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khanacademy | Given the equation $8 x^2-9 x-10 y^2-9 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-9 y+8 x^2-9 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-10 y^2-9 y+8 x^2-9 x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(8 x^2-9 x+\underline{\text{ }}\right)+\left(-10 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2-9 x+\underline{\text{ }}\right)=8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right)$}+\left(-10 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2-9 y+\underline{\text{ }}\right)=-10 \left(y^2+\frac{9 y}{10}+\underline{\text{ }}\right): \\
8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2+\frac{9 y}{10}+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{8}}{2}\right)^2=\frac{81}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{81}{256}=\frac{81}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6+\frac{81}{32}=\frac{273}{32}: \\
8 \left(x^2-\frac{9 x}{8}+\frac{81}{256}\right)-10 \left(y^2+\frac{9 y}{10}+\underline{\text{ }}\right)=\fbox{$\frac{273}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{9}{10}}{2}\right)^2=\frac{81}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{81}{400}=-\frac{81}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{273}{32}-\frac{81}{40}=\frac{1041}{160}: \\
8 \left(x^2-\frac{9 x}{8}+\frac{81}{256}\right)-10 \left(y^2+\frac{9 y}{10}+\frac{81}{400}\right)=\fbox{$\frac{1041}{160}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{9 x}{8}+\frac{81}{256}=\left(x-\frac{9}{16}\right)^2: \\
8 \fbox{$\left(x-\frac{9}{16}\right)^2$}-10 \left(y^2+\frac{9 y}{10}+\frac{81}{400}\right)=\frac{1041}{160} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{9 y}{10}+\frac{81}{400}=\left(y+\frac{9}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \left(x-\frac{9}{16}\right)^2-\text{10 }\fbox{$\left(y+\frac{9}{20}\right)^2$}=\frac{1041}{160} \\
\end{array}
| amps |
khanacademy | Given the equation $-8 x^2-2 x-7 y^2+6 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2+6 y-8 x^2-2 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
-7 y^2+6 y-8 x^2-2 x=2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2-2 x+\underline{\text{ }}\right)+\left(-7 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2-2 x+\underline{\text{ }}\right)=-8 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right)$}+\left(-7 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2+6 y+\underline{\text{ }}\right)=-7 \left(y^2-\frac{6 y}{7}+\underline{\text{ }}\right): \\
-8 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2-\frac{6 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{-8}{64}=-\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2-\frac{1}{8}=\frac{15}{8}: \\
-8 \left(x^2+\frac{x}{4}+\frac{1}{64}\right)-7 \left(y^2-\frac{6 y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{15}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-6}{7}}{2}\right)^2=\frac{9}{49} \text{on }\text{the }\text{left }\text{and }-7\times \frac{9}{49}=-\frac{9}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{15}{8}-\frac{9}{7}=\frac{33}{56}: \\
-8 \left(x^2+\frac{x}{4}+\frac{1}{64}\right)-7 \left(y^2-\frac{6 y}{7}+\frac{9}{49}\right)=\fbox{$\frac{33}{56}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{4}+\frac{1}{64}=\left(x+\frac{1}{8}\right)^2: \\
-8 \fbox{$\left(x+\frac{1}{8}\right)^2$}-7 \left(y^2-\frac{6 y}{7}+\frac{9}{49}\right)=\frac{33}{56} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{6 y}{7}+\frac{9}{49}=\left(y-\frac{3}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x+\frac{1}{8}\right)^2-7 \fbox{$\left(y-\frac{3}{7}\right)^2$}=\frac{33}{56} \\
\end{array}
| amps |
khanacademy | Given the equation $5 x^2-9 x+7 y^2-9 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2-9 y+5 x^2-9 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
7 y^2-9 y+5 x^2-9 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2-9 x+\underline{\text{ }}\right)+\left(7 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2-9 x+\underline{\text{ }}\right)=5 \left(x^2-\frac{9 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-\frac{9 x}{5}+\underline{\text{ }}\right)$}+\left(7 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2-9 y+\underline{\text{ }}\right)=7 \left(y^2-\frac{9 y}{7}+\underline{\text{ }}\right): \\
5 \left(x^2-\frac{9 x}{5}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2-\frac{9 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{5}}{2}\right)^2=\frac{81}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{81}{100}=\frac{81}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{81}{20}-1=\frac{61}{20}: \\
5 \left(x^2-\frac{9 x}{5}+\frac{81}{100}\right)+7 \left(y^2-\frac{9 y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{61}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{7}}{2}\right)^2=\frac{81}{196} \text{on }\text{the }\text{left }\text{and }7\times \frac{81}{196}=\frac{81}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{61}{20}+\frac{81}{28}=\frac{208}{35}: \\
5 \left(x^2-\frac{9 x}{5}+\frac{81}{100}\right)+7 \left(y^2-\frac{9 y}{7}+\frac{81}{196}\right)=\fbox{$\frac{208}{35}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{9 x}{5}+\frac{81}{100}=\left(x-\frac{9}{10}\right)^2: \\
5 \fbox{$\left(x-\frac{9}{10}\right)^2$}+7 \left(y^2-\frac{9 y}{7}+\frac{81}{196}\right)=\frac{208}{35} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{9 y}{7}+\frac{81}{196}=\left(y-\frac{9}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x-\frac{9}{10}\right)^2+7 \fbox{$\left(y-\frac{9}{14}\right)^2$}=\frac{208}{35} \\
\end{array}
| amps |
khanacademy | Given the equation $2 x^2+6 x+10 y^2-8 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2-8 y+2 x^2+6 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
10 y^2-8 y+2 x^2+6 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2+6 x+\underline{\text{ }}\right)+\left(10 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2+6 x+\underline{\text{ }}\right)=2 \left(x^2+3 x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2+3 x+\underline{\text{ }}\right)$}+\left(10 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2-8 y+\underline{\text{ }}\right)=10 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right): \\
2 \left(x^2+3 x+\underline{\text{ }}\right)+\fbox{$10 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }2\times \frac{9}{4}=\frac{9}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{2}-9=-\frac{9}{2}: \\
2 \left(x^2+3 x+\frac{9}{4}\right)+10 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{9}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }10\times \frac{4}{25}=\frac{8}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{8}{5}-\frac{9}{2}=-\frac{29}{10}: \\
2 \left(x^2+3 x+\frac{9}{4}\right)+10 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=\fbox{$-\frac{29}{10}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+3 x+\frac{9}{4}=\left(x+\frac{3}{2}\right)^2: \\
2 \fbox{$\left(x+\frac{3}{2}\right)^2$}+10 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=-\frac{29}{10} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{4 y}{5}+\frac{4}{25}=\left(y-\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x+\frac{3}{2}\right)^2+\text{10 }\fbox{$\left(y-\frac{2}{5}\right)^2$}=-\frac{29}{10} \\
\end{array}
| amps |
khanacademy | Given the equation $-3 x^2-2 x+2 y^2-7 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-7 y-3 x^2-2 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
2 y^2-7 y-3 x^2-2 x=2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2-2 x+\underline{\text{ }}\right)+\left(2 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2-2 x+\underline{\text{ }}\right)=-3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(2 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2-7 y+\underline{\text{ }}\right)=2 \left(y^2-\frac{7 y}{2}+\underline{\text{ }}\right): \\
-3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-\frac{7 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{-3}{9}=-\frac{1}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2-\frac{1}{3}=\frac{5}{3}: \\
-3 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+2 \left(y^2-\frac{7 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{5}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{2}}{2}\right)^2=\frac{49}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{49}{16}=\frac{49}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{5}{3}+\frac{49}{8}=\frac{187}{24}: \\
-3 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+2 \left(y^2-\frac{7 y}{2}+\frac{49}{16}\right)=\fbox{$\frac{187}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{3}+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2: \\
-3 \fbox{$\left(x+\frac{1}{3}\right)^2$}+2 \left(y^2-\frac{7 y}{2}+\frac{49}{16}\right)=\frac{187}{24} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{2}+\frac{49}{16}=\left(y-\frac{7}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x+\frac{1}{3}\right)^2+2 \fbox{$\left(y-\frac{7}{4}\right)^2$}=\frac{187}{24} \\
\end{array}
| amps |
khanacademy | Given the equation $-4 x^2+10 x+8 y^2+4 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2+4 y-4 x^2+10 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
8 y^2+4 y-4 x^2+10 x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-4 x^2+10 x+\underline{\text{ }}\right)+\left(8 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2+10 x+\underline{\text{ }}\right)=-4 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)$}+\left(8 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2+4 y+\underline{\text{ }}\right)=8 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right): \\
-4 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }-4\times \frac{25}{16}=-\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6-\frac{25}{4}=-\frac{1}{4}: \\
-4 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)+8 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{1}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{8}{16}=\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{2}-\frac{1}{4}=\frac{1}{4}: \\
-4 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)+8 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\fbox{$\frac{1}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{2}+\frac{25}{16}=\left(x-\frac{5}{4}\right)^2: \\
-4 \fbox{$\left(x-\frac{5}{4}\right)^2$}+8 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\frac{1}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{2}+\frac{1}{16}=\left(y+\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x-\frac{5}{4}\right)^2+8 \fbox{$\left(y+\frac{1}{4}\right)^2$}=\frac{1}{4} \\
\end{array}
| amps |
khanacademy | Given the equation $10 x^2-8 x-6 y^2-y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-y+10 x^2-8 x=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(10 x^2-8 x+\underline{\text{ }}\right)+\left(-6 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(10 x^2-8 x+\underline{\text{ }}\right)=10 \left(x^2-\frac{4 x}{5}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2-\frac{4 x}{5}+\underline{\text{ }}\right)$}+\left(-6 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 y^2-y+\underline{\text{ }}\right)=-6 \left(y^2+\frac{y}{6}+\underline{\text{ }}\right): \\
10 \left(x^2-\frac{4 x}{5}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{y}{6}+\underline{\text{ }}\right)$}=\underline{\text{ }}+0 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }10\times \frac{4}{25}=\frac{8}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{6}}{2}\right)^2=\frac{1}{144} \text{on }\text{the }\text{left }\text{and }\frac{-6}{144}=-\frac{1}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{8}{5}-\frac{1}{24}=\frac{187}{120}: \\
10 \left(x^2-\frac{4 x}{5}+\frac{4}{25}\right)-6 \left(y^2+\frac{y}{6}+\frac{1}{144}\right)=\fbox{$\frac{187}{120}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{4 x}{5}+\frac{4}{25}=\left(x-\frac{2}{5}\right)^2: \\
\text{10 }\fbox{$\left(x-\frac{2}{5}\right)^2$}-6 \left(y^2+\frac{y}{6}+\frac{1}{144}\right)=\frac{187}{120} \\
\end{array}
Step 9:
\begin{array}{l}
y^2+\frac{y}{6}+\frac{1}{144}=\left(y+\frac{1}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 10 \left(x-\frac{2}{5}\right)^2-6 \fbox{$\left(y+\frac{1}{12}\right)^2$}=\frac{187}{120} \\
\end{array}
| amps |
khanacademy | Given the equation $-8 x^2+3 x-5 y^2+2 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+2 y-8 x^2+3 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-5 y^2+2 y-8 x^2+3 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2+3 x+\underline{\text{ }}\right)+\left(-5 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2+3 x+\underline{\text{ }}\right)=-8 \left(x^2-\frac{3 x}{8}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2-\frac{3 x}{8}+\underline{\text{ }}\right)$}+\left(-5 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+2 y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{2 y}{5}+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{3 x}{8}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{2 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{8}}{2}\right)^2=\frac{9}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{9}{256}=-\frac{9}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-9-\frac{9}{32}=-\frac{297}{32}: \\
-8 \left(x^2-\frac{3 x}{8}+\frac{9}{256}\right)-5 \left(y^2-\frac{2 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{297}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{5}}{2}\right)^2=\frac{1}{25} \text{on }\text{the }\text{left }\text{and }\frac{-5}{25}=-\frac{1}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{297}{32}-\frac{1}{5}=-\frac{1517}{160}: \\
-8 \left(x^2-\frac{3 x}{8}+\frac{9}{256}\right)-5 \left(y^2-\frac{2 y}{5}+\frac{1}{25}\right)=\fbox{$-\frac{1517}{160}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{8}+\frac{9}{256}=\left(x-\frac{3}{16}\right)^2: \\
-8 \fbox{$\left(x-\frac{3}{16}\right)^2$}-5 \left(y^2-\frac{2 y}{5}+\frac{1}{25}\right)=-\frac{1517}{160} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{2 y}{5}+\frac{1}{25}=\left(y-\frac{1}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x-\frac{3}{16}\right)^2-5 \fbox{$\left(y-\frac{1}{5}\right)^2$}=-\frac{1517}{160} \\
\end{array}
| amps |
khanacademy | Given the equation $-2 x^2+5 x-2 y^2-6 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-6 y-2 x^2+5 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-2 y^2-6 y-2 x^2+5 x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-2 x^2+5 x+\underline{\text{ }}\right)+\left(-2 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 x^2+5 x+\underline{\text{ }}\right)=-2 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$-2 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)$}+\left(-2 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2-6 y+\underline{\text{ }}\right)=-2 \left(y^2+3 y+\underline{\text{ }}\right): \\
-2 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2+3 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }-2\times \frac{25}{16}=-\frac{25}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6-\frac{25}{8}=\frac{23}{8}: \\
-2 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)-2 \left(y^2+3 y+\underline{\text{ }}\right)=\fbox{$\frac{23}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }-2\times \frac{9}{4}=-\frac{9}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{23}{8}-\frac{9}{2}=-\frac{13}{8}: \\
-2 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)-2 \left(y^2+3 y+\frac{9}{4}\right)=\fbox{$-\frac{13}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{2}+\frac{25}{16}=\left(x-\frac{5}{4}\right)^2: \\
-2 \fbox{$\left(x-\frac{5}{4}\right)^2$}-2 \left(y^2+3 y+\frac{9}{4}\right)=-\frac{13}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+3 y+\frac{9}{4}=\left(y+\frac{3}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -2 \left(x-\frac{5}{4}\right)^2-2 \fbox{$\left(y+\frac{3}{2}\right)^2$}=-\frac{13}{8} \\
\end{array}
| amps |
khanacademy | Given the equation $7 x^2-3 x-6 y^2-3 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-3 y+7 x^2-3 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
-6 y^2-3 y+7 x^2-3 x=8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(7 x^2-3 x+\underline{\text{ }}\right)+\left(-6 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2-3 x+\underline{\text{ }}\right)=7 \left(x^2-\frac{3 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2-\frac{3 x}{7}+\underline{\text{ }}\right)$}+\left(-6 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2-3 y+\underline{\text{ }}\right)=-6 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right): \\
7 \left(x^2-\frac{3 x}{7}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{7}}{2}\right)^2=\frac{9}{196} \text{on }\text{the }\text{left }\text{and }7\times \frac{9}{196}=\frac{9}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8+\frac{9}{28}=\frac{233}{28}: \\
7 \left(x^2-\frac{3 x}{7}+\frac{9}{196}\right)-6 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{233}{28}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{-6}{16}=-\frac{3}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{233}{28}-\frac{3}{8}=\frac{445}{56}: \\
7 \left(x^2-\frac{3 x}{7}+\frac{9}{196}\right)-6 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\fbox{$\frac{445}{56}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{7}+\frac{9}{196}=\left(x-\frac{3}{14}\right)^2: \\
7 \fbox{$\left(x-\frac{3}{14}\right)^2$}-6 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\frac{445}{56} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{2}+\frac{1}{16}=\left(y+\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x-\frac{3}{14}\right)^2-6 \fbox{$\left(y+\frac{1}{4}\right)^2$}=\frac{445}{56} \\
\end{array}
| amps |
khanacademy | Given the equation $-3 x^2+8 x+9 y^2+8 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+8 y-3 x^2+8 x=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2+8 x+\underline{\text{ }}\right)+\left(9 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(-3 x^2+8 x+\underline{\text{ }}\right)=-3 \left(x^2-\frac{8 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2-\frac{8 x}{3}+\underline{\text{ }}\right)$}+\left(9 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 y^2+8 y+\underline{\text{ }}\right)=9 \left(y^2+\frac{8 y}{9}+\underline{\text{ }}\right): \\
-3 \left(x^2-\frac{8 x}{3}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2+\frac{8 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}+0 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-8}{3}}{2}\right)^2=\frac{16}{9} \text{on }\text{the }\text{left }\text{and }-3\times \frac{16}{9}=-\frac{16}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{9}}{2}\right)^2=\frac{16}{81} \text{on }\text{the }\text{left }\text{and }9\times \frac{16}{81}=\frac{16}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{16}{9}-\frac{16}{3}=-\frac{32}{9}: \\
-3 \left(x^2-\frac{8 x}{3}+\frac{16}{9}\right)+9 \left(y^2+\frac{8 y}{9}+\frac{16}{81}\right)=\fbox{$-\frac{32}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{8 x}{3}+\frac{16}{9}=\left(x-\frac{4}{3}\right)^2: \\
-3 \fbox{$\left(x-\frac{4}{3}\right)^2$}+9 \left(y^2+\frac{8 y}{9}+\frac{16}{81}\right)=-\frac{32}{9} \\
\end{array}
Step 9:
\begin{array}{l}
y^2+\frac{8 y}{9}+\frac{16}{81}=\left(y+\frac{4}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x-\frac{4}{3}\right)^2+9 \fbox{$\left(y+\frac{4}{9}\right)^2$}=-\frac{32}{9} \\
\end{array}
| amps |
khanacademy | Given the equation $3 x^2-2 x-7 y^2+10 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2+10 y+3 x^2-2 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
-7 y^2+10 y+3 x^2-2 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2-2 x+\underline{\text{ }}\right)+\left(-7 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2-2 x+\underline{\text{ }}\right)=3 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(-7 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2+10 y+\underline{\text{ }}\right)=-7 \left(y^2-\frac{10 y}{7}+\underline{\text{ }}\right): \\
3 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2-\frac{10 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}+3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{3}{9}=\frac{1}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3+\frac{1}{3}=\frac{10}{3}: \\
3 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)-7 \left(y^2-\frac{10 y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{10}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-10}{7}}{2}\right)^2=\frac{25}{49} \text{on }\text{the }\text{left }\text{and }-7\times \frac{25}{49}=-\frac{25}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{10}{3}-\frac{25}{7}=-\frac{5}{21}: \\
3 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)-7 \left(y^2-\frac{10 y}{7}+\frac{25}{49}\right)=\fbox{$-\frac{5}{21}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{2 x}{3}+\frac{1}{9}=\left(x-\frac{1}{3}\right)^2: \\
