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Given the equation $-5 x^2-6 x-10 y^2+10 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+10 y-5 x^2-6 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
-10 y^2+10 y-5 x^2-6 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(-5 x^2-6 x+\underline{\text{ }}\right)+\left(-10 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\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(-10 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2+10 y+\underline{\text{ }}\right)=-10 \left(y^2-y+\underline{\text{ }}\right): \\
-5 \left(x^2+\frac{6 x}{5}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2-y+\underline{\text{ }}\right)$}=\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 }-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}
-6-\frac{9}{5}=-\frac{39}{5}: \\
-5 \left(x^2+\frac{6 x}{5}+\frac{9}{25}\right)-10 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$-\frac{39}{5}$} \\
\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 }-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{39}{5}-\frac{5}{2}=-\frac{103}{10}: \\
-5 \left(x^2+\frac{6 x}{5}+\frac{9}{25}\right)-10 \left(y^2-y+\frac{1}{4}\right)=\fbox{$-\frac{103}{10}$} \\
\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{$\left(x+\frac{3}{5}\right)^2$}-10 \left(y^2-y+\frac{1}{4}\right)=-\frac{103}{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{} & -5 \left(x+\frac{3}{5}\right)^2-\text{10 }\fbox{$\left(y-\frac{1}{2}\right)^2$}=-\frac{103}{10} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2-7 x-6 y^2-6 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-6 y-4 x^2-7 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-6 y^2-6 y-4 x^2-7 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^2-7 x+\underline{\text{ }}\right)+\left(-6 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2-6 y+\underline{\text{ }}\right)=-6 \left(y^2+y+\underline{\text{ }}\right): \\
-4 \left(x^2+\frac{7 x}{4}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+y+\underline{\text{ }}\right)$}=\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 }-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}
-5-\frac{49}{16}=-\frac{129}{16}: \\
-4 \left(x^2+\frac{7 x}{4}+\frac{49}{64}\right)-6 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$-\frac{129}{16}$} \\
\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{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 9:
\begin{array}{l}
-\frac{129}{16}-\frac{3}{2}=-\frac{153}{16}: \\
-4 \left(x^2+\frac{7 x}{4}+\frac{49}{64}\right)-6 \left(y^2+y+\frac{1}{4}\right)=\fbox{$-\frac{153}{16}$} \\
\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+y+\frac{1}{4}\right)=-\frac{153}{16} \\
\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{7}{8}\right)^2-6 \fbox{$\left(y+\frac{1}{2}\right)^2$}=-\frac{153}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-2 x-10 y^2-5 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 x^2-10 y^2-2 x-5 y-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
4 x^2-10 y^2-2 x-5 y=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(4 x^2-2 x+\underline{\text{ }}\right)+\left(-10 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2-2 x+\underline{\text{ }}\right)=4 \left(x^2-\frac{x^2}{2}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)$}+\left(-10 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2-5 y+\underline{\text{ }}\right)=-10 \left(y^2+\frac{y^2}{2}+\underline{\text{ }}\right): \\
4 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)$}=\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 }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 7:
\begin{array}{l}
7+\frac{1}{4}=\frac{29}{4}: \\
4 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)-10 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{29}{4}$} \\
\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{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 9:
\begin{array}{l}
\frac{29}{4}-\frac{5}{8}=\frac{53}{8}: \\
4 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)-10 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\fbox{$\frac{53}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
4 \fbox{$\left(x-\frac{1}{4}\right)^2$}-10 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\frac{53}{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{} & 4 \left(x-\frac{1}{4}\right)^2-\text{10 }\fbox{$\left(y+\frac{1}{4}\right)^2$}=\frac{53}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2+3 x+7 y^2+y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2+y-4 x^2+3 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
7 y^2+y-4 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(-4 x^2+3 x+\underline{\text{ }}\right)+\left(7 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2+3 x+\underline{\text{ }}\right)=-4 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)$}+\left(7 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\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): \\
-4 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2+\frac{y}{7}+\underline{\text{ }}\right)$}=\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 }-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}
2-\frac{9}{16}=\frac{23}{16}: \\
-4 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)+7 \left(y^2+\frac{y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{23}{16}$} \\
\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{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{23}{16}+\frac{1}{28}=\frac{165}{112}: \\
-4 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)+7 \left(y^2+\frac{y}{7}+\frac{1}{196}\right)=\fbox{$\frac{165}{112}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{4}+\frac{9}{64}=\left(x-\frac{3}{8}\right)^2: \\
-4 \fbox{$\left(x-\frac{3}{8}\right)^2$}+7 \left(y^2+\frac{y}{7}+\frac{1}{196}\right)=\frac{165}{112} \\
\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{} & -4 \left(x-\frac{3}{8}\right)^2+7 \fbox{$\left(y+\frac{1}{14}\right)^2$}=\frac{165}{112} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2-x-8 y^2-10 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2-10 y-5 x^2-x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
-8 y^2-10 y-5 x^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(-5 x^2-x+\underline{\text{ }}\right)+\left(-8 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2-x+\underline{\text{ }}\right)=-5 \left(x^2+\frac{x}{5}+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2+\frac{x}{5}+\underline{\text{ }}\right)$}+\left(-8 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2-10 y+\underline{\text{ }}\right)=-8 \left(y^2+\frac{5 y}{4}+\underline{\text{ }}\right): \\
-5 \left(x^2+\frac{x}{5}+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2+\frac{5 y}{4}+\underline{\text{ }}\right)$}=\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 }-5 \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{-5}{100}=-\frac{1}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-10-\frac{1}{20}=-\frac{201}{20}: \\
-5 \left(x^2+\frac{x}{5}+\frac{1}{100}\right)-8 \left(y^2+\frac{5 y}{4}+\underline{\text{ }}\right)=\fbox{$-\frac{201}{20}$} \\
\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 }-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 9:
\begin{array}{l}
-\frac{201}{20}-\frac{25}{8}=-\frac{527}{40}: \\
-5 \left(x^2+\frac{x}{5}+\frac{1}{100}\right)-8 \left(y^2+\frac{5 y}{4}+\frac{25}{64}\right)=\fbox{$-\frac{527}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{5}+\frac{1}{100}=\left(x+\frac{1}{10}\right)^2: \\
-5 \fbox{$\left(x+\frac{1}{10}\right)^2$}-8 \left(y^2+\frac{5 y}{4}+\frac{25}{64}\right)=-\frac{527}{40} \\
\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{} & -5 \left(x+\frac{1}{10}\right)^2-8 \fbox{$\left(y+\frac{5}{8}\right)^2$}=-\frac{527}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2+x-3 y^2-6 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2-6 y-4 x^2+x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-3 y^2-6 y-4 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(-4 x^2+x+\underline{\text{ }}\right)+\left(-3 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\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(-3 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2-6 y+\underline{\text{ }}\right)=-3 \left(y^2+2 y+\underline{\text{ }}\right): \\
-4 \left(x^2-\frac{x}{4}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2+2 y+\underline{\text{ }}\right)$}=\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 }-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}
6-\frac{1}{16}=\frac{95}{16}: \\
-4 \left(x^2-\frac{x}{4}+\frac{1}{64}\right)-3 \left(y^2+2 y+\underline{\text{ }}\right)=\fbox{$\frac{95}{16}$} \\
\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 }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-3\times 1=-3 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{95}{16}-3=\frac{47}{16}: \\
-4 \left(x^2-\frac{x}{4}+\frac{1}{64}\right)-3 \left(y^2+2 y+1\right)=\fbox{$\frac{47}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{4}+\frac{1}{64}=\left(x-\frac{1}{8}\right)^2: \\
-4 \fbox{$\left(x-\frac{1}{8}\right)^2$}-3 \left(y^2+2 y+1\right)=\frac{47}{16} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+2 y+1=(y+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x-\frac{1}{8}\right)^2-3 \fbox{$(y+1)^2$}=\frac{47}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2-2 x-5 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 x^2-2 x+(-5 y-9)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 y+9 x^2+2 x+9 \text{to }\text{both }\text{sides}: \\
9 x^2+2 x+(5 y+9)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }5 y+9 \text{from }\text{both }\text{sides}: \\
9 x^2+2 x=-5 y-9 \\
\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 x^2+2 x+\underline{\text{ }}\right)=(-5 y-9)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 x^2+2 x+\underline{\text{ }}\right)=9 \left(x^2+\frac{2 x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2+\frac{2 x}{9}+\underline{\text{ }}\right)$}=(-5 y-9)+\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 }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 7:
\begin{array}{l}
(-5 y-9)+\frac{1}{9}=-5 y-\frac{80}{9}: \\
9 \left(x^2+\frac{2 x}{9}+\frac{1}{81}\right)=\fbox{$-5 y-\frac{80}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{2 x}{9}+\frac{1}{81}=\left(x+\frac{1}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \fbox{$\left(x+\frac{1}{9}\right)^2$}=-5 y-\frac{80}{9} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2-9 x-7 y^2-5 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2-5 y+x^2-9 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
-7 y^2-5 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(-7 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 y^2-5 y+\underline{\text{ }}\right)=-7 \left(y^2+\frac{5 y}{7}+\underline{\text{ }}\right): \\
\left(x^2-9 x+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2+\frac{5 y}{7}+\underline{\text{ }}\right)$}=\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{-9}{2}\right)^2=\frac{81}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
4+\frac{81}{4}=\frac{97}{4}: \\
\left(x^2-9 x+\frac{81}{4}\right)-7 \left(y^2+\frac{5 y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{97}{4}$} \\
\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 }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{7}}{2}\right)^2=\frac{25}{196} \text{on }\text{the }\text{left }\text{and }-7\times \frac{25}{196}=-\frac{25}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 8:
\begin{array}{l}
\frac{97}{4}-\frac{25}{28}=\frac{327}{14}: \\
\left(x^2-9 x+\frac{81}{4}\right)-7 \left(y^2+\frac{5 y}{7}+\frac{25}{196}\right)=\fbox{$\frac{327}{14}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-9 x+\frac{81}{4}=\left(x-\frac{9}{2}\right)^2: \\
\fbox{$\left(x-\frac{9}{2}\right)^2$}-7 \left(y^2+\frac{5 y}{7}+\frac{25}{196}\right)=\frac{327}{14} \\
\end{array}
Step 10:
\begin{array}{l}
y^2+\frac{5 y}{7}+\frac{25}{196}=\left(y+\frac{5}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \left(x-\frac{9}{2}\right)^2-7 \fbox{$\left(y+\frac{5}{14}\right)^2$}=\frac{327}{14} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2+7 x+6 y^2-4 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2-4 y+9 x^2+7 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
6 y^2-4 y+9 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(9 x^2+7 x+\underline{\text{ }}\right)+\left(6 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2+7 x+\underline{\text{ }}\right)=9 \left(x^2+\frac{7 x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2+\frac{7 x}{9}+\underline{\text{ }}\right)$}+\left(6 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\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+\frac{7 x}{9}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)$}=\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 }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 7:
\begin{array}{l}
\frac{49}{36}-3=-\frac{59}{36}: \\
9 \left(x^2+\frac{7 x}{9}+\frac{49}{324}\right)+6 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{59}{36}$} \\
\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{-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{2}{3}-\frac{59}{36}=-\frac{35}{36}: \\
9 \left(x^2+\frac{7 x}{9}+\frac{49}{324}\right)+6 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=\fbox{$-\frac{35}{36}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{9}+\frac{49}{324}=\left(x+\frac{7}{18}\right)^2: \\
9 \fbox{$\left(x+\frac{7}{18}\right)^2$}+6 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=-\frac{35}{36} \\
\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{7}{18}\right)^2+6 \fbox{$\left(y-\frac{1}{3}\right)^2$}=-\frac{35}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+5 x+10 y^2+9 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2+9 y-3 x^2+5 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
10 y^2+9 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(10 y^2+9 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(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): \\
-3 \left(x^2-\frac{5 x}{3}+\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 }-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{47}{12}: \\
-3 \left(x^2-\frac{5 x}{3}+\frac{25}{36}\right)+10 \left(y^2+\frac{9 y}{10}+\underline{\text{ }}\right)=\fbox{$\frac{47}{12}$} \\
\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 }\text{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{47}{12}+\frac{81}{40}=\frac{713}{120}: \\
-3 \left(x^2-\frac{5 x}{3}+\frac{25}{36}\right)+10 \left(y^2+\frac{9 y}{10}+\frac{81}{400}\right)=\fbox{$\frac{713}{120}$} \\
\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{$\left(x-\frac{5}{6}\right)^2$}+10 \left(y^2+\frac{9 y}{10}+\frac{81}{400}\right)=\frac{713}{120} \\
\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{} & -3 \left(x-\frac{5}{6}\right)^2+\text{10 }\fbox{$\left(y+\frac{9}{20}\right)^2$}=\frac{713}{120} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2-x-2 y^2-y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-y+8 x^2-x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-2 y^2-y+8 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(8 x^2-x+\underline{\text{ }}\right)+\left(-2 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\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-y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2-y+\underline{\text{ }}\right)=-2 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right): \\
8 \left(x^2-\frac{x}{8}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)$}=\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 }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}
\frac{1}{32}-7=-\frac{223}{32}: \\
8 \left(x^2-\frac{x}{8}+\frac{1}{256}\right)-2 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{223}{32}$} \\
\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{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{-2}{16}=-\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{223}{32}-\frac{1}{8}=-\frac{227}{32}: \\
8 \left(x^2-\frac{x}{8}+\frac{1}{256}\right)-2 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\fbox{$-\frac{227}{32}$} \\
\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{y}{2}+\frac{1}{16}\right)=-\frac{227}{32} \\
\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{1}{16}\right)^2-2 \fbox{$\left(y+\frac{1}{4}\right)^2$}=-\frac{227}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2+4 x-9 y^2+y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2+y-6 x^2+4 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-9 y^2+y-6 x^2+4 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(-6 x^2+4 x+\underline{\text{ }}\right)+\left(-9 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\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{ }}+6 \\
\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{ }}+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 }-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}
6-\frac{2}{3}=\frac{16}{3}: \\
-6 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)-9 \left(y^2-\frac{y}{9}+\underline{\text{ }}\right)=\fbox{$\frac{16}{3}$} \\
\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{16}{3}-\frac{1}{36}=\frac{191}{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{$\frac{191}{36}$} \\
\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{191}{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{191}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+8 x-7 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 x^2+8 x+(6-7 y)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-7 y-3 x^2+8 x+6 \text{from }\text{both }\text{sides}: \\
3 x^2-8 x+(7 y-6)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }7 y-6 \text{from }\text{both }\text{sides}: \\
3 x^2-8 x=6-7 y \\
