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Given the equation $-4 x^2+5 x+8 y^2-5 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-5 y-4 x^2+5 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
8 y^2-5 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(8 y^2-5 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(8 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-5 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{5 y}{8}+\underline{\text{ }}\right): \\
-4 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{5 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-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)+8 \left(y^2-\frac{5 y}{8}+\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 }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{8}}{2}\right)^2=\frac{25}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{25}{256}=\frac{25}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{25}{32}-\frac{89}{16}=-\frac{153}{32}: \\
-4 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)+8 \left(y^2-\frac{5 y}{8}+\frac{25}{256}\right)=\fbox{$-\frac{153}{32}$} \\
\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{$\left(x-\frac{5}{8}\right)^2$}+8 \left(y^2-\frac{5 y}{8}+\frac{25}{256}\right)=-\frac{153}{32} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{5 y}{8}+\frac{25}{256}=\left(y-\frac{5}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x-\frac{5}{8}\right)^2+8 \fbox{$\left(y-\frac{5}{16}\right)^2$}=-\frac{153}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2+x-8 y^2+3 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2+3 y+10 x^2+x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
-8 y^2+3 y+10 x^2+x=-8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(10 x^2+x+\underline{\text{ }}\right)+\left(-8 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\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(-8 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2+3 y+\underline{\text{ }}\right)=-8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right): \\
10 \left(x^2+\frac{x}{10}+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right)$}=\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 }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{10}}{2}\right)^2=\frac{1}{400} \text{on }\text{the }\text{left }\text{and }\frac{10}{400}=\frac{1}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{1}{40}-8=-\frac{319}{40}: \\
10 \left(x^2+\frac{x}{10}+\frac{1}{400}\right)-8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right)=\fbox{$-\frac{319}{40}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-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 9:
\begin{array}{l}
-\frac{319}{40}-\frac{9}{32}=-\frac{1321}{160}: \\
10 \left(x^2+\frac{x}{10}+\frac{1}{400}\right)-8 \left(y^2-\frac{3 y}{8}+\frac{9}{256}\right)=\fbox{$-\frac{1321}{160}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{10}+\frac{1}{400}=\left(x+\frac{1}{20}\right)^2: \\
\text{10 }\fbox{$\left(x+\frac{1}{20}\right)^2$}-8 \left(y^2-\frac{3 y}{8}+\frac{9}{256}\right)=-\frac{1321}{160} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{8}+\frac{9}{256}=\left(y-\frac{3}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 10 \left(x+\frac{1}{20}\right)^2-8 \fbox{$\left(y-\frac{3}{16}\right)^2$}=-\frac{1321}{160} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-8 x+9 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 x^2-8 x+(9 y-2)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 y-10 x^2-8 x-2 \text{from }\text{both }\text{sides}: \\
10 x^2+8 x+(2-9 y)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }2-9 y \text{from }\text{both }\text{sides}: \\
10 x^2+8 x=9 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(10 x^2+8 x+\underline{\text{ }}\right)=(9 y-2)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 x^2+8 x+\underline{\text{ }}\right)=10 \left(x^2+\frac{4 x}{5}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2+\frac{4 x}{5}+\underline{\text{ }}\right)$}=(9 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 }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }10\times \frac{4}{25}=\frac{8}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
(9 y-2)+\frac{8}{5}=9 y-\frac{2}{5}: \\
10 \left(x^2+\frac{4 x}{5}+\frac{4}{25}\right)=\fbox{$9 y-\frac{2}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{4 x}{5}+\frac{4}{25}=\left(x+\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \text{10 }\fbox{$\left(x+\frac{2}{5}\right)^2$}=9 y-\frac{2}{5} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2+7 x-2 y^2-4 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-4 y+10 x^2+7 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-2 y^2-4 y+10 x^2+7 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+7 x+\underline{\text{ }}\right)+\left(-2 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\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(-2 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2-4 y+\underline{\text{ }}\right)=-2 \left(y^2+2 y+\underline{\text{ }}\right): \\
10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2+2 y+\underline{\text{ }}\right)$}=\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 }\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 7:
\begin{array}{l}
6+\frac{49}{40}=\frac{289}{40}: \\
10 \left(x^2+\frac{7 x}{10}+\frac{49}{400}\right)-2 \left(y^2+2 y+\underline{\text{ }}\right)=\fbox{$\frac{289}{40}$} \\
\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 }-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 9:
\begin{array}{l}
\frac{289}{40}-2=\frac{209}{40}: \\
10 \left(x^2+\frac{7 x}{10}+\frac{49}{400}\right)-2 \left(y^2+2 y+1\right)=\fbox{$\frac{209}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{10}+\frac{49}{400}=\left(x+\frac{7}{20}\right)^2: \\
\text{10 }\fbox{$\left(x+\frac{7}{20}\right)^2$}-2 \left(y^2+2 y+1\right)=\frac{209}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+2 y+1=(y+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 10 \left(x+\frac{7}{20}\right)^2-2 \fbox{$(y+1)^2$}=\frac{209}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2+2 x-5 y^2-y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2-y+x^2+2 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
-5 y^2-y+x^2+2 x=8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(x^2+2 x+\underline{\text{ }}\right)+\left(-5 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 y^2-y+\underline{\text{ }}\right)=-5 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right): \\
\left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+8 \\
\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{2}{2}\right)^2=1 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
8+1=9: \\
\left(x^2+2 x+1\right)-5 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right)=\fbox{$9$} \\
\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 }-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 8:
\begin{array}{l}
9-\frac{1}{20}=\frac{179}{20}: \\
\left(x^2+2 x+1\right)-5 \left(y^2+\frac{y}{5}+\frac{1}{100}\right)=\fbox{$\frac{179}{20}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
\fbox{$(x+1)^2$}-5 \left(y^2+\frac{y}{5}+\frac{1}{100}\right)=\frac{179}{20} \\
\end{array}
Step 10:
\begin{array}{l}
y^2+\frac{y}{5}+\frac{1}{100}=\left(y+\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & (x+1)^2-5 \fbox{$\left(y+\frac{1}{10}\right)^2$}=\frac{179}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-3 x+8 y^2-3 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-3 y+5 x^2-3 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
8 y^2-3 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(8 y^2-3 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(8 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-3 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right): \\
5 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\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{189}{20}: \\
5 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)+8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right)=\fbox{$\frac{189}{20}$} \\
\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{-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 9:
\begin{array}{l}
\frac{189}{20}+\frac{9}{32}=\frac{1557}{160}: \\
5 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)+8 \left(y^2-\frac{3 y}{8}+\frac{9}{256}\right)=\fbox{$\frac{1557}{160}$} \\
\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{$\left(x-\frac{3}{10}\right)^2$}+8 \left(y^2-\frac{3 y}{8}+\frac{9}{256}\right)=\frac{1557}{160} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{8}+\frac{9}{256}=\left(y-\frac{3}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x-\frac{3}{10}\right)^2+8 \fbox{$\left(y-\frac{3}{16}\right)^2$}=\frac{1557}{160} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2-10 x+3 y^2-4 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2-4 y-x^2-10 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(-x^2-10 x+\underline{\text{ }}\right)+\left(3 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(-x^2-10 x+\underline{\text{ }}\right)=-\left(x^2+10 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2+10 x+\underline{\text{ }}\right)$}+\left(3 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 y^2-4 y+\underline{\text{ }}\right)=3 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right): \\
-\left(x^2+10 x+\underline{\text{ }}\right)+\fbox{$3 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+0 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-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 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{3}}{2}\right)^2=\frac{4}{9} \text{on }\text{the }\text{left }\text{and }3\times \frac{4}{9}=\frac{4}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{4}{3}-25=-\frac{71}{3}: \\
-\left(x^2+10 x+25\right)+3 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$-\frac{71}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+10 x+25=(x+5)^2: \\
-\fbox{$(x+5)^2$}+3 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=-\frac{71}{3} \\
\end{array}
Step 9:
\begin{array}{l}
y^2-\frac{4 y}{3}+\frac{4}{9}=\left(y-\frac{2}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -(x+5)^2+3 \fbox{$\left(y-\frac{2}{3}\right)^2$}=-\frac{71}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+6 x-2 y^2+4 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+4 y+8 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+4 y+8 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(8 x^2+6 x+\underline{\text{ }}\right)+\left(-2 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2+6 x+\underline{\text{ }}\right)=8 \left(x^2+\frac{3 x}{4}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+\frac{3 x}{4}+\underline{\text{ }}\right)$}+\left(-2 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+4 y+\underline{\text{ }}\right)=-2 \left(y^2-2 y+\underline{\text{ }}\right): \\
8 \left(x^2+\frac{3 x}{4}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-2 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }8\times \frac{9}{64}=\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{8}-9=-\frac{63}{8}: \\
8 \left(x^2+\frac{3 x}{4}+\frac{9}{64}\right)-2 \left(y^2-2 y+\underline{\text{ }}\right)=\fbox{$-\frac{63}{8}$} \\
\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{-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 9:
\begin{array}{l}
-\frac{63}{8}-2=-\frac{79}{8}: \\
8 \left(x^2+\frac{3 x}{4}+\frac{9}{64}\right)-2 \left(y^2-2 y+1\right)=\fbox{$-\frac{79}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{4}+\frac{9}{64}=\left(x+\frac{3}{8}\right)^2: \\
8 \fbox{$\left(x+\frac{3}{8}\right)^2$}-2 \left(y^2-2 y+1\right)=-\frac{79}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-2 y+1=(y-1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \left(x+\frac{3}{8}\right)^2-2 \fbox{$(y-1)^2$}=-\frac{79}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2-8 x+4 y^2-3 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2-3 y-9 x^2-8 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
4 y^2-3 y-9 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(-9 x^2-8 x+\underline{\text{ }}\right)+\left(4 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2-8 x+\underline{\text{ }}\right)=-9 \left(x^2+\frac{8 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2+\frac{8 x}{9}+\underline{\text{ }}\right)$}+\left(4 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2-3 y+\underline{\text{ }}\right)=4 \left(y^2-\frac{3 y}{4}+\underline{\text{ }}\right): \\
-9 \left(x^2+\frac{8 x}{9}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2-\frac{3 y}{4}+\underline{\text{ }}\right)$}=\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 }-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}
9-\frac{16}{9}=\frac{65}{9}: \\
-9 \left(x^2+\frac{8 x}{9}+\frac{16}{81}\right)+4 \left(y^2-\frac{3 y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{65}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }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 9:
\begin{array}{l}
\frac{65}{9}+\frac{9}{16}=\frac{1121}{144}: \\
-9 \left(x^2+\frac{8 x}{9}+\frac{16}{81}\right)+4 \left(y^2-\frac{3 y}{4}+\frac{9}{64}\right)=\fbox{$\frac{1121}{144}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{8 x}{9}+\frac{16}{81}=\left(x+\frac{4}{9}\right)^2: \\
-9 \fbox{$\left(x+\frac{4}{9}\right)^2$}+4 \left(y^2-\frac{3 y}{4}+\frac{9}{64}\right)=\frac{1121}{144} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{4}+\frac{9}{64}=\left(y-\frac{3}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x+\frac{4}{9}\right)^2+4 \fbox{$\left(y-\frac{3}{8}\right)^2$}=\frac{1121}{144} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-2 x+6 y^2+8 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2+8 y+5 x^2-2 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
6 y^2+8 y+5 x^2-2 x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2-2 x+\underline{\text{ }}\right)+\left(6 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2-2 x+\underline{\text{ }}\right)=5 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right)$}+\left(6 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2+8 y+\underline{\text{ }}\right)=6 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right): \\
5 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{5}}{2}\right)^2=\frac{1}{25} \text{on }\text{the }\text{left }\text{and }\frac{5}{25}=\frac{1}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{1}{5}-4=-\frac{19}{5}: \\
5 \left(x^2-\frac{2 x}{5}+\frac{1}{25}\right)+6 \left(y^2+\frac{4 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{19}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{3}}{2}\right)^2=\frac{4}{9} \text{on }\text{the }\text{left }\text{and }6\times \frac{4}{9}=\frac{8}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{8}{3}-\frac{19}{5}=-\frac{17}{15}: \\
5 \left(x^2-\frac{2 x}{5}+\frac{1}{25}\right)+6 \left(y^2+\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$-\frac{17}{15}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{2 x}{5}+\frac{1}{25}=\left(x-\frac{1}{5}\right)^2: \\
5 \fbox{$\left(x-\frac{1}{5}\right)^2$}+6 \left(y^2+\frac{4 y}{3}+\frac{4}{9}\right)=-\frac{17}{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{1}{5}\right)^2+6 \fbox{$\left(y+\frac{2}{3}\right)^2$}=-\frac{17}{15} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+4 x+9 y^2+3 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+3 y+8 x^2+4 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
9 y^2+3 y+8 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(8 x^2+4 x+\underline{\text{ }}\right)+\left(9 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2+4 x+\underline{\text{ }}\right)=8 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)$}+\left(9 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2+3 y+\underline{\text{ }}\right)=9 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right): \\
8 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)$}=\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 }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{8}{16}=\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8+\frac{1}{2}=\frac{17}{2}: \\
8 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)+9 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{17}{2}$} \\
\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 }9 \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{9}{36}=\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{17}{2}+\frac{1}{4}=\frac{35}{4}: \\
8 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)+9 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=\fbox{$\frac{35}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{2}+\frac{1}{16}=\left(x+\frac{1}{4}\right)^2: \\