3 \fbox{$\left(x-\frac{1}{3}\right)^2$}-7 \left(y^2-\frac{10 y}{7}+\frac{25}{49}\right)=-\frac{5}{21} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{10 y}{7}+\frac{25}{49}=\left(y-\frac{5}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x-\frac{1}{3}\right)^2-7 \fbox{$\left(y-\frac{5}{7}\right)^2$}=-\frac{5}{21} \\
\end{array}
| amps |
khanacademy | Given the equation $4 x^2+x+4 y^2+9 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2+9 y+4 x^2+x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
4 y^2+9 y+4 x^2+x=2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2+x+\underline{\text{ }}\right)+\left(4 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2+x+\underline{\text{ }}\right)=4 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right)$}+\left(4 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2+9 y+\underline{\text{ }}\right)=4 \left(y^2+\frac{9 y}{4}+\underline{\text{ }}\right): \\
4 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2+\frac{9 y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{4}{64}=\frac{1}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2+\frac{1}{16}=\frac{33}{16}: \\
4 \left(x^2+\frac{x}{4}+\frac{1}{64}\right)+4 \left(y^2+\frac{9 y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{33}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{9}{4}}{2}\right)^2=\frac{81}{64} \text{on }\text{the }\text{left }\text{and }4\times \frac{81}{64}=\frac{81}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{33}{16}+\frac{81}{16}=\frac{57}{8}: \\
4 \left(x^2+\frac{x}{4}+\frac{1}{64}\right)+4 \left(y^2+\frac{9 y}{4}+\frac{81}{64}\right)=\fbox{$\frac{57}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{4}+\frac{1}{64}=\left(x+\frac{1}{8}\right)^2: \\
4 \fbox{$\left(x+\frac{1}{8}\right)^2$}+4 \left(y^2+\frac{9 y}{4}+\frac{81}{64}\right)=\frac{57}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{9 y}{4}+\frac{81}{64}=\left(y+\frac{9}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x+\frac{1}{8}\right)^2+4 \fbox{$\left(y+\frac{9}{8}\right)^2$}=\frac{57}{8} \\
\end{array}
| amps |
khanacademy | Given the equation $-9 x^2-5 x+10 y^2+y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2+y-9 x^2-5 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
10 y^2+y-9 x^2-5 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2-5 x+\underline{\text{ }}\right)+\left(10 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2-5 x+\underline{\text{ }}\right)=-9 \left(x^2+\frac{5 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2+\frac{5 x}{9}+\underline{\text{ }}\right)$}+\left(10 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2+y+\underline{\text{ }}\right)=10 \left(y^2+\frac{y}{10}+\underline{\text{ }}\right): \\
-9 \left(x^2+\frac{5 x}{9}+\underline{\text{ }}\right)+\fbox{$10 \left(y^2+\frac{y}{10}+\underline{\text{ }}\right)$}=\underline{\text{ }}-1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{9}}{2}\right)^2=\frac{25}{324} \text{on }\text{the }\text{left }\text{and }-9\times \frac{25}{324}=-\frac{25}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-1-\frac{25}{36}=-\frac{61}{36}: \\
-9 \left(x^2+\frac{5 x}{9}+\frac{25}{324}\right)+10 \left(y^2+\frac{y}{10}+\underline{\text{ }}\right)=\fbox{$-\frac{61}{36}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{10}}{2}\right)^2=\frac{1}{400} \text{on }\text{the }\text{left }\text{and }\frac{10}{400}=\frac{1}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{40}-\frac{61}{36}=-\frac{601}{360}: \\
-9 \left(x^2+\frac{5 x}{9}+\frac{25}{324}\right)+10 \left(y^2+\frac{y}{10}+\frac{1}{400}\right)=\fbox{$-\frac{601}{360}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{9}+\frac{25}{324}=\left(x+\frac{5}{18}\right)^2: \\
-9 \fbox{$\left(x+\frac{5}{18}\right)^2$}+10 \left(y^2+\frac{y}{10}+\frac{1}{400}\right)=-\frac{601}{360} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{10}+\frac{1}{400}=\left(y+\frac{1}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x+\frac{5}{18}\right)^2+\text{10 }\fbox{$\left(y+\frac{1}{20}\right)^2$}=-\frac{601}{360} \\
\end{array}
| amps |
khanacademy | Given the equation $3 x^2+2 x-6 y^2-5 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-5 y+3 x^2+2 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
-6 y^2-5 y+3 x^2+2 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2+2 x+\underline{\text{ }}\right)+\left(-6 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2+2 x+\underline{\text{ }}\right)=3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(-6 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2-5 y+\underline{\text{ }}\right)=-6 \left(y^2+\frac{5 y}{6}+\underline{\text{ }}\right): \\
3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{5 y}{6}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{3}{9}=\frac{1}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9+\frac{1}{3}=\frac{28}{3}: \\
3 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)-6 \left(y^2+\frac{5 y}{6}+\underline{\text{ }}\right)=\fbox{$\frac{28}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{6}}{2}\right)^2=\frac{25}{144} \text{on }\text{the }\text{left }\text{and }-6\times \frac{25}{144}=-\frac{25}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{28}{3}-\frac{25}{24}=\frac{199}{24}: \\
3 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)-6 \left(y^2+\frac{5 y}{6}+\frac{25}{144}\right)=\fbox{$\frac{199}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{3}+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2: \\
3 \fbox{$\left(x+\frac{1}{3}\right)^2$}-6 \left(y^2+\frac{5 y}{6}+\frac{25}{144}\right)=\frac{199}{24} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{6}+\frac{25}{144}=\left(y+\frac{5}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x+\frac{1}{3}\right)^2-6 \fbox{$\left(y+\frac{5}{12}\right)^2$}=\frac{199}{24} \\
\end{array}
| amps |
khanacademy | Given the equation $8 x^2+4 x-4 y^2-10 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2-10 y+8 x^2+4 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
-4 y^2-10 y+8 x^2+4 x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(8 x^2+4 x+\underline{\text{ }}\right)+\left(-4 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2+4 x+\underline{\text{ }}\right)=8 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)$}+\left(-4 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2-10 y+\underline{\text{ }}\right)=-4 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right): \\
8 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{8}{16}=\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{1}{2}-4=-\frac{7}{2}: \\
8 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)-4 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{7}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }-4\times \frac{25}{16}=-\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{7}{2}-\frac{25}{4}=-\frac{39}{4}: \\
8 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)-4 \left(y^2+\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$-\frac{39}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{2}+\frac{1}{16}=\left(x+\frac{1}{4}\right)^2: \\
8 \fbox{$\left(x+\frac{1}{4}\right)^2$}-4 \left(y^2+\frac{5 y}{2}+\frac{25}{16}\right)=-\frac{39}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{2}+\frac{25}{16}=\left(y+\frac{5}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \left(x+\frac{1}{4}\right)^2-4 \fbox{$\left(y+\frac{5}{4}\right)^2$}=-\frac{39}{4} \\
\end{array}
| amps |
khanacademy | Given the equation $2 x^2-7 x-2 y^2-3 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-3 y+2 x^2-7 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
-2 y^2-3 y+2 x^2-7 x=-8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2-7 x+\underline{\text{ }}\right)+\left(-2 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2-7 x+\underline{\text{ }}\right)=2 \left(x^2-\frac{7 x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-\frac{7 x}{2}+\underline{\text{ }}\right)$}+\left(-2 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2-3 y+\underline{\text{ }}\right)=-2 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right): \\
2 \left(x^2-\frac{7 x}{2}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{2}}{2}\right)^2=\frac{49}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{49}{16}=\frac{49}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{49}{8}-8=-\frac{15}{8}: \\
2 \left(x^2-\frac{7 x}{2}+\frac{49}{16}\right)-2 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{15}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }-2\times \frac{9}{16}=-\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{15}{8}-\frac{9}{8}=-3: \\
2 \left(x^2-\frac{7 x}{2}+\frac{49}{16}\right)-2 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$-3$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{7 x}{2}+\frac{49}{16}=\left(x-\frac{7}{4}\right)^2: \\
2 \fbox{$\left(x-\frac{7}{4}\right)^2$}-2 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=-3 \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{2}+\frac{9}{16}=\left(y+\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x-\frac{7}{4}\right)^2-2 \fbox{$\left(y+\frac{3}{4}\right)^2$}=-3 \\
\end{array}
| amps |
khanacademy | Given the equation $3 x^2-9 x-4 y^2-5 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2-5 y+3 x^2-9 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
-4 y^2-5 y+3 x^2-9 x=-3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2-9 x+\underline{\text{ }}\right)+\left(-4 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2-9 x+\underline{\text{ }}\right)=3 \left(x^2-3 x+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2-3 x+\underline{\text{ }}\right)$}+\left(-4 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2-5 y+\underline{\text{ }}\right)=-4 \left(y^2+\frac{5 y}{4}+\underline{\text{ }}\right): \\
3 \left(x^2-3 x+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2+\frac{5 y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}-3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }3\times \frac{9}{4}=\frac{27}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{27}{4}-3=\frac{15}{4}: \\
3 \left(x^2-3 x+\frac{9}{4}\right)-4 \left(y^2+\frac{5 y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{15}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{4}}{2}\right)^2=\frac{25}{64} \text{on }\text{the }\text{left }\text{and }-4\times \frac{25}{64}=-\frac{25}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{15}{4}-\frac{25}{16}=\frac{35}{16}: \\
3 \left(x^2-3 x+\frac{9}{4}\right)-4 \left(y^2+\frac{5 y}{4}+\frac{25}{64}\right)=\fbox{$\frac{35}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-3 x+\frac{9}{4}=\left(x-\frac{3}{2}\right)^2: \\
3 \fbox{$\left(x-\frac{3}{2}\right)^2$}-4 \left(y^2+\frac{5 y}{4}+\frac{25}{64}\right)=\frac{35}{16} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{4}+\frac{25}{64}=\left(y+\frac{5}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x-\frac{3}{2}\right)^2-4 \fbox{$\left(y+\frac{5}{8}\right)^2$}=\frac{35}{16} \\
\end{array}
| amps |
khanacademy | Given the equation $3 x^2-2 x-2 y^2+8 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+8 y+3 x^2-2 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
-2 y^2+8 y+3 x^2-2 x=-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2-2 x+\underline{\text{ }}\right)+\left(-2 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2-2 x+\underline{\text{ }}\right)=3 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(-2 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+8 y+\underline{\text{ }}\right)=-2 \left(y^2-4 y+\underline{\text{ }}\right): \\
3 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-4 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{3}{9}=\frac{1}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{1}{3}-10=-\frac{29}{3}: \\
3 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)-2 \left(y^2-4 y+\underline{\text{ }}\right)=\fbox{$-\frac{29}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-4}{2}\right)^2=4 \text{on }\text{the }\text{left }\text{and }-2\times 4=-8 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{29}{3}-8=-\frac{53}{3}: \\
3 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)-2 \left(y^2-4 y+4\right)=\fbox{$-\frac{53}{3}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{2 x}{3}+\frac{1}{9}=\left(x-\frac{1}{3}\right)^2: \\
3 \fbox{$\left(x-\frac{1}{3}\right)^2$}-2 \left(y^2-4 y+4\right)=-\frac{53}{3} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-4 y+4=(y-2)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x-\frac{1}{3}\right)^2-2 \fbox{$(y-2)^2$}=-\frac{53}{3} \\
\end{array}
| amps |
khanacademy | Given the equation $x^2-10 x-6 y^2-6 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-6 y+x^2-10 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
-6 y^2-6 y+x^2-10 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(x^2-10 x+\underline{\text{ }}\right)+\left(-6 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 y^2-6 y+\underline{\text{ }}\right)=-6 \left(y^2+y+\underline{\text{ }}\right): \\
\left(x^2-10 x+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+y+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{-10}{2}\right)^2=25 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
9+25=34: \\
\left(x^2-10 x+25\right)-6 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$34$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-6}{4}=-\frac{3}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 8:
\begin{array}{l}
34-\frac{3}{2}=\frac{65}{2}: \\
\left(x^2-10 x+25\right)-6 \left(y^2+y+\frac{1}{4}\right)=\fbox{$\frac{65}{2}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-10 x+25=(x-5)^2: \\
\fbox{$(x-5)^2$}-6 \left(y^2+y+\frac{1}{4}\right)=\frac{65}{2} \\
\end{array}
Step 10:
\begin{array}{l}
y^2+y+\frac{1}{4}=\left(y+\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & (x-5)^2-6 \fbox{$\left(y+\frac{1}{2}\right)^2$}=\frac{65}{2} \\
\end{array}
| amps |
khanacademy | Given the equation $-4 x^2+7 x+6 y^2-7 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2-7 y-4 x^2+7 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
6 y^2-7 y-4 x^2+7 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-4 x^2+7 x+\underline{\text{ }}\right)+\left(6 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2+7 x+\underline{\text{ }}\right)=-4 \left(x^2-\frac{7 x}{4}+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2-\frac{7 x}{4}+\underline{\text{ }}\right)$}+\left(6 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2-7 y+\underline{\text{ }}\right)=6 \left(y^2-\frac{7 y}{6}+\underline{\text{ }}\right): \\
-4 \left(x^2-\frac{7 x}{4}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2-\frac{7 y}{6}+\underline{\text{ }}\right)$}=\underline{\text{ }}+3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{4}}{2}\right)^2=\frac{49}{64} \text{on }\text{the }\text{left }\text{and }-4\times \frac{49}{64}=-\frac{49}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3-\frac{49}{16}=-\frac{1}{16}: \\
-4 \left(x^2-\frac{7 x}{4}+\frac{49}{64}\right)+6 \left(y^2-\frac{7 y}{6}+\underline{\text{ }}\right)=\fbox{$-\frac{1}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{6}}{2}\right)^2=\frac{49}{144} \text{on }\text{the }\text{left }\text{and }6\times \frac{49}{144}=\frac{49}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{49}{24}-\frac{1}{16}=\frac{95}{48}: \\
-4 \left(x^2-\frac{7 x}{4}+\frac{49}{64}\right)+6 \left(y^2-\frac{7 y}{6}+\frac{49}{144}\right)=\fbox{$\frac{95}{48}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{7 x}{4}+\frac{49}{64}=\left(x-\frac{7}{8}\right)^2: \\
-4 \fbox{$\left(x-\frac{7}{8}\right)^2$}+6 \left(y^2-\frac{7 y}{6}+\frac{49}{144}\right)=\frac{95}{48} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{6}+\frac{49}{144}=\left(y-\frac{7}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x-\frac{7}{8}\right)^2+6 \fbox{$\left(y-\frac{7}{12}\right)^2$}=\frac{95}{48} \\
\end{array}
| amps |
khanacademy | Given the equation $-5 x^2-9 x-y^2-6 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2-6 y-5 x^2-9 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-y^2-6 y-5 x^2-9 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2-9 x+\underline{\text{ }}\right)+\left(-y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2-9 x+\underline{\text{ }}\right)=-5 \left(x^2+\frac{9 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2+\frac{9 x}{5}+\underline{\text{ }}\right)$}+\left(-y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2-6 y+\underline{\text{ }}\right)=-\left(y^2+6 y+\underline{\text{ }}\right): \\
-5 \left(x^2+\frac{9 x}{5}+\underline{\text{ }}\right)+\fbox{$-\left(y^2+6 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{9}{5}}{2}\right)^2=\frac{81}{100} \text{on }\text{the }\text{left }\text{and }-5\times \frac{81}{100}=-\frac{81}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5-\frac{81}{20}=\frac{19}{20}: \\
-5 \left(x^2+\frac{9 x}{5}+\frac{81}{100}\right)-\left(y^2+6 y+\underline{\text{ }}\right)=\fbox{$\frac{19}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{6}{2}\right)^2=9 \text{on }\text{the }\text{left }\text{and }-9=-9 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{19}{20}-9=-\frac{161}{20}: \\
-5 \left(x^2+\frac{9 x}{5}+\frac{81}{100}\right)-\left(y^2+6 y+9\right)=\fbox{$-\frac{161}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{9 x}{5}+\frac{81}{100}=\left(x+\frac{9}{10}\right)^2: \\
-5 \fbox{$\left(x+\frac{9}{10}\right)^2$}-\left(y^2+6 y+9\right)=-\frac{161}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+6 y+9=(y+3)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 \left(x+\frac{9}{10}\right)^2-\fbox{$(y+3)^2$}=-\frac{161}{20} \\
\end{array}
| amps |
khanacademy | Given the equation $5 x^2+2 x-3 y^2+10 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2+10 y+5 x^2+2 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
-3 y^2+10 y+5 x^2+2 x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2+2 x+\underline{\text{ }}\right)+\left(-3 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2+2 x+\underline{\text{ }}\right)=5 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right)$}+\left(-3 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2+10 y+\underline{\text{ }}\right)=-3 \left(y^2-\frac{10 y}{3}+\underline{\text{ }}\right): \\
5 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2-\frac{10 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{5}}{2}\right)^2=\frac{1}{25} \text{on }\text{the }\text{left }\text{and }\frac{5}{25}=\frac{1}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{1}{5}-4=-\frac{19}{5}: \\
5 \left(x^2+\frac{2 x}{5}+\frac{1}{25}\right)-3 \left(y^2-\frac{10 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{19}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-10}{3}}{2}\right)^2=\frac{25}{9} \text{on }\text{the }\text{left }\text{and }-3\times \frac{25}{9}=-\frac{25}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{19}{5}-\frac{25}{3}=-\frac{182}{15}: \\
5 \left(x^2+\frac{2 x}{5}+\frac{1}{25}\right)-3 \left(y^2-\frac{10 y}{3}+\frac{25}{9}\right)=\fbox{$-\frac{182}{15}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{5}+\frac{1}{25}=\left(x+\frac{1}{5}\right)^2: \\
5 \fbox{$\left(x+\frac{1}{5}\right)^2$}-3 \left(y^2-\frac{10 y}{3}+\frac{25}{9}\right)=-\frac{182}{15} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{10 y}{3}+\frac{25}{9}=\left(y-\frac{5}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x+\frac{1}{5}\right)^2-3 \fbox{$\left(y-\frac{5}{3}\right)^2$}=-\frac{182}{15} \\
\end{array}
| amps |
khanacademy | Given the equation $5 x^2+8 x-6 y^2-2 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-2 y+5 x^2+8 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
-6 y^2-2 y+5 x^2+8 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2+8 x+\underline{\text{ }}\right)+\left(-6 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2+8 x+\underline{\text{ }}\right)=5 \left(x^2+\frac{8 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+\frac{8 x}{5}+\underline{\text{ }}\right)$}+\left(-6 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2-2 y+\underline{\text{ }}\right)=-6 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right): \\
5 \left(x^2+\frac{8 x}{5}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{5}}{2}\right)^2=\frac{16}{25} \text{on }\text{the }\text{left }\text{and }5\times \frac{16}{25}=\frac{16}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9+\frac{16}{5}=\frac{61}{5}: \\
5 \left(x^2+\frac{8 x}{5}+\frac{16}{25}\right)-6 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{61}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{-6}{36}=-\frac{1}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{61}{5}-\frac{1}{6}=\frac{361}{30}: \\