\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^2-8 x+\underline{\text{ }}\right)=(6-7 y)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 x^2-8 x+\underline{\text{ }}\right)=3 \left(x^2-\frac{8 x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2-\frac{8 x}{3}+\underline{\text{ }}\right)$}=(6-7 y)+\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{-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 7:
\begin{array}{l}
(6-7 y)+\frac{16}{3}=\frac{34}{3}-7 y: \\
3 \left(x^2-\frac{8 x}{3}+\frac{16}{9}\right)=\fbox{$\frac{34}{3}-7 y$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{8 x}{3}+\frac{16}{9}=\left(x-\frac{4}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \fbox{$\left(x-\frac{4}{3}\right)^2$}=\frac{34}{3}-7 y \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+4 x-2 y^2-7 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-7 y+6 x^2+4 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
-2 y^2-7 y+6 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(6 x^2+4 x+\underline{\text{ }}\right)+\left(-2 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\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(-2 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\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): \\
6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2+\frac{7 y}{2}+\underline{\text{ }}\right)$}=\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 }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}
\frac{2}{3}-8=-\frac{22}{3}: \\
6 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)-2 \left(y^2+\frac{7 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{22}{3}$} \\
\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{22}{3}-\frac{49}{8}=-\frac{323}{24}: \\
6 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)-2 \left(y^2+\frac{7 y}{2}+\frac{49}{16}\right)=\fbox{$-\frac{323}{24}$} \\
\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{$\left(x+\frac{1}{3}\right)^2$}-2 \left(y^2+\frac{7 y}{2}+\frac{49}{16}\right)=-\frac{323}{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{} & 6 \left(x+\frac{1}{3}\right)^2-2 \fbox{$\left(y+\frac{7}{4}\right)^2$}=-\frac{323}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2-x-6 y^2-8 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-8 y+3 x^2-x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
-6 y^2-8 y+3 x^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-x+\underline{\text{ }}\right)+\left(-6 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2-x+\underline{\text{ }}\right)=3 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)$}+\left(-6 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\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): \\
3 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right)$}=\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{-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 7:
\begin{array}{l}
3+\frac{1}{12}=\frac{37}{12}: \\
3 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)-6 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{37}{12}$} \\
\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{37}{12}-\frac{8}{3}=\frac{5}{12}: \\
3 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)-6 \left(y^2+\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$\frac{5}{12}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{3}+\frac{1}{36}=\left(x-\frac{1}{6}\right)^2: \\
3 \fbox{$\left(x-\frac{1}{6}\right)^2$}-6 \left(y^2+\frac{4 y}{3}+\frac{4}{9}\right)=\frac{5}{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{1}{6}\right)^2-6 \fbox{$\left(y+\frac{2}{3}\right)^2$}=\frac{5}{12} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-6 x-4 y^2+6 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+6 y-3 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+6 y-3 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(-3 x^2-6 x+\underline{\text{ }}\right)+\left(-4 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2-6 x+\underline{\text{ }}\right)=-3 \left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+2 x+\underline{\text{ }}\right)$}+\left(-4 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2+6 y+\underline{\text{ }}\right)=-4 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right): \\
-3 \left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)$}=\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 }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-3\times 1=-3 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-10-3=-13: \\
-3 \left(x^2+2 x+1\right)-4 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$-13$} \\
\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{-3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }-4\times \frac{9}{16}=-\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-13-\frac{9}{4}=-\frac{61}{4}: \\
-3 \left(x^2+2 x+1\right)-4 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$-\frac{61}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
-3 \fbox{$(x+1)^2$}-4 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=-\frac{61}{4} \\
\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{} & -3 (x+1)^2-4 \fbox{$\left(y-\frac{3}{4}\right)^2$}=-\frac{61}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+6 x-4 y^2-y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2-y-3 x^2+6 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
-4 y^2-y-3 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(-3 x^2+6 x+\underline{\text{ }}\right)+\left(-4 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2+6 x+\underline{\text{ }}\right)=-3 \left(x^2-2 x+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2-2 x+\underline{\text{ }}\right)$}+\left(-4 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\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): \\
-3 \left(x^2-2 x+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)$}=\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{-2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-3\times 1=-3 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9-3=6: \\
-3 \left(x^2-2 x+1\right)-4 \left(y^2+\frac{y}{4}+\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 }-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}
6-\frac{1}{16}=\frac{95}{16}: \\
-3 \left(x^2-2 x+1\right)-4 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=\fbox{$\frac{95}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-2 x+1=(x-1)^2: \\
-3 \fbox{$(x-1)^2$}-4 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=\frac{95}{16} \\
\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{} & -3 (x-1)^2-4 \fbox{$\left(y+\frac{1}{8}\right)^2$}=\frac{95}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-9 x+3 y^2+7 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2+7 y+2 x^2-9 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
3 y^2+7 y+2 x^2-9 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^2-9 x+\underline{\text{ }}\right)+\left(3 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2-9 x+\underline{\text{ }}\right)=2 \left(x^2-\frac{9 x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-\frac{9 x}{2}+\underline{\text{ }}\right)$}+\left(3 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 y^2+7 y+\underline{\text{ }}\right)=3 \left(y^2+\frac{7 y}{3}+\underline{\text{ }}\right): \\
2 \left(x^2-\frac{9 x}{2}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2+\frac{7 y}{3}+\underline{\text{ }}\right)$}=\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{-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 7:
\begin{array}{l}
8+\frac{81}{8}=\frac{145}{8}: \\
2 \left(x^2-\frac{9 x}{2}+\frac{81}{16}\right)+3 \left(y^2+\frac{7 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{145}{8}$} \\
\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 }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 9:
\begin{array}{l}
\frac{145}{8}+\frac{49}{12}=\frac{533}{24}: \\
2 \left(x^2-\frac{9 x}{2}+\frac{81}{16}\right)+3 \left(y^2+\frac{7 y}{3}+\frac{49}{36}\right)=\fbox{$\frac{533}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{9 x}{2}+\frac{81}{16}=\left(x-\frac{9}{4}\right)^2: \\
2 \fbox{$\left(x-\frac{9}{4}\right)^2$}+3 \left(y^2+\frac{7 y}{3}+\frac{49}{36}\right)=\frac{533}{24} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{7 y}{3}+\frac{49}{36}=\left(y+\frac{7}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x-\frac{9}{4}\right)^2+3 \fbox{$\left(y+\frac{7}{6}\right)^2$}=\frac{533}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2+5 x-8 y^2+y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2+y+9 x^2+5 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
-8 y^2+y+9 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(9 x^2+5 x+\underline{\text{ }}\right)+\left(-8 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\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(-8 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2+y+\underline{\text{ }}\right)=-8 \left(y^2-\frac{y}{8}+\underline{\text{ }}\right): \\
9 \left(x^2+\frac{5 x}{9}+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2-\frac{y}{8}+\underline{\text{ }}\right)$}=\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 }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}
\frac{25}{36}-8=-\frac{263}{36}: \\
9 \left(x^2+\frac{5 x}{9}+\frac{25}{324}\right)-8 \left(y^2-\frac{y}{8}+\underline{\text{ }}\right)=\fbox{$-\frac{263}{36}$} \\
\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 }-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 9:
\begin{array}{l}
-\frac{263}{36}-\frac{1}{32}=-\frac{2113}{288}: \\
9 \left(x^2+\frac{5 x}{9}+\frac{25}{324}\right)-8 \left(y^2-\frac{y}{8}+\frac{1}{256}\right)=\fbox{$-\frac{2113}{288}$} \\
\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{$\left(x+\frac{5}{18}\right)^2$}-8 \left(y^2-\frac{y}{8}+\frac{1}{256}\right)=-\frac{2113}{288} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{8}+\frac{1}{256}=\left(y-\frac{1}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x+\frac{5}{18}\right)^2-8 \fbox{$\left(y-\frac{1}{16}\right)^2$}=-\frac{2113}{288} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2+3 x+5 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 x^2+3 x+(5 y+10)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 y+10 \text{from }\text{both }\text{sides}: \\
3 x^2+3 x=-5 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(3 x^2+3 x+\underline{\text{ }}\right)=(-5 y-10)+\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{$3 \left(x^2+x+\underline{\text{ }}\right)$}=(-5 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 }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}
(-5 y-10)+\frac{3}{4}=-5 y-\frac{37}{4}: \\
3 \left(x^2+x+\frac{1}{4}\right)=\fbox{$-5 y-\frac{37}{4}$} \\
\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{$\left(x+\frac{1}{2}\right)^2$}=-5 y-\frac{37}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2-2 x+6 y^2-8 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2-8 y+8 x^2-2 x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
6 y^2-8 y+8 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(8 x^2-2 x+\underline{\text{ }}\right)+\left(6 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\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(6 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\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): \\
8 \left(x^2-\frac{x}{4}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right)$}=\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 }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}
7+\frac{1}{8}=\frac{57}{8}: \\
8 \left(x^2-\frac{x}{4}+\frac{1}{64}\right)+6 \left(y^2-\frac{4 y}{3}+\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 }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{57}{8}+\frac{8}{3}=\frac{235}{24}: \\
8 \left(x^2-\frac{x}{4}+\frac{1}{64}\right)+6 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$\frac{235}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{4}+\frac{1}{64}=\left(x-\frac{1}{8}\right)^2: \\
8 \fbox{$\left(x-\frac{1}{8}\right)^2$}+6 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=\frac{235}{24} \\
\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{} & 8 \left(x-\frac{1}{8}\right)^2+6 \fbox{$\left(y-\frac{2}{3}\right)^2$}=\frac{235}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2+6 y^2-2 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2-2 y+\left(10 x^2-7\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
6 y^2-2 y+10 x^2=7 \\
\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^2-2 y+\underline{\text{ }}\right)+10 x^2=\underline{\text{ }}+7 \\
\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): \\
\fbox{$6 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right)$}+10 x^2=\underline{\text{ }}+7 \\
\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{\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 6:
\begin{array}{l}
7+\frac{1}{6}=\frac{43}{6}: \\
6 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)+10 x^2=\fbox{$\frac{43}{6}$} \\
\end{array}
Step 7:
\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{} & 6 \fbox{$\left(y-\frac{1}{6}\right)^2$}+10 x^2=\frac{43}{6} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2+2 x-10 y^2+y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+y+3 x^2+2 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
-10 y^2+y+3 x^2+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(3 x^2+2 x+\underline{\text{ }}\right)+\left(-10 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\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(-10 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\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): \\
3 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2-\frac{y}{10}+\underline{\text{ }}\right)$}=\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 }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}-6=-\frac{17}{3}: \\
3 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)-10 \left(y^2-\frac{y}{10}+\underline{\text{ }}\right)=\fbox{$-\frac{17}{3}$} \\
\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 }-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{17}{3}-\frac{1}{40}=-\frac{683}{120}: \\
3 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)-10 \left(y^2-\frac{y}{10}+\frac{1}{400}\right)=\fbox{$-\frac{683}{120}$} \\
\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{$\left(x+\frac{1}{3}\right)^2$}-10 \left(y^2-\frac{y}{10}+\frac{1}{400}\right)=-\frac{683}{120} \\
\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{} & 3 \left(x+\frac{1}{3}\right)^2-\text{10 }\fbox{$\left(y-\frac{1}{20}\right)^2$}=-\frac{683}{120} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-6 x-3 y^2-10 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2-10 y-10 x^2-6 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
-3 y^2-10 y-10 x^2-6 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^2-6 x+\underline{\text{ }}\right)+\left(-3 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2-6 x+\underline{\text{ }}\right)=-10 \left(x^2+\frac{3 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2+\frac{3 x}{5}+\underline{\text{ }}\right)$}+\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): \\
-10 \left(x^2+\frac{3 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 }-10 \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 }-10\times \frac{9}{100}=-\frac{9}{10} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-4-\frac{9}{10}=-\frac{49}{10}: \\
-10 \left(x^2+\frac{3 x}{5}+\frac{9}{100}\right)-3 \left(y^2+\frac{10 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{49}{10}$} \\
\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 }-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{49}{10}-\frac{25}{3}=-\frac{397}{30}: \\
-10 \left(x^2+\frac{3 x}{5}+\frac{9}{100}\right)-3 \left(y^2+\frac{10 y}{3}+\frac{25}{9}\right)=\fbox{$-\frac{397}{30}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{5}+\frac{9}{100}=\left(x+\frac{3}{10}\right)^2: \\
-10 \fbox{$\left(x+\frac{3}{10}\right)^2$}-3 \left(y^2+\frac{10 y}{3}+\frac{25}{9}\right)=-\frac{397}{30} \\
\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{} & -10 \left(x+\frac{3}{10}\right)^2-3 \fbox{$\left(y+\frac{5}{3}\right)^2$}=-\frac{397}{30} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+8 x+3 y^2-2 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2-2 y+6 x^2+8 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
3 y^2-2 y+6 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(6 x^2+8 x+\underline{\text{ }}\right)+\left(3 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\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(3 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 y^2-2 y+\underline{\text{ }}\right)=3 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right): \\
6 \left(x^2+\frac{4 x}{3}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)$}=\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 }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}
8+\frac{8}{3}=\frac{32}{3}: \\
6 \left(x^2+\frac{4 x}{3}+\frac{4}{9}\right)+3 \left(y^2-\frac{2 y}{3}+\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 }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 9:
\begin{array}{l}
\frac{32}{3}+\frac{1}{3}=11: \\
6 \left(x^2+\frac{4 x}{3}+\frac{4}{9}\right)+3 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=\fbox{$11$} \\
\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{$\left(x+\frac{2}{3}\right)^2$}+3 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=11 \\
\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+3 \fbox{$\left(y-\frac{1}{3}\right)^2$}=11 \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-3 x+9 y^2+7 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+7 y-10 x^2-3 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
9 y^2+7 y-10 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(-10 x^2-3 x+\underline{\text{ }}\right)+\left(9 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\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(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): \\