8 \fbox{$\left(x+\frac{1}{4}\right)^2$}+9 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=\frac{35}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{3}+\frac{1}{36}=\left(y+\frac{1}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \left(x+\frac{1}{4}\right)^2+9 \fbox{$\left(y+\frac{1}{6}\right)^2$}=\frac{35}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2+6 x+7 y^2-3 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2-3 y+5 x^2+6 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
7 y^2-3 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(7 y^2-3 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(7 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2-3 y+\underline{\text{ }}\right)=7 \left(y^2-\frac{3 y}{7}+\underline{\text{ }}\right): \\
5 \left(x^2+\frac{6 x}{5}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2-\frac{3 y}{7}+\underline{\text{ }}\right)$}=\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 }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{6}{5}}{2}\right)^2=\frac{9}{25} \text{on }\text{the }\text{left }\text{and }5\times \frac{9}{25}=\frac{9}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{5}-6=-\frac{21}{5}: \\
5 \left(x^2+\frac{6 x}{5}+\frac{9}{25}\right)+7 \left(y^2-\frac{3 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{21}{5}$} \\
\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{-3}{7}}{2}\right)^2=\frac{9}{196} \text{on }\text{the }\text{left }\text{and }7\times \frac{9}{196}=\frac{9}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{9}{28}-\frac{21}{5}=-\frac{543}{140}: \\
5 \left(x^2+\frac{6 x}{5}+\frac{9}{25}\right)+7 \left(y^2-\frac{3 y}{7}+\frac{9}{196}\right)=\fbox{$-\frac{543}{140}$} \\
\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{$\left(x+\frac{3}{5}\right)^2$}+7 \left(y^2-\frac{3 y}{7}+\frac{9}{196}\right)=-\frac{543}{140} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{7}+\frac{9}{196}=\left(y-\frac{3}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x+\frac{3}{5}\right)^2+7 \fbox{$\left(y-\frac{3}{14}\right)^2$}=-\frac{543}{140} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2+x+2 y^2+4 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+4 y-2 x^2+x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
2 y^2+4 y-2 x^2+x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-2 x^2+x+\underline{\text{ }}\right)+\left(2 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 x^2+x+\underline{\text{ }}\right)=-2 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$-2 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)$}+\left(2 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+4 y+\underline{\text{ }}\right)=2 \left(y^2+2 y+\underline{\text{ }}\right): \\
-2 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+2 y+\underline{\text{ }}\right)$}=\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 }-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 7:
\begin{array}{l}
-1-\frac{1}{8}=-\frac{9}{8}: \\
-2 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+2 \left(y^2+2 y+\underline{\text{ }}\right)=\fbox{$-\frac{9}{8}$} \\
\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{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 9:
\begin{array}{l}
2-\frac{9}{8}=\frac{7}{8}: \\
-2 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+2 \left(y^2+2 y+1\right)=\fbox{$\frac{7}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
-2 \fbox{$\left(x-\frac{1}{4}\right)^2$}+2 \left(y^2+2 y+1\right)=\frac{7}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+2 y+1=(y+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -2 \left(x-\frac{1}{4}\right)^2+2 \fbox{$(y+1)^2$}=\frac{7}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-10 x+3 y^2-4 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2-4 y+2 x^2-10 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
3 y^2-4 y+2 x^2-10 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2-10 x+\underline{\text{ }}\right)+\left(3 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2-10 x+\underline{\text{ }}\right)=2 \left(x^2-5 x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-5 x+\underline{\text{ }}\right)$}+\left(3 y^2-4 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): \\
2 \left(x^2-5 x+\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 }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }2\times \frac{25}{4}=\frac{25}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{25}{2}-1=\frac{23}{2}: \\
2 \left(x^2-5 x+\frac{25}{4}\right)+3 \left(y^2-\frac{4 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{23}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\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{23}{2}+\frac{4}{3}=\frac{77}{6}: \\
2 \left(x^2-5 x+\frac{25}{4}\right)+3 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=\fbox{$\frac{77}{6}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-5 x+\frac{25}{4}=\left(x-\frac{5}{2}\right)^2: \\
2 \fbox{$\left(x-\frac{5}{2}\right)^2$}+3 \left(y^2-\frac{4 y}{3}+\frac{4}{9}\right)=\frac{77}{6} \\
\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{} & 2 \left(x-\frac{5}{2}\right)^2+3 \fbox{$\left(y-\frac{2}{3}\right)^2$}=\frac{77}{6} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2+5 x+5 y^2+9 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+9 y-6 x^2+5 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
5 y^2+9 y-6 x^2+5 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2+5 x+\underline{\text{ }}\right)+\left(5 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\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(5 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\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): \\
-6 \left(x^2-\frac{5 x}{6}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2+\frac{9 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-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}
-9-\frac{25}{24}=-\frac{241}{24}: \\
-6 \left(x^2-\frac{5 x}{6}+\frac{25}{144}\right)+5 \left(y^2+\frac{9 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{241}{24}$} \\
\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 }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{241}{24}=-\frac{719}{120}: \\
-6 \left(x^2-\frac{5 x}{6}+\frac{25}{144}\right)+5 \left(y^2+\frac{9 y}{5}+\frac{81}{100}\right)=\fbox{$-\frac{719}{120}$} \\
\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{$\left(x-\frac{5}{12}\right)^2$}+5 \left(y^2+\frac{9 y}{5}+\frac{81}{100}\right)=-\frac{719}{120} \\
\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{} & -6 \left(x-\frac{5}{12}\right)^2+5 \fbox{$\left(y+\frac{9}{10}\right)^2$}=-\frac{719}{120} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2-8 x-y^2+10 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2+10 y-6 x^2-8 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
-y^2+10 y-6 x^2-8 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^2-8 x+\underline{\text{ }}\right)+\left(-y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\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(-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): \\
-6 \left(x^2+\frac{4 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 }-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}
-6-\frac{8}{3}=-\frac{26}{3}: \\
-6 \left(x^2+\frac{4 x}{3}+\frac{4}{9}\right)-\left(y^2-10 y+\underline{\text{ }}\right)=\fbox{$-\frac{26}{3}$} \\
\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{26}{3}-25=-\frac{101}{3}: \\
-6 \left(x^2+\frac{4 x}{3}+\frac{4}{9}\right)-\left(y^2-10 y+25\right)=\fbox{$-\frac{101}{3}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{4 x}{3}+\frac{4}{9}=\left(x+\frac{2}{3}\right)^2: \\
-6 \fbox{$\left(x+\frac{2}{3}\right)^2$}-\left(y^2-10 y+25\right)=-\frac{101}{3} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-10 y+25=(y-5)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x+\frac{2}{3}\right)^2-\fbox{$(y-5)^2$}=-\frac{101}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2-8 x-7 y^2-3 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2-3 y-x^2-8 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
-7 y^2-3 y-x^2-8 x=-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2-8 x+\underline{\text{ }}\right)+\left(-7 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2-8 x+\underline{\text{ }}\right)=-\left(x^2+8 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2+8 x+\underline{\text{ }}\right)$}+\left(-7 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2-3 y+\underline{\text{ }}\right)=-7 \left(y^2+\frac{3 y}{7}+\underline{\text{ }}\right): \\
-\left(x^2+8 x+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2+\frac{3 y}{7}+\underline{\text{ }}\right)$}=\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 }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{8}{2}\right)^2=16 \text{on }\text{the }\text{left }\text{and }-16=-16 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-10-16=-26: \\
-\left(x^2+8 x+16\right)-7 \left(y^2+\frac{3 y}{7}+\underline{\text{ }}\right)=\fbox{$-26$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{7}}{2}\right)^2=\frac{9}{196} \text{on }\text{the }\text{left }\text{and }-7\times \frac{9}{196}=-\frac{9}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-26-\frac{9}{28}=-\frac{737}{28}: \\
-\left(x^2+8 x+16\right)-7 \left(y^2+\frac{3 y}{7}+\frac{9}{196}\right)=\fbox{$-\frac{737}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+8 x+16=(x+4)^2: \\
-\fbox{$(x+4)^2$}-7 \left(y^2+\frac{3 y}{7}+\frac{9}{196}\right)=-\frac{737}{28} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{7}+\frac{9}{196}=\left(y+\frac{3}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -(x+4)^2-7 \fbox{$\left(y+\frac{3}{14}\right)^2$}=-\frac{737}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2-10 y^2-10 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-10 y+\left(-x^2-10\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }10 y^2+10 y+x^2+10 \text{to }\text{both }\text{sides}: \\
10 y^2+10 y+\left(x^2+10\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
10 y^2+10 y+x^2=-10 \\
\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(10 y^2+10 y+\underline{\text{ }}\right)+x^2=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2+10 y+\underline{\text{ }}\right)=10 \left(y^2+y+\underline{\text{ }}\right): \\
\fbox{$10 \left(y^2+y+\underline{\text{ }}\right)$}+x^2=\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 }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{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}-10=-\frac{15}{2}: \\
10 \left(y^2+y+\frac{1}{4}\right)+x^2=\fbox{$-\frac{15}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
y^2+y+\frac{1}{4}=\left(y+\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \text{10 }\fbox{$\left(y+\frac{1}{2}\right)^2$}+x^2=-\frac{15}{2} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2+2 x-5 y^2+8 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+8 y-4 x^2+2 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-5 y^2+8 y-4 x^2+2 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-4 x^2+2 x+\underline{\text{ }}\right)+\left(-5 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2+2 x+\underline{\text{ }}\right)=-4 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)$}+\left(-5 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+8 y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{8 y}{5}+\underline{\text{ }}\right): \\
-4 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{8 y}{5}+\underline{\text{ }}\right)$}=\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{-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}
-1-\frac{1}{4}=-\frac{5}{4}: \\
-4 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)-5 \left(y^2-\frac{8 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{5}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-8}{5}}{2}\right)^2=\frac{16}{25} \text{on }\text{the }\text{left }\text{and }-5\times \frac{16}{25}=-\frac{16}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{5}{4}-\frac{16}{5}=-\frac{89}{20}: \\
-4 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)-5 \left(y^2-\frac{8 y}{5}+\frac{16}{25}\right)=\fbox{$-\frac{89}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
-4 \fbox{$\left(x-\frac{1}{4}\right)^2$}-5 \left(y^2-\frac{8 y}{5}+\frac{16}{25}\right)=-\frac{89}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{8 y}{5}+\frac{16}{25}=\left(y-\frac{4}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x-\frac{1}{4}\right)^2-5 \fbox{$\left(y-\frac{4}{5}\right)^2$}=-\frac{89}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2+x+8 y^2-7 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-7 y+5 x^2+x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
8 y^2-7 y+5 x^2+x=1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2+x+\underline{\text{ }}\right)+\left(8 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\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^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-7 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{7 y}{8}+\underline{\text{ }}\right): \\
5 \left(x^2+\frac{x}{5}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{7 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}+1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{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}
1+\frac{1}{20}=\frac{21}{20}: \\
5 \left(x^2+\frac{x}{5}+\frac{1}{100}\right)+8 \left(y^2-\frac{7 y}{8}+\underline{\text{ }}\right)=\fbox{$\frac{21}{20}$} \\
\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{-7}{8}}{2}\right)^2=\frac{49}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{49}{256}=\frac{49}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{21}{20}+\frac{49}{32}=\frac{413}{160}: \\
5 \left(x^2+\frac{x}{5}+\frac{1}{100}\right)+8 \left(y^2-\frac{7 y}{8}+\frac{49}{256}\right)=\fbox{$\frac{413}{160}$} \\
\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{7 y}{8}+\frac{49}{256}\right)=\frac{413}{160} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{8}+\frac{49}{256}=\left(y-\frac{7}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x+\frac{1}{10}\right)^2+8 \fbox{$\left(y-\frac{7}{16}\right)^2$}=\frac{413}{160} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2+3 x+y^2+6 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2+6 y-9 x^2+3 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
y^2+6 y-9 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(-9 x^2+3 x+\underline{\text{ }}\right)+\left(y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2+3 x+\underline{\text{ }}\right)=-9 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)$}+\left(y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\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 }-9 \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{-9}{36}=-\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
-8-\frac{1}{4}=-\frac{33}{4}: \\
-9 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)+\left(y^2+6 y+\underline{\text{ }}\right)=\fbox{$-\frac{33}{4}$} \\
\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{Add }\left(\frac{6}{2}\right)^2=9 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 8:
\begin{array}{l}
9-\frac{33}{4}=\frac{3}{4}: \\
-9 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)+\left(y^2+6 y+9\right)=\fbox{$\frac{3}{4}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-\frac{x}{3}+\frac{1}{36}=\left(x-\frac{1}{6}\right)^2: \\
-9 \fbox{$\left(x-\frac{1}{6}\right)^2$}+\left(y^2+6 y+9\right)=\frac{3}{4} \\
\end{array}
Step 10:
\begin{array}{l}
y^2+6 y+9=(y+3)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x-\frac{1}{6}\right)^2+\fbox{$(y+3)^2$}=\frac{3}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2+5 y^2-4 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2-4 y+\left(-10 x^2-8\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
5 y^2-4 y-10 x^2=8 \\
\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 y^2-4 y+\underline{\text{ }}\right)-10 x^2=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 y^2-4 y+\underline{\text{ }}\right)=5 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)$}-10 x^2=\underline{\text{ }}+8 \\
\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 }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }5\times \frac{4}{25}=\frac{4}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
8+\frac{4}{5}=\frac{44}{5}: \\
5 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)-10 x^2=\fbox{$\frac{44}{5}$} \\
\end{array}
Step 7:
\begin{array}{l}
y^2-\frac{4 y}{5}+\frac{4}{25}=\left(y-\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$\left(y-\frac{2}{5}\right)^2$}-10 x^2=\frac{44}{5} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+5 y^2-y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2-y+\left(4 x^2+3\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