5 \left(x^2+\frac{8 x}{5}+\frac{16}{25}\right)-6 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=\fbox{$\frac{361}{30}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{8 x}{5}+\frac{16}{25}=\left(x+\frac{4}{5}\right)^2: \\
5 \fbox{$\left(x+\frac{4}{5}\right)^2$}-6 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=\frac{361}{30} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{3}+\frac{1}{36}=\left(y+\frac{1}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x+\frac{4}{5}\right)^2-6 \fbox{$\left(y+\frac{1}{6}\right)^2$}=\frac{361}{30} \\
\end{array}
| amps |
khanacademy | Given the equation $8 x^2+8 x-10 y^2+2 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+2 y+8 x^2+8 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
-10 y^2+2 y+8 x^2+8 x=8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(8 x^2+8 x+\underline{\text{ }}\right)+\left(-10 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2+8 x+\underline{\text{ }}\right)=8 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+x+\underline{\text{ }}\right)$}+\left(-10 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2+2 y+\underline{\text{ }}\right)=-10 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right): \\
8 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{8}{4}=2 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8+2=10: \\
8 \left(x^2+x+\frac{1}{4}\right)-10 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right)=\fbox{$10$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{-10}{100}=-\frac{1}{10} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
10-\frac{1}{10}=\frac{99}{10}: \\
8 \left(x^2+x+\frac{1}{4}\right)-10 \left(y^2-\frac{y}{5}+\frac{1}{100}\right)=\fbox{$\frac{99}{10}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
8 \fbox{$\left(x+\frac{1}{2}\right)^2$}-10 \left(y^2-\frac{y}{5}+\frac{1}{100}\right)=\frac{99}{10} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{5}+\frac{1}{100}=\left(y-\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \left(x+\frac{1}{2}\right)^2-\text{10 }\fbox{$\left(y-\frac{1}{10}\right)^2$}=\frac{99}{10} \\
\end{array}
| amps |
khanacademy | Given the equation $-9 x^2+4 x+7 y^2+y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2+y-9 x^2+4 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
7 y^2+y-9 x^2+4 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2+4 x+\underline{\text{ }}\right)+\left(7 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2+4 x+\underline{\text{ }}\right)=-9 \left(x^2-\frac{4 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2-\frac{4 x}{9}+\underline{\text{ }}\right)$}+\left(7 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2+y+\underline{\text{ }}\right)=7 \left(y^2+\frac{y}{7}+\underline{\text{ }}\right): \\
-9 \left(x^2-\frac{4 x}{9}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2+\frac{y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{9}}{2}\right)^2=\frac{4}{81} \text{on }\text{the }\text{left }\text{and }-9\times \frac{4}{81}=-\frac{4}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9-\frac{4}{9}=\frac{77}{9}: \\
-9 \left(x^2-\frac{4 x}{9}+\frac{4}{81}\right)+7 \left(y^2+\frac{y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{77}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{7}}{2}\right)^2=\frac{1}{196} \text{on }\text{the }\text{left }\text{and }\frac{7}{196}=\frac{1}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{77}{9}+\frac{1}{28}=\frac{2165}{252}: \\
-9 \left(x^2-\frac{4 x}{9}+\frac{4}{81}\right)+7 \left(y^2+\frac{y}{7}+\frac{1}{196}\right)=\fbox{$\frac{2165}{252}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{4 x}{9}+\frac{4}{81}=\left(x-\frac{2}{9}\right)^2: \\
-9 \fbox{$\left(x-\frac{2}{9}\right)^2$}+7 \left(y^2+\frac{y}{7}+\frac{1}{196}\right)=\frac{2165}{252} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{7}+\frac{1}{196}=\left(y+\frac{1}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x-\frac{2}{9}\right)^2+7 \fbox{$\left(y+\frac{1}{14}\right)^2$}=\frac{2165}{252} \\
\end{array}
| amps |
khanacademy | Given the equation $-9 x^2+10 x-8 y^2+4 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2+4 y-9 x^2+10 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-8 y^2+4 y-9 x^2+10 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2+10 x+\underline{\text{ }}\right)+\left(-8 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2+10 x+\underline{\text{ }}\right)=-9 \left(x^2-\frac{10 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2-\frac{10 x}{9}+\underline{\text{ }}\right)$}+\left(-8 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2+4 y+\underline{\text{ }}\right)=-8 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right): \\
-9 \left(x^2-\frac{10 x}{9}+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-10}{9}}{2}\right)^2=\frac{25}{81} \text{on }\text{the }\text{left }\text{and }-9\times \frac{25}{81}=-\frac{25}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-5-\frac{25}{9}=-\frac{70}{9}: \\
-9 \left(x^2-\frac{10 x}{9}+\frac{25}{81}\right)-8 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{70}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{-8}{16}=-\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{70}{9}-\frac{1}{2}=-\frac{149}{18}: \\
-9 \left(x^2-\frac{10 x}{9}+\frac{25}{81}\right)-8 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\fbox{$-\frac{149}{18}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{10 x}{9}+\frac{25}{81}=\left(x-\frac{5}{9}\right)^2: \\
-9 \fbox{$\left(x-\frac{5}{9}\right)^2$}-8 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=-\frac{149}{18} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{2}+\frac{1}{16}=\left(y-\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x-\frac{5}{9}\right)^2-8 \fbox{$\left(y-\frac{1}{4}\right)^2$}=-\frac{149}{18} \\
\end{array}
| amps |
khanacademy | Given the equation $-3 x^2-7 x+6 y^2-8 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 x^2+6 y^2-7 x-8 y-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
-3 x^2+6 y^2-7 x-8 y=4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2-7 x+\underline{\text{ }}\right)+\left(6 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2-7 x+\underline{\text{ }}\right)=-3 \left(x^2+\frac{7 x^2}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+\frac{7 x}{3}+\underline{\text{ }}\right)$}+\left(6 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2-8 y+\underline{\text{ }}\right)=6 \left(y^2-\frac{4 y^2}{3}+\underline{\text{ }}\right): \\
-3 \left(x^2+\frac{7 x}{3}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{3}}{2}\right)^2=\frac{49}{36} \text{on }\text{the }\text{left }\text{and }-3\times \frac{49}{36}=-\frac{49}{12} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
4-\frac{49}{12}=-\frac{1}{12}: \\
-3 \left(x^2+\frac{7 x}{3}+\frac{49}{36}\right)+6 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{-1}{12}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{3}}{2}\right)^2=\frac{4}{9} \text{on }\text{the }\text{left }\text{and }6\times \frac{4}{9}=\frac{8}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{8}{3}-\frac{1}{12}=\frac{31}{12}: \\
-3 \left(x^2+\frac{7 x}{3}+\frac{49}{36}\right)+6 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$\frac{31}{12}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{3}+\frac{49}{36}=\left(x+\frac{7}{6}\right)^2: \\
-3 \fbox{$\left(x+\frac{7}{6}\right)^2$}+6 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=\frac{31}{12} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{4 y}{3}+\frac{4}{9}=\left(y-\frac{2}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x+\frac{7}{6}\right)^2+6 \fbox{$\left(y-\frac{2}{3}\right)^2$}=\frac{31}{12} \\
\end{array}
| amps |
khanacademy | Given the equation $-2 x^2+8 x-5 y^2-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 x^2+8 x+\left(-5 y^2-4\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-5 y^2-2 x^2+8 x-4 \text{from }\text{both }\text{sides}: \\
2 x^2-8 x+\left(5 y^2+4\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
5 y^2+2 x^2-8 x=-4 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(2 x^2-8 x+\underline{\text{ }}\right)+5 y^2=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 x^2-8 x+\underline{\text{ }}\right)=2 \left(x^2-4 x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-4 x+\underline{\text{ }}\right)$}+5 y^2=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-4}{2}\right)^2=4 \text{on }\text{the }\text{left }\text{and }2\times 4=8 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8-4=4: \\
2 \left(x^2-4 x+4\right)+5 y^2=\fbox{$4$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-4 x+4=(x-2)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \fbox{$(x-2)^2$}+5 y^2=4 \\
\end{array}
| amps |
khanacademy | Given the equation $8 x^2+7 x-y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 x^2+7 x+(10-y)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }10-y \text{from }\text{both }\text{sides}: \\
8 x^2+7 x=y-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(8 x^2+7 x+\underline{\text{ }}\right)=(y-10)+\underline{\text{ }} \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2+7 x+\underline{\text{ }}\right)=8 \left(x^2+\frac{7 x}{8}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+\frac{7 x}{8}+\underline{\text{ }}\right)$}=(y-10)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{8}}{2}\right)^2=\frac{49}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{49}{256}=\frac{49}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
(y-10)+\frac{49}{32}=y-\frac{271}{32}: \\
8 \left(x^2+\frac{7 x}{8}+\frac{49}{256}\right)=\fbox{$y-\frac{271}{32}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2+\frac{7 x}{8}+\frac{49}{256}=\left(x+\frac{7}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \fbox{$\left(x+\frac{7}{16}\right)^2$}=y-\frac{271}{32} \\
\end{array}
| amps |
khanacademy | Given the equation $7 x^2+4 x-7 y^2+10 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2+10 y+7 x^2+4 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
-7 y^2+10 y+7 x^2+4 x=-8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(7 x^2+4 x+\underline{\text{ }}\right)+\left(-7 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2+4 x+\underline{\text{ }}\right)=7 \left(x^2+\frac{4 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2+\frac{4 x}{7}+\underline{\text{ }}\right)$}+\left(-7 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2+10 y+\underline{\text{ }}\right)=-7 \left(y^2-\frac{10 y}{7}+\underline{\text{ }}\right): \\
7 \left(x^2+\frac{4 x}{7}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2-\frac{10 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{7}}{2}\right)^2=\frac{4}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{4}{49}=\frac{4}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{4}{7}-8=-\frac{52}{7}: \\
7 \left(x^2+\frac{4 x}{7}+\frac{4}{49}\right)-7 \left(y^2-\frac{10 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{52}{7}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-10}{7}}{2}\right)^2=\frac{25}{49} \text{on }\text{the }\text{left }\text{and }-7\times \frac{25}{49}=-\frac{25}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{52}{7}-\frac{25}{7}=-11: \\
7 \left(x^2+\frac{4 x}{7}+\frac{4}{49}\right)-7 \left(y^2-\frac{10 y}{7}+\frac{25}{49}\right)=\fbox{$-11$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{4 x}{7}+\frac{4}{49}=\left(x+\frac{2}{7}\right)^2: \\
7 \fbox{$\left(x+\frac{2}{7}\right)^2$}-7 \left(y^2-\frac{10 y}{7}+\frac{25}{49}\right)=-11 \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{10 y}{7}+\frac{25}{49}=\left(y-\frac{5}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x+\frac{2}{7}\right)^2-7 \fbox{$\left(y-\frac{5}{7}\right)^2$}=-11 \\
\end{array}
| amps |
khanacademy | Given the equation $9 x^2+10 x-7 y^2+y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2+y+9 x^2+10 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-7 y^2+y+9 x^2+10 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(9 x^2+10 x+\underline{\text{ }}\right)+\left(-7 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2+10 x+\underline{\text{ }}\right)=9 \left(x^2+\frac{10 x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2+\frac{10 x}{9}+\underline{\text{ }}\right)$}+\left(-7 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2+y+\underline{\text{ }}\right)=-7 \left(y^2-\frac{y}{7}+\underline{\text{ }}\right): \\
9 \left(x^2+\frac{10 x}{9}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2-\frac{y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{10}{9}}{2}\right)^2=\frac{25}{81} \text{on }\text{the }\text{left }\text{and }9\times \frac{25}{81}=\frac{25}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{25}{9}-9=-\frac{56}{9}: \\
9 \left(x^2+\frac{10 x}{9}+\frac{25}{81}\right)-7 \left(y^2-\frac{y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{56}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{7}}{2}\right)^2=\frac{1}{196} \text{on }\text{the }\text{left }\text{and }\frac{-7}{196}=-\frac{1}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{56}{9}-\frac{1}{28}=-\frac{1577}{252}: \\
9 \left(x^2+\frac{10 x}{9}+\frac{25}{81}\right)-7 \left(y^2-\frac{y}{7}+\frac{1}{196}\right)=\fbox{$-\frac{1577}{252}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{10 x}{9}+\frac{25}{81}=\left(x+\frac{5}{9}\right)^2: \\
9 \fbox{$\left(x+\frac{5}{9}\right)^2$}-7 \left(y^2-\frac{y}{7}+\frac{1}{196}\right)=-\frac{1577}{252} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{7}+\frac{1}{196}=\left(y-\frac{1}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x+\frac{5}{9}\right)^2-7 \fbox{$\left(y-\frac{1}{14}\right)^2$}=-\frac{1577}{252} \\
\end{array}
| amps |
khanacademy | Given the equation $8 x^2+3 x-6 y^2+2 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+2 y+8 x^2+3 x=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(8 x^2+3 x+\underline{\text{ }}\right)+\left(-6 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(8 x^2+3 x+\underline{\text{ }}\right)=8 \left(x^2+\frac{3 x}{8}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+\frac{3 x}{8}+\underline{\text{ }}\right)$}+\left(-6 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 y^2+2 y+\underline{\text{ }}\right)=-6 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right): \\
8 \left(x^2+\frac{3 x}{8}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+0 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{8}}{2}\right)^2=\frac{9}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{9}{256}=\frac{9}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{-6}{36}=-\frac{1}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{32}-\frac{1}{6}=\frac{11}{96}: \\
8 \left(x^2+\frac{3 x}{8}+\frac{9}{256}\right)-6 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\fbox{$\frac{11}{96}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{3 x}{8}+\frac{9}{256}=\left(x+\frac{3}{16}\right)^2: \\
8 \fbox{$\left(x+\frac{3}{16}\right)^2$}-6 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\frac{11}{96} \\
\end{array}
Step 9:
\begin{array}{l}
y^2-\frac{y}{3}+\frac{1}{36}=\left(y-\frac{1}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \left(x+\frac{3}{16}\right)^2-6 \fbox{$\left(y-\frac{1}{6}\right)^2$}=\frac{11}{96} \\
\end{array}
| amps |
khanacademy | Given the equation $-3 x^2-5 x-5 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 x^2-5 x+(-5 y-1)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 y+3 x^2+5 x+1 \text{to }\text{both }\text{sides}: \\
3 x^2+5 x+(5 y+1)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }5 y+1 \text{from }\text{both }\text{sides}: \\
3 x^2+5 x=-5 y-1 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(3 x^2+5 x+\underline{\text{ }}\right)=(-5 y-1)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 x^2+5 x+\underline{\text{ }}\right)=3 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right)$}=(-5 y-1)+\underline{\text{ }} \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{3}}{2}\right)^2=\frac{25}{36} \text{on }\text{the }\text{left }\text{and }3\times \frac{25}{36}=\frac{25}{12} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
(-5 y-1)+\frac{25}{12}=\frac{13}{12}-5 y: \\
3 \left(x^2+\frac{5 x}{3}+\frac{25}{36}\right)=\fbox{$\frac{13}{12}-5 y$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{5 x}{3}+\frac{25}{36}=\left(x+\frac{5}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \fbox{$\left(x+\frac{5}{6}\right)^2$}=\frac{13}{12}-5 y \\
\end{array}
| amps |
khanacademy | Given the equation $-9 x^2-4 x-2 y^2-10 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-10 y-9 x^2-4 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-2 y^2-10 y-9 x^2-4 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2-4 x+\underline{\text{ }}\right)+\left(-2 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2-4 x+\underline{\text{ }}\right)=-9 \left(x^2+\frac{4 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2+\frac{4 x}{9}+\underline{\text{ }}\right)$}+\left(-2 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2-10 y+\underline{\text{ }}\right)=-2 \left(y^2+5 y+\underline{\text{ }}\right): \\
-9 \left(x^2+\frac{4 x}{9}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2+5 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{9}}{2}\right)^2=\frac{4}{81} \text{on }\text{the }\text{left }\text{and }-9\times \frac{4}{81}=-\frac{4}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5-\frac{4}{9}=\frac{41}{9}: \\
-9 \left(x^2+\frac{4 x}{9}+\frac{4}{81}\right)-2 \left(y^2+5 y+\underline{\text{ }}\right)=\fbox{$\frac{41}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }-2\times \frac{25}{4}=-\frac{25}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{41}{9}-\frac{25}{2}=-\frac{143}{18}: \\
-9 \left(x^2+\frac{4 x}{9}+\frac{4}{81}\right)-2 \left(y^2+5 y+\frac{25}{4}\right)=\fbox{$-\frac{143}{18}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{4 x}{9}+\frac{4}{81}=\left(x+\frac{2}{9}\right)^2: \\
-9 \fbox{$\left(x+\frac{2}{9}\right)^2$}-2 \left(y^2+5 y+\frac{25}{4}\right)=-\frac{143}{18} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+5 y+\frac{25}{4}=\left(y+\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x+\frac{2}{9}\right)^2-2 \fbox{$\left(y+\frac{5}{2}\right)^2$}=-\frac{143}{18} \\
\end{array}
| amps |
khanacademy | Given the equation $-5 x^2-8 x+7 y^2+10 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2+10 y-5 x^2-8 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
7 y^2+10 y-5 x^2-8 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2-8 x+\underline{\text{ }}\right)+\left(7 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2-8 x+\underline{\text{ }}\right)=-5 \left(x^2+\frac{8 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2+\frac{8 x}{5}+\underline{\text{ }}\right)$}+\left(7 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2+10 y+\underline{\text{ }}\right)=7 \left(y^2+\frac{10 y}{7}+\underline{\text{ }}\right): \\
-5 \left(x^2+\frac{8 x}{5}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2+\frac{10 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{5}}{2}\right)^2=\frac{16}{25} \text{on }\text{the }\text{left }\text{and }-5\times \frac{16}{25}=-\frac{16}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-5-\frac{16}{5}=-\frac{41}{5}: \\
-5 \left(x^2+\frac{8 x}{5}+\frac{16}{25}\right)+7 \left(y^2+\frac{10 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{41}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{10}{7}}{2}\right)^2=\frac{25}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{25}{49}=\frac{25}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{25}{7}-\frac{41}{5}=-\frac{162}{35}: \\
-5 \left(x^2+\frac{8 x}{5}+\frac{16}{25}\right)+7 \left(y^2+\frac{10 y}{7}+\frac{25}{49}\right)=\fbox{$-\frac{162}{35}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{8 x}{5}+\frac{16}{25}=\left(x+\frac{4}{5}\right)^2: \\
-5 \fbox{$\left(x+\frac{4}{5}\right)^2$}+7 \left(y^2+\frac{10 y}{7}+\frac{25}{49}\right)=-\frac{162}{35} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{10 y}{7}+\frac{25}{49}=\left(y+\frac{5}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 \left(x+\frac{4}{5}\right)^2+7 \fbox{$\left(y+\frac{5}{7}\right)^2$}=-\frac{162}{35} \\
\end{array}
| amps |
khanacademy | Given the equation $-10 x^2+3 x+8 y^2-9 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-9 y-10 x^2+3 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
8 y^2-9 y-10 x^2+3 x=-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-10 x^2+3 x+\underline{\text{ }}\right)+\left(8 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2+3 x+\underline{\text{ }}\right)=-10 \left(x^2-\frac{3 x}{10}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2-\frac{3 x}{10}+\underline{\text{ }}\right)$}+\left(8 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-9 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{9 y}{8}+\underline{\text{ }}\right): \\