-10 \left(x^2+\frac{3 x}{10}+\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 }-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}
-4-\frac{9}{40}=-\frac{169}{40}: \\
-10 \left(x^2+\frac{3 x}{10}+\frac{9}{400}\right)+9 \left(y^2+\frac{7 y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{169}{40}$} \\
\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 }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{169}{40}=-\frac{1031}{360}: \\
-10 \left(x^2+\frac{3 x}{10}+\frac{9}{400}\right)+9 \left(y^2+\frac{7 y}{9}+\frac{49}{324}\right)=\fbox{$-\frac{1031}{360}$} \\
\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{$\left(x+\frac{3}{20}\right)^2$}+9 \left(y^2+\frac{7 y}{9}+\frac{49}{324}\right)=-\frac{1031}{360} \\
\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{} & -10 \left(x+\frac{3}{20}\right)^2+9 \fbox{$\left(y+\frac{7}{18}\right)^2$}=-\frac{1031}{360} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-4 x+7 y^2+8 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2+8 y-10 x^2-4 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
7 y^2+8 y-10 x^2-4 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(-10 x^2-4 x+\underline{\text{ }}\right)+\left(7 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2-4 x+\underline{\text{ }}\right)=-10 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right)$}+\left(7 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\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): \\
-10 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right)$}=\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 }-10 \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{-10}{25}=-\frac{2}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6-\frac{2}{5}=\frac{28}{5}: \\
-10 \left(x^2+\frac{2 x}{5}+\frac{1}{25}\right)+7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{28}{5}$} \\
\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{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{28}{5}+\frac{16}{7}=\frac{276}{35}: \\
-10 \left(x^2+\frac{2 x}{5}+\frac{1}{25}\right)+7 \left(y^2+\frac{8 y}{7}+\frac{16}{49}\right)=\fbox{$\frac{276}{35}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{5}+\frac{1}{25}=\left(x+\frac{1}{5}\right)^2: \\
-10 \fbox{$\left(x+\frac{1}{5}\right)^2$}+7 \left(y^2+\frac{8 y}{7}+\frac{16}{49}\right)=\frac{276}{35} \\
\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{} & -10 \left(x+\frac{1}{5}\right)^2+7 \fbox{$\left(y+\frac{4}{7}\right)^2$}=\frac{276}{35} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-x-3 y^2+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 x^2-x+\left(4-3 y^2\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
-3 y^2+5 x^2-x=-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-x+\underline{\text{ }}\right)-3 y^2=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2-x+\underline{\text{ }}\right)=5 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right)$}-3 y^2=\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{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{-1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{5}{100}=\frac{1}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\frac{1}{20}-4=-\frac{79}{20}: \\
5 \left(x^2-\frac{x}{5}+\frac{1}{100}\right)-3 y^2=\fbox{$-\frac{79}{20}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2-\frac{x}{5}+\frac{1}{100}=\left(x-\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$\left(x-\frac{1}{10}\right)^2$}-3 y^2=-\frac{79}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+5 x+6 y^2+10 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2+10 y+6 x^2+5 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
6 y^2+10 y+6 x^2+5 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(6 x^2+5 x+\underline{\text{ }}\right)+\left(6 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\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(6 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2+10 y+\underline{\text{ }}\right)=6 \left(y^2+\frac{5 y}{3}+\underline{\text{ }}\right): \\
6 \left(x^2+\frac{5 x}{6}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2+\frac{5 y}{3}+\underline{\text{ }}\right)$}=\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 }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}
\frac{25}{24}-2=-\frac{23}{24}: \\
6 \left(x^2+\frac{5 x}{6}+\frac{25}{144}\right)+6 \left(y^2+\frac{5 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{23}{24}$} \\
\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 }6 \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 }6\times \frac{25}{36}=\frac{25}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{25}{6}-\frac{23}{24}=\frac{77}{24}: \\
6 \left(x^2+\frac{5 x}{6}+\frac{25}{144}\right)+6 \left(y^2+\frac{5 y}{3}+\frac{25}{36}\right)=\fbox{$\frac{77}{24}$} \\
\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{$\left(x+\frac{5}{12}\right)^2$}+6 \left(y^2+\frac{5 y}{3}+\frac{25}{36}\right)=\frac{77}{24} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{3}+\frac{25}{36}=\left(y+\frac{5}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x+\frac{5}{12}\right)^2+6 \fbox{$\left(y+\frac{5}{6}\right)^2$}=\frac{77}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2-7 x+8 y^2-8 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-8 y+7 x^2-7 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
8 y^2-8 y+7 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(7 x^2-7 x+\underline{\text{ }}\right)+\left(8 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2-7 x+\underline{\text{ }}\right)=7 \left(x^2-x+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2-x+\underline{\text{ }}\right)$}+\left(8 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-8 y+\underline{\text{ }}\right)=8 \left(y^2-y+\underline{\text{ }}\right): \\
7 \left(x^2-x+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-y+\underline{\text{ }}\right)$}=\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 }7 \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{7}{4}=\frac{7}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2+\frac{7}{4}=\frac{15}{4}: \\
7 \left(x^2-x+\frac{1}{4}\right)+8 \left(y^2-y+\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 }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}
\frac{15}{4}+2=\frac{23}{4}: \\
7 \left(x^2-x+\frac{1}{4}\right)+8 \left(y^2-y+\frac{1}{4}\right)=\fbox{$\frac{23}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2: \\
7 \fbox{$\left(x-\frac{1}{2}\right)^2$}+8 \left(y^2-y+\frac{1}{4}\right)=\frac{23}{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{} & 7 \left(x-\frac{1}{2}\right)^2+8 \fbox{$\left(y-\frac{1}{2}\right)^2$}=\frac{23}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-3 x-7 y^2+5 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2+5 y+2 x^2-3 x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
-7 y^2+5 y+2 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(2 x^2-3 x+\underline{\text{ }}\right)+\left(-7 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\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(-7 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2+5 y+\underline{\text{ }}\right)=-7 \left(y^2-\frac{5 y}{7}+\underline{\text{ }}\right): \\
2 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2-\frac{5 y}{7}+\underline{\text{ }}\right)$}=\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 }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}
7+\frac{9}{8}=\frac{65}{8}: \\
2 \left(x^2-\frac{3 x}{2}+\frac{9}{16}\right)-7 \left(y^2-\frac{5 y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{65}{8}$} \\
\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{-5}{7}}{2}\right)^2=\frac{25}{196} \text{on }\text{the }\text{left }\text{and }-7\times \frac{25}{196}=-\frac{25}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{65}{8}-\frac{25}{28}=\frac{405}{56}: \\
2 \left(x^2-\frac{3 x}{2}+\frac{9}{16}\right)-7 \left(y^2-\frac{5 y}{7}+\frac{25}{196}\right)=\fbox{$\frac{405}{56}$} \\
\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{$\left(x-\frac{3}{4}\right)^2$}-7 \left(y^2-\frac{5 y}{7}+\frac{25}{196}\right)=\frac{405}{56} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{5 y}{7}+\frac{25}{196}=\left(y-\frac{5}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x-\frac{3}{4}\right)^2-7 \fbox{$\left(y-\frac{5}{14}\right)^2$}=\frac{405}{56} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2+8 x-3 y^2+4 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2+4 y+5 x^2+8 x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
-3 y^2+4 y+5 x^2+8 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+8 x+\underline{\text{ }}\right)+\left(-3 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\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(-3 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2+4 y+\underline{\text{ }}\right)=-3 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right): \\
5 \left(x^2+\frac{8 x}{5}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right)$}=\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{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}
7+\frac{16}{5}=\frac{51}{5}: \\
5 \left(x^2+\frac{8 x}{5}+\frac{16}{25}\right)-3 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{51}{5}$} \\
\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{-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 9:
\begin{array}{l}
\frac{51}{5}-\frac{4}{3}=\frac{133}{15}: \\
5 \left(x^2+\frac{8 x}{5}+\frac{16}{25}\right)-3 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$\frac{133}{15}$} \\
\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{$\left(x+\frac{4}{5}\right)^2$}-3 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=\frac{133}{15} \\
\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{} & 5 \left(x+\frac{4}{5}\right)^2-3 \fbox{$\left(y-\frac{2}{3}\right)^2$}=\frac{133}{15} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-5 x-9 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 x^2-5 x+(-9 y-7)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 y+7 \text{to }\text{both }\text{sides}: \\
2 x^2-5 x=9 y+7 \\
\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(2 x^2-5 x+\underline{\text{ }}\right)=(9 y+7)+\underline{\text{ }} \\
\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{$2 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)$}=(9 y+7)+\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 }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 6:
\begin{array}{l}
(9 y+7)+\frac{25}{8}=9 y+\frac{81}{8}: \\
2 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)=\fbox{$9 y+\frac{81}{8}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2-\frac{5 x}{2}+\frac{25}{16}=\left(x-\frac{5}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \fbox{$\left(x-\frac{5}{4}\right)^2$}=9 y+\frac{81}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2-9 x+6 y^2+10 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2+10 y-2 x^2-9 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
6 y^2+10 y-2 x^2-9 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^2-9 x+\underline{\text{ }}\right)+\left(6 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 x^2-9 x+\underline{\text{ }}\right)=-2 \left(x^2+\frac{9 x}{2}+\underline{\text{ }}\right): \\
\fbox{$-2 \left(x^2+\frac{9 x}{2}+\underline{\text{ }}\right)$}+\left(6 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2+10 y+\underline{\text{ }}\right)=6 \left(y^2+\frac{5 y}{3}+\underline{\text{ }}\right): \\
-2 \left(x^2+\frac{9 x}{2}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2+\frac{5 y}{3}+\underline{\text{ }}\right)$}=\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 }-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 7:
\begin{array}{l}
-9-\frac{81}{8}=-\frac{153}{8}: \\
-2 \left(x^2+\frac{9 x}{2}+\frac{81}{16}\right)+6 \left(y^2+\frac{5 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{153}{8}$} \\
\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 }6 \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 }6\times \frac{25}{36}=\frac{25}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{25}{6}-\frac{153}{8}=-\frac{359}{24}: \\
-2 \left(x^2+\frac{9 x}{2}+\frac{81}{16}\right)+6 \left(y^2+\frac{5 y}{3}+\frac{25}{36}\right)=\fbox{$-\frac{359}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{9 x}{2}+\frac{81}{16}=\left(x+\frac{9}{4}\right)^2: \\
-2 \fbox{$\left(x+\frac{9}{4}\right)^2$}+6 \left(y^2+\frac{5 y}{3}+\frac{25}{36}\right)=-\frac{359}{24} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{3}+\frac{25}{36}=\left(y+\frac{5}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -2 \left(x+\frac{9}{4}\right)^2+6 \fbox{$\left(y+\frac{5}{6}\right)^2$}=-\frac{359}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+4 x-3 y^2-5 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2-5 y-3 x^2+4 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
-3 y^2-5 y-3 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(-3 x^2+4 x+\underline{\text{ }}\right)+\left(-3 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2+4 x+\underline{\text{ }}\right)=-3 \left(x^2-\frac{4 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2-\frac{4 x}{3}+\underline{\text{ }}\right)$}+\left(-3 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2-5 y+\underline{\text{ }}\right)=-3 \left(y^2+\frac{5 y}{3}+\underline{\text{ }}\right): \\
-3 \left(x^2-\frac{4 x}{3}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2+\frac{5 y}{3}+\underline{\text{ }}\right)$}=\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 }-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}
-8-\frac{4}{3}=-\frac{28}{3}: \\
-3 \left(x^2-\frac{4 x}{3}+\frac{4}{9}\right)-3 \left(y^2+\frac{5 y}{3}+\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 }-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 9:
\begin{array}{l}
-\frac{28}{3}-\frac{25}{12}=-\frac{137}{12}: \\
-3 \left(x^2-\frac{4 x}{3}+\frac{4}{9}\right)-3 \left(y^2+\frac{5 y}{3}+\frac{25}{36}\right)=\fbox{$-\frac{137}{12}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{4 x}{3}+\frac{4}{9}=\left(x-\frac{2}{3}\right)^2: \\
-3 \fbox{$\left(x-\frac{2}{3}\right)^2$}-3 \left(y^2+\frac{5 y}{3}+\frac{25}{36}\right)=-\frac{137}{12} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{3}+\frac{25}{36}=\left(y+\frac{5}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x-\frac{2}{3}\right)^2-3 \fbox{$\left(y+\frac{5}{6}\right)^2$}=-\frac{137}{12} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+x-6 y^2+4 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+4 y-3 x^2+x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
-6 y^2+4 y-3 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(-3 x^2+x+\underline{\text{ }}\right)+\left(-6 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2+x+\underline{\text{ }}\right)=-3 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)$}+\left(-6 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\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): \\
-3 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)$}=\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{-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 7:
\begin{array}{l}
-2-\frac{1}{12}=-\frac{25}{12}: \\
-3 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)-6 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{25}{12}$} \\
\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{-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{25}{12}-\frac{2}{3}=-\frac{11}{4}: \\
-3 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)-6 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=\fbox{$-\frac{11}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{3}+\frac{1}{36}=\left(x-\frac{1}{6}\right)^2: \\
-3 \fbox{$\left(x-\frac{1}{6}\right)^2$}-6 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)=-\frac{11}{4} \\
\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{} & -3 \left(x-\frac{1}{6}\right)^2-6 \fbox{$\left(y-\frac{1}{3}\right)^2$}=-\frac{11}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-7 x+9 y^2-5 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2-5 y-10 x^2-7 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
9 y^2-5 y-10 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(-10 x^2-7 x+\underline{\text{ }}\right)+\left(9 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2-7 x+\underline{\text{ }}\right)=-10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right)$}+\left(9 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\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): \\
-10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2-\frac{5 y}{9}+\underline{\text{ }}\right)$}=\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 }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{10}}{2}\right)^2=\frac{49}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{49}{400}=-\frac{49}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-1-\frac{49}{40}=-\frac{89}{40}: \\
-10 \left(x^2+\frac{7 x}{10}+\frac{49}{400}\right)+9 \left(y^2-\frac{5 y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{89}{40}$} \\
\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 }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{25}{36}-\frac{89}{40}=-\frac{551}{360}: \\
-10 \left(x^2+\frac{7 x}{10}+\frac{49}{400}\right)+9 \left(y^2-\frac{5 y}{9}+\frac{25}{324}\right)=\fbox{$-\frac{551}{360}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{10}+\frac{49}{400}=\left(x+\frac{7}{20}\right)^2: \\
-10 \fbox{$\left(x+\frac{7}{20}\right)^2$}+9 \left(y^2-\frac{5 y}{9}+\frac{25}{324}\right)=-\frac{551}{360} \\
\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{} & -10 \left(x+\frac{7}{20}\right)^2+9 \fbox{$\left(y-\frac{5}{18}\right)^2$}=-\frac{551}{360} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-5 x+9 y^2+y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+y+4 x^2-5 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