5 y^2-y+4 x^2=-3 \\
\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 y^2-y+\underline{\text{ }}\right)+4 x^2=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 y^2-y+\underline{\text{ }}\right)=5 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right)$}+4 x^2=\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 }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{-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}-3=-\frac{59}{20}: \\
5 \left(y^2-\frac{y}{5}+\frac{1}{100}\right)+4 x^2=\fbox{$-\frac{59}{20}$} \\
\end{array}
Step 7:
\begin{array}{l}
y^2-\frac{y}{5}+\frac{1}{100}=\left(y-\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$\left(y-\frac{1}{10}\right)^2$}+4 x^2=-\frac{59}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2-8 x-2 y^2-3 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-3 y+x^2-8 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-2 y^2-3 y+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(x^2-8 x+\underline{\text{ }}\right)+\left(-2 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\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): \\
\left(x^2-8 x+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{-8}{2}\right)^2=16 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
16-9=7: \\
\left(x^2-8 x+16\right)-2 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$7$} \\
\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 }-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 8:
\begin{array}{l}
7-\frac{9}{8}=\frac{47}{8}: \\
\left(x^2-8 x+16\right)-2 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$\frac{47}{8}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-8 x+16=(x-4)^2: \\
\fbox{$(x-4)^2$}-2 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\frac{47}{8} \\
\end{array}
Step 10:
\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{} & (x-4)^2-2 \fbox{$\left(y+\frac{3}{4}\right)^2$}=\frac{47}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-10 x+4 y^2-7 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2-7 y+4 x^2-10 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
4 y^2-7 y+4 x^2-10 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2-10 x+\underline{\text{ }}\right)+\left(4 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\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^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2-7 y+\underline{\text{ }}\right)=4 \left(y^2-\frac{7 y}{4}+\underline{\text{ }}\right): \\
4 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2-\frac{7 y}{4}+\underline{\text{ }}\right)$}=\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{-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}
\frac{25}{4}-1=\frac{21}{4}: \\
4 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)+4 \left(y^2-\frac{7 y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{21}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{4}}{2}\right)^2=\frac{49}{64} \text{on }\text{the }\text{left }\text{and }4\times \frac{49}{64}=\frac{49}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{21}{4}+\frac{49}{16}=\frac{133}{16}: \\
4 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)+4 \left(y^2-\frac{7 y}{4}+\frac{49}{64}\right)=\fbox{$\frac{133}{16}$} \\
\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-\frac{7 y}{4}+\frac{49}{64}\right)=\frac{133}{16} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{4}+\frac{49}{64}=\left(y-\frac{7}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x-\frac{5}{4}\right)^2+4 \fbox{$\left(y-\frac{7}{8}\right)^2$}=\frac{133}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2+9 x-9 y^2-5 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-5 y-6 x^2+9 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
-9 y^2-5 y-6 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(-6 x^2+9 x+\underline{\text{ }}\right)+\left(-9 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\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(-9 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\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): \\
-6 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2+\frac{5 y}{9}+\underline{\text{ }}\right)$}=\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 }-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}
-4-\frac{27}{8}=-\frac{59}{8}: \\
-6 \left(x^2-\frac{3 x}{2}+\frac{9}{16}\right)-9 \left(y^2+\frac{5 y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{59}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{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{59}{8}-\frac{25}{36}=-\frac{581}{72}: \\
-6 \left(x^2-\frac{3 x}{2}+\frac{9}{16}\right)-9 \left(y^2+\frac{5 y}{9}+\frac{25}{324}\right)=\fbox{$-\frac{581}{72}$} \\
\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{$\left(x-\frac{3}{4}\right)^2$}-9 \left(y^2+\frac{5 y}{9}+\frac{25}{324}\right)=-\frac{581}{72} \\
\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{} & -6 \left(x-\frac{3}{4}\right)^2-9 \fbox{$\left(y+\frac{5}{18}\right)^2$}=-\frac{581}{72} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2-5 x-6 y^2-4 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-4 y+3 x^2-5 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-6 y^2-4 y+3 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(3 x^2-5 x+\underline{\text{ }}\right)+\left(-6 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\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(-6 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\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{5 x}{3}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right)$}=\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 }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}
\frac{25}{12}-1=\frac{13}{12}: \\
3 \left(x^2-\frac{5 x}{3}+\frac{25}{36}\right)-6 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{13}{12}$} \\
\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{13}{12}-\frac{2}{3}=\frac{5}{12}: \\
3 \left(x^2-\frac{5 x}{3}+\frac{25}{36}\right)-6 \left(y^2+\frac{2 y}{3}+\frac{1}{9}\right)=\fbox{$\frac{5}{12}$} \\
\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{$\left(x-\frac{5}{6}\right)^2$}-6 \left(y^2+\frac{2 y}{3}+\frac{1}{9}\right)=\frac{5}{12} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{2 y}{3}+\frac{1}{9}=\left(y+\frac{1}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x-\frac{5}{6}\right)^2-6 \fbox{$\left(y+\frac{1}{3}\right)^2$}=\frac{5}{12} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2+6 x+y^2-5 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2-5 y-x^2+6 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
y^2-5 y-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(-x^2+6 x+\underline{\text{ }}\right)+\left(y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2+6 x+\underline{\text{ }}\right)=-\left(x^2-6 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2-6 x+\underline{\text{ }}\right)$}+\left(y^2-5 y+\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{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-6}{2}\right)^2=9 \text{on }\text{the }\text{left }\text{and }-9=-9 \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
4-9=-5: \\
-\left(x^2-6 x+9\right)+\left(y^2-5 y+\underline{\text{ }}\right)=\fbox{$-5$} \\
\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{Add }\left(\frac{-5}{2}\right)^2=\frac{25}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 8:
\begin{array}{l}
\frac{25}{4}-5=\frac{5}{4}: \\
-\left(x^2-6 x+9\right)+\left(y^2-5 y+\frac{25}{4}\right)=\fbox{$\frac{5}{4}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-6 x+9=(x-3)^2: \\
-\fbox{$(x-3)^2$}+\left(y^2-5 y+\frac{25}{4}\right)=\frac{5}{4} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-5 y+\frac{25}{4}=\left(y-\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -(x-3)^2+\fbox{$\left(y-\frac{5}{2}\right)^2$}=\frac{5}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2+8 x+7 y^2-8 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2-8 y-7 x^2+8 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
7 y^2-8 y-7 x^2+8 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+8 x+\underline{\text{ }}\right)+\left(7 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 x^2+8 x+\underline{\text{ }}\right)=-7 \left(x^2-\frac{8 x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2-\frac{8 x}{7}+\underline{\text{ }}\right)$}+\left(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): \\
-7 \left(x^2-\frac{8 x}{7}+\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 }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-8}{7}}{2}\right)^2=\frac{16}{49} \text{on }\text{the }\text{left }\text{and }-7\times \frac{16}{49}=-\frac{16}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-3-\frac{16}{7}=-\frac{37}{7}: \\
-7 \left(x^2-\frac{8 x}{7}+\frac{16}{49}\right)+7 \left(y^2-\frac{8 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{37}{7}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-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{37}{7}=-3: \\
-7 \left(x^2-\frac{8 x}{7}+\frac{16}{49}\right)+7 \left(y^2-\frac{8 y}{7}+\frac{16}{49}\right)=\fbox{$-3$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{8 x}{7}+\frac{16}{49}=\left(x-\frac{4}{7}\right)^2: \\
-7 \fbox{$\left(x-\frac{4}{7}\right)^2$}+7 \left(y^2-\frac{8 y}{7}+\frac{16}{49}\right)=-3 \\
\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{} & -7 \left(x-\frac{4}{7}\right)^2+7 \fbox{$\left(y-\frac{4}{7}\right)^2$}=-3 \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2-7 x-10 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 x^2-7 x+(-10 y-7)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }10 y+2 x^2+7 x+7 \text{to }\text{both }\text{sides}: \\
2 x^2+7 x+(10 y+7)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }10 y+7 \text{from }\text{both }\text{sides}: \\
2 x^2+7 x=-10 y-7 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(2 x^2+7 x+\underline{\text{ }}\right)=(-10 y-7)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 x^2+7 x+\underline{\text{ }}\right)=2 \left(x^2+\frac{7 x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2+\frac{7 x}{2}+\underline{\text{ }}\right)$}=(-10 y-7)+\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 }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 y-7)+\frac{49}{8}=-10 y-\frac{7}{8}: \\
2 \left(x^2+\frac{7 x}{2}+\frac{49}{16}\right)=\fbox{$-10 y-\frac{7}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{7 x}{2}+\frac{49}{16}=\left(x+\frac{7}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \fbox{$\left(x+\frac{7}{4}\right)^2$}=-10 y-\frac{7}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2+5 x+3 y^2-8 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2-8 y-7 x^2+5 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
3 y^2-8 y-7 x^2+5 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-7 x^2+5 x+\underline{\text{ }}\right)+\left(3 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\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(3 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\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{5 x}{7}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right)$}=\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 }-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}
3-\frac{25}{28}=\frac{59}{28}: \\
-7 \left(x^2-\frac{5 x}{7}+\frac{25}{196}\right)+3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{59}{28}$} \\
\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{59}{28}+\frac{16}{3}=\frac{625}{84}: \\
-7 \left(x^2-\frac{5 x}{7}+\frac{25}{196}\right)+3 \left(y^2-\frac{8 y}{3}+\frac{16}{9}\right)=\fbox{$\frac{625}{84}$} \\
\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{$\left(x-\frac{5}{14}\right)^2$}+3 \left(y^2-\frac{8 y}{3}+\frac{16}{9}\right)=\frac{625}{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{5}{14}\right)^2+3 \fbox{$\left(y-\frac{4}{3}\right)^2$}=\frac{625}{84} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+6 x-10 y^2+7 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+7 y-8 x^2+6 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-10 y^2+7 y-8 x^2+6 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(-8 x^2+6 x+\underline{\text{ }}\right)+\left(-10 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2+6 x+\underline{\text{ }}\right)=-8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)$}+\left(-10 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\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): \\
-8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2-\frac{7 y}{10}+\underline{\text{ }}\right)$}=\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 }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }-8\times \frac{9}{64}=-\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5-\frac{9}{8}=\frac{31}{8}: \\
-8 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)-10 \left(y^2-\frac{7 y}{10}+\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 }-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{31}{8}-\frac{49}{40}=\frac{53}{20}: \\
-8 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)-10 \left(y^2-\frac{7 y}{10}+\frac{49}{400}\right)=\fbox{$\frac{53}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{4}+\frac{9}{64}=\left(x-\frac{3}{8}\right)^2: \\
-8 \fbox{$\left(x-\frac{3}{8}\right)^2$}-10 \left(y^2-\frac{7 y}{10}+\frac{49}{400}\right)=\frac{53}{20} \\
\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{} & -8 \left(x-\frac{3}{8}\right)^2-\text{10 }\fbox{$\left(y-\frac{7}{20}\right)^2$}=\frac{53}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2+6 x-2 y^2-7 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-7 y-9 x^2+6 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
-2 y^2-7 y-9 x^2+6 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+6 x+\underline{\text{ }}\right)+\left(-2 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2+6 x+\underline{\text{ }}\right)=-9 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right)$}+\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): \\
-9 \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 }-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 7:
\begin{array}{l}
8-1=7: \\
-9 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)-2 \left(y^2+\frac{7 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 }-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}
7-\frac{49}{8}=\frac{7}{8}: \\
-9 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)-2 \left(y^2+\frac{7 y}{2}+\frac{49}{16}\right)=\fbox{$\frac{7}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{2 x}{3}+\frac{1}{9}=\left(x-\frac{1}{3}\right)^2: \\
-9 \fbox{$\left(x-\frac{1}{3}\right)^2$}-2 \left(y^2+\frac{7 y}{2}+\frac{49}{16}\right)=\frac{7}{8} \\
\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{} & -9 \left(x-\frac{1}{3}\right)^2-2 \fbox{$\left(y+\frac{7}{4}\right)^2$}=\frac{7}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-9 x+10 y^2+3 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2+3 y-10 x^2-9 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
10 y^2+3 y-10 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(-10 x^2-9 x+\underline{\text{ }}\right)+\left(10 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\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{$-10 \left(x^2+\frac{9 x}{10}+\underline{\text{ }}\right)$}+\left(10 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2+3 y+\underline{\text{ }}\right)=10 \left(y^2+\frac{3 y}{10}+\underline{\text{ }}\right): \\
-10 \left(x^2+\frac{9 x}{10}+\underline{\text{ }}\right)+\fbox{$10 \left(y^2+\frac{3 y}{10}+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-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 7:
\begin{array}{l}
4-\frac{81}{40}=\frac{79}{40}: \\
-10 \left(x^2+\frac{9 x}{10}+\frac{81}{400}\right)+10 \left(y^2+\frac{3 y}{10}+\underline{\text{ }}\right)=\fbox{$\frac{79}{40}$} \\
\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{3}{10}}{2}\right)^2=\frac{9}{400} \text{on }\text{the }\text{left }\text{and }10\times \frac{9}{400}=\frac{9}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{79}{40}+\frac{9}{40}=\frac{11}{5}: \\
-10 \left(x^2+\frac{9 x}{10}+\frac{81}{400}\right)+10 \left(y^2+\frac{3 y}{10}+\frac{9}{400}\right)=\fbox{$\frac{11}{5}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{9 x}{10}+\frac{81}{400}=\left(x+\frac{9}{20}\right)^2: \\
-10 \fbox{$\left(x+\frac{9}{20}\right)^2$}+10 \left(y^2+\frac{3 y}{10}+\frac{9}{400}\right)=\frac{11}{5} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{10}+\frac{9}{400}=\left(y+\frac{3}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x+\frac{9}{20}\right)^2+\text{10 }\fbox{$\left(y+\frac{3}{20}\right)^2$}=\frac{11}{5} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2-4 x-y^2-5 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2-5 y+8 x^2-4 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