-10 \left(x^2-\frac{3 x}{10}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{9 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}-10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{10}}{2}\right)^2=\frac{9}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{9}{400}=-\frac{9}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-10-\frac{9}{40}=-\frac{409}{40}: \\
-10 \left(x^2-\frac{3 x}{10}+\frac{9}{400}\right)+8 \left(y^2-\frac{9 y}{8}+\underline{\text{ }}\right)=\fbox{$-\frac{409}{40}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{8}}{2}\right)^2=\frac{81}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{81}{256}=\frac{81}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{81}{32}-\frac{409}{40}=-\frac{1231}{160}: \\
-10 \left(x^2-\frac{3 x}{10}+\frac{9}{400}\right)+8 \left(y^2-\frac{9 y}{8}+\frac{81}{256}\right)=\fbox{$-\frac{1231}{160}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{10}+\frac{9}{400}=\left(x-\frac{3}{20}\right)^2: \\
-10 \fbox{$\left(x-\frac{3}{20}\right)^2$}+8 \left(y^2-\frac{9 y}{8}+\frac{81}{256}\right)=-\frac{1231}{160} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{9 y}{8}+\frac{81}{256}=\left(y-\frac{9}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x-\frac{3}{20}\right)^2+8 \fbox{$\left(y-\frac{9}{16}\right)^2$}=-\frac{1231}{160} \\
\end{array}
| amps |
khanacademy | Given the equation $-x^2-x+8 y^2+5 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2+5 y-x^2-x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
8 y^2+5 y-x^2-x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2-x+\underline{\text{ }}\right)+\left(8 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2-x+\underline{\text{ }}\right)=-\left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2+x+\underline{\text{ }}\right)$}+\left(8 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2+5 y+\underline{\text{ }}\right)=8 \left(y^2+\frac{5 y}{8}+\underline{\text{ }}\right): \\
-\left(x^2+x+\underline{\text{ }}\right)+\fbox{$8 \left(y^2+\frac{5 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-1}{4}=-\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6-\frac{1}{4}=\frac{23}{4}: \\
-\left(x^2+x+\frac{1}{4}\right)+8 \left(y^2+\frac{5 y}{8}+\underline{\text{ }}\right)=\fbox{$\frac{23}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{8}}{2}\right)^2=\frac{25}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{25}{256}=\frac{25}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{23}{4}+\frac{25}{32}=\frac{209}{32}: \\
-\left(x^2+x+\frac{1}{4}\right)+8 \left(y^2+\frac{5 y}{8}+\frac{25}{256}\right)=\fbox{$\frac{209}{32}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
-\fbox{$\left(x+\frac{1}{2}\right)^2$}+8 \left(y^2+\frac{5 y}{8}+\frac{25}{256}\right)=\frac{209}{32} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{8}+\frac{25}{256}=\left(y+\frac{5}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -\left(x+\frac{1}{2}\right)^2+8 \fbox{$\left(y+\frac{5}{16}\right)^2$}=\frac{209}{32} \\
\end{array}
| amps |
khanacademy | Given the equation $-5 x^2-9 x-y^2-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 x^2-9 x+\left(-y^2-1\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }y^2+5 x^2+9 x+1 \text{to }\text{both }\text{sides}: \\
5 x^2+9 x+\left(y^2+1\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
y^2+5 x^2+9 x=-1 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(5 x^2+9 x+\underline{\text{ }}\right)+y^2=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 x^2+9 x+\underline{\text{ }}\right)=5 \left(x^2+\frac{9 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+\frac{9 x}{5}+\underline{\text{ }}\right)$}+y^2=\underline{\text{ }}-1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{9}{5}}{2}\right)^2=\frac{81}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{81}{100}=\frac{81}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{81}{20}-1=\frac{61}{20}: \\
5 \left(x^2+\frac{9 x}{5}+\frac{81}{100}\right)+y^2=\fbox{$\frac{61}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{9 x}{5}+\frac{81}{100}=\left(x+\frac{9}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$\left(x+\frac{9}{10}\right)^2$}+y^2=\frac{61}{20} \\
\end{array}
| amps |
khanacademy | Given the equation $6 x^2+4 x+10 y^2+2 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2+2 y+6 x^2+4 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
10 y^2+2 y+6 x^2+4 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(6 x^2+4 x+\underline{\text{ }}\right)+\left(10 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(6 x^2+4 x+\underline{\text{ }}\right)=6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(10 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2+2 y+\underline{\text{ }}\right)=10 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right): \\
6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$10 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{6}{9}=\frac{2}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10+\frac{2}{3}=\frac{32}{3}: \\
6 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+10 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{32}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{10}{100}=\frac{1}{10} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{32}{3}+\frac{1}{10}=\frac{323}{30}: \\
6 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+10 \left(y^2+\frac{y}{5}+\frac{1}{100}\right)=\fbox{$\frac{323}{30}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{3}+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2: \\
6 \fbox{$\left(x+\frac{1}{3}\right)^2$}+10 \left(y^2+\frac{y}{5}+\frac{1}{100}\right)=\frac{323}{30} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{5}+\frac{1}{100}=\left(y+\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x+\frac{1}{3}\right)^2+\text{10 }\fbox{$\left(y+\frac{1}{10}\right)^2$}=\frac{323}{30} \\
\end{array}
| amps |
khanacademy | Given the equation $3 x^2-3 x-10 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 x^2-3 x+(6-10 y)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6-10 y \text{from }\text{both }\text{sides}: \\
3 x^2-3 x=10 y-6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(3 x^2-3 x+\underline{\text{ }}\right)=(10 y-6)+\underline{\text{ }} \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2-3 x+\underline{\text{ }}\right)=3 \left(x^2-x+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2-x+\underline{\text{ }}\right)$}=(10 y-6)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{3}{4}=\frac{3}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
(10 y-6)+\frac{3}{4}=10 y-\frac{21}{4}: \\
3 \left(x^2-x+\frac{1}{4}\right)=\fbox{$10 y-\frac{21}{4}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \fbox{$\left(x-\frac{1}{2}\right)^2$}=10 y-\frac{21}{4} \\
\end{array}
| amps |
khanacademy | Given the equation $-2 x^2-5 x+2 y^2-9 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-9 y-2 x^2-5 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
2 y^2-9 y-2 x^2-5 x=-3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-2 x^2-5 x+\underline{\text{ }}\right)+\left(2 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 x^2-5 x+\underline{\text{ }}\right)=-2 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$-2 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)$}+\left(2 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2-9 y+\underline{\text{ }}\right)=2 \left(y^2-\frac{9 y}{2}+\underline{\text{ }}\right): \\
-2 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-\frac{9 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }-2\times \frac{25}{16}=-\frac{25}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-3-\frac{25}{8}=-\frac{49}{8}: \\
-2 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)+2 \left(y^2-\frac{9 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{49}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{2}}{2}\right)^2=\frac{81}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{81}{16}=\frac{81}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{81}{8}-\frac{49}{8}=4: \\
-2 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)+2 \left(y^2-\frac{9 y}{2}+\frac{81}{16}\right)=\fbox{$4$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{2}+\frac{25}{16}=\left(x+\frac{5}{4}\right)^2: \\
-2 \fbox{$\left(x+\frac{5}{4}\right)^2$}+2 \left(y^2-\frac{9 y}{2}+\frac{81}{16}\right)=4 \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{9 y}{2}+\frac{81}{16}=\left(y-\frac{9}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -2 \left(x+\frac{5}{4}\right)^2+2 \fbox{$\left(y-\frac{9}{4}\right)^2$}=4 \\
\end{array}
| amps |
khanacademy | Given the equation $x^2+4 x-8 y^2+7 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2+7 y+x^2+4 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
-8 y^2+7 y+x^2+4 x=4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(x^2+4 x+\underline{\text{ }}\right)+\left(-8 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 y^2+7 y+\underline{\text{ }}\right)=-8 \left(y^2-\frac{7 y}{8}+\underline{\text{ }}\right): \\
\left(x^2+4 x+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2-\frac{7 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{4}{2}\right)^2=4 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
4+4=8: \\
\left(x^2+4 x+4\right)-8 \left(y^2-\frac{7 y}{8}+\underline{\text{ }}\right)=\fbox{$8$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{8}}{2}\right)^2=\frac{49}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{49}{256}=-\frac{49}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 8:
\begin{array}{l}
8-\frac{49}{32}=\frac{207}{32}: \\
\left(x^2+4 x+4\right)-8 \left(y^2-\frac{7 y}{8}+\frac{49}{256}\right)=\fbox{$\frac{207}{32}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+4 x+4=(x+2)^2: \\
\fbox{$(x+2)^2$}-8 \left(y^2-\frac{7 y}{8}+\frac{49}{256}\right)=\frac{207}{32} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-\frac{7 y}{8}+\frac{49}{256}=\left(y-\frac{7}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & (x+2)^2-8 \fbox{$\left(y-\frac{7}{16}\right)^2$}=\frac{207}{32} \\
\end{array}
| amps |
khanacademy | Given the equation $8 x^2-5 x-5 y^2+8 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+8 y+8 x^2-5 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
-5 y^2+8 y+8 x^2-5 x=-8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(8 x^2-5 x+\underline{\text{ }}\right)+\left(-5 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2-5 x+\underline{\text{ }}\right)=8 \left(x^2-\frac{5 x}{8}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2-\frac{5 x}{8}+\underline{\text{ }}\right)$}+\left(-5 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+8 y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{8 y}{5}+\underline{\text{ }}\right): \\
8 \left(x^2-\frac{5 x}{8}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{8 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{8}}{2}\right)^2=\frac{25}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{25}{256}=\frac{25}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{25}{32}-8=-\frac{231}{32}: \\
8 \left(x^2-\frac{5 x}{8}+\frac{25}{256}\right)-5 \left(y^2-\frac{8 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{231}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-8}{5}}{2}\right)^2=\frac{16}{25} \text{on }\text{the }\text{left }\text{and }-5\times \frac{16}{25}=-\frac{16}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{231}{32}-\frac{16}{5}=-\frac{1667}{160}: \\
8 \left(x^2-\frac{5 x}{8}+\frac{25}{256}\right)-5 \left(y^2-\frac{8 y}{5}+\frac{16}{25}\right)=\fbox{$-\frac{1667}{160}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{8}+\frac{25}{256}=\left(x-\frac{5}{16}\right)^2: \\
8 \fbox{$\left(x-\frac{5}{16}\right)^2$}-5 \left(y^2-\frac{8 y}{5}+\frac{16}{25}\right)=-\frac{1667}{160} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{8 y}{5}+\frac{16}{25}=\left(y-\frac{4}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \left(x-\frac{5}{16}\right)^2-5 \fbox{$\left(y-\frac{4}{5}\right)^2$}=-\frac{1667}{160} \\
\end{array}
| amps |
khanacademy | Given the equation $4 x^2-x-2 y^2+7 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+7 y+4 x^2-x=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2-x+\underline{\text{ }}\right)+\left(-2 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(4 x^2-x+\underline{\text{ }}\right)=4 \left(x^2-\frac{x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2-\frac{x}{4}+\underline{\text{ }}\right)$}+\left(-2 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 y^2+7 y+\underline{\text{ }}\right)=-2 \left(y^2-\frac{7 y}{2}+\underline{\text{ }}\right): \\
4 \left(x^2-\frac{x}{4}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-\frac{7 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+0 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{4}{64}=\frac{1}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{2}}{2}\right)^2=\frac{49}{16} \text{on }\text{the }\text{left }\text{and }-2\times \frac{49}{16}=-\frac{49}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{1}{16}-\frac{49}{8}=-\frac{97}{16}: \\
4 \left(x^2-\frac{x}{4}+\frac{1}{64}\right)-2 \left(y^2-\frac{7 y}{2}+\frac{49}{16}\right)=\fbox{$-\frac{97}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{x}{4}+\frac{1}{64}=\left(x-\frac{1}{8}\right)^2: \\
4 \fbox{$\left(x-\frac{1}{8}\right)^2$}-2 \left(y^2-\frac{7 y}{2}+\frac{49}{16}\right)=-\frac{97}{16} \\
\end{array}
Step 9:
\begin{array}{l}
y^2-\frac{7 y}{2}+\frac{49}{16}=\left(y-\frac{7}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x-\frac{1}{8}\right)^2-2 \fbox{$\left(y-\frac{7}{4}\right)^2$}=-\frac{97}{16} \\
\end{array}
| amps |
khanacademy | Given the equation $3 x^2-10 x-5 y^2+9 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+9 y+3 x^2-10 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-5 y^2+9 y+3 x^2-10 x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2-10 x+\underline{\text{ }}\right)+\left(-5 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2-10 x+\underline{\text{ }}\right)=3 \left(x^2-\frac{10 x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2-\frac{10 x}{3}+\underline{\text{ }}\right)$}+\left(-5 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+9 y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{9 y}{5}+\underline{\text{ }}\right): \\
3 \left(x^2-\frac{10 x}{3}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{9 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-10}{3}}{2}\right)^2=\frac{25}{9} \text{on }\text{the }\text{left }\text{and }3\times \frac{25}{9}=\frac{25}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{25}{3}-7=\frac{4}{3}: \\
3 \left(x^2-\frac{10 x}{3}+\frac{25}{9}\right)-5 \left(y^2-\frac{9 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{4}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{5}}{2}\right)^2=\frac{81}{100} \text{on }\text{the }\text{left }\text{and }-5\times \frac{81}{100}=-\frac{81}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{4}{3}-\frac{81}{20}=-\frac{163}{60}: \\
3 \left(x^2-\frac{10 x}{3}+\frac{25}{9}\right)-5 \left(y^2-\frac{9 y}{5}+\frac{81}{100}\right)=\fbox{$-\frac{163}{60}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{10 x}{3}+\frac{25}{9}=\left(x-\frac{5}{3}\right)^2: \\
3 \fbox{$\left(x-\frac{5}{3}\right)^2$}-5 \left(y^2-\frac{9 y}{5}+\frac{81}{100}\right)=-\frac{163}{60} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{9 y}{5}+\frac{81}{100}=\left(y-\frac{9}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x-\frac{5}{3}\right)^2-5 \fbox{$\left(y-\frac{9}{10}\right)^2$}=-\frac{163}{60} \\
\end{array}
| amps |
khanacademy | Given the equation $-10 x^2-2 y^2-4 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-4 y+\left(-10 x^2-2\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 y^2+4 y+10 x^2+2 \text{to }\text{both }\text{sides}: \\
2 y^2+4 y+\left(10 x^2+2\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
2 y^2+4 y+10 x^2=-2 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(2 y^2+4 y+\underline{\text{ }}\right)+10 x^2=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+4 y+\underline{\text{ }}\right)=2 \left(y^2+2 y+\underline{\text{ }}\right): \\
\fbox{$2 \left(y^2+2 y+\underline{\text{ }}\right)$}+10 x^2=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }2\times 1=2 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2-2=0: \\
2 \left(y^2+2 y+1\right)+10 x^2=\fbox{$0$} \\
\end{array}
Step 8:
\begin{array}{l}
y^2+2 y+1=(y+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \fbox{$(y+1)^2$}+10 x^2=0 \\
\end{array}
| amps |
khanacademy | Given the equation $-8 x^2-x+3 y^2+y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2+y-8 x^2-x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
3 y^2+y-8 x^2-x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2-x+\underline{\text{ }}\right)+\left(3 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2-x+\underline{\text{ }}\right)=-8 \left(x^2+\frac{x}{8}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2+\frac{x}{8}+\underline{\text{ }}\right)$}+\left(3 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 y^2+y+\underline{\text{ }}\right)=3 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right): \\
-8 \left(x^2+\frac{x}{8}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{8}}{2}\right)^2=\frac{1}{256} \text{on }\text{the }\text{left }\text{and }\frac{-8}{256}=-\frac{1}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9-\frac{1}{32}=\frac{287}{32}: \\
-8 \left(x^2+\frac{x}{8}+\frac{1}{256}\right)+3 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{287}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{3}{36}=\frac{1}{12} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{287}{32}+\frac{1}{12}=\frac{869}{96}: \\
-8 \left(x^2+\frac{x}{8}+\frac{1}{256}\right)+3 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=\fbox{$\frac{869}{96}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{8}+\frac{1}{256}=\left(x+\frac{1}{16}\right)^2: \\
-8 \fbox{$\left(x+\frac{1}{16}\right)^2$}+3 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=\frac{869}{96} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{3}+\frac{1}{36}=\left(y+\frac{1}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x+\frac{1}{16}\right)^2+3 \fbox{$\left(y+\frac{1}{6}\right)^2$}=\frac{869}{96} \\
\end{array}
| amps |
khanacademy | Given the equation $4 x^2+5 x+7 y^2-9 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2-9 y+4 x^2+5 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
7 y^2-9 y+4 x^2+5 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2+5 x+\underline{\text{ }}\right)+\left(7 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2+5 x+\underline{\text{ }}\right)=4 \left(x^2+\frac{5 x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2+\frac{5 x}{4}+\underline{\text{ }}\right)$}+\left(7 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2-9 y+\underline{\text{ }}\right)=7 \left(y^2-\frac{9 y}{7}+\underline{\text{ }}\right): \\
4 \left(x^2+\frac{5 x}{4}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2-\frac{9 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{4}}{2}\right)^2=\frac{25}{64} \text{on }\text{the }\text{left }\text{and }4\times \frac{25}{64}=\frac{25}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{25}{16}-5=-\frac{55}{16}: \\
4 \left(x^2+\frac{5 x}{4}+\frac{25}{64}\right)+7 \left(y^2-\frac{9 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{55}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{7}}{2}\right)^2=\frac{81}{196} \text{on }\text{the }\text{left }\text{and }7\times \frac{81}{196}=\frac{81}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{81}{28}-\frac{55}{16}=-\frac{61}{112}: \\
4 \left(x^2+\frac{5 x}{4}+\frac{25}{64}\right)+7 \left(y^2-\frac{9 y}{7}+\frac{81}{196}\right)=\fbox{$-\frac{61}{112}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{4}+\frac{25}{64}=\left(x+\frac{5}{8}\right)^2: \\
4 \fbox{$\left(x+\frac{5}{8}\right)^2$}+7 \left(y^2-\frac{9 y}{7}+\frac{81}{196}\right)=-\frac{61}{112} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{9 y}{7}+\frac{81}{196}=\left(y-\frac{9}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x+\frac{5}{8}\right)^2+7 \fbox{$\left(y-\frac{9}{14}\right)^2$}=-\frac{61}{112} \\
\end{array}
| amps |
khanacademy | Given the equation $5 x^2+3 x+9 y^2+4 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+4 y+5 x^2+3 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
9 y^2+4 y+5 x^2+3 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2+3 x+\underline{\text{ }}\right)+\left(9 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2+3 x+\underline{\text{ }}\right)=5 \left(x^2+\frac{3 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+\frac{3 x}{5}+\underline{\text{ }}\right)$}+\left(9 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2+4 y+\underline{\text{ }}\right)=9 \left(y^2+\frac{4 y}{9}+\underline{\text{ }}\right): \\