9 y^2+y+4 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(4 x^2-5 x+\underline{\text{ }}\right)+\left(9 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\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(9 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\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): \\
4 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2+\frac{y}{9}+\underline{\text{ }}\right)$}=\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 }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}
4+\frac{25}{16}=\frac{89}{16}: \\
4 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)+9 \left(y^2+\frac{y}{9}+\underline{\text{ }}\right)=\fbox{$\frac{89}{16}$} \\
\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 }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{89}{16}+\frac{1}{36}=\frac{805}{144}: \\
4 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)+9 \left(y^2+\frac{y}{9}+\frac{1}{324}\right)=\fbox{$\frac{805}{144}$} \\
\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{$\left(x-\frac{5}{8}\right)^2$}+9 \left(y^2+\frac{y}{9}+\frac{1}{324}\right)=\frac{805}{144} \\
\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{} & 4 \left(x-\frac{5}{8}\right)^2+9 \fbox{$\left(y+\frac{1}{18}\right)^2$}=\frac{805}{144} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2-9 x+2 y^2+6 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+6 y-4 x^2-9 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
2 y^2+6 y-4 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(-4 x^2-9 x+\underline{\text{ }}\right)+\left(2 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2-9 x+\underline{\text{ }}\right)=-4 \left(x^2+\frac{9 x}{4}+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2+\frac{9 x}{4}+\underline{\text{ }}\right)$}+\left(2 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\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): \\
-4 \left(x^2+\frac{9 x}{4}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+3 y+\underline{\text{ }}\right)$}=\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 }-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 7:
\begin{array}{l}
1-\frac{81}{16}=-\frac{65}{16}: \\
-4 \left(x^2+\frac{9 x}{4}+\frac{81}{64}\right)+2 \left(y^2+3 y+\underline{\text{ }}\right)=\fbox{$-\frac{65}{16}$} \\
\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 }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{9}{2}-\frac{65}{16}=\frac{7}{16}: \\
-4 \left(x^2+\frac{9 x}{4}+\frac{81}{64}\right)+2 \left(y^2+3 y+\frac{9}{4}\right)=\fbox{$\frac{7}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{9 x}{4}+\frac{81}{64}=\left(x+\frac{9}{8}\right)^2: \\
-4 \fbox{$\left(x+\frac{9}{8}\right)^2$}+2 \left(y^2+3 y+\frac{9}{4}\right)=\frac{7}{16} \\
\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{} & -4 \left(x+\frac{9}{8}\right)^2+2 \fbox{$\left(y+\frac{3}{2}\right)^2$}=\frac{7}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2+7 x-9 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
x^2+7 x+(9-9 y)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9-9 y \text{from }\text{both }\text{sides}: \\
x^2+7 x=9 y-9 \\
\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(x^2+7 x+\underline{\text{ }}\right)=(9 y-9)+\underline{\text{ }} \\
\end{array}
Step 4:
\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{7}{2}\right)^2=\frac{49}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 5:
\begin{array}{l}
(9 y-9)+\frac{49}{4}=9 y+\frac{13}{4}: \\
\left(x^2+7 x+\frac{49}{4}\right)=\fbox{$9 y+\frac{13}{4}$} \\
\end{array}
Step 6:
\begin{array}{l}
x^2+7 x+\frac{49}{4}=\left(x+\frac{7}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \fbox{$\left(x+\frac{7}{2}\right)^2$}=9 y+\frac{13}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2+9 x-4 y^2+2 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+2 y-4 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+2 y-4 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(-4 x^2+9 x+\underline{\text{ }}\right)+\left(-4 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2+9 x+\underline{\text{ }}\right)=-4 \left(x^2-\frac{9 x}{4}+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2-\frac{9 x}{4}+\underline{\text{ }}\right)$}+\left(-4 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\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): \\
-4 \left(x^2-\frac{9 x}{4}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}=\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 }-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 7:
\begin{array}{l}
-3-\frac{81}{16}=-\frac{129}{16}: \\
-4 \left(x^2-\frac{9 x}{4}+\frac{81}{64}\right)-4 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{129}{16}$} \\
\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{-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{129}{16}-\frac{1}{4}=-\frac{133}{16}: \\
-4 \left(x^2-\frac{9 x}{4}+\frac{81}{64}\right)-4 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\fbox{$-\frac{133}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{9 x}{4}+\frac{81}{64}=\left(x-\frac{9}{8}\right)^2: \\
-4 \fbox{$\left(x-\frac{9}{8}\right)^2$}-4 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=-\frac{133}{16} \\
\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{9}{8}\right)^2-4 \fbox{$\left(y-\frac{1}{4}\right)^2$}=-\frac{133}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2-4 x+7 y^2+8 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2+8 y+10 x^2-4 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
7 y^2+8 y+10 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(10 x^2-4 x+\underline{\text{ }}\right)+\left(7 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(10 x^2-4 x+\underline{\text{ }}\right)=10 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right)$}+\left(7 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\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): \\
10 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right)$}=\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 }\text{10 }\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{10}{25}=\frac{2}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{2}{5}-3=-\frac{13}{5}: \\
10 \left(x^2-\frac{2 x}{5}+\frac{1}{25}\right)+7 \left(y^2+\frac{8 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{13}{5}$} \\
\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{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{16}{7}-\frac{13}{5}=-\frac{11}{35}: \\
10 \left(x^2-\frac{2 x}{5}+\frac{1}{25}\right)+7 \left(y^2+\frac{8 y}{7}+\frac{16}{49}\right)=\fbox{$-\frac{11}{35}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{2 x}{5}+\frac{1}{25}=\left(x-\frac{1}{5}\right)^2: \\
\text{10 }\fbox{$\left(x-\frac{1}{5}\right)^2$}+7 \left(y^2+\frac{8 y}{7}+\frac{16}{49}\right)=-\frac{11}{35} \\
\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{} & 10 \left(x-\frac{1}{5}\right)^2+7 \fbox{$\left(y+\frac{4}{7}\right)^2$}=-\frac{11}{35} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-3 x-2 y^2+5 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+5 y+5 x^2-3 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
-2 y^2+5 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(-2 y^2+5 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(-2 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+5 y+\underline{\text{ }}\right)=-2 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right): \\
5 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)$}=\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 }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}
2+\frac{9}{20}=\frac{49}{20}: \\
5 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)-2 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{49}{20}$} \\
\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{\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 9:
\begin{array}{l}
\frac{49}{20}-\frac{25}{8}=-\frac{27}{40}: \\
5 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)-2 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$-\frac{27}{40}$} \\
\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{$\left(x-\frac{3}{10}\right)^2$}-2 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=-\frac{27}{40} \\
\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{} & 5 \left(x-\frac{3}{10}\right)^2-2 \fbox{$\left(y-\frac{5}{4}\right)^2$}=-\frac{27}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2-4 x-3 y^2-7 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2-7 y-7 x^2-4 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-3 y^2-7 y-7 x^2-4 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(-7 x^2-4 x+\underline{\text{ }}\right)+\left(-3 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\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(-3 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2-7 y+\underline{\text{ }}\right)=-3 \left(y^2+\frac{7 y}{3}+\underline{\text{ }}\right): \\
-7 \left(x^2+\frac{4 x}{7}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2+\frac{7 y}{3}+\underline{\text{ }}\right)$}=\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 }-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}
6-\frac{4}{7}=\frac{38}{7}: \\
-7 \left(x^2+\frac{4 x}{7}+\frac{4}{49}\right)-3 \left(y^2+\frac{7 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{38}{7}$} \\
\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 }-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 9:
\begin{array}{l}
\frac{38}{7}-\frac{49}{12}=\frac{113}{84}: \\
-7 \left(x^2+\frac{4 x}{7}+\frac{4}{49}\right)-3 \left(y^2+\frac{7 y}{3}+\frac{49}{36}\right)=\fbox{$\frac{113}{84}$} \\
\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{$\left(x+\frac{2}{7}\right)^2$}-3 \left(y^2+\frac{7 y}{3}+\frac{49}{36}\right)=\frac{113}{84} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{7 y}{3}+\frac{49}{36}=\left(y+\frac{7}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x+\frac{2}{7}\right)^2-3 \fbox{$\left(y+\frac{7}{6}\right)^2$}=\frac{113}{84} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-x-9 y^2-4 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-4 y-10 x^2-x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-9 y^2-4 y-10 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(-10 x^2-x+\underline{\text{ }}\right)+\left(-9 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\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(-9 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\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): \\
-10 \left(x^2+\frac{x}{10}+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2+\frac{4 y}{9}+\underline{\text{ }}\right)$}=\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 }-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}
-7-\frac{1}{40}=-\frac{281}{40}: \\
-10 \left(x^2+\frac{x}{10}+\frac{1}{400}\right)-9 \left(y^2+\frac{4 y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{281}{40}$} \\
\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 }-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{281}{40}-\frac{4}{9}=-\frac{2689}{360}: \\
-10 \left(x^2+\frac{x}{10}+\frac{1}{400}\right)-9 \left(y^2+\frac{4 y}{9}+\frac{4}{81}\right)=\fbox{$-\frac{2689}{360}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{10}+\frac{1}{400}=\left(x+\frac{1}{20}\right)^2: \\
-10 \fbox{$\left(x+\frac{1}{20}\right)^2$}-9 \left(y^2+\frac{4 y}{9}+\frac{4}{81}\right)=-\frac{2689}{360} \\
\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{} & -10 \left(x+\frac{1}{20}\right)^2-9 \fbox{$\left(y+\frac{2}{9}\right)^2$}=-\frac{2689}{360} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2+3 x-7 y^2-4 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2-4 y-5 x^2+3 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
-7 y^2-4 y-5 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(-5 x^2+3 x+\underline{\text{ }}\right)+\left(-7 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\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(-7 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\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): \\
-5 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2+\frac{4 y}{7}+\underline{\text{ }}\right)$}=\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 }-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}
9-\frac{9}{20}=\frac{171}{20}: \\
-5 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)-7 \left(y^2+\frac{4 y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{171}{20}$} \\
\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{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{171}{20}-\frac{4}{7}=\frac{1117}{140}: \\
-5 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)-7 \left(y^2+\frac{4 y}{7}+\frac{4}{49}\right)=\fbox{$\frac{1117}{140}$} \\
\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{$\left(x-\frac{3}{10}\right)^2$}-7 \left(y^2+\frac{4 y}{7}+\frac{4}{49}\right)=\frac{1117}{140} \\
\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{} & -5 \left(x-\frac{3}{10}\right)^2-7 \fbox{$\left(y+\frac{2}{7}\right)^2$}=\frac{1117}{140} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2+10 x+8 y^2+9 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2+9 y-4 x^2+10 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
8 y^2+9 y-4 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(-4 x^2+10 x+\underline{\text{ }}\right)+\left(8 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\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+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\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): \\
-4 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2+\frac{9 y}{8}+\underline{\text{ }}\right)$}=\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{-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}
-2-\frac{25}{4}=-\frac{33}{4}: \\
-4 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)+8 \left(y^2+\frac{9 y}{8}+\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 }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{33}{4}=-\frac{183}{32}: \\
-4 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)+8 \left(y^2+\frac{9 y}{8}+\frac{81}{256}\right)=\fbox{$-\frac{183}{32}$} \\
\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{9 y}{8}+\frac{81}{256}\right)=-\frac{183}{32} \\
\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{} & -4 \left(x-\frac{5}{4}\right)^2+8 \fbox{$\left(y+\frac{9}{16}\right)^2$}=-\frac{183}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2+10 x-4 y^2-8 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2-8 y-4 x^2+10 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
-4 y^2-8 y-4 x^2+10 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(-4 x^2+10 x+\underline{\text{ }}\right)+\left(-4 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\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(-4 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2-8 y+\underline{\text{ }}\right)=-4 \left(y^2+2 y+\underline{\text{ }}\right): \\
-4 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2+2 y+\underline{\text{ }}\right)$}=\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 }-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}
10-\frac{25}{4}=\frac{15}{4}: \\
-4 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)-4 \left(y^2+2 y+\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{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-4\times 1=-4 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{15}{4}-4=-\frac{1}{4}: \\
-4 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)-4 \left(y^2+2 y+1\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{$\left(x-\frac{5}{4}\right)^2$}-4 \left(y^2+2 y+1\right)=-\frac{1}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+2 y+1=(y+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x-\frac{5}{4}\right)^2-4 \fbox{$(y+1)^2$}=-\frac{1}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-7 x-6 y^2-3 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-3 y-3 x^2-7 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
-6 y^2-3 y-3 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(-3 x^2-7 x+\underline{\text{ }}\right)+\left(-6 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2-7 x+\underline{\text{ }}\right)=-3 \left(x^2+\frac{7 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+\frac{7 x}{3}+\underline{\text{ }}\right)$}+\left(-6 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\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): \\
-3 \left(x^2+\frac{7 x}{3}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)$}=\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 }-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}
-3-\frac{49}{12}=-\frac{85}{12}: \\
-3 \left(x^2+\frac{7 x}{3}+\frac{49}{36}\right)-6 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{85}{12}$} \\
\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{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{85}{12}-\frac{3}{8}=-\frac{179}{24}: \\
-3 \left(x^2+\frac{7 x}{3}+\frac{49}{36}\right)-6 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\fbox{$-\frac{179}{24}$} \\
\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{y}{2}+\frac{1}{16}\right)=-\frac{179}{24} \\
\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{} & -3 \left(x+\frac{7}{6}\right)^2-6 \fbox{$\left(y+\frac{1}{4}\right)^2$}=-\frac{179}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+7 x+7 y^2+7 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 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}: \\
7 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(7 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(7 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2+7 y+\underline{\text{ }}\right)=7 \left(y^2+y+\underline{\text{ }}\right): \\