-y^2-5 y+8 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(8 x^2-4 x+\underline{\text{ }}\right)+\left(-y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2-4 x+\underline{\text{ }}\right)=8 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)$}+\left(-y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2-5 y+\underline{\text{ }}\right)=-\left(y^2+5 y+\underline{\text{ }}\right): \\
8 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$-\left(y^2+5 y+\underline{\text{ }}\right)$}=\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{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{8}{16}=\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3+\frac{1}{2}=\frac{7}{2}: \\
8 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)-\left(y^2+5 y+\underline{\text{ }}\right)=\fbox{$\frac{7}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }-\frac{25}{4}=-\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{7}{2}-\frac{25}{4}=-\frac{11}{4}: \\
8 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)-\left(y^2+5 y+\frac{25}{4}\right)=\fbox{$-\frac{11}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
8 \fbox{$\left(x-\frac{1}{4}\right)^2$}-\left(y^2+5 y+\frac{25}{4}\right)=-\frac{11}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+5 y+\frac{25}{4}=\left(y+\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \left(x-\frac{1}{4}\right)^2-\fbox{$\left(y+\frac{5}{2}\right)^2$}=-\frac{11}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2+6 x+7 y^2-5 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2-5 y-9 x^2+6 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(-9 x^2+6 x+\underline{\text{ }}\right)+\left(7 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(-9 x^2+6 x+\underline{\text{ }}\right)=-9 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(7 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\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): \\
-9 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$7 \left(y^2-\frac{5 y}{7}+\underline{\text{ }}\right)$}=\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 }-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}
\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 7:
\begin{array}{l}
\frac{25}{28}-1=-\frac{3}{28}: \\
-9 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)+7 \left(y^2-\frac{5 y}{7}+\frac{25}{196}\right)=\fbox{$-\frac{3}{28}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{2 x}{3}+\frac{1}{9}=\left(x-\frac{1}{3}\right)^2: \\
-9 \fbox{$\left(x-\frac{1}{3}\right)^2$}+7 \left(y^2-\frac{5 y}{7}+\frac{25}{196}\right)=-\frac{3}{28} \\
\end{array}
Step 9:
\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{} & -9 \left(x-\frac{1}{3}\right)^2+7 \fbox{$\left(y-\frac{5}{14}\right)^2$}=-\frac{3}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+10 x+3 y^2-8 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2-8 y-8 x^2+10 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
3 y^2-8 y-8 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(-8 x^2+10 x+\underline{\text{ }}\right)+\left(3 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\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(3 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\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): \\
-8 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2-\frac{8 y}{3}+\underline{\text{ }}\right)$}=\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 }-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}
-4-\frac{25}{8}=-\frac{57}{8}: \\
-8 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)+3 \left(y^2-\frac{8 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 }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{57}{8}=-\frac{43}{24}: \\
-8 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)+3 \left(y^2-\frac{8 y}{3}+\frac{16}{9}\right)=\fbox{$-\frac{43}{24}$} \\
\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{$\left(x-\frac{5}{8}\right)^2$}+3 \left(y^2-\frac{8 y}{3}+\frac{16}{9}\right)=-\frac{43}{24} \\
\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{} & -8 \left(x-\frac{5}{8}\right)^2+3 \fbox{$\left(y-\frac{4}{3}\right)^2$}=-\frac{43}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+8 x+6 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 x^2+8 x+(6 y+9)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 y-8 x^2+8 x+9 \text{from }\text{both }\text{sides}: \\
8 x^2-8 x+(-6 y-9)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }6 y+9 \text{to }\text{both }\text{sides}: \\
8 x^2-8 x=6 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(8 x^2-8 x+\underline{\text{ }}\right)=(6 y+9)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 x^2-8 x+\underline{\text{ }}\right)=8 \left(x^2-x+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2-x+\underline{\text{ }}\right)$}=(6 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 }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{8}{4}=2 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
(6 y+9)+2=6 y+11: \\
8 \left(x^2-x+\frac{1}{4}\right)=\fbox{$6 y+11$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \fbox{$\left(x-\frac{1}{2}\right)^2$}=6 y+11 \\
\end{array}
| khanacademy | amps |
Given the equation $x^2-5 x-8 y^2+10 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2+10 y+x^2-5 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
-8 y^2+10 y+x^2-5 x=-3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(x^2-5 x+\underline{\text{ }}\right)+\left(-8 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 y^2+10 y+\underline{\text{ }}\right)=-8 \left(y^2-\frac{5 y}{4}+\underline{\text{ }}\right): \\
\left(x^2-5 x+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2-\frac{5 y}{4}+\underline{\text{ }}\right)$}=\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{-5}{2}\right)^2=\frac{25}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
\frac{25}{4}-3=\frac{13}{4}: \\
\left(x^2-5 x+\frac{25}{4}\right)-8 \left(y^2-\frac{5 y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{13}{4}$} \\
\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 }-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 8:
\begin{array}{l}
\frac{13}{4}-\frac{25}{8}=\frac{1}{8}: \\
\left(x^2-5 x+\frac{25}{4}\right)-8 \left(y^2-\frac{5 y}{4}+\frac{25}{64}\right)=\fbox{$\frac{1}{8}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-5 x+\frac{25}{4}=\left(x-\frac{5}{2}\right)^2: \\
\fbox{$\left(x-\frac{5}{2}\right)^2$}-8 \left(y^2-\frac{5 y}{4}+\frac{25}{64}\right)=\frac{1}{8} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-\frac{5 y}{4}+\frac{25}{64}=\left(y-\frac{5}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \left(x-\frac{5}{2}\right)^2-8 \fbox{$\left(y-\frac{5}{8}\right)^2$}=\frac{1}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2+10 x+10 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 x^2+10 x+(10 y+8)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }10 y-2 x^2+10 x+8 \text{from }\text{both }\text{sides}: \\
2 x^2-10 x+(-10 y-8)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }10 y+8 \text{to }\text{both }\text{sides}: \\
2 x^2-10 x=10 y+8 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(2 x^2-10 x+\underline{\text{ }}\right)=(10 y+8)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 x^2-10 x+\underline{\text{ }}\right)=2 \left(x^2-5 x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-5 x+\underline{\text{ }}\right)$}=(10 y+8)+\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 }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }2\times \frac{25}{4}=\frac{25}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
(10 y+8)+\frac{25}{2}=10 y+\frac{41}{2}: \\
2 \left(x^2-5 x+\frac{25}{4}\right)=\fbox{$10 y+\frac{41}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-5 x+\frac{25}{4}=\left(x-\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \fbox{$\left(x-\frac{5}{2}\right)^2$}=10 y+\frac{41}{2} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-9 y^2-3 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-3 y+\left(2 x^2-2\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-9 y^2-3 y+2 x^2-2 \text{from }\text{both }\text{sides}: \\
9 y^2+3 y+\left(2-2 x^2\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
9 y^2+3 y-2 x^2=-2 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(9 y^2+3 y+\underline{\text{ }}\right)-2 x^2=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2+3 y+\underline{\text{ }}\right)=9 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right): \\
\fbox{$9 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)$}-2 x^2=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \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{9}{36}=\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{1}{4}-2=-\frac{7}{4}: \\
9 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)-2 x^2=\fbox{$-\frac{7}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
y^2+\frac{y}{3}+\frac{1}{36}=\left(y+\frac{1}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \fbox{$\left(y+\frac{1}{6}\right)^2$}-2 x^2=-\frac{7}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2+3 x-5 y^2+4 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+4 y-10 x^2+3 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
-5 y^2+4 y-10 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(-10 x^2+3 x+\underline{\text{ }}\right)+\left(-5 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2+3 x+\underline{\text{ }}\right)=-10 \left(x^2-\frac{3 x}{10}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2-\frac{3 x}{10}+\underline{\text{ }}\right)$}+\left(-5 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+4 y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right): \\
-10 \left(x^2-\frac{3 x}{10}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-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}
-8-\frac{9}{40}=-\frac{329}{40}: \\
-10 \left(x^2-\frac{3 x}{10}+\frac{9}{400}\right)-5 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{329}{40}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }-5\times \frac{4}{25}=-\frac{4}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{329}{40}-\frac{4}{5}=-\frac{361}{40}: \\
-10 \left(x^2-\frac{3 x}{10}+\frac{9}{400}\right)-5 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=\fbox{$-\frac{361}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{10}+\frac{9}{400}=\left(x-\frac{3}{20}\right)^2: \\
-10 \fbox{$\left(x-\frac{3}{20}\right)^2$}-5 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=-\frac{361}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{4 y}{5}+\frac{4}{25}=\left(y-\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x-\frac{3}{20}\right)^2-5 \fbox{$\left(y-\frac{2}{5}\right)^2$}=-\frac{361}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-5 y^2-y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2-y+\left(5 x^2+7\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-5 y^2-y+5 x^2+7 \text{from }\text{both }\text{sides}: \\
5 y^2+y+\left(-5 x^2-7\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
5 y^2+y-5 x^2=7 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(5 y^2+y+\underline{\text{ }}\right)-5 x^2=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2+y+\underline{\text{ }}\right)=5 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right)$}-5 x^2=\underline{\text{ }}+7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }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}
7+\frac{1}{20}=\frac{141}{20}: \\
5 \left(y^2+\frac{y}{5}+\frac{1}{100}\right)-5 x^2=\fbox{$\frac{141}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
y^2+\frac{y}{5}+\frac{1}{100}=\left(y+\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$\left(y+\frac{1}{10}\right)^2$}-5 x^2=\frac{141}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2+x-6 y^2+5 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+5 y+2 x^2+x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-6 y^2+5 y+2 x^2+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+x+\underline{\text{ }}\right)+\left(-6 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2+x+\underline{\text{ }}\right)=2 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)$}+\left(-6 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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): \\
2 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{5 y}{6}+\underline{\text{ }}\right)$}=\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{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 7:
\begin{array}{l}
\frac{1}{8}-5=-\frac{39}{8}: \\
2 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)-6 \left(y^2-\frac{5 y}{6}+\underline{\text{ }}\right)=\fbox{$-\frac{39}{8}$} \\
\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}{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{39}{8}-\frac{25}{24}=-\frac{71}{12}: \\
2 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)-6 \left(y^2-\frac{5 y}{6}+\frac{25}{144}\right)=\fbox{$-\frac{71}{12}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{2}+\frac{1}{16}=\left(x+\frac{1}{4}\right)^2: \\
2 \fbox{$\left(x+\frac{1}{4}\right)^2$}-6 \left(y^2-\frac{5 y}{6}+\frac{25}{144}\right)=-\frac{71}{12} \\
\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{} & 2 \left(x+\frac{1}{4}\right)^2-6 \fbox{$\left(y-\frac{5}{12}\right)^2$}=-\frac{71}{12} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2+4 x+3 y^2-10 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2-10 y-9 x^2+4 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
3 y^2-10 y-9 x^2+4 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2+4 x+\underline{\text{ }}\right)+\left(3 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2+4 x+\underline{\text{ }}\right)=-9 \left(x^2-\frac{4 x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2-\frac{4 x}{9}+\underline{\text{ }}\right)$}+\left(3 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\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): \\
-9 \left(x^2-\frac{4 x}{9}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2-\frac{10 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{9}}{2}\right)^2=\frac{4}{81} \text{on }\text{the }\text{left }\text{and }-9\times \frac{4}{81}=-\frac{4}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10-\frac{4}{9}=\frac{86}{9}: \\
-9 \left(x^2-\frac{4 x}{9}+\frac{4}{81}\right)+3 \left(y^2-\frac{10 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{86}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }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{86}{9}+\frac{25}{3}=\frac{161}{9}: \\
-9 \left(x^2-\frac{4 x}{9}+\frac{4}{81}\right)+3 \left(y^2-\frac{10 y}{3}+\frac{25}{9}\right)=\fbox{$\frac{161}{9}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{4 x}{9}+\frac{4}{81}=\left(x-\frac{2}{9}\right)^2: \\
-9 \fbox{$\left(x-\frac{2}{9}\right)^2$}+3 \left(y^2-\frac{10 y}{3}+\frac{25}{9}\right)=\frac{161}{9} \\
\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{} & -9 \left(x-\frac{2}{9}\right)^2+3 \fbox{$\left(y-\frac{5}{3}\right)^2$}=\frac{161}{9} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2-4 x+3 y^2+7 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2+7 y+6 x^2-4 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
3 y^2+7 y+6 x^2-4 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(6 x^2-4 x+\underline{\text{ }}\right)+\left(3 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\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(3 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\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): \\
6 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2+\frac{7 y}{3}+\underline{\text{ }}\right)$}=\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 }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}
5+\frac{2}{3}=\frac{17}{3}: \\
6 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)+3 \left(y^2+\frac{7 y}{3}+\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 }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{17}{3}+\frac{49}{12}=\frac{39}{4}: \\
6 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)+3 \left(y^2+\frac{7 y}{3}+\frac{49}{36}\right)=\fbox{$\frac{39}{4}$} \\
\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{$\left(x-\frac{1}{3}\right)^2$}+3 \left(y^2+\frac{7 y}{3}+\frac{49}{36}\right)=\frac{39}{4} \\
\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{} & 6 \left(x-\frac{1}{3}\right)^2+3 \fbox{$\left(y+\frac{7}{6}\right)^2$}=\frac{39}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2+6 x+5 y^2-5 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2-5 y+x^2+6 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