5 \left(x^2+\frac{3 x}{5}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2+\frac{4 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{5}}{2}\right)^2=\frac{9}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{9}{100}=\frac{9}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{20}-2=-\frac{31}{20}: \\
5 \left(x^2+\frac{3 x}{5}+\frac{9}{100}\right)+9 \left(y^2+\frac{4 y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{31}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{9}}{2}\right)^2=\frac{4}{81} \text{on }\text{the }\text{left }\text{and }9\times \frac{4}{81}=\frac{4}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{4}{9}-\frac{31}{20}=-\frac{199}{180}: \\
5 \left(x^2+\frac{3 x}{5}+\frac{9}{100}\right)+9 \left(y^2+\frac{4 y}{9}+\frac{4}{81}\right)=\fbox{$-\frac{199}{180}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{5}+\frac{9}{100}=\left(x+\frac{3}{10}\right)^2: \\
5 \fbox{$\left(x+\frac{3}{10}\right)^2$}+9 \left(y^2+\frac{4 y}{9}+\frac{4}{81}\right)=-\frac{199}{180} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{4 y}{9}+\frac{4}{81}=\left(y+\frac{2}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x+\frac{3}{10}\right)^2+9 \fbox{$\left(y+\frac{2}{9}\right)^2$}=-\frac{199}{180} \\
\end{array}
| amps |
khanacademy | Given the equation $-2 x^2-10 x-3 y^2-3 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2-3 y-2 x^2-10 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
-3 y^2-3 y-2 x^2-10 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-2 x^2-10 x+\underline{\text{ }}\right)+\left(-3 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 x^2-10 x+\underline{\text{ }}\right)=-2 \left(x^2+5 x+\underline{\text{ }}\right): \\
\fbox{$-2 \left(x^2+5 x+\underline{\text{ }}\right)$}+\left(-3 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2-3 y+\underline{\text{ }}\right)=-3 \left(y^2+y+\underline{\text{ }}\right): \\
-2 \left(x^2+5 x+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2+y+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }-2\times \frac{25}{4}=-\frac{25}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-2-\frac{25}{2}=-\frac{29}{2}: \\
-2 \left(x^2+5 x+\frac{25}{4}\right)-3 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$-\frac{29}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-3}{4}=-\frac{3}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{29}{2}-\frac{3}{4}=-\frac{61}{4}: \\
-2 \left(x^2+5 x+\frac{25}{4}\right)-3 \left(y^2+y+\frac{1}{4}\right)=\fbox{$-\frac{61}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+5 x+\frac{25}{4}=\left(x+\frac{5}{2}\right)^2: \\
-2 \fbox{$\left(x+\frac{5}{2}\right)^2$}-3 \left(y^2+y+\frac{1}{4}\right)=-\frac{61}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+y+\frac{1}{4}=\left(y+\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -2 \left(x+\frac{5}{2}\right)^2-3 \fbox{$\left(y+\frac{1}{2}\right)^2$}=-\frac{61}{4} \\
\end{array}
| amps |
khanacademy | Given the equation $5 x^2+10 x-6 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 x^2+10 x+(4-6 y)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4-6 y \text{from }\text{both }\text{sides}: \\
5 x^2+10 x=6 y-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(5 x^2+10 x+\underline{\text{ }}\right)=(6 y-4)+\underline{\text{ }} \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2+10 x+\underline{\text{ }}\right)=5 \left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+2 x+\underline{\text{ }}\right)$}=(6 y-4)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }5\times 1=5 \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
(6 y-4)+5=6 y+1: \\
5 \left(x^2+2 x+1\right)=\fbox{$6 y+1$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$(x+1)^2$}=6 y+1 \\
\end{array}
| amps |
khanacademy | Given the equation $7 x^2+8 x-5 y^2+5 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+5 y+7 x^2+8 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-5 y^2+5 y+7 x^2+8 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(7 x^2+8 x+\underline{\text{ }}\right)+\left(-5 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2+8 x+\underline{\text{ }}\right)=7 \left(x^2+\frac{8 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2+\frac{8 x}{7}+\underline{\text{ }}\right)$}+\left(-5 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+5 y+\underline{\text{ }}\right)=-5 \left(y^2-y+\underline{\text{ }}\right): \\
7 \left(x^2+\frac{8 x}{7}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-y+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{7}}{2}\right)^2=\frac{16}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{16}{49}=\frac{16}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{16}{7}-5=-\frac{19}{7}: \\
7 \left(x^2+\frac{8 x}{7}+\frac{16}{49}\right)-5 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$-\frac{19}{7}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-5}{4}=-\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{19}{7}-\frac{5}{4}=-\frac{111}{28}: \\
7 \left(x^2+\frac{8 x}{7}+\frac{16}{49}\right)-5 \left(y^2-y+\frac{1}{4}\right)=\fbox{$-\frac{111}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{8 x}{7}+\frac{16}{49}=\left(x+\frac{4}{7}\right)^2: \\
7 \fbox{$\left(x+\frac{4}{7}\right)^2$}-5 \left(y^2-y+\frac{1}{4}\right)=-\frac{111}{28} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x+\frac{4}{7}\right)^2-5 \fbox{$\left(y-\frac{1}{2}\right)^2$}=-\frac{111}{28} \\
\end{array}
| amps |
khanacademy | Given the equation $-8 x^2+9 x+5 y^2+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 x^2+9 x+\left(5 y^2+7\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 y^2-8 x^2+9 x+7 \text{from }\text{both }\text{sides}: \\
8 x^2-9 x+\left(-5 y^2-7\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
-5 y^2+8 x^2-9 x=7 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(8 x^2-9 x+\underline{\text{ }}\right)-5 y^2=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 x^2-9 x+\underline{\text{ }}\right)=8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right)$}-5 y^2=\underline{\text{ }}+7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{8}}{2}\right)^2=\frac{81}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{81}{256}=\frac{81}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
7+\frac{81}{32}=\frac{305}{32}: \\
8 \left(x^2-\frac{9 x}{8}+\frac{81}{256}\right)-5 y^2=\fbox{$\frac{305}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{9 x}{8}+\frac{81}{256}=\left(x-\frac{9}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \fbox{$\left(x-\frac{9}{16}\right)^2$}-5 y^2=\frac{305}{32} \\
\end{array}
| amps |
khanacademy | Given the equation $-5 x^2-2 x+9 y^2-5 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2-5 y-5 x^2-2 x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
9 y^2-5 y-5 x^2-2 x=7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2-2 x+\underline{\text{ }}\right)+\left(9 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2-2 x+\underline{\text{ }}\right)=-5 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right)$}+\left(9 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2-5 y+\underline{\text{ }}\right)=9 \left(y^2-\frac{5 y}{9}+\underline{\text{ }}\right): \\
-5 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2-\frac{5 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}+7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{5}}{2}\right)^2=\frac{1}{25} \text{on }\text{the }\text{left }\text{and }\frac{-5}{25}=-\frac{1}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
7-\frac{1}{5}=\frac{34}{5}: \\
-5 \left(x^2+\frac{2 x}{5}+\frac{1}{25}\right)+9 \left(y^2-\frac{5 y}{9}+\underline{\text{ }}\right)=\fbox{$\frac{34}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{9}}{2}\right)^2=\frac{25}{324} \text{on }\text{the }\text{left }\text{and }9\times \frac{25}{324}=\frac{25}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{34}{5}+\frac{25}{36}=\frac{1349}{180}: \\
-5 \left(x^2+\frac{2 x}{5}+\frac{1}{25}\right)+9 \left(y^2-\frac{5 y}{9}+\frac{25}{324}\right)=\fbox{$\frac{1349}{180}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{5}+\frac{1}{25}=\left(x+\frac{1}{5}\right)^2: \\
-5 \fbox{$\left(x+\frac{1}{5}\right)^2$}+9 \left(y^2-\frac{5 y}{9}+\frac{25}{324}\right)=\frac{1349}{180} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{5 y}{9}+\frac{25}{324}=\left(y-\frac{5}{18}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 \left(x+\frac{1}{5}\right)^2+9 \fbox{$\left(y-\frac{5}{18}\right)^2$}=\frac{1349}{180} \\
\end{array}
| amps |
khanacademy | Given the equation $6 x^2-9 y^2+5 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2+5 y+\left(6 x^2+7\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-9 y^2+5 y+6 x^2+7 \text{from }\text{both }\text{sides}: \\
9 y^2-5 y+\left(-6 x^2-7\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
9 y^2-5 y-6 x^2=7 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(9 y^2-5 y+\underline{\text{ }}\right)-6 x^2=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2-5 y+\underline{\text{ }}\right)=9 \left(y^2-\frac{5 y}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(y^2-\frac{5 y}{9}+\underline{\text{ }}\right)$}-6 x^2=\underline{\text{ }}+7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{9}}{2}\right)^2=\frac{25}{324} \text{on }\text{the }\text{left }\text{and }9\times \frac{25}{324}=\frac{25}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
7+\frac{25}{36}=\frac{277}{36}: \\
9 \left(y^2-\frac{5 y}{9}+\frac{25}{324}\right)-6 x^2=\fbox{$\frac{277}{36}$} \\
\end{array}
Step 8:
\begin{array}{l}
y^2-\frac{5 y}{9}+\frac{25}{324}=\left(y-\frac{5}{18}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \fbox{$\left(y-\frac{5}{18}\right)^2$}-6 x^2=\frac{277}{36} \\
\end{array}
| amps |
khanacademy | Given the equation $-x^2-7 x-10 y^2-3 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-3 y-x^2-7 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
-10 y^2-3 y-x^2-7 x=4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2-7 x+\underline{\text{ }}\right)+\left(-10 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2-7 x+\underline{\text{ }}\right)=-\left(x^2+7 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2+7 x+\underline{\text{ }}\right)$}+\left(-10 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2-3 y+\underline{\text{ }}\right)=-10 \left(y^2+\frac{3 y}{10}+\underline{\text{ }}\right): \\
-\left(x^2+7 x+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2+\frac{3 y}{10}+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{7}{2}\right)^2=\frac{49}{4} \text{on }\text{the }\text{left }\text{and }-\frac{49}{4}=-\frac{49}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
4-\frac{49}{4}=-\frac{33}{4}: \\
-\left(x^2+7 x+\frac{49}{4}\right)-10 \left(y^2+\frac{3 y}{10}+\underline{\text{ }}\right)=\fbox{$-\frac{33}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{10}}{2}\right)^2=\frac{9}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{9}{400}=-\frac{9}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{33}{4}-\frac{9}{40}=-\frac{339}{40}: \\
-\left(x^2+7 x+\frac{49}{4}\right)-10 \left(y^2+\frac{3 y}{10}+\frac{9}{400}\right)=\fbox{$-\frac{339}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+7 x+\frac{49}{4}=\left(x+\frac{7}{2}\right)^2: \\
-\fbox{$\left(x+\frac{7}{2}\right)^2$}-10 \left(y^2+\frac{3 y}{10}+\frac{9}{400}\right)=-\frac{339}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{10}+\frac{9}{400}=\left(y+\frac{3}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -\left(x+\frac{7}{2}\right)^2-\text{10 }\fbox{$\left(y+\frac{3}{20}\right)^2$}=-\frac{339}{40} \\
\end{array}
| amps |
khanacademy | Given the equation $-9 x^2-x-6 y^2-8 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-8 y-9 x^2-x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-6 y^2-8 y-9 x^2-x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2-x+\underline{\text{ }}\right)+\left(-6 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2-x+\underline{\text{ }}\right)=-9 \left(x^2+\frac{x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2+\frac{x}{9}+\underline{\text{ }}\right)$}+\left(-6 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2-8 y+\underline{\text{ }}\right)=-6 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right): \\
-9 \left(x^2+\frac{x}{9}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{9}}{2}\right)^2=\frac{1}{324} \text{on }\text{the }\text{left }\text{and }\frac{-9}{324}=-\frac{1}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-1-\frac{1}{36}=-\frac{37}{36}: \\
-9 \left(x^2+\frac{x}{9}+\frac{1}{324}\right)-6 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{37}{36}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{3}}{2}\right)^2=\frac{4}{9} \text{on }\text{the }\text{left }\text{and }-6\times \frac{4}{9}=-\frac{8}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{37}{36}-\frac{8}{3}=-\frac{133}{36}: \\
-9 \left(x^2+\frac{x}{9}+\frac{1}{324}\right)-6 \left(y^2+\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$-\frac{133}{36}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{9}+\frac{1}{324}=\left(x+\frac{1}{18}\right)^2: \\
-9 \fbox{$\left(x+\frac{1}{18}\right)^2$}-6 \left(y^2+\frac{4 y}{3}+\frac{4}{9}\right)=-\frac{133}{36} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{4 y}{3}+\frac{4}{9}=\left(y+\frac{2}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x+\frac{1}{18}\right)^2-6 \fbox{$\left(y+\frac{2}{3}\right)^2$}=-\frac{133}{36} \\
\end{array}
| amps |
khanacademy | Given the equation $-6 x^2-5 x+4 y^2-y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2-y-6 x^2-5 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
4 y^2-y-6 x^2-5 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2-5 x+\underline{\text{ }}\right)+\left(4 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2-5 x+\underline{\text{ }}\right)=-6 \left(x^2+\frac{5 x}{6}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2+\frac{5 x}{6}+\underline{\text{ }}\right)$}+\left(4 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2-y+\underline{\text{ }}\right)=4 \left(y^2-\frac{y}{4}+\underline{\text{ }}\right): \\
-6 \left(x^2+\frac{5 x}{6}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2-\frac{y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{6}}{2}\right)^2=\frac{25}{144} \text{on }\text{the }\text{left }\text{and }-6\times \frac{25}{144}=-\frac{25}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5-\frac{25}{24}=\frac{95}{24}: \\
-6 \left(x^2+\frac{5 x}{6}+\frac{25}{144}\right)+4 \left(y^2-\frac{y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{95}{24}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{4}{64}=\frac{1}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{95}{24}+\frac{1}{16}=\frac{193}{48}: \\
-6 \left(x^2+\frac{5 x}{6}+\frac{25}{144}\right)+4 \left(y^2-\frac{y}{4}+\frac{1}{64}\right)=\fbox{$\frac{193}{48}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{6}+\frac{25}{144}=\left(x+\frac{5}{12}\right)^2: \\
-6 \fbox{$\left(x+\frac{5}{12}\right)^2$}+4 \left(y^2-\frac{y}{4}+\frac{1}{64}\right)=\frac{193}{48} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{4}+\frac{1}{64}=\left(y-\frac{1}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x+\frac{5}{12}\right)^2+4 \fbox{$\left(y-\frac{1}{8}\right)^2$}=\frac{193}{48} \\
\end{array}
| amps |
khanacademy | Given the equation $-6 x^2-8 x+9 y^2+6 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+6 y-6 x^2-8 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
9 y^2+6 y-6 x^2-8 x=-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2-8 x+\underline{\text{ }}\right)+\left(9 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2-8 x+\underline{\text{ }}\right)=-6 \left(x^2+\frac{4 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2+\frac{4 x}{3}+\underline{\text{ }}\right)$}+\left(9 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2+6 y+\underline{\text{ }}\right)=9 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right): \\
-6 \left(x^2+\frac{4 x}{3}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{3}}{2}\right)^2=\frac{4}{9} \text{on }\text{the }\text{left }\text{and }-6\times \frac{4}{9}=-\frac{8}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-10-\frac{8}{3}=-\frac{38}{3}: \\
-6 \left(x^2+\frac{4 x}{3}+\frac{4}{9}\right)+9 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{38}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{9}{9}=1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
1-\frac{38}{3}=-\frac{35}{3}: \\
-6 \left(x^2+\frac{4 x}{3}+\frac{4}{9}\right)+9 \left(y^2+\frac{2 y}{3}+\frac{1}{9}\right)=\fbox{$-\frac{35}{3}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{4 x}{3}+\frac{4}{9}=\left(x+\frac{2}{3}\right)^2: \\
-6 \fbox{$\left(x+\frac{2}{3}\right)^2$}+9 \left(y^2+\frac{2 y}{3}+\frac{1}{9}\right)=-\frac{35}{3} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{2 y}{3}+\frac{1}{9}=\left(y+\frac{1}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x+\frac{2}{3}\right)^2+9 \fbox{$\left(y+\frac{1}{3}\right)^2$}=-\frac{35}{3} \\
\end{array}
| amps |
khanacademy | Given the equation $-x^2-x-6 y^2-7 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-7 y-x^2-x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
-6 y^2-7 y-x^2-x=7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2-x+\underline{\text{ }}\right)+\left(-6 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2-x+\underline{\text{ }}\right)=-\left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2+x+\underline{\text{ }}\right)$}+\left(-6 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2-7 y+\underline{\text{ }}\right)=-6 \left(y^2+\frac{7 y}{6}+\underline{\text{ }}\right): \\
-\left(x^2+x+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{7 y}{6}+\underline{\text{ }}\right)$}=\underline{\text{ }}+7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-1}{4}=-\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
7-\frac{1}{4}=\frac{27}{4}: \\
-\left(x^2+x+\frac{1}{4}\right)-6 \left(y^2+\frac{7 y}{6}+\underline{\text{ }}\right)=\fbox{$\frac{27}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{6}}{2}\right)^2=\frac{49}{144} \text{on }\text{the }\text{left }\text{and }-6\times \frac{49}{144}=-\frac{49}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{27}{4}-\frac{49}{24}=\frac{113}{24}: \\
-\left(x^2+x+\frac{1}{4}\right)-6 \left(y^2+\frac{7 y}{6}+\frac{49}{144}\right)=\fbox{$\frac{113}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
-\fbox{$\left(x+\frac{1}{2}\right)^2$}-6 \left(y^2+\frac{7 y}{6}+\frac{49}{144}\right)=\frac{113}{24} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{7 y}{6}+\frac{49}{144}=\left(y+\frac{7}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -\left(x+\frac{1}{2}\right)^2-6 \fbox{$\left(y+\frac{7}{12}\right)^2$}=\frac{113}{24} \\
\end{array}
| amps |
khanacademy | Given the equation $2 x^2-10 x+2 y^2+6 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+6 y+2 x^2-10 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
2 y^2+6 y+2 x^2-10 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2-10 x+\underline{\text{ }}\right)+\left(2 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2-10 x+\underline{\text{ }}\right)=2 \left(x^2-5 x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-5 x+\underline{\text{ }}\right)$}+\left(2 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+6 y+\underline{\text{ }}\right)=2 \left(y^2+3 y+\underline{\text{ }}\right): \\
2 \left(x^2-5 x+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+3 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }2\times \frac{25}{4}=\frac{25}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5+\frac{25}{2}=\frac{35}{2}: \\
2 \left(x^2-5 x+\frac{25}{4}\right)+2 \left(y^2+3 y+\underline{\text{ }}\right)=\fbox{$\frac{35}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }2\times \frac{9}{4}=\frac{9}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{35}{2}+\frac{9}{2}=22: \\
2 \left(x^2-5 x+\frac{25}{4}\right)+2 \left(y^2+3 y+\frac{9}{4}\right)=\fbox{$22$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-5 x+\frac{25}{4}=\left(x-\frac{5}{2}\right)^2: \\
2 \fbox{$\left(x-\frac{5}{2}\right)^2$}+2 \left(y^2+3 y+\frac{9}{4}\right)=22 \\
\end{array}
Step 11:
\begin{array}{l}
y^2+3 y+\frac{9}{4}=\left(y+\frac{3}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x-\frac{5}{2}\right)^2+2 \fbox{$\left(y+\frac{3}{2}\right)^2$}=22 \\
\end{array}
| amps |
khanacademy | Given the equation $5 x^2-5 x+8 y^2+5 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2+5 y+5 x^2-5 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
8 y^2+5 y+5 x^2-5 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2-5 x+\underline{\text{ }}\right)+\left(8 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2-5 x+\underline{\text{ }}\right)=5 \left(x^2-x+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-x+\underline{\text{ }}\right)$}+\left(8 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2+5 y+\underline{\text{ }}\right)=8 \left(y^2+\frac{5 y}{8}+\underline{\text{ }}\right): \\