4 \left(x^2+\frac{7 x}{4}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2+y+\underline{\text{ }}\right)$}=\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{97}{16}: \\
4 \left(x^2+\frac{7 x}{4}+\frac{49}{64}\right)+7 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$\frac{97}{16}$} \\
\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{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{7}{4}=\frac{7}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{97}{16}+\frac{7}{4}=\frac{125}{16}: \\
4 \left(x^2+\frac{7 x}{4}+\frac{49}{64}\right)+7 \left(y^2+y+\frac{1}{4}\right)=\fbox{$\frac{125}{16}$} \\
\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{$\left(x+\frac{7}{8}\right)^2$}+7 \left(y^2+y+\frac{1}{4}\right)=\frac{125}{16} \\
\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{7}{8}\right)^2+7 \fbox{$\left(y+\frac{1}{2}\right)^2$}=\frac{125}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2+5 x-6 y^2+5 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+5 y+7 x^2+5 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-6 y^2+5 y+7 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(7 x^2+5 x+\underline{\text{ }}\right)+\left(-6 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2+5 x+\underline{\text{ }}\right)=7 \left(x^2+\frac{5 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2+\frac{5 x}{7}+\underline{\text{ }}\right)$}+\left(-6 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\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): \\
7 \left(x^2+\frac{5 x}{7}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{5 y}{6}+\underline{\text{ }}\right)$}=\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 }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{7}}{2}\right)^2=\frac{25}{196} \text{on }\text{the }\text{left }\text{and }7\times \frac{25}{196}=\frac{25}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6+\frac{25}{28}=\frac{193}{28}: \\
7 \left(x^2+\frac{5 x}{7}+\frac{25}{196}\right)-6 \left(y^2-\frac{5 y}{6}+\underline{\text{ }}\right)=\fbox{$\frac{193}{28}$} \\
\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{-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{193}{28}-\frac{25}{24}=\frac{983}{168}: \\
7 \left(x^2+\frac{5 x}{7}+\frac{25}{196}\right)-6 \left(y^2-\frac{5 y}{6}+\frac{25}{144}\right)=\fbox{$\frac{983}{168}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{7}+\frac{25}{196}=\left(x+\frac{5}{14}\right)^2: \\
7 \fbox{$\left(x+\frac{5}{14}\right)^2$}-6 \left(y^2-\frac{5 y}{6}+\frac{25}{144}\right)=\frac{983}{168} \\
\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{} & 7 \left(x+\frac{5}{14}\right)^2-6 \fbox{$\left(y-\frac{5}{12}\right)^2$}=\frac{983}{168} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2-5 x-2 y^2-3 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-3 y-6 x^2-5 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
-2 y^2-3 y-6 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(-6 x^2-5 x+\underline{\text{ }}\right)+\left(-2 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\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(-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): \\
-6 \left(x^2+\frac{5 x}{6}+\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 }-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}
8-\frac{25}{24}=\frac{167}{24}: \\
-6 \left(x^2+\frac{5 x}{6}+\frac{25}{144}\right)-2 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{167}{24}$} \\
\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 }-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{167}{24}-\frac{9}{8}=\frac{35}{6}: \\
-6 \left(x^2+\frac{5 x}{6}+\frac{25}{144}\right)-2 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$\frac{35}{6}$} \\
\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{$\left(x+\frac{5}{12}\right)^2$}-2 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\frac{35}{6} \\
\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{} & -6 \left(x+\frac{5}{12}\right)^2-2 \fbox{$\left(y+\frac{3}{4}\right)^2$}=\frac{35}{6} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2+5 x-4 y^2+10 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+10 y-7 x^2+5 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-4 y^2+10 y-7 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(-7 x^2+5 x+\underline{\text{ }}\right)+\left(-4 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 x^2+5 x+\underline{\text{ }}\right)=-7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right)$}+\left(-4 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\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): \\
-7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)$}=\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 }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{7}}{2}\right)^2=\frac{25}{196} \text{on }\text{the }\text{left }\text{and }-7\times \frac{25}{196}=-\frac{25}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-1-\frac{25}{28}=-\frac{53}{28}: \\
-7 \left(x^2-\frac{5 x}{7}+\frac{25}{196}\right)-4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{53}{28}$} \\
\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 }-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{53}{28}-\frac{25}{4}=-\frac{57}{7}: \\
-7 \left(x^2-\frac{5 x}{7}+\frac{25}{196}\right)-4 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$-\frac{57}{7}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{7}+\frac{25}{196}=\left(x-\frac{5}{14}\right)^2: \\
-7 \fbox{$\left(x-\frac{5}{14}\right)^2$}-4 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=-\frac{57}{7} \\
\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{} & -7 \left(x-\frac{5}{14}\right)^2-4 \fbox{$\left(y-\frac{5}{4}\right)^2$}=-\frac{57}{7} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+2 y^2+3 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+3 y+\left(4 x^2+6\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
2 y^2+3 y+4 x^2=-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(2 y^2+3 y+\underline{\text{ }}\right)+4 x^2=\underline{\text{ }}-6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 y^2+3 y+\underline{\text{ }}\right)=2 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)$}+4 x^2=\underline{\text{ }}-6 \\
\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 }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 6:
\begin{array}{l}
\frac{9}{8}-6=-\frac{39}{8}: \\
2 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)+4 x^2=\fbox{$-\frac{39}{8}$} \\
\end{array}
Step 7:
\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 \fbox{$\left(y+\frac{3}{4}\right)^2$}+4 x^2=-\frac{39}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2-9 x+2 y^2+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 x^2-9 x+\left(2 y^2+7\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
2 y^2+10 x^2-9 x=-7 \\
\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(10 x^2-9 x+\underline{\text{ }}\right)+2 y^2=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(10 x^2-9 x+\underline{\text{ }}\right)=10 \left(x^2-\frac{9 x}{10}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2-\frac{9 x}{10}+\underline{\text{ }}\right)$}+2 y^2=\underline{\text{ }}-7 \\
\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{-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 6:
\begin{array}{l}
\frac{81}{40}-7=-\frac{199}{40}: \\
10 \left(x^2-\frac{9 x}{10}+\frac{81}{400}\right)+2 y^2=\fbox{$-\frac{199}{40}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2-\frac{9 x}{10}+\frac{81}{400}=\left(x-\frac{9}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \text{10 }\fbox{$\left(x-\frac{9}{20}\right)^2$}+2 y^2=-\frac{199}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-8 x+4 y^2-9 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2-9 y-10 x^2-8 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
4 y^2-9 y-10 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(-10 x^2-8 x+\underline{\text{ }}\right)+\left(4 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\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(4 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\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): \\
-10 \left(x^2+\frac{4 x}{5}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2-\frac{9 y}{4}+\underline{\text{ }}\right)$}=\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 }-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 7:
\begin{array}{l}
-8-\frac{8}{5}=-\frac{48}{5}: \\
-10 \left(x^2+\frac{4 x}{5}+\frac{4}{25}\right)+4 \left(y^2-\frac{9 y}{4}+\underline{\text{ }}\right)=\fbox{$-\frac{48}{5}$} \\
\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 }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{81}{16}-\frac{48}{5}=-\frac{363}{80}: \\
-10 \left(x^2+\frac{4 x}{5}+\frac{4}{25}\right)+4 \left(y^2-\frac{9 y}{4}+\frac{81}{64}\right)=\fbox{$-\frac{363}{80}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{4 x}{5}+\frac{4}{25}=\left(x+\frac{2}{5}\right)^2: \\
-10 \fbox{$\left(x+\frac{2}{5}\right)^2$}+4 \left(y^2-\frac{9 y}{4}+\frac{81}{64}\right)=-\frac{363}{80} \\
\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{} & -10 \left(x+\frac{2}{5}\right)^2+4 \fbox{$\left(y-\frac{9}{8}\right)^2$}=-\frac{363}{80} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2+9 y^2-6 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2-6 y+\left(2-6 x^2\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
9 y^2-6 y-6 x^2=-2 \\
\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(9 y^2-6 y+\underline{\text{ }}\right)-6 x^2=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 y^2-6 y+\underline{\text{ }}\right)=9 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right): \\
\fbox{$9 \left(y^2-\frac{2 y}{3}+\underline{\text{ }}\right)$}-6 x^2=\underline{\text{ }}-2 \\
\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 }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 6:
\begin{array}{l}
1-2=-1: \\
9 \left(y^2-\frac{2 y}{3}+\frac{1}{9}\right)-6 x^2=\fbox{$-1$} \\
\end{array}
Step 7:
\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 \fbox{$\left(y-\frac{1}{3}\right)^2$}-6 x^2=-1 \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+10 x-5 y^2-6 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2-6 y+6 x^2+10 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
-5 y^2-6 y+6 x^2+10 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+10 x+\underline{\text{ }}\right)+\left(-5 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(6 x^2+10 x+\underline{\text{ }}\right)=6 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right)$}+\left(-5 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\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): \\
6 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2+\frac{6 y}{5}+\underline{\text{ }}\right)$}=\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{5}{3}}{2}\right)^2=\frac{25}{36} \text{on }\text{the }\text{left }\text{and }6\times \frac{25}{36}=\frac{25}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{25}{6}-10=-\frac{35}{6}: \\
6 \left(x^2+\frac{5 x}{3}+\frac{25}{36}\right)-5 \left(y^2+\frac{6 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{35}{6}$} \\
\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 }-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{35}{6}-\frac{9}{5}=-\frac{229}{30}: \\
6 \left(x^2+\frac{5 x}{3}+\frac{25}{36}\right)-5 \left(y^2+\frac{6 y}{5}+\frac{9}{25}\right)=\fbox{$-\frac{229}{30}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{3}+\frac{25}{36}=\left(x+\frac{5}{6}\right)^2: \\
6 \fbox{$\left(x+\frac{5}{6}\right)^2$}-5 \left(y^2+\frac{6 y}{5}+\frac{9}{25}\right)=-\frac{229}{30} \\
\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{} & 6 \left(x+\frac{5}{6}\right)^2-5 \fbox{$\left(y+\frac{3}{5}\right)^2$}=-\frac{229}{30} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2+10 x-2 y^2+5 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+5 y+10 x^2+10 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
-2 y^2+5 y+10 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(10 x^2+10 x+\underline{\text{ }}\right)+\left(-2 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(10 x^2+10 x+\underline{\text{ }}\right)=10 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2+x+\underline{\text{ }}\right)$}+\left(-2 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+5 y+\underline{\text{ }}\right)=-2 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right): \\
10 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)$}=\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 }\text{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 7:
\begin{array}{l}
\frac{5}{2}-4=-\frac{3}{2}: \\
10 \left(x^2+x+\frac{1}{4}\right)-2 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{3}{2}$} \\
\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{\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 9:
\begin{array}{l}
-\frac{3}{2}-\frac{25}{8}=-\frac{37}{8}: \\
10 \left(x^2+x+\frac{1}{4}\right)-2 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$-\frac{37}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
\text{10 }\fbox{$\left(x+\frac{1}{2}\right)^2$}-2 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=-\frac{37}{8} \\
\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{} & 10 \left(x+\frac{1}{2}\right)^2-2 \fbox{$\left(y-\frac{5}{4}\right)^2$}=-\frac{37}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2+3 x-2 y^2+5 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+5 y-x^2+3 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-2 y^2+5 y-x^2+3 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+3 x+\underline{\text{ }}\right)+\left(-2 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2+3 x+\underline{\text{ }}\right)=-\left(x^2-3 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2-3 x+\underline{\text{ }}\right)$}+\left(-2 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+5 y+\underline{\text{ }}\right)=-2 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right): \\
-\left(x^2-3 x+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)$}=\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 }-1 \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 }-\frac{9}{4}=-\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6-\frac{9}{4}=\frac{15}{4}: \\
-\left(x^2-3 x+\frac{9}{4}\right)-2 \left(y^2-\frac{5 y}{2}+\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 }-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 9:
\begin{array}{l}
\frac{15}{4}-\frac{25}{8}=\frac{5}{8}: \\
-\left(x^2-3 x+\frac{9}{4}\right)-2 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$\frac{5}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-3 x+\frac{9}{4}=\left(x-\frac{3}{2}\right)^2: \\
-\fbox{$\left(x-\frac{3}{2}\right)^2$}-2 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\frac{5}{8} \\
\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{} & -\left(x-\frac{3}{2}\right)^2-2 \fbox{$\left(y-\frac{5}{4}\right)^2$}=\frac{5}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+3 x+8 y^2+10 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2+10 y-3 x^2+3 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
8 y^2+10 y-3 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(-3 x^2+3 x+\underline{\text{ }}\right)+\left(8 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\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{$-3 \left(x^2-x+\underline{\text{ }}\right)$}+\left(8 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2+10 y+\underline{\text{ }}\right)=8 \left(y^2+\frac{5 y}{4}+\underline{\text{ }}\right): \\
-3 \left(x^2-x+\underline{\text{ }}\right)+\fbox{$8 \left(y^2+\frac{5 y}{4}+\underline{\text{ }}\right)$}=\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 }-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 7:
\begin{array}{l}
-8-\frac{3}{4}=-\frac{35}{4}: \\
-3 \left(x^2-x+\frac{1}{4}\right)+8 \left(y^2+\frac{5 y}{4}+\underline{\text{ }}\right)=\fbox{$-\frac{35}{4}$} \\
\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}{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 9:
\begin{array}{l}
\frac{25}{8}-\frac{35}{4}=-\frac{45}{8}: \\
-3 \left(x^2-x+\frac{1}{4}\right)+8 \left(y^2+\frac{5 y}{4}+\frac{25}{64}\right)=\fbox{$-\frac{45}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2: \\
-3 \fbox{$\left(x-\frac{1}{2}\right)^2$}+8 \left(y^2+\frac{5 y}{4}+\frac{25}{64}\right)=-\frac{45}{8} \\
\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{1}{2}\right)^2+8 \fbox{$\left(y+\frac{5}{8}\right)^2$}=-\frac{45}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2-3 x+3 y^2-4 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2-4 y+6 x^2-3 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
3 y^2-4 y+6 x^2-3 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(6 x^2-3 x+\underline{\text{ }}\right)+\left(3 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(6 x^2-3 x+\underline{\text{ }}\right)=6 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)$}+\left(3 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 y^2-4 y+\underline{\text{ }}\right)=3 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right): \\
6 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right)$}=\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 }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 7:
\begin{array}{l}
1+\frac{3}{8}=\frac{11}{8}: \\
6 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+3 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{11}{8}$} \\
\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 }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 9:
\begin{array}{l}
\frac{11}{8}+\frac{4}{3}=\frac{65}{24}: \\
6 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+3 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$\frac{65}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
6 \fbox{$\left(x-\frac{1}{4}\right)^2$}+3 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=\frac{65}{24} \\