5 y^2-5 y+x^2+6 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(x^2+6 x+\underline{\text{ }}\right)+\left(5 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 y^2-5 y+\underline{\text{ }}\right)=5 \left(y^2-y+\underline{\text{ }}\right): \\
\left(x^2+6 x+\underline{\text{ }}\right)+\fbox{$5 \left(y^2-y+\underline{\text{ }}\right)$}=\underline{\text{ }}+8 \\
\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{6}{2}\right)^2=9 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
8+9=17: \\
\left(x^2+6 x+9\right)+5 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$17$} \\
\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 }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{5}{4}=\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 8:
\begin{array}{l}
17+\frac{5}{4}=\frac{73}{4}: \\
\left(x^2+6 x+9\right)+5 \left(y^2-y+\frac{1}{4}\right)=\fbox{$\frac{73}{4}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+6 x+9=(x+3)^2: \\
\fbox{$(x+3)^2$}+5 \left(y^2-y+\frac{1}{4}\right)=\frac{73}{4} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & (x+3)^2+5 \fbox{$\left(y-\frac{1}{2}\right)^2$}=\frac{73}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2-4 x-5 y^2+2 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+2 y-6 x^2-4 x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
-5 y^2+2 y-6 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(-6 x^2-4 x+\underline{\text{ }}\right)+\left(-5 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\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(-5 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+2 y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{2 y}{5}+\underline{\text{ }}\right): \\
-6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{2 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-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}
7-\frac{2}{3}=\frac{19}{3}: \\
-6 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)-5 \left(y^2-\frac{2 y}{5}+\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 }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{5}}{2}\right)^2=\frac{1}{25} \text{on }\text{the }\text{left }\text{and }\frac{-5}{25}=-\frac{1}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{19}{3}-\frac{1}{5}=\frac{92}{15}: \\
-6 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)-5 \left(y^2-\frac{2 y}{5}+\frac{1}{25}\right)=\fbox{$\frac{92}{15}$} \\
\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{$\left(x+\frac{1}{3}\right)^2$}-5 \left(y^2-\frac{2 y}{5}+\frac{1}{25}\right)=\frac{92}{15} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{2 y}{5}+\frac{1}{25}=\left(y-\frac{1}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x+\frac{1}{3}\right)^2-5 \fbox{$\left(y-\frac{1}{5}\right)^2$}=\frac{92}{15} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2-6 x-5 y^2-7 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2-7 y-x^2-6 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
-5 y^2-7 y-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(-x^2-6 x+\underline{\text{ }}\right)+\left(-5 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2-6 x+\underline{\text{ }}\right)=-\left(x^2+6 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2+6 x+\underline{\text{ }}\right)$}+\left(-5 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2-7 y+\underline{\text{ }}\right)=-5 \left(y^2+\frac{7 y}{5}+\underline{\text{ }}\right): \\
-\left(x^2+6 x+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2+\frac{7 y}{5}+\underline{\text{ }}\right)$}=\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 }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{6}{2}\right)^2=9 \text{on }\text{the }\text{left }\text{and }-9=-9 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3-9=-6: \\
-\left(x^2+6 x+9\right)-5 \left(y^2+\frac{7 y}{5}+\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 }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{5}}{2}\right)^2=\frac{49}{100} \text{on }\text{the }\text{left }\text{and }-5\times \frac{49}{100}=-\frac{49}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-6-\frac{49}{20}=-\frac{169}{20}: \\
-\left(x^2+6 x+9\right)-5 \left(y^2+\frac{7 y}{5}+\frac{49}{100}\right)=\fbox{$-\frac{169}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+6 x+9=(x+3)^2: \\
-\fbox{$(x+3)^2$}-5 \left(y^2+\frac{7 y}{5}+\frac{49}{100}\right)=-\frac{169}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{7 y}{5}+\frac{49}{100}=\left(y+\frac{7}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -(x+3)^2-5 \fbox{$\left(y+\frac{7}{10}\right)^2$}=-\frac{169}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2-8 x-7 y^2+8 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2+8 y+6 x^2-8 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
-7 y^2+8 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(-7 y^2+8 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(-7 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\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): \\
6 \left(x^2-\frac{4 x}{3}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2-\frac{8 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }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)-7 \left(y^2-\frac{8 y}{7}+\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 }-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{20}{3}-\frac{16}{7}=\frac{92}{21}: \\
6 \left(x^2-\frac{4 x}{3}+\frac{4}{9}\right)-7 \left(y^2-\frac{8 y}{7}+\frac{16}{49}\right)=\fbox{$\frac{92}{21}$} \\
\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{$\left(x-\frac{2}{3}\right)^2$}-7 \left(y^2-\frac{8 y}{7}+\frac{16}{49}\right)=\frac{92}{21} \\
\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{} & 6 \left(x-\frac{2}{3}\right)^2-7 \fbox{$\left(y-\frac{4}{7}\right)^2$}=\frac{92}{21} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2-6 x-4 y^2-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 x^2-6 x+\left(-4 y^2-2\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 y^2+8 x^2+6 x+2 \text{to }\text{both }\text{sides}: \\
8 x^2+6 x+\left(4 y^2+2\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
4 y^2+8 x^2+6 x=-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(8 x^2+6 x+\underline{\text{ }}\right)+4 y^2=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 x^2+6 x+\underline{\text{ }}\right)=8 \left(x^2+\frac{3 x}{4}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+\frac{3 x}{4}+\underline{\text{ }}\right)$}+4 y^2=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }8\times \frac{9}{64}=\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{8}-2=-\frac{7}{8}: \\
8 \left(x^2+\frac{3 x}{4}+\frac{9}{64}\right)+4 y^2=\fbox{$-\frac{7}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{3 x}{4}+\frac{9}{64}=\left(x+\frac{3}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \fbox{$\left(x+\frac{3}{8}\right)^2$}+4 y^2=-\frac{7}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-10 y^2-8 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-8 y+\left(5 x^2+1\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-10 y^2-8 y+5 x^2+1 \text{from }\text{both }\text{sides}: \\
10 y^2+8 y+\left(-5 x^2-1\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
10 y^2+8 y-5 x^2=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(10 y^2+8 y+\underline{\text{ }}\right)-5 x^2=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2+8 y+\underline{\text{ }}\right)=10 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right): \\
\fbox{$10 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right)$}-5 x^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 }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }10\times \frac{4}{25}=\frac{8}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
1+\frac{8}{5}=\frac{13}{5}: \\
10 \left(y^2+\frac{4 y}{5}+\frac{4}{25}\right)-5 x^2=\fbox{$\frac{13}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
y^2+\frac{4 y}{5}+\frac{4}{25}=\left(y+\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \text{10 }\fbox{$\left(y+\frac{2}{5}\right)^2$}-5 x^2=\frac{13}{5} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+6 x-3 y^2-3 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2-3 y-8 x^2+6 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-3 y^2-3 y-8 x^2+6 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+6 x+\underline{\text{ }}\right)+\left(-3 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2+6 x+\underline{\text{ }}\right)=-8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)$}+\left(-3 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2-3 y+\underline{\text{ }}\right)=-3 \left(y^2+y+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2+y+\underline{\text{ }}\right)$}=\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 }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }-8\times \frac{9}{64}=-\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-1-\frac{9}{8}=-\frac{17}{8}: \\
-8 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)-3 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$-\frac{17}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-3}{4}=-\frac{3}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{17}{8}-\frac{3}{4}=-\frac{23}{8}: \\
-8 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)-3 \left(y^2+y+\frac{1}{4}\right)=\fbox{$-\frac{23}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{4}+\frac{9}{64}=\left(x-\frac{3}{8}\right)^2: \\
-8 \fbox{$\left(x-\frac{3}{8}\right)^2$}-3 \left(y^2+y+\frac{1}{4}\right)=-\frac{23}{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{3}{8}\right)^2-3 \fbox{$\left(y+\frac{1}{2}\right)^2$}=-\frac{23}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2+3 x-6 y^2+10 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+10 y-6 x^2+3 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-6 y^2+10 y-6 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(-6 x^2+3 x+\underline{\text{ }}\right)+\left(-6 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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(-6 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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{x}{2}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{5 y}{3}+\underline{\text{ }}\right)$}=\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 }-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}
-5-\frac{3}{8}=-\frac{43}{8}: \\
-6 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)-6 \left(y^2-\frac{5 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{43}{8}$} \\
\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{43}{8}-\frac{25}{6}=-\frac{229}{24}: \\
-6 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)-6 \left(y^2-\frac{5 y}{3}+\frac{25}{36}\right)=\fbox{$-\frac{229}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
-6 \fbox{$\left(x-\frac{1}{4}\right)^2$}-6 \left(y^2-\frac{5 y}{3}+\frac{25}{36}\right)=-\frac{229}{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{1}{4}\right)^2-6 \fbox{$\left(y-\frac{5}{6}\right)^2$}=-\frac{229}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2-4 x+2 y^2+2 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+2 y+3 x^2-4 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
2 y^2+2 y+3 x^2-4 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2-4 x+\underline{\text{ }}\right)+\left(2 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\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(2 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+2 y+\underline{\text{ }}\right)=2 \left(y^2+y+\underline{\text{ }}\right): \\
3 \left(x^2-\frac{4 x}{3}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+y+\underline{\text{ }}\right)$}=\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 }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}
10+\frac{4}{3}=\frac{34}{3}: \\
3 \left(x^2-\frac{4 x}{3}+\frac{4}{9}\right)+2 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$\frac{34}{3}$} \\
\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{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{34}{3}+\frac{1}{2}=\frac{71}{6}: \\
3 \left(x^2-\frac{4 x}{3}+\frac{4}{9}\right)+2 \left(y^2+y+\frac{1}{4}\right)=\fbox{$\frac{71}{6}$} \\
\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{$\left(x-\frac{2}{3}\right)^2$}+2 \left(y^2+y+\frac{1}{4}\right)=\frac{71}{6} \\
\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{2}{3}\right)^2+2 \fbox{$\left(y+\frac{1}{2}\right)^2$}=\frac{71}{6} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2+x-2 y^2-10 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-10 y+5 x^2+x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
-2 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(-2 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(-2 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2-10 y+\underline{\text{ }}\right)=-2 \left(y^2+5 y+\underline{\text{ }}\right): \\
5 \left(x^2+\frac{x}{5}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2+5 y+\underline{\text{ }}\right)$}=\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 }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)-2 \left(y^2+5 y+\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 }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }-2\times \frac{25}{4}=-\frac{25}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{201}{20}-\frac{25}{2}=-\frac{49}{20}: \\
5 \left(x^2+\frac{x}{5}+\frac{1}{100}\right)-2 \left(y^2+5 y+\frac{25}{4}\right)=\fbox{$-\frac{49}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{5}+\frac{1}{100}=\left(x+\frac{1}{10}\right)^2: \\
5 \fbox{$\left(x+\frac{1}{10}\right)^2$}-2 \left(y^2+5 y+\frac{25}{4}\right)=-\frac{49}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+5 y+\frac{25}{4}=\left(y+\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x+\frac{1}{10}\right)^2-2 \fbox{$\left(y+\frac{5}{2}\right)^2$}=-\frac{49}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2+8 x+y^2-4 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2-4 y+2 x^2+8 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
y^2-4 y+2 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(2 x^2+8 x+\underline{\text{ }}\right)+\left(y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2+8 x+\underline{\text{ }}\right)=2 \left(x^2+4 x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2+4 x+\underline{\text{ }}\right)$}+\left(y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\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{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 6:
\begin{array}{l}
5+8=13: \\
2 \left(x^2+4 x+4\right)+\left(y^2-4 y+\underline{\text{ }}\right)=\fbox{$13$} \\
\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{Add }\left(\frac{-4}{2}\right)^2=4 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 8:
\begin{array}{l}
13+4=17: \\
2 \left(x^2+4 x+4\right)+\left(y^2-4 y+4\right)=\fbox{$17$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+4 x+4=(x+2)^2: \\
2 \fbox{$(x+2)^2$}+\left(y^2-4 y+4\right)=17 \\
\end{array}
Step 10:
\begin{array}{l}
y^2-4 y+4=(y-2)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 (x+2)^2+\fbox{$(y-2)^2$}=17 \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2+10 x+8 y^2-5 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-5 y+5 x^2+10 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
8 y^2-5 y+5 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(5 x^2+10 x+\underline{\text{ }}\right)+\left(8 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2+10 x+\underline{\text{ }}\right)=5 \left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+2 x+\underline{\text{ }}\right)$}+\left(8 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-5 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{5 y}{8}+\underline{\text{ }}\right): \\
5 \left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{5 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}-10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }5\times 1=5 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5-10=-5: \\
5 \left(x^2+2 x+1\right)+8 \left(y^2-\frac{5 y}{8}+\underline{\text{ }}\right)=\fbox{$-5$} \\
\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 }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{8}}{2}\right)^2=\frac{25}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{25}{256}=\frac{25}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{25}{32}-5=-\frac{135}{32}: \\
5 \left(x^2+2 x+1\right)+8 \left(y^2-\frac{5 y}{8}+\frac{25}{256}\right)=\fbox{$-\frac{135}{32}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