5 \left(x^2-x+\underline{\text{ }}\right)+\fbox{$8 \left(y^2+\frac{5 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{5}{4}=\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9+\frac{5}{4}=\frac{41}{4}: \\
5 \left(x^2-x+\frac{1}{4}\right)+8 \left(y^2+\frac{5 y}{8}+\underline{\text{ }}\right)=\fbox{$\frac{41}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{8}}{2}\right)^2=\frac{25}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{25}{256}=\frac{25}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{41}{4}+\frac{25}{32}=\frac{353}{32}: \\
5 \left(x^2-x+\frac{1}{4}\right)+8 \left(y^2+\frac{5 y}{8}+\frac{25}{256}\right)=\fbox{$\frac{353}{32}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2: \\
5 \fbox{$\left(x-\frac{1}{2}\right)^2$}+8 \left(y^2+\frac{5 y}{8}+\frac{25}{256}\right)=\frac{353}{32} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{8}+\frac{25}{256}=\left(y+\frac{5}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x-\frac{1}{2}\right)^2+8 \fbox{$\left(y+\frac{5}{16}\right)^2$}=\frac{353}{32} \\
\end{array}
| amps |
khanacademy | Given the equation $-6 x^2+5 x-7 y^2+3 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2+3 y-6 x^2+5 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-7 y^2+3 y-6 x^2+5 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2+5 x+\underline{\text{ }}\right)+\left(-7 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2+5 x+\underline{\text{ }}\right)=-6 \left(x^2-\frac{5 x}{6}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2-\frac{5 x}{6}+\underline{\text{ }}\right)$}+\left(-7 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2+3 y+\underline{\text{ }}\right)=-7 \left(y^2-\frac{3 y}{7}+\underline{\text{ }}\right): \\
-6 \left(x^2-\frac{5 x}{6}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2-\frac{3 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{6}}{2}\right)^2=\frac{25}{144} \text{on }\text{the }\text{left }\text{and }-6\times \frac{25}{144}=-\frac{25}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5-\frac{25}{24}=\frac{95}{24}: \\
-6 \left(x^2-\frac{5 x}{6}+\frac{25}{144}\right)-7 \left(y^2-\frac{3 y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{95}{24}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{7}}{2}\right)^2=\frac{9}{196} \text{on }\text{the }\text{left }\text{and }-7\times \frac{9}{196}=-\frac{9}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{95}{24}-\frac{9}{28}=\frac{611}{168}: \\
-6 \left(x^2-\frac{5 x}{6}+\frac{25}{144}\right)-7 \left(y^2-\frac{3 y}{7}+\frac{9}{196}\right)=\fbox{$\frac{611}{168}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{6}+\frac{25}{144}=\left(x-\frac{5}{12}\right)^2: \\
-6 \fbox{$\left(x-\frac{5}{12}\right)^2$}-7 \left(y^2-\frac{3 y}{7}+\frac{9}{196}\right)=\frac{611}{168} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{7}+\frac{9}{196}=\left(y-\frac{3}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x-\frac{5}{12}\right)^2-7 \fbox{$\left(y-\frac{3}{14}\right)^2$}=\frac{611}{168} \\
\end{array}
| amps |
khanacademy | Given the equation $x^2+3 x-10 y^2+5 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+5 y+x^2+3 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
-10 y^2+5 y+x^2+3 x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(x^2+3 x+\underline{\text{ }}\right)+\left(-10 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 y^2+5 y+\underline{\text{ }}\right)=-10 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right): \\
\left(x^2+3 x+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{3}{2}\right)^2=\frac{9}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
\frac{9}{4}-4=-\frac{7}{4}: \\
\left(x^2+3 x+\frac{9}{4}\right)-10 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{7}{4}$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{-10}{16}=-\frac{5}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 8:
\begin{array}{l}
-\frac{7}{4}-\frac{5}{8}=-\frac{19}{8}: \\
\left(x^2+3 x+\frac{9}{4}\right)-10 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\fbox{$-\frac{19}{8}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+3 x+\frac{9}{4}=\left(x+\frac{3}{2}\right)^2: \\
\fbox{$\left(x+\frac{3}{2}\right)^2$}-10 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=-\frac{19}{8} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-\frac{y}{2}+\frac{1}{16}=\left(y-\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \left(x+\frac{3}{2}\right)^2-\text{10 }\fbox{$\left(y-\frac{1}{4}\right)^2$}=-\frac{19}{8} \\
\end{array}
| amps |
khanacademy | Given the equation $-7 x^2-x-10 y^2-10 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-10 y-7 x^2-x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
-10 y^2-10 y-7 x^2-x=8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-7 x^2-x+\underline{\text{ }}\right)+\left(-10 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 x^2-x+\underline{\text{ }}\right)=-7 \left(x^2+\frac{x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2+\frac{x}{7}+\underline{\text{ }}\right)$}+\left(-10 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2-10 y+\underline{\text{ }}\right)=-10 \left(y^2+y+\underline{\text{ }}\right): \\
-7 \left(x^2+\frac{x}{7}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2+y+\underline{\text{ }}\right)$}=\underline{\text{ }}+8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{7}}{2}\right)^2=\frac{1}{196} \text{on }\text{the }\text{left }\text{and }\frac{-7}{196}=-\frac{1}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8-\frac{1}{28}=\frac{223}{28}: \\
-7 \left(x^2+\frac{x}{7}+\frac{1}{196}\right)-10 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$\frac{223}{28}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-10}{4}=-\frac{5}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{223}{28}-\frac{5}{2}=\frac{153}{28}: \\
-7 \left(x^2+\frac{x}{7}+\frac{1}{196}\right)-10 \left(y^2+y+\frac{1}{4}\right)=\fbox{$\frac{153}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{7}+\frac{1}{196}=\left(x+\frac{1}{14}\right)^2: \\
-7 \fbox{$\left(x+\frac{1}{14}\right)^2$}-10 \left(y^2+y+\frac{1}{4}\right)=\frac{153}{28} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+y+\frac{1}{4}=\left(y+\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x+\frac{1}{14}\right)^2-\text{10 }\fbox{$\left(y+\frac{1}{2}\right)^2$}=\frac{153}{28} \\
\end{array}
| amps |
khanacademy | Given the equation $3 x^2+9 x-8 y^2-7 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2-7 y+3 x^2+9 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
-8 y^2-7 y+3 x^2+9 x=4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2+9 x+\underline{\text{ }}\right)+\left(-8 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2+9 x+\underline{\text{ }}\right)=3 \left(x^2+3 x+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2+3 x+\underline{\text{ }}\right)$}+\left(-8 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2-7 y+\underline{\text{ }}\right)=-8 \left(y^2+\frac{7 y}{8}+\underline{\text{ }}\right): \\
3 \left(x^2+3 x+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2+\frac{7 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }3\times \frac{9}{4}=\frac{27}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
4+\frac{27}{4}=\frac{43}{4}: \\
3 \left(x^2+3 x+\frac{9}{4}\right)-8 \left(y^2+\frac{7 y}{8}+\underline{\text{ }}\right)=\fbox{$\frac{43}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{8}}{2}\right)^2=\frac{49}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{49}{256}=-\frac{49}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{43}{4}-\frac{49}{32}=\frac{295}{32}: \\
3 \left(x^2+3 x+\frac{9}{4}\right)-8 \left(y^2+\frac{7 y}{8}+\frac{49}{256}\right)=\fbox{$\frac{295}{32}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+3 x+\frac{9}{4}=\left(x+\frac{3}{2}\right)^2: \\
3 \fbox{$\left(x+\frac{3}{2}\right)^2$}-8 \left(y^2+\frac{7 y}{8}+\frac{49}{256}\right)=\frac{295}{32} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{7 y}{8}+\frac{49}{256}=\left(y+\frac{7}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x+\frac{3}{2}\right)^2-8 \fbox{$\left(y+\frac{7}{16}\right)^2$}=\frac{295}{32} \\
\end{array}
| amps |
khanacademy | Given the equation $-9 x^2-9 x-6 y^2-4 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-4 y-9 x^2-9 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
-6 y^2-4 y-9 x^2-9 x=-6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2-9 x+\underline{\text{ }}\right)+\left(-6 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2-9 x+\underline{\text{ }}\right)=-9 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2+x+\underline{\text{ }}\right)$}+\left(-6 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2-4 y+\underline{\text{ }}\right)=-6 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right): \\
-9 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-9}{4}=-\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-6-\frac{9}{4}=-\frac{33}{4}: \\
-9 \left(x^2+x+\frac{1}{4}\right)-6 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{33}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{-6}{9}=-\frac{2}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{33}{4}-\frac{2}{3}=-\frac{107}{12}: \\
-9 \left(x^2+x+\frac{1}{4}\right)-6 \left(y^2+\frac{2 y}{3}+\frac{1}{9}\right)=\fbox{$-\frac{107}{12}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
-9 \fbox{$\left(x+\frac{1}{2}\right)^2$}-6 \left(y^2+\frac{2 y}{3}+\frac{1}{9}\right)=-\frac{107}{12} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{2 y}{3}+\frac{1}{9}=\left(y+\frac{1}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x+\frac{1}{2}\right)^2-6 \fbox{$\left(y+\frac{1}{3}\right)^2$}=-\frac{107}{12} \\
\end{array}
| amps |
khanacademy | Given the equation $5 x^2-6 x-6 y^2-9 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-9 y+5 x^2-6 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-6 y^2-9 y+5 x^2-6 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2-6 x+\underline{\text{ }}\right)+\left(-6 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2-6 x+\underline{\text{ }}\right)=5 \left(x^2-\frac{6 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-\frac{6 x}{5}+\underline{\text{ }}\right)$}+\left(-6 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2-9 y+\underline{\text{ }}\right)=-6 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right): \\
5 \left(x^2-\frac{6 x}{5}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-6}{5}}{2}\right)^2=\frac{9}{25} \text{on }\text{the }\text{left }\text{and }5\times \frac{9}{25}=\frac{9}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{5}-9=-\frac{36}{5}: \\
5 \left(x^2-\frac{6 x}{5}+\frac{9}{25}\right)-6 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{36}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }-6\times \frac{9}{16}=-\frac{27}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{36}{5}-\frac{27}{8}=-\frac{423}{40}: \\
5 \left(x^2-\frac{6 x}{5}+\frac{9}{25}\right)-6 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$-\frac{423}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{6 x}{5}+\frac{9}{25}=\left(x-\frac{3}{5}\right)^2: \\
5 \fbox{$\left(x-\frac{3}{5}\right)^2$}-6 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=-\frac{423}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{2}+\frac{9}{16}=\left(y+\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x-\frac{3}{5}\right)^2-6 \fbox{$\left(y+\frac{3}{4}\right)^2$}=-\frac{423}{40} \\
\end{array}
| amps |
khanacademy | Given the equation $-10 x^2-3 x-5 y^2+10 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+10 y-10 x^2-3 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
-5 y^2+10 y-10 x^2-3 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-10 x^2-3 x+\underline{\text{ }}\right)+\left(-5 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2-3 x+\underline{\text{ }}\right)=-10 \left(x^2+\frac{3 x}{10}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2+\frac{3 x}{10}+\underline{\text{ }}\right)$}+\left(-5 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+10 y+\underline{\text{ }}\right)=-5 \left(y^2-2 y+\underline{\text{ }}\right): \\
-10 \left(x^2+\frac{3 x}{10}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-2 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{10}}{2}\right)^2=\frac{9}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{9}{400}=-\frac{9}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9-\frac{9}{40}=\frac{351}{40}: \\
-10 \left(x^2+\frac{3 x}{10}+\frac{9}{400}\right)-5 \left(y^2-2 y+\underline{\text{ }}\right)=\fbox{$\frac{351}{40}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-5\times 1=-5 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{351}{40}-5=\frac{151}{40}: \\
-10 \left(x^2+\frac{3 x}{10}+\frac{9}{400}\right)-5 \left(y^2-2 y+1\right)=\fbox{$\frac{151}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{10}+\frac{9}{400}=\left(x+\frac{3}{20}\right)^2: \\
-10 \fbox{$\left(x+\frac{3}{20}\right)^2$}-5 \left(y^2-2 y+1\right)=\frac{151}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-2 y+1=(y-1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x+\frac{3}{20}\right)^2-5 \fbox{$(y-1)^2$}=\frac{151}{40} \\
\end{array}
| amps |
khanacademy | Given the equation $-3 x^2-2 x+3 y^2+4 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2+4 y-3 x^2-2 x=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2-2 x+\underline{\text{ }}\right)+\left(3 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(-3 x^2-2 x+\underline{\text{ }}\right)=-3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(3 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 y^2+4 y+\underline{\text{ }}\right)=3 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right): \\
-3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+0 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{-3}{9}=-\frac{1}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{3}}{2}\right)^2=\frac{4}{9} \text{on }\text{the }\text{left }\text{and }3\times \frac{4}{9}=\frac{4}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{4}{3}-\frac{1}{3}=1: \\
-3 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+3 \left(y^2+\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$1$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{2 x}{3}+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2: \\
-3 \fbox{$\left(x+\frac{1}{3}\right)^2$}+3 \left(y^2+\frac{4 y}{3}+\frac{4}{9}\right)=1 \\
\end{array}
Step 9:
\begin{array}{l}
y^2+\frac{4 y}{3}+\frac{4}{9}=\left(y+\frac{2}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x+\frac{1}{3}\right)^2+3 \fbox{$\left(y+\frac{2}{3}\right)^2$}=1 \\
\end{array}
| amps |
khanacademy | Given the equation $9 x^2+x+6 y^2+7 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2+7 y+9 x^2+x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
6 y^2+7 y+9 x^2+x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(9 x^2+x+\underline{\text{ }}\right)+\left(6 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2+x+\underline{\text{ }}\right)=9 \left(x^2+\frac{x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2+\frac{x}{9}+\underline{\text{ }}\right)$}+\left(6 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2+7 y+\underline{\text{ }}\right)=6 \left(y^2+\frac{7 y}{6}+\underline{\text{ }}\right): \\
9 \left(x^2+\frac{x}{9}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2+\frac{7 y}{6}+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{9}}{2}\right)^2=\frac{1}{324} \text{on }\text{the }\text{left }\text{and }\frac{9}{324}=\frac{1}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{1}{36}-7=-\frac{251}{36}: \\
9 \left(x^2+\frac{x}{9}+\frac{1}{324}\right)+6 \left(y^2+\frac{7 y}{6}+\underline{\text{ }}\right)=\fbox{$-\frac{251}{36}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{6}}{2}\right)^2=\frac{49}{144} \text{on }\text{the }\text{left }\text{and }6\times \frac{49}{144}=\frac{49}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{49}{24}-\frac{251}{36}=-\frac{355}{72}: \\
9 \left(x^2+\frac{x}{9}+\frac{1}{324}\right)+6 \left(y^2+\frac{7 y}{6}+\frac{49}{144}\right)=\fbox{$-\frac{355}{72}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{9}+\frac{1}{324}=\left(x+\frac{1}{18}\right)^2: \\
9 \fbox{$\left(x+\frac{1}{18}\right)^2$}+6 \left(y^2+\frac{7 y}{6}+\frac{49}{144}\right)=-\frac{355}{72} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{7 y}{6}+\frac{49}{144}=\left(y+\frac{7}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x+\frac{1}{18}\right)^2+6 \fbox{$\left(y+\frac{7}{12}\right)^2$}=-\frac{355}{72} \\
\end{array}
| amps |
khanacademy | Given the equation $-3 x^2-2 x+9 y^2+10 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+10 y-3 x^2-2 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
9 y^2+10 y-3 x^2-2 x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2-2 x+\underline{\text{ }}\right)+\left(9 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2-2 x+\underline{\text{ }}\right)=-3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(9 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2+10 y+\underline{\text{ }}\right)=9 \left(y^2+\frac{10 y}{9}+\underline{\text{ }}\right): \\
-3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2+\frac{10 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{-3}{9}=-\frac{1}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-4-\frac{1}{3}=-\frac{13}{3}: \\
-3 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+9 \left(y^2+\frac{10 y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{13}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{10}{9}}{2}\right)^2=\frac{25}{81} \text{on }\text{the }\text{left }\text{and }9\times \frac{25}{81}=\frac{25}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{25}{9}-\frac{13}{3}=-\frac{14}{9}: \\
-3 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+9 \left(y^2+\frac{10 y}{9}+\frac{25}{81}\right)=\fbox{$-\frac{14}{9}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{3}+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2: \\
-3 \fbox{$\left(x+\frac{1}{3}\right)^2$}+9 \left(y^2+\frac{10 y}{9}+\frac{25}{81}\right)=-\frac{14}{9} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{10 y}{9}+\frac{25}{81}=\left(y+\frac{5}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x+\frac{1}{3}\right)^2+9 \fbox{$\left(y+\frac{5}{9}\right)^2$}=-\frac{14}{9} \\
\end{array}
| amps |
khanacademy | Given the equation $4 x^2+10 x-5 y^2+5 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+5 y+4 x^2+10 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
-5 y^2+5 y+4 x^2+10 x=4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2+10 x+\underline{\text{ }}\right)+\left(-5 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2+10 x+\underline{\text{ }}\right)=4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)$}+\left(-5 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+5 y+\underline{\text{ }}\right)=-5 \left(y^2-y+\underline{\text{ }}\right): \\
4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-y+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }4\times \frac{25}{16}=\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
4+\frac{25}{4}=\frac{41}{4}: \\
4 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)-5 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$\frac{41}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-5}{4}=-\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{41}{4}-\frac{5}{4}=9: \\
4 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)-5 \left(y^2-y+\frac{1}{4}\right)=\fbox{$9$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{2}+\frac{25}{16}=\left(x+\frac{5}{4}\right)^2: \\
4 \fbox{$\left(x+\frac{5}{4}\right)^2$}-5 \left(y^2-y+\frac{1}{4}\right)=9 \\
\end{array}
Step 11:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x+\frac{5}{4}\right)^2-5 \fbox{$\left(y-\frac{1}{2}\right)^2$}=9 \\
\end{array}
| amps |
khanacademy | Given the equation $-6 x^2-4 x+9 y^2-2 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2-2 y-6 x^2-4 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
9 y^2-2 y-6 x^2-4 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2-4 x+\underline{\text{ }}\right)+\left(9 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2-4 x+\underline{\text{ }}\right)=-6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(9 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2-2 y+\underline{\text{ }}\right)=9 \left(y^2-\frac{2 y}{9}+\underline{\text{ }}\right): \\
-6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2-\frac{2 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{-6}{9}=-\frac{2}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9-\frac{2}{3}=\frac{25}{3}: \\
-6 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+9 \left(y^2-\frac{2 y}{9}+\underline{\text{ }}\right)=\fbox{$\frac{25}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{9}}{2}\right)^2=\frac{1}{81} \text{on }\text{the }\text{left }\text{and }\frac{9}{81}=\frac{1}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{25}{3}+\frac{1}{9}=\frac{76}{9}: \\