\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{} & 6 \left(x-\frac{1}{4}\right)^2+3 \fbox{$\left(y-\frac{2}{3}\right)^2$}=\frac{65}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2+x+9 y^2+8 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+8 y+7 x^2+x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
9 y^2+8 y+7 x^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(7 x^2+x+\underline{\text{ }}\right)+\left(9 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\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(9 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\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): \\
7 \left(x^2+\frac{x}{7}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2+\frac{8 y}{9}+\underline{\text{ }}\right)$}=\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 }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}
3+\frac{1}{28}=\frac{85}{28}: \\
7 \left(x^2+\frac{x}{7}+\frac{1}{196}\right)+9 \left(y^2+\frac{8 y}{9}+\underline{\text{ }}\right)=\fbox{$\frac{85}{28}$} \\
\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 }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{85}{28}+\frac{16}{9}=\frac{1213}{252}: \\
7 \left(x^2+\frac{x}{7}+\frac{1}{196}\right)+9 \left(y^2+\frac{8 y}{9}+\frac{16}{81}\right)=\fbox{$\frac{1213}{252}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{7}+\frac{1}{196}=\left(x+\frac{1}{14}\right)^2: \\
7 \fbox{$\left(x+\frac{1}{14}\right)^2$}+9 \left(y^2+\frac{8 y}{9}+\frac{16}{81}\right)=\frac{1213}{252} \\
\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{} & 7 \left(x+\frac{1}{14}\right)^2+9 \fbox{$\left(y+\frac{4}{9}\right)^2$}=\frac{1213}{252} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2+x-3 y^2+8 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2+8 y-7 x^2+x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
-3 y^2+8 y-7 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(-7 x^2+x+\underline{\text{ }}\right)+\left(-3 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\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(-3 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2+8 y+\underline{\text{ }}\right)=-3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right): \\
-7 \left(x^2-\frac{x}{7}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right)$}=\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 }-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}
-6-\frac{1}{28}=-\frac{169}{28}: \\
-7 \left(x^2-\frac{x}{7}+\frac{1}{196}\right)-3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{169}{28}$} \\
\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 }-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 9:
\begin{array}{l}
-\frac{169}{28}-\frac{16}{3}=-\frac{955}{84}: \\
-7 \left(x^2-\frac{x}{7}+\frac{1}{196}\right)-3 \left(y^2-\frac{8 y}{3}+\frac{16}{9}\right)=\fbox{$-\frac{955}{84}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{7}+\frac{1}{196}=\left(x-\frac{1}{14}\right)^2: \\
-7 \fbox{$\left(x-\frac{1}{14}\right)^2$}-3 \left(y^2-\frac{8 y}{3}+\frac{16}{9}\right)=-\frac{955}{84} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{8 y}{3}+\frac{16}{9}=\left(y-\frac{4}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x-\frac{1}{14}\right)^2-3 \fbox{$\left(y-\frac{4}{3}\right)^2$}=-\frac{955}{84} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2-x+6 y^2+y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2+y+x^2-x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
6 y^2+y+x^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(x^2-x+\underline{\text{ }}\right)+\left(6 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\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): \\
\left(x^2-x+\underline{\text{ }}\right)+\fbox{$6 \left(y^2+\frac{y}{6}+\underline{\text{ }}\right)$}=\underline{\text{ }}-3 \\
\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{-1}{2}\right)^2=\frac{1}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
\frac{1}{4}-3=-\frac{11}{4}: \\
\left(x^2-x+\frac{1}{4}\right)+6 \left(y^2+\frac{y}{6}+\underline{\text{ }}\right)=\fbox{$-\frac{11}{4}$} \\
\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 }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 8:
\begin{array}{l}
\frac{1}{24}-\frac{11}{4}=-\frac{65}{24}: \\
\left(x^2-x+\frac{1}{4}\right)+6 \left(y^2+\frac{y}{6}+\frac{1}{144}\right)=\fbox{$-\frac{65}{24}$} \\
\end{array}
Step 9:
\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{y}{6}+\frac{1}{144}\right)=-\frac{65}{24} \\
\end{array}
Step 10:
\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{} & \left(x-\frac{1}{2}\right)^2+6 \fbox{$\left(y+\frac{1}{12}\right)^2$}=-\frac{65}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-x-y^2-y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2-y-3 x^2-x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-y^2-y-3 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(-3 x^2-x+\underline{\text{ }}\right)+\left(-y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2-x+\underline{\text{ }}\right)=-3 \left(x^2+\frac{x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+\frac{x}{3}+\underline{\text{ }}\right)$}+\left(-y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2-y+\underline{\text{ }}\right)=-\left(y^2+y+\underline{\text{ }}\right): \\
-3 \left(x^2+\frac{x}{3}+\underline{\text{ }}\right)+\fbox{$-\left(y^2+y+\underline{\text{ }}\right)$}=\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 }-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 7:
\begin{array}{l}
-7-\frac{1}{12}=-\frac{85}{12}: \\
-3 \left(x^2+\frac{x}{3}+\frac{1}{36}\right)-\left(y^2+y+\underline{\text{ }}\right)=\fbox{$-\frac{85}{12}$} \\
\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 }-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 9:
\begin{array}{l}
-\frac{85}{12}-\frac{1}{4}=-\frac{22}{3}: \\
-3 \left(x^2+\frac{x}{3}+\frac{1}{36}\right)-\left(y^2+y+\frac{1}{4}\right)=\fbox{$-\frac{22}{3}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{3}+\frac{1}{36}=\left(x+\frac{1}{6}\right)^2: \\
-3 \fbox{$\left(x+\frac{1}{6}\right)^2$}-\left(y^2+y+\frac{1}{4}\right)=-\frac{22}{3} \\
\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{} & -3 \left(x+\frac{1}{6}\right)^2-\fbox{$\left(y+\frac{1}{2}\right)^2$}=-\frac{22}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2+6 x+6 y^2+2 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2+2 y+3 x^2+6 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
6 y^2+2 y+3 x^2+6 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+6 x+\underline{\text{ }}\right)+\left(6 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2+6 x+\underline{\text{ }}\right)=3 \left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2+2 x+\underline{\text{ }}\right)$}+\left(6 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\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): \\
3 \left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$6 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)$}=\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{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }3\times 1=3 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3-3=0: \\
3 \left(x^2+2 x+1\right)+6 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)=\fbox{$0$} \\
\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}
x^2+2 x+1=(x+1)^2: \\
3 \fbox{$(x+1)^2$}+6 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=\frac{1}{6} \\
\end{array}
Step 10:
\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{} & 3 (x+1)^2+6 \fbox{$\left(y+\frac{1}{6}\right)^2$}=\frac{1}{6} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-6 x-2 y^2+5 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+5 y+2 x^2-6 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-2 y^2+5 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(-2 y^2+5 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(-2 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+5 y+\underline{\text{ }}\right)=-2 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right): \\
2 \left(x^2-3 x+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)$}=\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 }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)-2 \left(y^2-\frac{5 y}{2}+\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 }-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 9:
\begin{array}{l}
-\frac{9}{2}-\frac{25}{8}=-\frac{61}{8}: \\
2 \left(x^2-3 x+\frac{9}{4}\right)-2 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$-\frac{61}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-3 x+\frac{9}{4}=\left(x-\frac{3}{2}\right)^2: \\
2 \fbox{$\left(x-\frac{3}{2}\right)^2$}-2 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=-\frac{61}{8} \\
\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{} & 2 \left(x-\frac{3}{2}\right)^2-2 \fbox{$\left(y-\frac{5}{4}\right)^2$}=-\frac{61}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2+10 x+8 y^2-8 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-8 y-4 x^2+10 x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
8 y^2-8 y-4 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(-4 x^2+10 x+\underline{\text{ }}\right)+\left(8 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\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^2-8 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): \\
-4 \left(x^2-\frac{5 x}{2}+\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 }-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}
7-\frac{25}{4}=\frac{3}{4}: \\
-4 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)+8 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$\frac{3}{4}$} \\
\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{-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}
\frac{3}{4}+2=\frac{11}{4}: \\
-4 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)+8 \left(y^2-y+\frac{1}{4}\right)=\fbox{$\frac{11}{4}$} \\
\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-y+\frac{1}{4}\right)=\frac{11}{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{} & -4 \left(x-\frac{5}{4}\right)^2+8 \fbox{$\left(y-\frac{1}{2}\right)^2$}=\frac{11}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2-3 x-5 y^2+3 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+3 y-7 x^2-3 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
-5 y^2+3 y-7 x^2-3 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(-7 x^2-3 x+\underline{\text{ }}\right)+\left(-5 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\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(-5 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+3 y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{3 y}{5}+\underline{\text{ }}\right): \\
-7 \left(x^2+\frac{3 x}{7}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{3 y}{5}+\underline{\text{ }}\right)$}=\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 }-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}
3-\frac{9}{28}=\frac{75}{28}: \\
-7 \left(x^2+\frac{3 x}{7}+\frac{9}{196}\right)-5 \left(y^2-\frac{3 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{75}{28}$} \\
\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 }-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 9:
\begin{array}{l}
\frac{75}{28}-\frac{9}{20}=\frac{78}{35}: \\
-7 \left(x^2+\frac{3 x}{7}+\frac{9}{196}\right)-5 \left(y^2-\frac{3 y}{5}+\frac{9}{100}\right)=\fbox{$\frac{78}{35}$} \\
\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{$\left(x+\frac{3}{14}\right)^2$}-5 \left(y^2-\frac{3 y}{5}+\frac{9}{100}\right)=\frac{78}{35} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{5}+\frac{9}{100}=\left(y-\frac{3}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x+\frac{3}{14}\right)^2-5 \fbox{$\left(y-\frac{3}{10}\right)^2$}=\frac{78}{35} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2-3 x+5 y^2+9 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+9 y+7 x^2-3 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
5 y^2+9 y+7 x^2-3 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(7 x^2-3 x+\underline{\text{ }}\right)+\left(5 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\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(5 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\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): \\
7 \left(x^2-\frac{3 x}{7}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2+\frac{9 y}{5}+\underline{\text{ }}\right)$}=\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 }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}
\frac{9}{28}-6=-\frac{159}{28}: \\
7 \left(x^2-\frac{3 x}{7}+\frac{9}{196}\right)+5 \left(y^2+\frac{9 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{159}{28}$} \\
\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 }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{81}{20}-\frac{159}{28}=-\frac{57}{35}: \\
7 \left(x^2-\frac{3 x}{7}+\frac{9}{196}\right)+5 \left(y^2+\frac{9 y}{5}+\frac{81}{100}\right)=\fbox{$-\frac{57}{35}$} \\
\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{$\left(x-\frac{3}{14}\right)^2$}+5 \left(y^2+\frac{9 y}{5}+\frac{81}{100}\right)=-\frac{57}{35} \\
\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{} & 7 \left(x-\frac{3}{14}\right)^2+5 \fbox{$\left(y+\frac{9}{10}\right)^2$}=-\frac{57}{35} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2+2 x+3 y^2+10 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2+10 y+7 x^2+2 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
3 y^2+10 y+7 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(7 x^2+2 x+\underline{\text{ }}\right)+\left(3 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\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(3 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\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): \\
7 \left(x^2+\frac{2 x}{7}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2+\frac{10 y}{3}+\underline{\text{ }}\right)$}=\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 }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}
2+\frac{1}{7}=\frac{15}{7}: \\
7 \left(x^2+\frac{2 x}{7}+\frac{1}{49}\right)+3 \left(y^2+\frac{10 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{15}{7}$} \\
\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 }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{15}{7}+\frac{25}{3}=\frac{220}{21}: \\
7 \left(x^2+\frac{2 x}{7}+\frac{1}{49}\right)+3 \left(y^2+\frac{10 y}{3}+\frac{25}{9}\right)=\fbox{$\frac{220}{21}$} \\
\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{$\left(x+\frac{1}{7}\right)^2$}+3 \left(y^2+\frac{10 y}{3}+\frac{25}{9}\right)=\frac{220}{21} \\
\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{} & 7 \left(x+\frac{1}{7}\right)^2+3 \fbox{$\left(y+\frac{5}{3}\right)^2$}=\frac{220}{21} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2-9 x-8 y^2-7 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2-7 y-5 x^2-9 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(-5 x^2-9 x+\underline{\text{ }}\right)+\left(-8 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\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(-8 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\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): \\
-5 \left(x^2+\frac{9 x}{5}+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2+\frac{7 y}{8}+\underline{\text{ }}\right)$}=\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 }-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 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{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 7:
\begin{array}{l}
-\frac{81}{20}-\frac{49}{32}=-\frac{893}{160}: \\
-5 \left(x^2+\frac{9 x}{5}+\frac{81}{100}\right)-8 \left(y^2+\frac{7 y}{8}+\frac{49}{256}\right)=\fbox{$-\frac{893}{160}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{9 x}{5}+\frac{81}{100}=\left(x+\frac{9}{10}\right)^2: \\
-5 \fbox{$\left(x+\frac{9}{10}\right)^2$}-8 \left(y^2+\frac{7 y}{8}+\frac{49}{256}\right)=-\frac{893}{160} \\
\end{array}
Step 9:
\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{} & -5 \left(x+\frac{9}{10}\right)^2-8 \fbox{$\left(y+\frac{7}{16}\right)^2$}=-\frac{893}{160} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-6 x-4 y^2+6 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+6 y-3 x^2-6 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
-4 y^2+6 y-3 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(-3 x^2-6 x+\underline{\text{ }}\right)+\left(-4 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2-6 x+\underline{\text{ }}\right)=-3 \left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+2 x+\underline{\text{ }}\right)$}+\left(-4 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2+6 y+\underline{\text{ }}\right)=-4 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right): \\
-3 \left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)$}=\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 }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-3\times 1=-3 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10-3=7: \\
-3 \left(x^2+2 x+1\right)-4 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$7$} \\
\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 }-4 \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 }-4\times \frac{9}{16}=-\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
7-\frac{9}{4}=\frac{19}{4}: \\
-3 \left(x^2+2 x+1\right)-4 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$\frac{19}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
-3 \fbox{$(x+1)^2$}-4 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\frac{19}{4} \\