5 \fbox{$(x+1)^2$}+8 \left(y^2-\frac{5 y}{8}+\frac{25}{256}\right)=-\frac{135}{32} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{5 y}{8}+\frac{25}{256}=\left(y-\frac{5}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 (x+1)^2+8 \fbox{$\left(y-\frac{5}{16}\right)^2$}=-\frac{135}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-x+4 y^2+y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2+y+2 x^2-x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
4 y^2+y+2 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(2 x^2-x+\underline{\text{ }}\right)+\left(4 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2-x+\underline{\text{ }}\right)=2 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)$}+\left(4 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\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): \\
2 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }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 7:
\begin{array}{l}
\frac{1}{8}-2=-\frac{15}{8}: \\
2 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+4 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)=\fbox{$-\frac{15}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{4}{64}=\frac{1}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{16}-\frac{15}{8}=-\frac{29}{16}: \\
2 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)+4 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=\fbox{$-\frac{29}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
2 \fbox{$\left(x-\frac{1}{4}\right)^2$}+4 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=-\frac{29}{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{} & 2 \left(x-\frac{1}{4}\right)^2+4 \fbox{$\left(y+\frac{1}{8}\right)^2$}=-\frac{29}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2-3 x-7 y^2+4 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2+4 y-2 x^2-3 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-7 y^2+4 y-2 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(-2 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(-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+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): \\
-2 \left(x^2+\frac{3 x}{2}+\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 }-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}
-9-\frac{9}{8}=-\frac{81}{8}: \\
-2 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)-7 \left(y^2-\frac{4 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{81}{8}$} \\
\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{-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{81}{8}-\frac{4}{7}=-\frac{599}{56}: \\
-2 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)-7 \left(y^2-\frac{4 y}{7}+\frac{4}{49}\right)=\fbox{$-\frac{599}{56}$} \\
\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{4 y}{7}+\frac{4}{49}\right)=-\frac{599}{56} \\
\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{} & -2 \left(x+\frac{3}{4}\right)^2-7 \fbox{$\left(y-\frac{2}{7}\right)^2$}=-\frac{599}{56} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2-9 x-8 y^2+2 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2+2 y-8 x^2-9 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
-8 y^2+2 y-8 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(-8 x^2-9 x+\underline{\text{ }}\right)+\left(-8 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2-9 x+\underline{\text{ }}\right)=-8 \left(x^2+\frac{9 x}{8}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2+\frac{9 x}{8}+\underline{\text{ }}\right)$}+\left(-8 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2+2 y+\underline{\text{ }}\right)=-8 \left(y^2-\frac{y}{4}+\underline{\text{ }}\right): \\
-8 \left(x^2+\frac{9 x}{8}+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2-\frac{y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}-3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{9}{8}}{2}\right)^2=\frac{81}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{81}{256}=-\frac{81}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-3-\frac{81}{32}=-\frac{177}{32}: \\
-8 \left(x^2+\frac{9 x}{8}+\frac{81}{256}\right)-8 \left(y^2-\frac{y}{4}+\underline{\text{ }}\right)=\fbox{$-\frac{177}{32}$} \\
\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 }-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 9:
\begin{array}{l}
-\frac{177}{32}-\frac{1}{8}=-\frac{181}{32}: \\
-8 \left(x^2+\frac{9 x}{8}+\frac{81}{256}\right)-8 \left(y^2-\frac{y}{4}+\frac{1}{64}\right)=\fbox{$-\frac{181}{32}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{9 x}{8}+\frac{81}{256}=\left(x+\frac{9}{16}\right)^2: \\
-8 \fbox{$\left(x+\frac{9}{16}\right)^2$}-8 \left(y^2-\frac{y}{4}+\frac{1}{64}\right)=-\frac{181}{32} \\
\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{} & -8 \left(x+\frac{9}{16}\right)^2-8 \fbox{$\left(y-\frac{1}{8}\right)^2$}=-\frac{181}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2-8 x+10 y^2+4 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2+4 y-8 x^2-8 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
10 y^2+4 y-8 x^2-8 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2-8 x+\underline{\text{ }}\right)+\left(10 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2-8 x+\underline{\text{ }}\right)=-8 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2+x+\underline{\text{ }}\right)$}+\left(10 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2+4 y+\underline{\text{ }}\right)=10 \left(y^2+\frac{2 y}{5}+\underline{\text{ }}\right): \\
-8 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$10 \left(y^2+\frac{2 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-8}{4}=-2 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-2-2=-4: \\
-8 \left(x^2+x+\frac{1}{4}\right)+10 \left(y^2+\frac{2 y}{5}+\underline{\text{ }}\right)=\fbox{$-4$} \\
\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 }\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 9:
\begin{array}{l}
\frac{2}{5}-4=-\frac{18}{5}: \\
-8 \left(x^2+x+\frac{1}{4}\right)+10 \left(y^2+\frac{2 y}{5}+\frac{1}{25}\right)=\fbox{$-\frac{18}{5}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
-8 \fbox{$\left(x+\frac{1}{2}\right)^2$}+10 \left(y^2+\frac{2 y}{5}+\frac{1}{25}\right)=-\frac{18}{5} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{2 y}{5}+\frac{1}{25}=\left(y+\frac{1}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x+\frac{1}{2}\right)^2+\text{10 }\fbox{$\left(y+\frac{1}{5}\right)^2$}=-\frac{18}{5} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2+8 x-5 y^2+5 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+5 y-2 x^2+8 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-5 y^2+5 y-2 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(-2 x^2+8 x+\underline{\text{ }}\right)+\left(-5 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 x^2+8 x+\underline{\text{ }}\right)=-2 \left(x^2-4 x+\underline{\text{ }}\right): \\
\fbox{$-2 \left(x^2-4 x+\underline{\text{ }}\right)$}+\left(-5 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+5 y+\underline{\text{ }}\right)=-5 \left(y^2-y+\underline{\text{ }}\right): \\
-2 \left(x^2-4 x+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-y+\underline{\text{ }}\right)$}=\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 }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-4}{2}\right)^2=4 \text{on }\text{the }\text{left }\text{and }-2\times 4=-8 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-7-8=-15: \\
-2 \left(x^2-4 x+4\right)-5 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$-15$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-5}{4}=-\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-15-\frac{5}{4}=-\frac{65}{4}: \\
-2 \left(x^2-4 x+4\right)-5 \left(y^2-y+\frac{1}{4}\right)=\fbox{$-\frac{65}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-4 x+4=(x-2)^2: \\
-2 \fbox{$(x-2)^2$}-5 \left(y^2-y+\frac{1}{4}\right)=-\frac{65}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -2 (x-2)^2-5 \fbox{$\left(y-\frac{1}{2}\right)^2$}=-\frac{65}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+9 x+3 y^2+10 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2+10 y+8 x^2+9 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
3 y^2+10 y+8 x^2+9 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+9 x+\underline{\text{ }}\right)+\left(3 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2+9 x+\underline{\text{ }}\right)=8 \left(x^2+\frac{9 x}{8}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+\frac{9 x}{8}+\underline{\text{ }}\right)$}+\left(3 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\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): \\
8 \left(x^2+\frac{9 x}{8}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2+\frac{10 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{9}{8}}{2}\right)^2=\frac{81}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{81}{256}=\frac{81}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10+\frac{81}{32}=\frac{401}{32}: \\
8 \left(x^2+\frac{9 x}{8}+\frac{81}{256}\right)+3 \left(y^2+\frac{10 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{401}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{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{401}{32}+\frac{25}{3}=\frac{2003}{96}: \\
8 \left(x^2+\frac{9 x}{8}+\frac{81}{256}\right)+3 \left(y^2+\frac{10 y}{3}+\frac{25}{9}\right)=\fbox{$\frac{2003}{96}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{9 x}{8}+\frac{81}{256}=\left(x+\frac{9}{16}\right)^2: \\
8 \fbox{$\left(x+\frac{9}{16}\right)^2$}+3 \left(y^2+\frac{10 y}{3}+\frac{25}{9}\right)=\frac{2003}{96} \\
\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{} & 8 \left(x+\frac{9}{16}\right)^2+3 \fbox{$\left(y+\frac{5}{3}\right)^2$}=\frac{2003}{96} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2-2 x-6 y^2+7 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+7 y-8 x^2-2 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
-6 y^2+7 y-8 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(-8 x^2-2 x+\underline{\text{ }}\right)+\left(-6 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\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+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2+7 y+\underline{\text{ }}\right)=-6 \left(y^2-\frac{7 y}{6}+\underline{\text{ }}\right): \\
-8 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{7 y}{6}+\underline{\text{ }}\right)$}=\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 }-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}
-6-\frac{1}{8}=-\frac{49}{8}: \\
-8 \left(x^2+\frac{x}{4}+\frac{1}{64}\right)-6 \left(y^2-\frac{7 y}{6}+\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 }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{6}}{2}\right)^2=\frac{49}{144} \text{on }\text{the }\text{left }\text{and }-6\times \frac{49}{144}=-\frac{49}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{49}{8}-\frac{49}{24}=-\frac{49}{6}: \\
-8 \left(x^2+\frac{x}{4}+\frac{1}{64}\right)-6 \left(y^2-\frac{7 y}{6}+\frac{49}{144}\right)=\fbox{$-\frac{49}{6}$} \\
\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{7 y}{6}+\frac{49}{144}\right)=-\frac{49}{6} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{6}+\frac{49}{144}=\left(y-\frac{7}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x+\frac{1}{8}\right)^2-6 \fbox{$\left(y-\frac{7}{12}\right)^2$}=-\frac{49}{6} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2-9 y^2+8 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2+8 y+\left(x^2-6\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-9 y^2+8 y+x^2-6 \text{from }\text{both }\text{sides}: \\
9 y^2-8 y+\left(6-x^2\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
9 y^2-8 y-x^2=-6 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(9 y^2-8 y+\underline{\text{ }}\right)-x^2=\underline{\text{ }}-6 \\
\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): \\
\fbox{$9 \left(y^2-\frac{8 y}{9}+\underline{\text{ }}\right)$}-x^2=\underline{\text{ }}-6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-8}{9}}{2}\right)^2=\frac{16}{81} \text{on }\text{the }\text{left }\text{and }9\times \frac{16}{81}=\frac{16}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{16}{9}-6=-\frac{38}{9}: \\
9 \left(y^2-\frac{8 y}{9}+\frac{16}{81}\right)-x^2=\fbox{$-\frac{38}{9}$} \\
\end{array}
Step 8:
\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{$\left(y-\frac{4}{9}\right)^2$}-x^2=-\frac{38}{9} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2-2 x-9 y^2-6 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-6 y-x^2-2 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
-9 y^2-6 y-x^2-2 x=-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2-2 x+\underline{\text{ }}\right)+\left(-9 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2-2 x+\underline{\text{ }}\right)=-\left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2+2 x+\underline{\text{ }}\right)$}+\left(-9 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2-6 y+\underline{\text{ }}\right)=-9 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right): \\
-\left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-1=-1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-10-1=-11: \\
-\left(x^2+2 x+1\right)-9 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right)=\fbox{$-11$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{-9}{9}=-1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-11-1=-12: \\
-\left(x^2+2 x+1\right)-9 \left(y^2+\frac{2 y}{3}+\frac{1}{9}\right)=\fbox{$-12$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
-\fbox{$(x+1)^2$}-9 \left(y^2+\frac{2 y}{3}+\frac{1}{9}\right)=-12 \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{2 y}{3}+\frac{1}{9}=\left(y+\frac{1}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -(x+1)^2-9 \fbox{$\left(y+\frac{1}{3}\right)^2$}=-12 \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2+x-10 y^2+10 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+10 y-9 x^2+x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-10 y^2+10 y-9 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(-9 x^2+x+\underline{\text{ }}\right)+\left(-10 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2+x+\underline{\text{ }}\right)=-9 \left(x^2-\frac{x}{9}+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2-\frac{x}{9}+\underline{\text{ }}\right)$}+\left(-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): \\
-9 \left(x^2-\frac{x}{9}+\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 }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{9}}{2}\right)^2=\frac{1}{324} \text{on }\text{the }\text{left }\text{and }\frac{-9}{324}=-\frac{1}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6-\frac{1}{36}=\frac{215}{36}: \\
-9 \left(x^2-\frac{x}{9}+\frac{1}{324}\right)-10 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$\frac{215}{36}$} \\
\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 }-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{215}{36}-\frac{5}{2}=\frac{125}{36}: \\
-9 \left(x^2-\frac{x}{9}+\frac{1}{324}\right)-10 \left(y^2-y+\frac{1}{4}\right)=\fbox{$\frac{125}{36}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{9}+\frac{1}{324}=\left(x-\frac{1}{18}\right)^2: \\
-9 \fbox{$\left(x-\frac{1}{18}\right)^2$}-10 \left(y^2-y+\frac{1}{4}\right)=\frac{125}{36} \\
\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{} & -9 \left(x-\frac{1}{18}\right)^2-\text{10 }\fbox{$\left(y-\frac{1}{2}\right)^2$}=\frac{125}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-7 x-6 y^2-y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-y-10 x^2-7 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-6 y^2-y-10 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(-10 x^2-7 x+\underline{\text{ }}\right)+\left(-6 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\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(-6 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2-y+\underline{\text{ }}\right)=-6 \left(y^2+\frac{y}{6}+\underline{\text{ }}\right): \\
-10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{y}{6}+\underline{\text{ }}\right)$}=\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 }-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}
5-\frac{49}{40}=\frac{151}{40}: \\
-10 \left(x^2+\frac{7 x}{10}+\frac{49}{400}\right)-6 \left(y^2+\frac{y}{6}+\underline{\text{ }}\right)=\fbox{$\frac{151}{40}$} \\
\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 }-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 9:
\begin{array}{l}
\frac{151}{40}-\frac{1}{24}=\frac{56}{15}: \\
-10 \left(x^2+\frac{7 x}{10}+\frac{49}{400}\right)-6 \left(y^2+\frac{y}{6}+\frac{1}{144}\right)=\fbox{$\frac{56}{15}$} \\
\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{$\left(x+\frac{7}{20}\right)^2$}-6 \left(y^2+\frac{y}{6}+\frac{1}{144}\right)=\frac{56}{15} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{6}+\frac{1}{144}=\left(y+\frac{1}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x+\frac{7}{20}\right)^2-6 \fbox{$\left(y+\frac{1}{12}\right)^2$}=\frac{56}{15} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2+2 x-y^2-9 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2-9 y-6 x^2+2 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