-6 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+9 \left(y^2-\frac{2 y}{9}+\frac{1}{81}\right)=\fbox{$\frac{76}{9}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{3}+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2: \\
-6 \fbox{$\left(x+\frac{1}{3}\right)^2$}+9 \left(y^2-\frac{2 y}{9}+\frac{1}{81}\right)=\frac{76}{9} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{2 y}{9}+\frac{1}{81}=\left(y-\frac{1}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x+\frac{1}{3}\right)^2+9 \fbox{$\left(y-\frac{1}{9}\right)^2$}=\frac{76}{9} \\
\end{array}
| amps |
khanacademy | Given the equation $-x^2+9 x+9 y^2+7 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+7 y-x^2+9 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
9 y^2+7 y-x^2+9 x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2+9 x+\underline{\text{ }}\right)+\left(9 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2+9 x+\underline{\text{ }}\right)=-\left(x^2-9 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2-9 x+\underline{\text{ }}\right)$}+\left(9 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2+7 y+\underline{\text{ }}\right)=9 \left(y^2+\frac{7 y}{9}+\underline{\text{ }}\right): \\
-\left(x^2-9 x+\underline{\text{ }}\right)+\fbox{$9 \left(y^2+\frac{7 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-9}{2}\right)^2=\frac{81}{4} \text{on }\text{the }\text{left }\text{and }-\frac{81}{4}=-\frac{81}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-4-\frac{81}{4}=-\frac{97}{4}: \\
-\left(x^2-9 x+\frac{81}{4}\right)+9 \left(y^2+\frac{7 y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{97}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{9}}{2}\right)^2=\frac{49}{324} \text{on }\text{the }\text{left }\text{and }9\times \frac{49}{324}=\frac{49}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{49}{36}-\frac{97}{4}=-\frac{206}{9}: \\
-\left(x^2-9 x+\frac{81}{4}\right)+9 \left(y^2+\frac{7 y}{9}+\frac{49}{324}\right)=\fbox{$-\frac{206}{9}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-9 x+\frac{81}{4}=\left(x-\frac{9}{2}\right)^2: \\
-\fbox{$\left(x-\frac{9}{2}\right)^2$}+9 \left(y^2+\frac{7 y}{9}+\frac{49}{324}\right)=-\frac{206}{9} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{7 y}{9}+\frac{49}{324}=\left(y+\frac{7}{18}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -\left(x-\frac{9}{2}\right)^2+9 \fbox{$\left(y+\frac{7}{18}\right)^2$}=-\frac{206}{9} \\
\end{array}
| amps |
khanacademy | Given the equation $-3 x^2-5 x+5 y^2+8 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+8 y-3 x^2-5 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
5 y^2+8 y-3 x^2-5 x=-6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2-5 x+\underline{\text{ }}\right)+\left(5 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2-5 x+\underline{\text{ }}\right)=-3 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right)$}+\left(5 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2+8 y+\underline{\text{ }}\right)=5 \left(y^2+\frac{8 y}{5}+\underline{\text{ }}\right): \\
-3 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2+\frac{8 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{3}}{2}\right)^2=\frac{25}{36} \text{on }\text{the }\text{left }\text{and }-3\times \frac{25}{36}=-\frac{25}{12} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-6-\frac{25}{12}=-\frac{97}{12}: \\
-3 \left(x^2+\frac{5 x}{3}+\frac{25}{36}\right)+5 \left(y^2+\frac{8 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{97}{12}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{5}}{2}\right)^2=\frac{16}{25} \text{on }\text{the }\text{left }\text{and }5\times \frac{16}{25}=\frac{16}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{16}{5}-\frac{97}{12}=-\frac{293}{60}: \\
-3 \left(x^2+\frac{5 x}{3}+\frac{25}{36}\right)+5 \left(y^2+\frac{8 y}{5}+\frac{16}{25}\right)=\fbox{$-\frac{293}{60}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{3}+\frac{25}{36}=\left(x+\frac{5}{6}\right)^2: \\
-3 \fbox{$\left(x+\frac{5}{6}\right)^2$}+5 \left(y^2+\frac{8 y}{5}+\frac{16}{25}\right)=-\frac{293}{60} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{8 y}{5}+\frac{16}{25}=\left(y+\frac{4}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x+\frac{5}{6}\right)^2+5 \fbox{$\left(y+\frac{4}{5}\right)^2$}=-\frac{293}{60} \\
\end{array}
| amps |
khanacademy | Given the equation $-8 x^2-x+2 y^2+7 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+7 y-8 x^2-x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
2 y^2+7 y-8 x^2-x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2-x+\underline{\text{ }}\right)+\left(2 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2-x+\underline{\text{ }}\right)=-8 \left(x^2+\frac{x}{8}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2+\frac{x}{8}+\underline{\text{ }}\right)$}+\left(2 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+7 y+\underline{\text{ }}\right)=2 \left(y^2+\frac{7 y}{2}+\underline{\text{ }}\right): \\
-8 \left(x^2+\frac{x}{8}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+\frac{7 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{8}}{2}\right)^2=\frac{1}{256} \text{on }\text{the }\text{left }\text{and }\frac{-8}{256}=-\frac{1}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-4-\frac{1}{32}=-\frac{129}{32}: \\
-8 \left(x^2+\frac{x}{8}+\frac{1}{256}\right)+2 \left(y^2+\frac{7 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{129}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{2}}{2}\right)^2=\frac{49}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{49}{16}=\frac{49}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{49}{8}-\frac{129}{32}=\frac{67}{32}: \\
-8 \left(x^2+\frac{x}{8}+\frac{1}{256}\right)+2 \left(y^2+\frac{7 y}{2}+\frac{49}{16}\right)=\fbox{$\frac{67}{32}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{8}+\frac{1}{256}=\left(x+\frac{1}{16}\right)^2: \\
-8 \fbox{$\left(x+\frac{1}{16}\right)^2$}+2 \left(y^2+\frac{7 y}{2}+\frac{49}{16}\right)=\frac{67}{32} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{7 y}{2}+\frac{49}{16}=\left(y+\frac{7}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x+\frac{1}{16}\right)^2+2 \fbox{$\left(y+\frac{7}{4}\right)^2$}=\frac{67}{32} \\
\end{array}
| amps |
khanacademy | Given the equation $6 x^2-4 x+9 y^2+y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+y+6 x^2-4 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
9 y^2+y+6 x^2-4 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(6 x^2-4 x+\underline{\text{ }}\right)+\left(9 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(6 x^2-4 x+\underline{\text{ }}\right)=6 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(9 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2+y+\underline{\text{ }}\right)=9 \left(y^2+\frac{y}{9}+\underline{\text{ }}\right): \\
6 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2+\frac{y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}+3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{6}{9}=\frac{2}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3+\frac{2}{3}=\frac{11}{3}: \\
6 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)+9 \left(y^2+\frac{y}{9}+\underline{\text{ }}\right)=\fbox{$\frac{11}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{9}}{2}\right)^2=\frac{1}{324} \text{on }\text{the }\text{left }\text{and }\frac{9}{324}=\frac{1}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{11}{3}+\frac{1}{36}=\frac{133}{36}: \\
6 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)+9 \left(y^2+\frac{y}{9}+\frac{1}{324}\right)=\fbox{$\frac{133}{36}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{2 x}{3}+\frac{1}{9}=\left(x-\frac{1}{3}\right)^2: \\
6 \fbox{$\left(x-\frac{1}{3}\right)^2$}+9 \left(y^2+\frac{y}{9}+\frac{1}{324}\right)=\frac{133}{36} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{9}+\frac{1}{324}=\left(y+\frac{1}{18}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x-\frac{1}{3}\right)^2+9 \fbox{$\left(y+\frac{1}{18}\right)^2$}=\frac{133}{36} \\
\end{array}
| amps |
khanacademy | Given the equation $2 x^2-3 x-5 y^2-6 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2-6 y+2 x^2-3 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-5 y^2-6 y+2 x^2-3 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2-3 x+\underline{\text{ }}\right)+\left(-5 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2-3 x+\underline{\text{ }}\right)=2 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right)$}+\left(-5 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2-6 y+\underline{\text{ }}\right)=-5 \left(y^2+\frac{6 y}{5}+\underline{\text{ }}\right): \\
2 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2+\frac{6 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{9}{16}=\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5+\frac{9}{8}=\frac{49}{8}: \\
2 \left(x^2-\frac{3 x}{2}+\frac{9}{16}\right)-5 \left(y^2+\frac{6 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{49}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{6}{5}}{2}\right)^2=\frac{9}{25} \text{on }\text{the }\text{left }\text{and }-5\times \frac{9}{25}=-\frac{9}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{49}{8}-\frac{9}{5}=\frac{173}{40}: \\
2 \left(x^2-\frac{3 x}{2}+\frac{9}{16}\right)-5 \left(y^2+\frac{6 y}{5}+\frac{9}{25}\right)=\fbox{$\frac{173}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{2}+\frac{9}{16}=\left(x-\frac{3}{4}\right)^2: \\
2 \fbox{$\left(x-\frac{3}{4}\right)^2$}-5 \left(y^2+\frac{6 y}{5}+\frac{9}{25}\right)=\frac{173}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{6 y}{5}+\frac{9}{25}=\left(y+\frac{3}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x-\frac{3}{4}\right)^2-5 \fbox{$\left(y+\frac{3}{5}\right)^2$}=\frac{173}{40} \\
\end{array}
| amps |
khanacademy | Given the equation $-8 x^2+10 x-7 y^2-8 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2-8 y-8 x^2+10 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
-7 y^2-8 y-8 x^2+10 x=1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2+10 x+\underline{\text{ }}\right)+\left(-7 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2+10 x+\underline{\text{ }}\right)=-8 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right)$}+\left(-7 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2-8 y+\underline{\text{ }}\right)=-7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}+1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{4}}{2}\right)^2=\frac{25}{64} \text{on }\text{the }\text{left }\text{and }-8\times \frac{25}{64}=-\frac{25}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
1-\frac{25}{8}=-\frac{17}{8}: \\
-8 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)-7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{17}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{7}}{2}\right)^2=\frac{16}{49} \text{on }\text{the }\text{left }\text{and }-7\times \frac{16}{49}=-\frac{16}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{17}{8}-\frac{16}{7}=-\frac{247}{56}: \\
-8 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)-7 \left(y^2+\frac{8 y}{7}+\frac{16}{49}\right)=\fbox{$-\frac{247}{56}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{4}+\frac{25}{64}=\left(x-\frac{5}{8}\right)^2: \\
-8 \fbox{$\left(x-\frac{5}{8}\right)^2$}-7 \left(y^2+\frac{8 y}{7}+\frac{16}{49}\right)=-\frac{247}{56} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{8 y}{7}+\frac{16}{49}=\left(y+\frac{4}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x-\frac{5}{8}\right)^2-7 \fbox{$\left(y+\frac{4}{7}\right)^2$}=-\frac{247}{56} \\
\end{array}
| amps |
khanacademy | Given the equation $7 x^2+2 x-2 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 x^2+2 x+(10-2 y)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }10-2 y \text{from }\text{both }\text{sides}: \\
7 x^2+2 x=2 y-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(7 x^2+2 x+\underline{\text{ }}\right)=(2 y-10)+\underline{\text{ }} \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2+2 x+\underline{\text{ }}\right)=7 \left(x^2+\frac{2 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2+\frac{2 x}{7}+\underline{\text{ }}\right)$}=(2 y-10)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{7}}{2}\right)^2=\frac{1}{49} \text{on }\text{the }\text{left }\text{and }\frac{7}{49}=\frac{1}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
(2 y-10)+\frac{1}{7}=2 y-\frac{69}{7}: \\
7 \left(x^2+\frac{2 x}{7}+\frac{1}{49}\right)=\fbox{$2 y-\frac{69}{7}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2+\frac{2 x}{7}+\frac{1}{49}=\left(x+\frac{1}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \fbox{$\left(x+\frac{1}{7}\right)^2$}=2 y-\frac{69}{7} \\
\end{array}
| amps |
khanacademy | Given the equation $-10 x^2-x+7 y^2-4 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2-4 y-10 x^2-x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
7 y^2-4 y-10 x^2-x=4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-10 x^2-x+\underline{\text{ }}\right)+\left(7 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2-x+\underline{\text{ }}\right)=-10 \left(x^2+\frac{x}{10}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2+\frac{x}{10}+\underline{\text{ }}\right)$}+\left(7 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2-4 y+\underline{\text{ }}\right)=7 \left(y^2-\frac{4 y}{7}+\underline{\text{ }}\right): \\
-10 \left(x^2+\frac{x}{10}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2-\frac{4 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{10}}{2}\right)^2=\frac{1}{400} \text{on }\text{the }\text{left }\text{and }\frac{-10}{400}=-\frac{1}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
4-\frac{1}{40}=\frac{159}{40}: \\
-10 \left(x^2+\frac{x}{10}+\frac{1}{400}\right)+7 \left(y^2-\frac{4 y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{159}{40}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{7}}{2}\right)^2=\frac{4}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{4}{49}=\frac{4}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{159}{40}+\frac{4}{7}=\frac{1273}{280}: \\
-10 \left(x^2+\frac{x}{10}+\frac{1}{400}\right)+7 \left(y^2-\frac{4 y}{7}+\frac{4}{49}\right)=\fbox{$\frac{1273}{280}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{10}+\frac{1}{400}=\left(x+\frac{1}{20}\right)^2: \\
-10 \fbox{$\left(x+\frac{1}{20}\right)^2$}+7 \left(y^2-\frac{4 y}{7}+\frac{4}{49}\right)=\frac{1273}{280} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{4 y}{7}+\frac{4}{49}=\left(y-\frac{2}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x+\frac{1}{20}\right)^2+7 \fbox{$\left(y-\frac{2}{7}\right)^2$}=\frac{1273}{280} \\
\end{array}
| amps |
khanacademy | Given the equation $-10 x^2+2 x-y^2+5 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2+5 y-10 x^2+2 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
-y^2+5 y-10 x^2+2 x=2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-10 x^2+2 x+\underline{\text{ }}\right)+\left(-y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2+2 x+\underline{\text{ }}\right)=-10 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right)$}+\left(-y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2+5 y+\underline{\text{ }}\right)=-\left(y^2-5 y+\underline{\text{ }}\right): \\
-10 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right)+\fbox{$-\left(y^2-5 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{-10}{100}=-\frac{1}{10} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2-\frac{1}{10}=\frac{19}{10}: \\
-10 \left(x^2-\frac{x}{5}+\frac{1}{100}\right)-\left(y^2-5 y+\underline{\text{ }}\right)=\fbox{$\frac{19}{10}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }-\frac{25}{4}=-\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{19}{10}-\frac{25}{4}=-\frac{87}{20}: \\
-10 \left(x^2-\frac{x}{5}+\frac{1}{100}\right)-\left(y^2-5 y+\frac{25}{4}\right)=\fbox{$-\frac{87}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{5}+\frac{1}{100}=\left(x-\frac{1}{10}\right)^2: \\
-10 \fbox{$\left(x-\frac{1}{10}\right)^2$}-\left(y^2-5 y+\frac{25}{4}\right)=-\frac{87}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-5 y+\frac{25}{4}=\left(y-\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x-\frac{1}{10}\right)^2-\fbox{$\left(y-\frac{5}{2}\right)^2$}=-\frac{87}{20} \\
\end{array}
| amps |
khanacademy | Given the equation $-9 x^2-5 x-9 y^2-2 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-2 y-9 x^2-5 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-9 y^2-2 y-9 x^2-5 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2-5 x+\underline{\text{ }}\right)+\left(-9 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2-5 x+\underline{\text{ }}\right)=-9 \left(x^2+\frac{5 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2+\frac{5 x}{9}+\underline{\text{ }}\right)$}+\left(-9 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2-2 y+\underline{\text{ }}\right)=-9 \left(y^2+\frac{2 y}{9}+\underline{\text{ }}\right): \\
-9 \left(x^2+\frac{5 x}{9}+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2+\frac{2 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{9}}{2}\right)^2=\frac{25}{324} \text{on }\text{the }\text{left }\text{and }-9\times \frac{25}{324}=-\frac{25}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-5-\frac{25}{36}=-\frac{205}{36}: \\
-9 \left(x^2+\frac{5 x}{9}+\frac{25}{324}\right)-9 \left(y^2+\frac{2 y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{205}{36}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{9}}{2}\right)^2=\frac{1}{81} \text{on }\text{the }\text{left }\text{and }\frac{-9}{81}=-\frac{1}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{205}{36}-\frac{1}{9}=-\frac{209}{36}: \\
-9 \left(x^2+\frac{5 x}{9}+\frac{25}{324}\right)-9 \left(y^2+\frac{2 y}{9}+\frac{1}{81}\right)=\fbox{$-\frac{209}{36}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{9}+\frac{25}{324}=\left(x+\frac{5}{18}\right)^2: \\
-9 \fbox{$\left(x+\frac{5}{18}\right)^2$}-9 \left(y^2+\frac{2 y}{9}+\frac{1}{81}\right)=-\frac{209}{36} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{2 y}{9}+\frac{1}{81}=\left(y+\frac{1}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x+\frac{5}{18}\right)^2-9 \fbox{$\left(y+\frac{1}{9}\right)^2$}=-\frac{209}{36} \\
\end{array}
| amps |
khanacademy | Given the equation $-5 x^2+3 x+4 y^2-7 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2-7 y-5 x^2+3 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
4 y^2-7 y-5 x^2+3 x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2+3 x+\underline{\text{ }}\right)+\left(4 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2+3 x+\underline{\text{ }}\right)=-5 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right)$}+\left(4 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2-7 y+\underline{\text{ }}\right)=4 \left(y^2-\frac{7 y}{4}+\underline{\text{ }}\right): \\
-5 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2-\frac{7 y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{5}}{2}\right)^2=\frac{9}{100} \text{on }\text{the }\text{left }\text{and }-5\times \frac{9}{100}=-\frac{9}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-7-\frac{9}{20}=-\frac{149}{20}: \\
-5 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)+4 \left(y^2-\frac{7 y}{4}+\underline{\text{ }}\right)=\fbox{$-\frac{149}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{4}}{2}\right)^2=\frac{49}{64} \text{on }\text{the }\text{left }\text{and }4\times \frac{49}{64}=\frac{49}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{49}{16}-\frac{149}{20}=-\frac{351}{80}: \\
-5 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)+4 \left(y^2-\frac{7 y}{4}+\frac{49}{64}\right)=\fbox{$-\frac{351}{80}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{5}+\frac{9}{100}=\left(x-\frac{3}{10}\right)^2: \\
-5 \fbox{$\left(x-\frac{3}{10}\right)^2$}+4 \left(y^2-\frac{7 y}{4}+\frac{49}{64}\right)=-\frac{351}{80} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{4}+\frac{49}{64}=\left(y-\frac{7}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 \left(x-\frac{3}{10}\right)^2+4 \fbox{$\left(y-\frac{7}{8}\right)^2$}=-\frac{351}{80} \\
\end{array}
| amps |
khanacademy | Given the equation $-9 x^2-5 x+3 y^2+y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2+y-9 x^2-5 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
3 y^2+y-9 x^2-5 x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2-5 x+\underline{\text{ }}\right)+\left(3 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2-5 x+\underline{\text{ }}\right)=-9 \left(x^2+\frac{5 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2+\frac{5 x}{9}+\underline{\text{ }}\right)$}+\left(3 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 y^2+y+\underline{\text{ }}\right)=3 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right): \\