\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{} & -3 (x+1)^2-4 \fbox{$\left(y-\frac{3}{4}\right)^2$}=\frac{19}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-3 x-y^2+3 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2+3 y+2 x^2-3 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-y^2+3 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(-y^2+3 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(-y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2+3 y+\underline{\text{ }}\right)=-\left(y^2-3 y+\underline{\text{ }}\right): \\
2 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right)+\fbox{$-\left(y^2-3 y+\underline{\text{ }}\right)$}=\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{\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}
\frac{9}{8}-5=-\frac{31}{8}: \\
2 \left(x^2-\frac{3 x}{2}+\frac{9}{16}\right)-\left(y^2-3 y+\underline{\text{ }}\right)=\fbox{$-\frac{31}{8}$} \\
\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 }-1 \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 }-\frac{9}{4}=-\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{31}{8}-\frac{9}{4}=-\frac{49}{8}: \\
2 \left(x^2-\frac{3 x}{2}+\frac{9}{16}\right)-\left(y^2-3 y+\frac{9}{4}\right)=\fbox{$-\frac{49}{8}$} \\
\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{$\left(x-\frac{3}{4}\right)^2$}-\left(y^2-3 y+\frac{9}{4}\right)=-\frac{49}{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{3}{4}\right)^2-\fbox{$\left(y-\frac{3}{2}\right)^2$}=-\frac{49}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+9 x+6 y^2-6 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2-6 y+6 x^2+9 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
6 y^2-6 y+6 x^2+9 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+9 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 x^2+9 x+\underline{\text{ }}\right)=6 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right)$}+\left(6 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2-6 y+\underline{\text{ }}\right)=6 \left(y^2-y+\underline{\text{ }}\right): \\
6 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2-y+\underline{\text{ }}\right)$}=\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 }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 7:
\begin{array}{l}
\frac{27}{8}-9=-\frac{45}{8}: \\
6 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)+6 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$-\frac{45}{8}$} \\
\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 }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 9:
\begin{array}{l}
\frac{3}{2}-\frac{45}{8}=-\frac{33}{8}: \\
6 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)+6 \left(y^2-y+\frac{1}{4}\right)=\fbox{$-\frac{33}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{2}+\frac{9}{16}=\left(x+\frac{3}{4}\right)^2: \\
6 \fbox{$\left(x+\frac{3}{4}\right)^2$}+6 \left(y^2-y+\frac{1}{4}\right)=-\frac{33}{8} \\
\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{} & 6 \left(x+\frac{3}{4}\right)^2+6 \fbox{$\left(y-\frac{1}{2}\right)^2$}=-\frac{33}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2+9 x+4 y^2-y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2-y-6 x^2+9 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(-6 x^2+9 x+\underline{\text{ }}\right)+\left(4 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(-6 x^2+9 x+\underline{\text{ }}\right)=-6 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right)$}+\left(4 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\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{3 x}{2}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2-\frac{y}{4}+\underline{\text{ }}\right)$}=\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 }-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 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{-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}
\frac{1}{16}-\frac{27}{8}=-\frac{53}{16}: \\
-6 \left(x^2-\frac{3 x}{2}+\frac{9}{16}\right)+4 \left(y^2-\frac{y}{4}+\frac{1}{64}\right)=\fbox{$-\frac{53}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{3 x}{2}+\frac{9}{16}=\left(x-\frac{3}{4}\right)^2: \\
-6 \fbox{$\left(x-\frac{3}{4}\right)^2$}+4 \left(y^2-\frac{y}{4}+\frac{1}{64}\right)=-\frac{53}{16} \\
\end{array}
Step 9:
\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{3}{4}\right)^2+4 \fbox{$\left(y-\frac{1}{8}\right)^2$}=-\frac{53}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2-10 x+2 y^2-2 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-2 y+8 x^2-10 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
2 y^2-2 y+8 x^2-10 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(8 x^2-10 x+\underline{\text{ }}\right)+\left(2 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\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(2 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2-2 y+\underline{\text{ }}\right)=2 \left(y^2-y+\underline{\text{ }}\right): \\
8 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-y+\underline{\text{ }}\right)$}=\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 }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}
3+\frac{25}{8}=\frac{49}{8}: \\
8 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)+2 \left(y^2-y+\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{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{2}{4}=\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{49}{8}+\frac{1}{2}=\frac{53}{8}: \\
8 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)+2 \left(y^2-y+\frac{1}{4}\right)=\fbox{$\frac{53}{8}$} \\
\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{$\left(x-\frac{5}{8}\right)^2$}+2 \left(y^2-y+\frac{1}{4}\right)=\frac{53}{8} \\
\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{} & 8 \left(x-\frac{5}{8}\right)^2+2 \fbox{$\left(y-\frac{1}{2}\right)^2$}=\frac{53}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2-2 x+8 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 x^2-2 x+(8 y-2)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 y-7 x^2-2 x-2 \text{from }\text{both }\text{sides}: \\
7 x^2+2 x+(2-8 y)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }2-8 y \text{from }\text{both }\text{sides}: \\
7 x^2+2 x=8 y-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(7 x^2+2 x+\underline{\text{ }}\right)=(8 y-2)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 x^2+2 x+\underline{\text{ }}\right)=7 \left(x^2+\frac{2 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2+\frac{2 x}{7}+\underline{\text{ }}\right)$}=(8 y-2)+\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 }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}
(8 y-2)+\frac{1}{7}=8 y-\frac{13}{7}: \\
7 \left(x^2+\frac{2 x}{7}+\frac{1}{49}\right)=\fbox{$8 y-\frac{13}{7}$} \\
\end{array}
Step 8:
\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{$\left(x+\frac{1}{7}\right)^2$}=8 y-\frac{13}{7} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+2 x-y^2-10 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2-10 y-8 x^2+2 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-y^2-10 y-8 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(-8 x^2+2 x+\underline{\text{ }}\right)+\left(-y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\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(-y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2-10 y+\underline{\text{ }}\right)=-\left(y^2+10 y+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{x}{4}+\underline{\text{ }}\right)+\fbox{$-\left(y^2+10 y+\underline{\text{ }}\right)$}=\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 }-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}
-7-\frac{1}{8}=-\frac{57}{8}: \\
-8 \left(x^2-\frac{x}{4}+\frac{1}{64}\right)-\left(y^2+10 y+\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 }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{10}{2}\right)^2=25 \text{on }\text{the }\text{left }\text{and }-25=-25 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{57}{8}-25=-\frac{257}{8}: \\
-8 \left(x^2-\frac{x}{4}+\frac{1}{64}\right)-\left(y^2+10 y+25\right)=\fbox{$-\frac{257}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{4}+\frac{1}{64}=\left(x-\frac{1}{8}\right)^2: \\
-8 \fbox{$\left(x-\frac{1}{8}\right)^2$}-\left(y^2+10 y+25\right)=-\frac{257}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+10 y+25=(y+5)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x-\frac{1}{8}\right)^2-\fbox{$(y+5)^2$}=-\frac{257}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+3 x-7 y^2+9 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2+9 y-8 x^2+3 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
-7 y^2+9 y-8 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(-8 x^2+3 x+\underline{\text{ }}\right)+\left(-7 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\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(-7 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\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): \\
-8 \left(x^2-\frac{3 x}{8}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2-\frac{9 y}{7}+\underline{\text{ }}\right)$}=\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}{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}
-10-\frac{9}{32}=-\frac{329}{32}: \\
-8 \left(x^2-\frac{3 x}{8}+\frac{9}{256}\right)-7 \left(y^2-\frac{9 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{329}{32}$} \\
\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 }-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{329}{32}-\frac{81}{28}=-\frac{2951}{224}: \\
-8 \left(x^2-\frac{3 x}{8}+\frac{9}{256}\right)-7 \left(y^2-\frac{9 y}{7}+\frac{81}{196}\right)=\fbox{$-\frac{2951}{224}$} \\
\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{$\left(x-\frac{3}{16}\right)^2$}-7 \left(y^2-\frac{9 y}{7}+\frac{81}{196}\right)=-\frac{2951}{224} \\
\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{} & -8 \left(x-\frac{3}{16}\right)^2-7 \fbox{$\left(y-\frac{9}{14}\right)^2$}=-\frac{2951}{224} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2+7 x-2 y^2+8 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+8 y+2 x^2+7 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
-2 y^2+8 y+2 x^2+7 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+7 x+\underline{\text{ }}\right)+\left(-2 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\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+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): \\
2 \left(x^2+\frac{7 x}{2}+\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 }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}
10+\frac{49}{8}=\frac{129}{8}: \\
2 \left(x^2+\frac{7 x}{2}+\frac{49}{16}\right)-2 \left(y^2-4 y+\underline{\text{ }}\right)=\fbox{$\frac{129}{8}$} \\
\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{-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{129}{8}-8=\frac{65}{8}: \\
2 \left(x^2+\frac{7 x}{2}+\frac{49}{16}\right)-2 \left(y^2-4 y+4\right)=\fbox{$\frac{65}{8}$} \\
\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^2-4 y+4\right)=\frac{65}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-4 y+4=(y-2)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x+\frac{7}{4}\right)^2-2 \fbox{$(y-2)^2$}=\frac{65}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2+6 x-10 y^2+7 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+7 y-4 x^2+6 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-10 y^2+7 y-4 x^2+6 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(-4 x^2+6 x+\underline{\text{ }}\right)+\left(-10 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2+6 x+\underline{\text{ }}\right)=-4 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right)$}+\left(-10 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2+7 y+\underline{\text{ }}\right)=-10 \left(y^2-\frac{7 y}{10}+\underline{\text{ }}\right): \\
-4 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2-\frac{7 y}{10}+\underline{\text{ }}\right)$}=\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 }-4 \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 }-4\times \frac{9}{16}=-\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-7-\frac{9}{4}=-\frac{37}{4}: \\
-4 \left(x^2-\frac{3 x}{2}+\frac{9}{16}\right)-10 \left(y^2-\frac{7 y}{10}+\underline{\text{ }}\right)=\fbox{$-\frac{37}{4}$} \\
\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{-7}{10}}{2}\right)^2=\frac{49}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{49}{400}=-\frac{49}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{37}{4}-\frac{49}{40}=-\frac{419}{40}: \\
-4 \left(x^2-\frac{3 x}{2}+\frac{9}{16}\right)-10 \left(y^2-\frac{7 y}{10}+\frac{49}{400}\right)=\fbox{$-\frac{419}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{2}+\frac{9}{16}=\left(x-\frac{3}{4}\right)^2: \\
-4 \fbox{$\left(x-\frac{3}{4}\right)^2$}-10 \left(y^2-\frac{7 y}{10}+\frac{49}{400}\right)=-\frac{419}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{10}+\frac{49}{400}=\left(y-\frac{7}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x-\frac{3}{4}\right)^2-\text{10 }\fbox{$\left(y-\frac{7}{20}\right)^2$}=-\frac{419}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+8 x-3 y^2+y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2+y+6 x^2+8 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
-3 y^2+y+6 x^2+8 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(6 x^2+8 x+\underline{\text{ }}\right)+\left(-3 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\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(-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): \\
6 \left(x^2+\frac{4 x}{3}+\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 }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}
4+\frac{8}{3}=\frac{20}{3}: \\
6 \left(x^2+\frac{4 x}{3}+\frac{4}{9}\right)-3 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{20}{3}$} \\
\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 }-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{20}{3}-\frac{1}{12}=\frac{79}{12}: \\
6 \left(x^2+\frac{4 x}{3}+\frac{4}{9}\right)-3 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\fbox{$\frac{79}{12}$} \\
\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{$\left(x+\frac{2}{3}\right)^2$}-3 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\frac{79}{12} \\
\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{} & 6 \left(x+\frac{2}{3}\right)^2-3 \fbox{$\left(y-\frac{1}{6}\right)^2$}=\frac{79}{12} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+9 y^2-8 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2-8 y+\left(-8 x^2-9\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
9 y^2-8 y-8 x^2=9 \\
\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(9 y^2-8 y+\underline{\text{ }}\right)-8 x^2=\underline{\text{ }}+9 \\
\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): \\
\fbox{$9 \left(y^2-\frac{8 y}{9}+\underline{\text{ }}\right)$}-8 x^2=\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 }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 6:
\begin{array}{l}
9+\frac{16}{9}=\frac{97}{9}: \\
9 \left(y^2-\frac{8 y}{9}+\frac{16}{81}\right)-8 x^2=\fbox{$\frac{97}{9}$} \\
\end{array}
Step 7:
\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{} & 9 \fbox{$\left(y-\frac{4}{9}\right)^2$}-8 x^2=\frac{97}{9} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+8 x+2 y^2+y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+y+4 x^2+8 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
2 y^2+y+4 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(4 x^2+8 x+\underline{\text{ }}\right)+\left(2 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2+8 x+\underline{\text{ }}\right)=4 \left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2+2 x+\underline{\text{ }}\right)$}+\left(2 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+y+\underline{\text{ }}\right)=2 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right): \\
4 \left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)$}=\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 }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }4\times 1=4 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9+4=13: \\
4 \left(x^2+2 x+1\right)+2 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$13$} \\
\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{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{2}{16}=\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
13+\frac{1}{8}=\frac{105}{8}: \\
4 \left(x^2+2 x+1\right)+2 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\fbox{$\frac{105}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
4 \fbox{$(x+1)^2$}+2 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\frac{105}{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{} & 4 (x+1)^2+2 \fbox{$\left(y+\frac{1}{4}\right)^2$}=\frac{105}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+3 x+2 y^2-6 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-6 y+8 x^2+3 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
2 y^2-6 y+8 x^2+3 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+3 x+\underline{\text{ }}\right)+\left(2 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\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(2 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\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): \\
8 \left(x^2+\frac{3 x}{8}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-3 y+\underline{\text{ }}\right)$}=\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 }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}
1+\frac{9}{32}=\frac{41}{32}: \\
8 \left(x^2+\frac{3 x}{8}+\frac{9}{256}\right)+2 \left(y^2-3 y+\underline{\text{ }}\right)=\fbox{$\frac{41}{32}$} \\
\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{-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{41}{32}+\frac{9}{2}=\frac{185}{32}: \\
8 \left(x^2+\frac{3 x}{8}+\frac{9}{256}\right)+2 \left(y^2-3 y+\frac{9}{4}\right)=\fbox{$\frac{185}{32}$} \\
\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{$\left(x+\frac{3}{16}\right)^2$}+2 \left(y^2-3 y+\frac{9}{4}\right)=\frac{185}{32} \\