-y^2-9 y-6 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(-6 x^2+2 x+\underline{\text{ }}\right)+\left(-y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2+2 x+\underline{\text{ }}\right)=-6 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)$}+\left(-y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2-9 y+\underline{\text{ }}\right)=-\left(y^2+9 y+\underline{\text{ }}\right): \\
-6 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)+\fbox{$-\left(y^2+9 y+\underline{\text{ }}\right)$}=\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 }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{-6}{36}=-\frac{1}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-2-\frac{1}{6}=-\frac{13}{6}: \\
-6 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)-\left(y^2+9 y+\underline{\text{ }}\right)=\fbox{$-\frac{13}{6}$} \\
\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 }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{9}{2}\right)^2=\frac{81}{4} \text{on }\text{the }\text{left }\text{and }-\frac{81}{4}=-\frac{81}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{13}{6}-\frac{81}{4}=-\frac{269}{12}: \\
-6 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)-\left(y^2+9 y+\frac{81}{4}\right)=\fbox{$-\frac{269}{12}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{3}+\frac{1}{36}=\left(x-\frac{1}{6}\right)^2: \\
-6 \fbox{$\left(x-\frac{1}{6}\right)^2$}-\left(y^2+9 y+\frac{81}{4}\right)=-\frac{269}{12} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+9 y+\frac{81}{4}=\left(y+\frac{9}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x-\frac{1}{6}\right)^2-\fbox{$\left(y+\frac{9}{2}\right)^2$}=-\frac{269}{12} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2+5 x-6 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 x^2+5 x+(7-6 y)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-6 y-2 x^2+5 x+7 \text{from }\text{both }\text{sides}: \\
2 x^2-5 x+(6 y-7)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }6 y-7 \text{from }\text{both }\text{sides}: \\
2 x^2-5 x=7-6 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(2 x^2-5 x+\underline{\text{ }}\right)=(7-6 y)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 x^2-5 x+\underline{\text{ }}\right)=2 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)$}=(7-6 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 }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{25}{16}=\frac{25}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
(7-6 y)+\frac{25}{8}=\frac{81}{8}-6 y: \\
2 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)=\fbox{$\frac{81}{8}-6 y$} \\
\end{array}
Step 8:
\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{$\left(x-\frac{5}{4}\right)^2$}=\frac{81}{8}-6 y \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2-7 x-10 y^2+7 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+7 y+9 x^2-7 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
-10 y^2+7 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(-10 y^2+7 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(-10 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\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): \\
9 \left(x^2-\frac{7 x}{9}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2-\frac{7 y}{10}+\underline{\text{ }}\right)$}=\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}
3+\frac{49}{36}=\frac{157}{36}: \\
9 \left(x^2-\frac{7 x}{9}+\frac{49}{324}\right)-10 \left(y^2-\frac{7 y}{10}+\underline{\text{ }}\right)=\fbox{$\frac{157}{36}$} \\
\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 }-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{157}{36}-\frac{49}{40}=\frac{1129}{360}: \\
9 \left(x^2-\frac{7 x}{9}+\frac{49}{324}\right)-10 \left(y^2-\frac{7 y}{10}+\frac{49}{400}\right)=\fbox{$\frac{1129}{360}$} \\
\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{$\left(x-\frac{7}{18}\right)^2$}-10 \left(y^2-\frac{7 y}{10}+\frac{49}{400}\right)=\frac{1129}{360} \\
\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{} & 9 \left(x-\frac{7}{18}\right)^2-\text{10 }\fbox{$\left(y-\frac{7}{20}\right)^2$}=\frac{1129}{360} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2-2 x+3 y^2-5 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2-5 y+x^2-2 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
3 y^2-5 y+x^2-2 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(x^2-2 x+\underline{\text{ }}\right)+\left(3 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 y^2-5 y+\underline{\text{ }}\right)=3 \left(y^2-\frac{5 y}{3}+\underline{\text{ }}\right): \\
\left(x^2-2 x+\underline{\text{ }}\right)+\fbox{$3 \left(y^2-\frac{5 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\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{-2}{2}\right)^2=1 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
1-5=-4: \\
\left(x^2-2 x+1\right)+3 \left(y^2-\frac{5 y}{3}+\underline{\text{ }}\right)=\fbox{$-4$} \\
\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 }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 8:
\begin{array}{l}
\frac{25}{12}-4=-\frac{23}{12}: \\
\left(x^2-2 x+1\right)+3 \left(y^2-\frac{5 y}{3}+\frac{25}{36}\right)=\fbox{$-\frac{23}{12}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-2 x+1=(x-1)^2: \\
\fbox{$(x-1)^2$}+3 \left(y^2-\frac{5 y}{3}+\frac{25}{36}\right)=-\frac{23}{12} \\
\end{array}
Step 10:
\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{} & (x-1)^2+3 \fbox{$\left(y-\frac{5}{6}\right)^2$}=-\frac{23}{12} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-8 x-y^2-5 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2-5 y+2 x^2-8 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
-y^2-5 y+2 x^2-8 x=1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2-8 x+\underline{\text{ }}\right)+\left(-y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2-8 x+\underline{\text{ }}\right)=2 \left(x^2-4 x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-4 x+\underline{\text{ }}\right)$}+\left(-y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2-5 y+\underline{\text{ }}\right)=-\left(y^2+5 y+\underline{\text{ }}\right): \\
2 \left(x^2-4 x+\underline{\text{ }}\right)+\fbox{$-\left(y^2+5 y+\underline{\text{ }}\right)$}=\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 }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-4}{2}\right)^2=4 \text{on }\text{the }\text{left }\text{and }2\times 4=8 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
1+8=9: \\
2 \left(x^2-4 x+4\right)-\left(y^2+5 y+\underline{\text{ }}\right)=\fbox{$9$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }-\frac{25}{4}=-\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
9-\frac{25}{4}=\frac{11}{4}: \\
2 \left(x^2-4 x+4\right)-\left(y^2+5 y+\frac{25}{4}\right)=\fbox{$\frac{11}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-4 x+4=(x-2)^2: \\
2 \fbox{$(x-2)^2$}-\left(y^2+5 y+\frac{25}{4}\right)=\frac{11}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+5 y+\frac{25}{4}=\left(y+\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 (x-2)^2-\fbox{$\left(y+\frac{5}{2}\right)^2$}=\frac{11}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2+8 x-2 y^2+2 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+2 y+7 x^2+8 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
-2 y^2+2 y+7 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(7 x^2+8 x+\underline{\text{ }}\right)+\left(-2 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2+8 x+\underline{\text{ }}\right)=7 \left(x^2+\frac{8 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2+\frac{8 x}{7}+\underline{\text{ }}\right)$}+\left(-2 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+2 y+\underline{\text{ }}\right)=-2 \left(y^2-y+\underline{\text{ }}\right): \\
7 \left(x^2+\frac{8 x}{7}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-y+\underline{\text{ }}\right)$}=\underline{\text{ }}+8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{7}}{2}\right)^2=\frac{16}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{16}{49}=\frac{16}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8+\frac{16}{7}=\frac{72}{7}: \\
7 \left(x^2+\frac{8 x}{7}+\frac{16}{49}\right)-2 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$\frac{72}{7}$} \\
\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 }-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{72}{7}-\frac{1}{2}=\frac{137}{14}: \\
7 \left(x^2+\frac{8 x}{7}+\frac{16}{49}\right)-2 \left(y^2-y+\frac{1}{4}\right)=\fbox{$\frac{137}{14}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{8 x}{7}+\frac{16}{49}=\left(x+\frac{4}{7}\right)^2: \\
7 \fbox{$\left(x+\frac{4}{7}\right)^2$}-2 \left(y^2-y+\frac{1}{4}\right)=\frac{137}{14} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x+\frac{4}{7}\right)^2-2 \fbox{$\left(y-\frac{1}{2}\right)^2$}=\frac{137}{14} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2-x-3 y^2-3 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2-3 y+8 x^2-x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-3 y^2-3 y+8 x^2-x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(8 x^2-x+\underline{\text{ }}\right)+\left(-3 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2-x+\underline{\text{ }}\right)=8 \left(x^2-\frac{x}{8}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2-\frac{x}{8}+\underline{\text{ }}\right)$}+\left(-3 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2-3 y+\underline{\text{ }}\right)=-3 \left(y^2+y+\underline{\text{ }}\right): \\
8 \left(x^2-\frac{x}{8}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2+y+\underline{\text{ }}\right)$}=\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 }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}-1=-\frac{31}{32}: \\
8 \left(x^2-\frac{x}{8}+\frac{1}{256}\right)-3 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$-\frac{31}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-3}{4}=-\frac{3}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{31}{32}-\frac{3}{4}=-\frac{55}{32}: \\
8 \left(x^2-\frac{x}{8}+\frac{1}{256}\right)-3 \left(y^2+y+\frac{1}{4}\right)=\fbox{$-\frac{55}{32}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{8}+\frac{1}{256}=\left(x-\frac{1}{16}\right)^2: \\
8 \fbox{$\left(x-\frac{1}{16}\right)^2$}-3 \left(y^2+y+\frac{1}{4}\right)=-\frac{55}{32} \\
\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{1}{16}\right)^2-3 \fbox{$\left(y+\frac{1}{2}\right)^2$}=-\frac{55}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2+5 x-6 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 x^2+5 x+(7-6 y)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-6 y-5 x^2+5 x+7 \text{from }\text{both }\text{sides}: \\
5 x^2-5 x+(6 y-7)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }6 y-7 \text{from }\text{both }\text{sides}: \\
5 x^2-5 x=7-6 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(5 x^2-5 x+\underline{\text{ }}\right)=(7-6 y)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 x^2-5 x+\underline{\text{ }}\right)=5 \left(x^2-x+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-x+\underline{\text{ }}\right)$}=(7-6 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 }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{5}{4}=\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
(7-6 y)+\frac{5}{4}=\frac{33}{4}-6 y: \\
5 \left(x^2-x+\frac{1}{4}\right)=\fbox{$\frac{33}{4}-6 y$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$\left(x-\frac{1}{2}\right)^2$}=\frac{33}{4}-6 y \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+8 x+9 y^2+2 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+2 y-3 x^2+8 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
9 y^2+2 y-3 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(-3 x^2+8 x+\underline{\text{ }}\right)+\left(9 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\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(9 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2+2 y+\underline{\text{ }}\right)=9 \left(y^2+\frac{2 y}{9}+\underline{\text{ }}\right): \\
-3 \left(x^2-\frac{8 x}{3}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2+\frac{2 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-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}
4-\frac{16}{3}=-\frac{4}{3}: \\
-3 \left(x^2-\frac{8 x}{3}+\frac{16}{9}\right)+9 \left(y^2+\frac{2 y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{4}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{9}}{2}\right)^2=\frac{1}{81} \text{on }\text{the }\text{left }\text{and }\frac{9}{81}=\frac{1}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{9}-\frac{4}{3}=-\frac{11}{9}: \\
-3 \left(x^2-\frac{8 x}{3}+\frac{16}{9}\right)+9 \left(y^2+\frac{2 y}{9}+\frac{1}{81}\right)=\fbox{$-\frac{11}{9}$} \\
\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{$\left(x-\frac{4}{3}\right)^2$}+9 \left(y^2+\frac{2 y}{9}+\frac{1}{81}\right)=-\frac{11}{9} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{2 y}{9}+\frac{1}{81}=\left(y+\frac{1}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x-\frac{4}{3}\right)^2+9 \fbox{$\left(y+\frac{1}{9}\right)^2$}=-\frac{11}{9} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-9 x-5 y^2+8 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+8 y+2 x^2-9 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
-5 y^2+8 y+2 x^2-9 x=2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2-9 x+\underline{\text{ }}\right)+\left(-5 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\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(-5 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+8 y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{8 y}{5}+\underline{\text{ }}\right): \\
2 \left(x^2-\frac{9 x}{2}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{8 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\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}
2+\frac{81}{8}=\frac{97}{8}: \\
2 \left(x^2-\frac{9 x}{2}+\frac{81}{16}\right)-5 \left(y^2-\frac{8 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{97}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-8}{5}}{2}\right)^2=\frac{16}{25} \text{on }\text{the }\text{left }\text{and }-5\times \frac{16}{25}=-\frac{16}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{97}{8}-\frac{16}{5}=\frac{357}{40}: \\
2 \left(x^2-\frac{9 x}{2}+\frac{81}{16}\right)-5 \left(y^2-\frac{8 y}{5}+\frac{16}{25}\right)=\fbox{$\frac{357}{40}$} \\
\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{$\left(x-\frac{9}{4}\right)^2$}-5 \left(y^2-\frac{8 y}{5}+\frac{16}{25}\right)=\frac{357}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{8 y}{5}+\frac{16}{25}=\left(y-\frac{4}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x-\frac{9}{4}\right)^2-5 \fbox{$\left(y-\frac{4}{5}\right)^2$}=\frac{357}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+3 x-4 y^2+10 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+10 y+4 x^2+3 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
-4 y^2+10 y+4 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(4 x^2+3 x+\underline{\text{ }}\right)+\left(-4 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\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(-4 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\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): \\
4 \left(x^2+\frac{3 x}{4}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)$}=\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{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}
\frac{9}{16}-3=-\frac{39}{16}: \\
4 \left(x^2+\frac{3 x}{4}+\frac{9}{64}\right)-4 \left(y^2-\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{39}{16}$} \\
\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{-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{39}{16}-\frac{25}{4}=-\frac{139}{16}: \\
4 \left(x^2+\frac{3 x}{4}+\frac{9}{64}\right)-4 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$-\frac{139}{16}$} \\
\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{$\left(x+\frac{3}{8}\right)^2$}-4 \left(y^2-\frac{5 y}{2}+\frac{25}{16}\right)=-\frac{139}{16} \\
\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{} & 4 \left(x+\frac{3}{8}\right)^2-4 \fbox{$\left(y-\frac{5}{4}\right)^2$}=-\frac{139}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+x+2 y^2+9 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+9 y+8 x^2+x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
2 y^2+9 y+8 x^2+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(8 x^2+x+\underline{\text{ }}\right)+\left(2 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+9 y+\underline{\text{ }}\right)=2 \left(y^2+\frac{9 y}{2}+\underline{\text{ }}\right): \\