-9 \left(x^2+\frac{5 x}{9}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{9}}{2}\right)^2=\frac{25}{324} \text{on }\text{the }\text{left }\text{and }-9\times \frac{25}{324}=-\frac{25}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-4-\frac{25}{36}=-\frac{169}{36}: \\
-9 \left(x^2+\frac{5 x}{9}+\frac{25}{324}\right)+3 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{169}{36}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{3}{36}=\frac{1}{12} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{12}-\frac{169}{36}=-\frac{83}{18}: \\
-9 \left(x^2+\frac{5 x}{9}+\frac{25}{324}\right)+3 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=\fbox{$-\frac{83}{18}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{9}+\frac{25}{324}=\left(x+\frac{5}{18}\right)^2: \\
-9 \fbox{$\left(x+\frac{5}{18}\right)^2$}+3 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=-\frac{83}{18} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{3}+\frac{1}{36}=\left(y+\frac{1}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x+\frac{5}{18}\right)^2+3 \fbox{$\left(y+\frac{1}{6}\right)^2$}=-\frac{83}{18} \\
\end{array}
| amps |
khanacademy | Given the equation $5 x^2-7 x-2 y^2+6 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+6 y+5 x^2-7 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
-2 y^2+6 y+5 x^2-7 x=2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2-7 x+\underline{\text{ }}\right)+\left(-2 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2-7 x+\underline{\text{ }}\right)=5 \left(x^2-\frac{7 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-\frac{7 x}{5}+\underline{\text{ }}\right)$}+\left(-2 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+6 y+\underline{\text{ }}\right)=-2 \left(y^2-3 y+\underline{\text{ }}\right): \\
5 \left(x^2-\frac{7 x}{5}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-3 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{5}}{2}\right)^2=\frac{49}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{49}{100}=\frac{49}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2+\frac{49}{20}=\frac{89}{20}: \\
5 \left(x^2-\frac{7 x}{5}+\frac{49}{100}\right)-2 \left(y^2-3 y+\underline{\text{ }}\right)=\fbox{$\frac{89}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }-2\times \frac{9}{4}=-\frac{9}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{89}{20}-\frac{9}{2}=-\frac{1}{20}: \\
5 \left(x^2-\frac{7 x}{5}+\frac{49}{100}\right)-2 \left(y^2-3 y+\frac{9}{4}\right)=\fbox{$-\frac{1}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{7 x}{5}+\frac{49}{100}=\left(x-\frac{7}{10}\right)^2: \\
5 \fbox{$\left(x-\frac{7}{10}\right)^2$}-2 \left(y^2-3 y+\frac{9}{4}\right)=-\frac{1}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-3 y+\frac{9}{4}=\left(y-\frac{3}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x-\frac{7}{10}\right)^2-2 \fbox{$\left(y-\frac{3}{2}\right)^2$}=-\frac{1}{20} \\
\end{array}
| amps |
khanacademy | Given the equation $-7 x^2-2 x-7 y^2-y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2-y-7 x^2-2 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-7 y^2-y-7 x^2-2 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-7 x^2-2 x+\underline{\text{ }}\right)+\left(-7 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 x^2-2 x+\underline{\text{ }}\right)=-7 \left(x^2+\frac{2 x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2+\frac{2 x}{7}+\underline{\text{ }}\right)$}+\left(-7 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2-y+\underline{\text{ }}\right)=-7 \left(y^2+\frac{y}{7}+\underline{\text{ }}\right): \\
-7 \left(x^2+\frac{2 x}{7}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2+\frac{y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{7}}{2}\right)^2=\frac{1}{49} \text{on }\text{the }\text{left }\text{and }\frac{-7}{49}=-\frac{1}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-9-\frac{1}{7}=-\frac{64}{7}: \\
-7 \left(x^2+\frac{2 x}{7}+\frac{1}{49}\right)-7 \left(y^2+\frac{y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{64}{7}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{7}}{2}\right)^2=\frac{1}{196} \text{on }\text{the }\text{left }\text{and }\frac{-7}{196}=-\frac{1}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{64}{7}-\frac{1}{28}=-\frac{257}{28}: \\
-7 \left(x^2+\frac{2 x}{7}+\frac{1}{49}\right)-7 \left(y^2+\frac{y}{7}+\frac{1}{196}\right)=\fbox{$-\frac{257}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{7}+\frac{1}{49}=\left(x+\frac{1}{7}\right)^2: \\
-7 \fbox{$\left(x+\frac{1}{7}\right)^2$}-7 \left(y^2+\frac{y}{7}+\frac{1}{196}\right)=-\frac{257}{28} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{7}+\frac{1}{196}=\left(y+\frac{1}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x+\frac{1}{7}\right)^2-7 \fbox{$\left(y+\frac{1}{14}\right)^2$}=-\frac{257}{28} \\
\end{array}
| amps |
khanacademy | Given the equation $-2 x^2-3 x-9 y^2-8 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-8 y-2 x^2-3 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
-9 y^2-8 y-2 x^2-3 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-2 x^2-3 x+\underline{\text{ }}\right)+\left(-9 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 x^2-3 x+\underline{\text{ }}\right)=-2 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right): \\
\fbox{$-2 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right)$}+\left(-9 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2-8 y+\underline{\text{ }}\right)=-9 \left(y^2+\frac{8 y}{9}+\underline{\text{ }}\right): \\
-2 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2+\frac{8 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }-2\times \frac{9}{16}=-\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10-\frac{9}{8}=\frac{71}{8}: \\
-2 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)-9 \left(y^2+\frac{8 y}{9}+\underline{\text{ }}\right)=\fbox{$\frac{71}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{9}}{2}\right)^2=\frac{16}{81} \text{on }\text{the }\text{left }\text{and }-9\times \frac{16}{81}=-\frac{16}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{71}{8}-\frac{16}{9}=\frac{511}{72}: \\
-2 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)-9 \left(y^2+\frac{8 y}{9}+\frac{16}{81}\right)=\fbox{$\frac{511}{72}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{2}+\frac{9}{16}=\left(x+\frac{3}{4}\right)^2: \\
-2 \fbox{$\left(x+\frac{3}{4}\right)^2$}-9 \left(y^2+\frac{8 y}{9}+\frac{16}{81}\right)=\frac{511}{72} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{8 y}{9}+\frac{16}{81}=\left(y+\frac{4}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -2 \left(x+\frac{3}{4}\right)^2-9 \fbox{$\left(y+\frac{4}{9}\right)^2$}=\frac{511}{72} \\
\end{array}
| amps |
khanacademy | Given the equation $-x^2-5 y^2-4 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2-4 y+\left(-x^2-6\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 y^2+4 y+x^2+6 \text{to }\text{both }\text{sides}: \\
5 y^2+4 y+\left(x^2+6\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
5 y^2+4 y+x^2=-6 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(5 y^2+4 y+\underline{\text{ }}\right)+x^2=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2+4 y+\underline{\text{ }}\right)=5 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right)$}+x^2=\underline{\text{ }}-6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }5\times \frac{4}{25}=\frac{4}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{4}{5}-6=-\frac{26}{5}: \\
5 \left(y^2+\frac{4 y}{5}+\frac{4}{25}\right)+x^2=\fbox{$-\frac{26}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
y^2+\frac{4 y}{5}+\frac{4}{25}=\left(y+\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$\left(y+\frac{2}{5}\right)^2$}+x^2=-\frac{26}{5} \\
\end{array}
| amps |
khanacademy | Given the equation $-7 x^2+2 x+8 y^2-6 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-6 y-7 x^2+2 x=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-7 x^2+2 x+\underline{\text{ }}\right)+\left(8 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(-7 x^2+2 x+\underline{\text{ }}\right)=-7 \left(x^2-\frac{2 x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2-\frac{2 x}{7}+\underline{\text{ }}\right)$}+\left(8 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 y^2-6 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{3 y}{4}+\underline{\text{ }}\right): \\
-7 \left(x^2-\frac{2 x}{7}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{3 y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}+0 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{7}}{2}\right)^2=\frac{1}{49} \text{on }\text{the }\text{left }\text{and }\frac{-7}{49}=-\frac{1}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }8\times \frac{9}{64}=\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{8}-\frac{1}{7}=\frac{55}{56}: \\
-7 \left(x^2-\frac{2 x}{7}+\frac{1}{49}\right)+8 \left(y^2-\frac{3 y}{4}+\frac{9}{64}\right)=\fbox{$\frac{55}{56}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{2 x}{7}+\frac{1}{49}=\left(x-\frac{1}{7}\right)^2: \\
-7 \fbox{$\left(x-\frac{1}{7}\right)^2$}+8 \left(y^2-\frac{3 y}{4}+\frac{9}{64}\right)=\frac{55}{56} \\
\end{array}
Step 9:
\begin{array}{l}
y^2-\frac{3 y}{4}+\frac{9}{64}=\left(y-\frac{3}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x-\frac{1}{7}\right)^2+8 \fbox{$\left(y-\frac{3}{8}\right)^2$}=\frac{55}{56} \\
\end{array}
| amps |
khanacademy | Given the equation $x^2+3 x-4 y^2+5 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+5 y+x^2+3 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
-4 y^2+5 y+x^2+3 x=-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(x^2+3 x+\underline{\text{ }}\right)+\left(-4 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 y^2+5 y+\underline{\text{ }}\right)=-4 \left(y^2-\frac{5 y}{4}+\underline{\text{ }}\right): \\
\left(x^2+3 x+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2-\frac{5 y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{3}{2}\right)^2=\frac{9}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
\frac{9}{4}-10=-\frac{31}{4}: \\
\left(x^2+3 x+\frac{9}{4}\right)-4 \left(y^2-\frac{5 y}{4}+\underline{\text{ }}\right)=\fbox{$-\frac{31}{4}$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{4}}{2}\right)^2=\frac{25}{64} \text{on }\text{the }\text{left }\text{and }-4\times \frac{25}{64}=-\frac{25}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 8:
\begin{array}{l}
-\frac{31}{4}-\frac{25}{16}=-\frac{149}{16}: \\
\left(x^2+3 x+\frac{9}{4}\right)-4 \left(y^2-\frac{5 y}{4}+\frac{25}{64}\right)=\fbox{$-\frac{149}{16}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+3 x+\frac{9}{4}=\left(x+\frac{3}{2}\right)^2: \\
\fbox{$\left(x+\frac{3}{2}\right)^2$}-4 \left(y^2-\frac{5 y}{4}+\frac{25}{64}\right)=-\frac{149}{16} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-\frac{5 y}{4}+\frac{25}{64}=\left(y-\frac{5}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \left(x+\frac{3}{2}\right)^2-4 \fbox{$\left(y-\frac{5}{8}\right)^2$}=-\frac{149}{16} \\
\end{array}
| amps |
khanacademy | Given the equation $6 x^2-4 y^2+3 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+3 y+\left(6 x^2+6\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-4 y^2+3 y+6 x^2+6 \text{from }\text{both }\text{sides}: \\
4 y^2-3 y+\left(-6 x^2-6\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
4 y^2-3 y-6 x^2=6 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(4 y^2-3 y+\underline{\text{ }}\right)-6 x^2=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2-3 y+\underline{\text{ }}\right)=4 \left(y^2-\frac{3 y}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(y^2-\frac{3 y}{4}+\underline{\text{ }}\right)$}-6 x^2=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }4\times \frac{9}{64}=\frac{9}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6+\frac{9}{16}=\frac{105}{16}: \\
4 \left(y^2-\frac{3 y}{4}+\frac{9}{64}\right)-6 x^2=\fbox{$\frac{105}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
y^2-\frac{3 y}{4}+\frac{9}{64}=\left(y-\frac{3}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \fbox{$\left(y-\frac{3}{8}\right)^2$}-6 x^2=\frac{105}{16} \\
\end{array}
| amps |
khanacademy | Given the equation $5 x^2+10 x-7 y^2-3 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2-3 y+5 x^2+10 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
-7 y^2-3 y+5 x^2+10 x=1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2+10 x+\underline{\text{ }}\right)+\left(-7 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2+10 x+\underline{\text{ }}\right)=5 \left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+2 x+\underline{\text{ }}\right)$}+\left(-7 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2-3 y+\underline{\text{ }}\right)=-7 \left(y^2+\frac{3 y}{7}+\underline{\text{ }}\right): \\
5 \left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2+\frac{3 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}+1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }5\times 1=5 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
1+5=6: \\
5 \left(x^2+2 x+1\right)-7 \left(y^2+\frac{3 y}{7}+\underline{\text{ }}\right)=\fbox{$6$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{7}}{2}\right)^2=\frac{9}{196} \text{on }\text{the }\text{left }\text{and }-7\times \frac{9}{196}=-\frac{9}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
6-\frac{9}{28}=\frac{159}{28}: \\
5 \left(x^2+2 x+1\right)-7 \left(y^2+\frac{3 y}{7}+\frac{9}{196}\right)=\fbox{$\frac{159}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
5 \fbox{$(x+1)^2$}-7 \left(y^2+\frac{3 y}{7}+\frac{9}{196}\right)=\frac{159}{28} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{7}+\frac{9}{196}=\left(y+\frac{3}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 (x+1)^2-7 \fbox{$\left(y+\frac{3}{14}\right)^2$}=\frac{159}{28} \\
\end{array}
| amps |
khanacademy | Given the equation $-8 x^2+6 x+4 y^2-2 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2-2 y-8 x^2+6 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
4 y^2-2 y-8 x^2+6 x=-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2+6 x+\underline{\text{ }}\right)+\left(4 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2+6 x+\underline{\text{ }}\right)=-8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)$}+\left(4 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2-2 y+\underline{\text{ }}\right)=4 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }-8\times \frac{9}{64}=-\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-10-\frac{9}{8}=-\frac{89}{8}: \\
-8 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)+4 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{89}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{4}{16}=\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{4}-\frac{89}{8}=-\frac{87}{8}: \\
-8 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)+4 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\fbox{$-\frac{87}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{4}+\frac{9}{64}=\left(x-\frac{3}{8}\right)^2: \\
-8 \fbox{$\left(x-\frac{3}{8}\right)^2$}+4 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=-\frac{87}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{2}+\frac{1}{16}=\left(y-\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x-\frac{3}{8}\right)^2+4 \fbox{$\left(y-\frac{1}{4}\right)^2$}=-\frac{87}{8} \\
\end{array}
| amps |
khanacademy | Given the equation $2 x^2-7 x-9 y^2-2 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-2 y+2 x^2-7 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
-9 y^2-2 y+2 x^2-7 x=1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2-7 x+\underline{\text{ }}\right)+\left(-9 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2-7 x+\underline{\text{ }}\right)=2 \left(x^2-\frac{7 x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-\frac{7 x}{2}+\underline{\text{ }}\right)$}+\left(-9 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2-2 y+\underline{\text{ }}\right)=-9 \left(y^2+\frac{2 y}{9}+\underline{\text{ }}\right): \\
2 \left(x^2-\frac{7 x}{2}+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2+\frac{2 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}+1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{2}}{2}\right)^2=\frac{49}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{49}{16}=\frac{49}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
1+\frac{49}{8}=\frac{57}{8}: \\
2 \left(x^2-\frac{7 x}{2}+\frac{49}{16}\right)-9 \left(y^2+\frac{2 y}{9}+\underline{\text{ }}\right)=\fbox{$\frac{57}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{9}}{2}\right)^2=\frac{1}{81} \text{on }\text{the }\text{left }\text{and }\frac{-9}{81}=-\frac{1}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{57}{8}-\frac{1}{9}=\frac{505}{72}: \\
2 \left(x^2-\frac{7 x}{2}+\frac{49}{16}\right)-9 \left(y^2+\frac{2 y}{9}+\frac{1}{81}\right)=\fbox{$\frac{505}{72}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{7 x}{2}+\frac{49}{16}=\left(x-\frac{7}{4}\right)^2: \\
2 \fbox{$\left(x-\frac{7}{4}\right)^2$}-9 \left(y^2+\frac{2 y}{9}+\frac{1}{81}\right)=\frac{505}{72} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{2 y}{9}+\frac{1}{81}=\left(y+\frac{1}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x-\frac{7}{4}\right)^2-9 \fbox{$\left(y+\frac{1}{9}\right)^2$}=\frac{505}{72} \\
\end{array}
| amps |
khanacademy | Given the equation $-x^2-2 x-8 y^2+8 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2+8 y-x^2-2 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-8 y^2+8 y-x^2-2 x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2-2 x+\underline{\text{ }}\right)+\left(-8 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2-2 x+\underline{\text{ }}\right)=-\left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2+2 x+\underline{\text{ }}\right)$}+\left(-8 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2+8 y+\underline{\text{ }}\right)=-8 \left(y^2-y+\underline{\text{ }}\right): \\
-\left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2-y+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-1=-1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-7-1=-8: \\
-\left(x^2+2 x+1\right)-8 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$-8$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-8}{4}=-2 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-8-2=-10: \\
-\left(x^2+2 x+1\right)-8 \left(y^2-y+\frac{1}{4}\right)=\fbox{$-10$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
-\fbox{$(x+1)^2$}-8 \left(y^2-y+\frac{1}{4}\right)=-10 \\
\end{array}
Step 11:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -(x+1)^2-8 \fbox{$\left(y-\frac{1}{2}\right)^2$}=-10 \\
\end{array}
| amps |
khanacademy | Given the equation $5 x^2-10 x+3 y^2-9 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2-9 y+5 x^2-10 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
3 y^2-9 y+5 x^2-10 x=1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2-10 x+\underline{\text{ }}\right)+\left(3 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2-10 x+\underline{\text{ }}\right)=5 \left(x^2-2 x+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-2 x+\underline{\text{ }}\right)$}+\left(3 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 y^2-9 y+\underline{\text{ }}\right)=3 \left(y^2-3 y+\underline{\text{ }}\right): \\
5 \left(x^2-2 x+\underline{\text{ }}\right)+\fbox{$3 \left(y^2-3 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }5\times 1=5 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
1+5=6: \\
5 \left(x^2-2 x+1\right)+3 \left(y^2-3 y+\underline{\text{ }}\right)=\fbox{$6$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }3\times \frac{9}{4}=\frac{27}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
6+\frac{27}{4}=\frac{51}{4}: \\
5 \left(x^2-2 x+1\right)+3 \left(y^2-3 y+\frac{9}{4}\right)=\fbox{$\frac{51}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-2 x+1=(x-1)^2: \\
5 \fbox{$(x-1)^2$}+3 \left(y^2-3 y+\frac{9}{4}\right)=\frac{51}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-3 y+\frac{9}{4}=\left(y-\frac{3}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 (x-1)^2+3 \fbox{$\left(y-\frac{3}{2}\right)^2$}=\frac{51}{4} \\
\end{array}
| amps |
khanacademy | Given the equation $3 x^2-6 y^2+6 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+6 y+3 x^2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-6 y^2+6 y+3 x^2 \text{from }\text{both }\text{sides}: \\
6 y^2-6 y-3 x^2=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(6 y^2-6 y+\underline{\text{ }}\right)-3 x^2=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(6 y^2-6 y+\underline{\text{ }}\right)=6 \left(y^2-y+\underline{\text{ }}\right): \\
\fbox{$6 \left(y^2-y+\underline{\text{ }}\right)$}-3 x^2=\underline{\text{ }}+0 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{6}{4}=\frac{3}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \fbox{$\left(y-\frac{1}{2}\right)^2$}-3 x^2=\frac{3}{2} \\
\end{array}
| amps |