\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{} & 8 \left(x+\frac{3}{16}\right)^2+2 \fbox{$\left(y-\frac{3}{2}\right)^2$}=\frac{185}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2+4 x+y^2-10 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2-10 y+x^2+4 x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
y^2-10 y+x^2+4 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+4 x+\underline{\text{ }}\right)+\left(y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 4:
\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 5:
\begin{array}{l}
7+4=11: \\
\left(x^2+4 x+4\right)+\left(y^2-10 y+\underline{\text{ }}\right)=\fbox{$11$} \\
\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{Add }\left(\frac{-10}{2}\right)^2=25 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 7:
\begin{array}{l}
11+25=36: \\
\left(x^2+4 x+4\right)+\left(y^2-10 y+25\right)=\fbox{$36$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+4 x+4=(x+2)^2: \\
\fbox{$(x+2)^2$}+\left(y^2-10 y+25\right)=36 \\
\end{array}
Step 9:
\begin{array}{l}
y^2-10 y+25=(y-5)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & (x+2)^2+\fbox{$(y-5)^2$}=36 \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2+10 x+8 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 x^2+10 x+(8 y-6)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 y-9 x^2+10 x-6 \text{from }\text{both }\text{sides}: \\
9 x^2-10 x+(6-8 y)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }6-8 y \text{from }\text{both }\text{sides}: \\
9 x^2-10 x=8 y-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(9 x^2-10 x+\underline{\text{ }}\right)=(8 y-6)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 x^2-10 x+\underline{\text{ }}\right)=9 \left(x^2-\frac{10 x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2-\frac{10 x}{9}+\underline{\text{ }}\right)$}=(8 y-6)+\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 }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}
(8 y-6)+\frac{25}{9}=8 y-\frac{29}{9}: \\
9 \left(x^2-\frac{10 x}{9}+\frac{25}{81}\right)=\fbox{$8 y-\frac{29}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{10 x}{9}+\frac{25}{81}=\left(x-\frac{5}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \fbox{$\left(x-\frac{5}{9}\right)^2$}=8 y-\frac{29}{9} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2+4 x+6 y^2-5 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2-5 y-10 x^2+4 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
6 y^2-5 y-10 x^2+4 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+4 x+\underline{\text{ }}\right)+\left(6 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2+4 x+\underline{\text{ }}\right)=-10 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right)$}+\left(6 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\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): \\
-10 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2-\frac{5 y}{6}+\underline{\text{ }}\right)$}=\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{-2}{5}}{2}\right)^2=\frac{1}{25} \text{on }\text{the }\text{left }\text{and }\frac{-10}{25}=-\frac{2}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2-\frac{2}{5}=\frac{8}{5}: \\
-10 \left(x^2-\frac{2 x}{5}+\frac{1}{25}\right)+6 \left(y^2-\frac{5 y}{6}+\underline{\text{ }}\right)=\fbox{$\frac{8}{5}$} \\
\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{-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{8}{5}+\frac{25}{24}=\frac{317}{120}: \\
-10 \left(x^2-\frac{2 x}{5}+\frac{1}{25}\right)+6 \left(y^2-\frac{5 y}{6}+\frac{25}{144}\right)=\fbox{$\frac{317}{120}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{2 x}{5}+\frac{1}{25}=\left(x-\frac{1}{5}\right)^2: \\
-10 \fbox{$\left(x-\frac{1}{5}\right)^2$}+6 \left(y^2-\frac{5 y}{6}+\frac{25}{144}\right)=\frac{317}{120} \\
\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{} & -10 \left(x-\frac{1}{5}\right)^2+6 \fbox{$\left(y-\frac{5}{12}\right)^2$}=\frac{317}{120} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2-2 x-y^2-10 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2-10 y+3 x^2-2 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-y^2-10 y+3 x^2-2 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(3 x^2-2 x+\underline{\text{ }}\right)+\left(-y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\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(-y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2-10 y+\underline{\text{ }}\right)=-\left(y^2+10 y+\underline{\text{ }}\right): \\
3 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$-\left(y^2+10 y+\underline{\text{ }}\right)$}=\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 }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}
6+\frac{1}{3}=\frac{19}{3}: \\
3 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)-\left(y^2+10 y+\underline{\text{ }}\right)=\fbox{$\frac{19}{3}$} \\
\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 }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{10}{2}\right)^2=25 \text{on }\text{the }\text{left }\text{and }-25=-25 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{19}{3}-25=-\frac{56}{3}: \\
3 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)-\left(y^2+10 y+25\right)=\fbox{$-\frac{56}{3}$} \\
\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{$\left(x-\frac{1}{3}\right)^2$}-\left(y^2+10 y+25\right)=-\frac{56}{3} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+10 y+25=(y+5)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x-\frac{1}{3}\right)^2-\fbox{$(y+5)^2$}=-\frac{56}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-6 x+4 y^2-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 x^2-6 x+\left(4 y^2-9\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
4 y^2+4 x^2-6 x=9 \\
\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(4 x^2-6 x+\underline{\text{ }}\right)+4 y^2=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2-6 x+\underline{\text{ }}\right)=4 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right)$}+4 y^2=\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{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}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }4\times \frac{9}{16}=\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
9+\frac{9}{4}=\frac{45}{4}: \\
4 \left(x^2-\frac{3 x}{2}+\frac{9}{16}\right)+4 y^2=\fbox{$\frac{45}{4}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2-\frac{3 x}{2}+\frac{9}{16}=\left(x-\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \fbox{$\left(x-\frac{3}{4}\right)^2$}+4 y^2=\frac{45}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2+4 x-y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 x^2+4 x+(-y-9)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }y+9 \text{to }\text{both }\text{sides}: \\
2 x^2+4 x=y+9 \\
\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(2 x^2+4 x+\underline{\text{ }}\right)=(y+9)+\underline{\text{ }} \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2+4 x+\underline{\text{ }}\right)=2 \left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2+2 x+\underline{\text{ }}\right)$}=(y+9)+\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 }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 6:
\begin{array}{l}
(y+9)+2=y+11: \\
2 \left(x^2+2 x+1\right)=\fbox{$y+11$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \fbox{$(x+1)^2$}=y+11 \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+5 x+10 y^2+7 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2+7 y+6 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+7 y+6 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(6 x^2+5 x+\underline{\text{ }}\right)+\left(10 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\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(10 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2+7 y+\underline{\text{ }}\right)=10 \left(y^2+\frac{7 y}{10}+\underline{\text{ }}\right): \\
6 \left(x^2+\frac{5 x}{6}+\underline{\text{ }}\right)+\fbox{$10 \left(y^2+\frac{7 y}{10}+\underline{\text{ }}\right)$}=\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 }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}
\frac{25}{24}-1=\frac{1}{24}: \\
6 \left(x^2+\frac{5 x}{6}+\frac{25}{144}\right)+10 \left(y^2+\frac{7 y}{10}+\underline{\text{ }}\right)=\fbox{$\frac{1}{24}$} \\
\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 }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{10}}{2}\right)^2=\frac{49}{400} \text{on }\text{the }\text{left }\text{and }10\times \frac{49}{400}=\frac{49}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{24}+\frac{49}{40}=\frac{19}{15}: \\
6 \left(x^2+\frac{5 x}{6}+\frac{25}{144}\right)+10 \left(y^2+\frac{7 y}{10}+\frac{49}{400}\right)=\fbox{$\frac{19}{15}$} \\
\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{$\left(x+\frac{5}{12}\right)^2$}+10 \left(y^2+\frac{7 y}{10}+\frac{49}{400}\right)=\frac{19}{15} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{7 y}{10}+\frac{49}{400}=\left(y+\frac{7}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x+\frac{5}{12}\right)^2+\text{10 }\fbox{$\left(y+\frac{7}{20}\right)^2$}=\frac{19}{15} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+9 x+4 y^2-8 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2-8 y+4 x^2+9 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
4 y^2-8 y+4 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(4 x^2+9 x+\underline{\text{ }}\right)+\left(4 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2+9 x+\underline{\text{ }}\right)=4 \left(x^2+\frac{9 x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2+\frac{9 x}{4}+\underline{\text{ }}\right)$}+\left(4 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2-8 y+\underline{\text{ }}\right)=4 \left(y^2-2 y+\underline{\text{ }}\right): \\
4 \left(x^2+\frac{9 x}{4}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2-2 y+\underline{\text{ }}\right)$}=\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 }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 7:
\begin{array}{l}
\frac{81}{16}-1=\frac{65}{16}: \\
4 \left(x^2+\frac{9 x}{4}+\frac{81}{64}\right)+4 \left(y^2-2 y+\underline{\text{ }}\right)=\fbox{$\frac{65}{16}$} \\
\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{-2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }4\times 1=4 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{65}{16}+4=\frac{129}{16}: \\
4 \left(x^2+\frac{9 x}{4}+\frac{81}{64}\right)+4 \left(y^2-2 y+1\right)=\fbox{$\frac{129}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{9 x}{4}+\frac{81}{64}=\left(x+\frac{9}{8}\right)^2: \\
4 \fbox{$\left(x+\frac{9}{8}\right)^2$}+4 \left(y^2-2 y+1\right)=\frac{129}{16} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-2 y+1=(y-1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x+\frac{9}{8}\right)^2+4 \fbox{$(y-1)^2$}=\frac{129}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2-9 x+9 y^2+9 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+9 y-7 x^2-9 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^2-9 x+\underline{\text{ }}\right)+\left(9 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(-7 x^2-9 x+\underline{\text{ }}\right)=-7 \left(x^2+\frac{9 x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2+\frac{9 x}{7}+\underline{\text{ }}\right)$}+\left(9 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 y^2+9 y+\underline{\text{ }}\right)=9 \left(y^2+y+\underline{\text{ }}\right): \\
-7 \left(x^2+\frac{9 x}{7}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2+y+\underline{\text{ }}\right)$}=\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{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 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{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}
\frac{9}{4}-\frac{81}{28}=-\frac{9}{14}: \\
-7 \left(x^2+\frac{9 x}{7}+\frac{81}{196}\right)+9 \left(y^2+y+\frac{1}{4}\right)=\fbox{$-\frac{9}{14}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{9 x}{7}+\frac{81}{196}=\left(x+\frac{9}{14}\right)^2: \\
-7 \fbox{$\left(x+\frac{9}{14}\right)^2$}+9 \left(y^2+y+\frac{1}{4}\right)=-\frac{9}{14} \\
\end{array}
Step 9:
\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{9}{14}\right)^2+9 \fbox{$\left(y+\frac{1}{2}\right)^2$}=-\frac{9}{14} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2+8 x+6 y^2+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 x^2+8 x+\left(6 y^2+1\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 y^2-9 x^2+8 x+1 \text{from }\text{both }\text{sides}: \\
9 x^2-8 x+\left(-6 y^2-1\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
-6 y^2+9 x^2-8 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(9 x^2-8 x+\underline{\text{ }}\right)-6 y^2=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 x^2-8 x+\underline{\text{ }}\right)=9 \left(x^2-\frac{8 x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2-\frac{8 x}{9}+\underline{\text{ }}\right)$}-6 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 }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}
1+\frac{16}{9}=\frac{25}{9}: \\
9 \left(x^2-\frac{8 x}{9}+\frac{16}{81}\right)-6 y^2=\fbox{$\frac{25}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{8 x}{9}+\frac{16}{81}=\left(x-\frac{4}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \fbox{$\left(x-\frac{4}{9}\right)^2$}-6 y^2=\frac{25}{9} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-8 x+3 y^2-8 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2-8 y-3 x^2-8 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
3 y^2-8 y-3 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(-3 x^2-8 x+\underline{\text{ }}\right)+\left(3 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\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(3 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 y^2-8 y+\underline{\text{ }}\right)=3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right): \\
-3 \left(x^2+\frac{8 x}{3}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right)$}=\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 }-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 7:
\begin{array}{l}
-5-\frac{16}{3}=-\frac{31}{3}: \\
-3 \left(x^2+\frac{8 x}{3}+\frac{16}{9}\right)+3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{31}{3}$} \\
\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 }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 9:
\begin{array}{l}
\frac{16}{3}-\frac{31}{3}=-5: \\
-3 \left(x^2+\frac{8 x}{3}+\frac{16}{9}\right)+3 \left(y^2-\frac{8 y}{3}+\frac{16}{9}\right)=\fbox{$-5$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{8 x}{3}+\frac{16}{9}=\left(x+\frac{4}{3}\right)^2: \\
-3 \fbox{$\left(x+\frac{4}{3}\right)^2$}+3 \left(y^2-\frac{8 y}{3}+\frac{16}{9}\right)=-5 \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{8 y}{3}+\frac{16}{9}=\left(y-\frac{4}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x+\frac{4}{3}\right)^2+3 \fbox{$\left(y-\frac{4}{3}\right)^2$}=-5 \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+2 x+2 y^2+9 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+9 y-8 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(-8 x^2+2 x+\underline{\text{ }}\right)+\left(2 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\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(2 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 y^2+9 y+\underline{\text{ }}\right)=2 \left(y^2+\frac{9 y}{2}+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{x}{4}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+\frac{9 y}{2}+\underline{\text{ }}\right)$}=\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 }-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 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{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 7:
\begin{array}{l}
\frac{81}{8}-\frac{1}{8}=10: \\
-8 \left(x^2-\frac{x}{4}+\frac{1}{64}\right)+2 \left(y^2+\frac{9 y}{2}+\frac{81}{16}\right)=\fbox{$10$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{x}{4}+\frac{1}{64}=\left(x-\frac{1}{8}\right)^2: \\
-8 \fbox{$\left(x-\frac{1}{8}\right)^2$}+2 \left(y^2+\frac{9 y}{2}+\frac{81}{16}\right)=10 \\
\end{array}
Step 9:
\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{} & -8 \left(x-\frac{1}{8}\right)^2+2 \fbox{$\left(y+\frac{9}{4}\right)^2$}=10 \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-7 x+10 y^2+9 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2+9 y-10 x^2-7 x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
10 y^2+9 y-10 x^2-7 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(-10 x^2-7 x+\underline{\text{ }}\right)+\left(10 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2-7 x+\underline{\text{ }}\right)=-10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right)$}+\left(10 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\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): \\
-10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right)+\fbox{$10 \left(y^2+\frac{9 y}{10}+\underline{\text{ }}\right)$}=\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 }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{10}}{2}\right)^2=\frac{49}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{49}{400}=-\frac{49}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
7-\frac{49}{40}=\frac{231}{40}: \\
-10 \left(x^2+\frac{7 x}{10}+\frac{49}{400}\right)+10 \left(y^2+\frac{9 y}{10}+\underline{\text{ }}\right)=\fbox{$\frac{231}{40}$} \\
\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 }\text{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{231}{40}+\frac{81}{40}=\frac{39}{5}: \\
-10 \left(x^2+\frac{7 x}{10}+\frac{49}{400}\right)+10 \left(y^2+\frac{9 y}{10}+\frac{81}{400}\right)=\fbox{$\frac{39}{5}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{10}+\frac{49}{400}=\left(x+\frac{7}{20}\right)^2: \\
-10 \fbox{$\left(x+\frac{7}{20}\right)^2$}+10 \left(y^2+\frac{9 y}{10}+\frac{81}{400}\right)=\frac{39}{5} \\
\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{} & -10 \left(x+\frac{7}{20}\right)^2+\text{10 }\fbox{$\left(y+\frac{9}{20}\right)^2$}=\frac{39}{5} \\
\end{array}
| khanacademy | amps |