8 \left(x^2+\frac{x}{8}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+\frac{9 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }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}-5=-\frac{159}{32}: \\
8 \left(x^2+\frac{x}{8}+\frac{1}{256}\right)+2 \left(y^2+\frac{9 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{159}{32}$} \\
\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{9}{2}}{2}\right)^2=\frac{81}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{81}{16}=\frac{81}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{81}{8}-\frac{159}{32}=\frac{165}{32}: \\
8 \left(x^2+\frac{x}{8}+\frac{1}{256}\right)+2 \left(y^2+\frac{9 y}{2}+\frac{81}{16}\right)=\fbox{$\frac{165}{32}$} \\
\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{9 y}{2}+\frac{81}{16}\right)=\frac{165}{32} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{9 y}{2}+\frac{81}{16}=\left(y+\frac{9}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \left(x+\frac{1}{16}\right)^2+2 \fbox{$\left(y+\frac{9}{4}\right)^2$}=\frac{165}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+9 x+5 y^2+4 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+4 y-8 x^2+9 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
5 y^2+4 y-8 x^2+9 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2+9 x+\underline{\text{ }}\right)+\left(5 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2+9 x+\underline{\text{ }}\right)=-8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right)$}+\left(5 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2+4 y+\underline{\text{ }}\right)=5 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{8}}{2}\right)^2=\frac{81}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{81}{256}=-\frac{81}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-2-\frac{81}{32}=-\frac{145}{32}: \\
-8 \left(x^2-\frac{9 x}{8}+\frac{81}{256}\right)+5 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{145}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }5\times \frac{4}{25}=\frac{4}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{4}{5}-\frac{145}{32}=-\frac{597}{160}: \\
-8 \left(x^2-\frac{9 x}{8}+\frac{81}{256}\right)+5 \left(y^2+\frac{4 y}{5}+\frac{4}{25}\right)=\fbox{$-\frac{597}{160}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{9 x}{8}+\frac{81}{256}=\left(x-\frac{9}{16}\right)^2: \\
-8 \fbox{$\left(x-\frac{9}{16}\right)^2$}+5 \left(y^2+\frac{4 y}{5}+\frac{4}{25}\right)=-\frac{597}{160} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{4 y}{5}+\frac{4}{25}=\left(y+\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x-\frac{9}{16}\right)^2+5 \fbox{$\left(y+\frac{2}{5}\right)^2$}=-\frac{597}{160} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2-x+2 y^2+8 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+8 y-5 x^2-x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
2 y^2+8 y-5 x^2-x=1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2-x+\underline{\text{ }}\right)+\left(2 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\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(2 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\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): \\
-5 \left(x^2+\frac{x}{5}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+4 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\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}
1-\frac{1}{20}=\frac{19}{20}: \\
-5 \left(x^2+\frac{x}{5}+\frac{1}{100}\right)+2 \left(y^2+4 y+\underline{\text{ }}\right)=\fbox{$\frac{19}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }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{19}{20}+8=\frac{179}{20}: \\
-5 \left(x^2+\frac{x}{5}+\frac{1}{100}\right)+2 \left(y^2+4 y+4\right)=\fbox{$\frac{179}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{5}+\frac{1}{100}=\left(x+\frac{1}{10}\right)^2: \\
-5 \fbox{$\left(x+\frac{1}{10}\right)^2$}+2 \left(y^2+4 y+4\right)=\frac{179}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+4 y+4=(y+2)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 \left(x+\frac{1}{10}\right)^2+2 \fbox{$(y+2)^2$}=\frac{179}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2-9 x-7 y^2-6 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2-6 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(-7 y^2-6 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(-7 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 y^2-6 y+\underline{\text{ }}\right)=-7 \left(y^2+\frac{6 y}{7}+\underline{\text{ }}\right): \\
-5 \left(x^2+\frac{9 x}{5}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2+\frac{6 y}{7}+\underline{\text{ }}\right)$}=\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 }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{6}{7}}{2}\right)^2=\frac{9}{49} \text{on }\text{the }\text{left }\text{and }-7\times \frac{9}{49}=-\frac{9}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-\frac{81}{20}-\frac{9}{7}=-\frac{747}{140}: \\
-5 \left(x^2+\frac{9 x}{5}+\frac{81}{100}\right)-7 \left(y^2+\frac{6 y}{7}+\frac{9}{49}\right)=\fbox{$-\frac{747}{140}$} \\
\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{$\left(x+\frac{9}{10}\right)^2$}-7 \left(y^2+\frac{6 y}{7}+\frac{9}{49}\right)=-\frac{747}{140} \\
\end{array}
Step 9:
\begin{array}{l}
y^2+\frac{6 y}{7}+\frac{9}{49}=\left(y+\frac{3}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 \left(x+\frac{9}{10}\right)^2-7 \fbox{$\left(y+\frac{3}{7}\right)^2$}=-\frac{747}{140} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2-7 x+2 y^2+7 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+7 y+3 x^2-7 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
2 y^2+7 y+3 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(3 x^2-7 x+\underline{\text{ }}\right)+\left(2 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\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(2 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+7 y+\underline{\text{ }}\right)=2 \left(y^2+\frac{7 y}{2}+\underline{\text{ }}\right): \\
3 \left(x^2-\frac{7 x}{3}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+\frac{7 y}{2}+\underline{\text{ }}\right)$}=\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 }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}
\frac{49}{12}-1=\frac{37}{12}: \\
3 \left(x^2-\frac{7 x}{3}+\frac{49}{36}\right)+2 \left(y^2+\frac{7 y}{2}+\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 }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{37}{12}+\frac{49}{8}=\frac{221}{24}: \\
3 \left(x^2-\frac{7 x}{3}+\frac{49}{36}\right)+2 \left(y^2+\frac{7 y}{2}+\frac{49}{16}\right)=\fbox{$\frac{221}{24}$} \\
\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{$\left(x-\frac{7}{6}\right)^2$}+2 \left(y^2+\frac{7 y}{2}+\frac{49}{16}\right)=\frac{221}{24} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{7 y}{2}+\frac{49}{16}=\left(y+\frac{7}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x-\frac{7}{6}\right)^2+2 \fbox{$\left(y+\frac{7}{4}\right)^2$}=\frac{221}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2+4 x-9 y^2+3 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2+3 y+2 x^2+4 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-9 y^2+3 y+2 x^2+4 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2+4 x+\underline{\text{ }}\right)+\left(-9 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\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{$2 \left(x^2+2 x+\underline{\text{ }}\right)$}+\left(-9 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2+3 y+\underline{\text{ }}\right)=-9 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right): \\
2 \left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2-\frac{y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }2\times 1=2 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2-9=-7: \\
2 \left(x^2+2 x+1\right)-9 \left(y^2-\frac{y}{3}+\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 }-9 \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{-9}{36}=-\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-7-\frac{1}{4}=-\frac{29}{4}: \\
2 \left(x^2+2 x+1\right)-9 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\fbox{$-\frac{29}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
2 \fbox{$(x+1)^2$}-9 \left(y^2-\frac{y}{3}+\frac{1}{36}\right)=-\frac{29}{4} \\
\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{} & 2 (x+1)^2-9 \fbox{$\left(y-\frac{1}{6}\right)^2$}=-\frac{29}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2-5 x+3 y^2+5 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2+5 y+7 x^2-5 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
3 y^2+5 y+7 x^2-5 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(7 x^2-5 x+\underline{\text{ }}\right)+\left(3 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\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(3 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\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): \\
7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2+\frac{5 y}{3}+\underline{\text{ }}\right)$}=\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 }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}
3+\frac{25}{28}=\frac{109}{28}: \\
7 \left(x^2-\frac{5 x}{7}+\frac{25}{196}\right)+3 \left(y^2+\frac{5 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{109}{28}$} \\
\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{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{109}{28}+\frac{25}{12}=\frac{251}{42}: \\
7 \left(x^2-\frac{5 x}{7}+\frac{25}{196}\right)+3 \left(y^2+\frac{5 y}{3}+\frac{25}{36}\right)=\fbox{$\frac{251}{42}$} \\
\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{$\left(x-\frac{5}{14}\right)^2$}+3 \left(y^2+\frac{5 y}{3}+\frac{25}{36}\right)=\frac{251}{42} \\
\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{} & 7 \left(x-\frac{5}{14}\right)^2+3 \fbox{$\left(y+\frac{5}{6}\right)^2$}=\frac{251}{42} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2-x-10 y^2+7 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+7 y+7 x^2-x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-10 y^2+7 y+7 x^2-x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(7 x^2-x+\underline{\text{ }}\right)+\left(-10 y^2+7 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2-x+\underline{\text{ }}\right)=7 \left(x^2-\frac{x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2-\frac{x}{7}+\underline{\text{ }}\right)$}+\left(-10 y^2+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): \\
7 \left(x^2-\frac{x}{7}+\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 }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}
\frac{1}{28}-1=-\frac{27}{28}: \\
7 \left(x^2-\frac{x}{7}+\frac{1}{196}\right)-10 \left(y^2-\frac{7 y}{10}+\underline{\text{ }}\right)=\fbox{$-\frac{27}{28}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\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{27}{28}-\frac{49}{40}=-\frac{613}{280}: \\
7 \left(x^2-\frac{x}{7}+\frac{1}{196}\right)-10 \left(y^2-\frac{7 y}{10}+\frac{49}{400}\right)=\fbox{$-\frac{613}{280}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{7}+\frac{1}{196}=\left(x-\frac{1}{14}\right)^2: \\
7 \fbox{$\left(x-\frac{1}{14}\right)^2$}-10 \left(y^2-\frac{7 y}{10}+\frac{49}{400}\right)=-\frac{613}{280} \\
\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{} & 7 \left(x-\frac{1}{14}\right)^2-\text{10 }\fbox{$\left(y-\frac{7}{20}\right)^2$}=-\frac{613}{280} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2+5 y^2-5 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2-5 y+\left(-7 x^2-5\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
5 y^2-5 y-7 x^2=5 \\
\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 y^2-5 y+\underline{\text{ }}\right)-7 x^2=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 y^2-5 y+\underline{\text{ }}\right)=5 \left(y^2-y+\underline{\text{ }}\right): \\
\fbox{$5 \left(y^2-y+\underline{\text{ }}\right)$}-7 x^2=\underline{\text{ }}+5 \\
\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 }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{5}{4}=\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
5+\frac{5}{4}=\frac{25}{4}: \\
5 \left(y^2-y+\frac{1}{4}\right)-7 x^2=\fbox{$\frac{25}{4}$} \\
\end{array}
Step 7:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$\left(y-\frac{1}{2}\right)^2$}-7 x^2=\frac{25}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+8 x-6 y^2-3 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-3 y+4 x^2+8 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
-6 y^2-3 y+4 x^2+8 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+8 x+\underline{\text{ }}\right)+\left(-6 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\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(-6 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\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): \\
4 \left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)$}=\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 }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}
4-2=2: \\
4 \left(x^2+2 x+1\right)-6 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$2$} \\
\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 }-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}
2-\frac{3}{8}=\frac{13}{8}: \\
4 \left(x^2+2 x+1\right)-6 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\fbox{$\frac{13}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
4 \fbox{$(x+1)^2$}-6 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\frac{13}{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)^2-6 \fbox{$\left(y+\frac{1}{4}\right)^2$}=\frac{13}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2-3 x+8 y^2+4 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2+4 y+9 x^2-3 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
8 y^2+4 y+9 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(9 x^2-3 x+\underline{\text{ }}\right)+\left(8 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2-3 x+\underline{\text{ }}\right)=9 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)$}+\left(8 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2+4 y+\underline{\text{ }}\right)=8 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right): \\
9 \left(x^2-\frac{x}{3}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \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{9}{36}=\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6+\frac{1}{4}=\frac{25}{4}: \\
9 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)+8 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{25}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{8}{16}=\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{25}{4}+\frac{1}{2}=\frac{27}{4}: \\
9 \left(x^2-\frac{x}{3}+\frac{1}{36}\right)+8 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\fbox{$\frac{27}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{3}+\frac{1}{36}=\left(x-\frac{1}{6}\right)^2: \\
9 \fbox{$\left(x-\frac{1}{6}\right)^2$}+8 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\frac{27}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{2}+\frac{1}{16}=\left(y+\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x-\frac{1}{6}\right)^2+8 \fbox{$\left(y+\frac{1}{4}\right)^2$}=\frac{27}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2+2 x-9 y^2-y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-y-10 x^2+2 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
-9 y^2-y-10 x^2+2 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-10 x^2+2 x+\underline{\text{ }}\right)+\left(-9 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2+2 x+\underline{\text{ }}\right)=-10 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right)$}+\left(-9 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\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): \\
-10 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2+\frac{y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{-10}{100}=-\frac{1}{10} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10-\frac{1}{10}=\frac{99}{10}: \\
-10 \left(x^2-\frac{x}{5}+\frac{1}{100}\right)-9 \left(y^2+\frac{y}{9}+\underline{\text{ }}\right)=\fbox{$\frac{99}{10}$} \\
\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 }-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{99}{10}-\frac{1}{36}=\frac{1777}{180}: \\
-10 \left(x^2-\frac{x}{5}+\frac{1}{100}\right)-9 \left(y^2+\frac{y}{9}+\frac{1}{324}\right)=\fbox{$\frac{1777}{180}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{5}+\frac{1}{100}=\left(x-\frac{1}{10}\right)^2: \\
-10 \fbox{$\left(x-\frac{1}{10}\right)^2$}-9 \left(y^2+\frac{y}{9}+\frac{1}{324}\right)=\frac{1777}{180} \\
\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{} & -10 \left(x-\frac{1}{10}\right)^2-9 \fbox{$\left(y+\frac{1}{18}\right)^2$}=\frac{1777}{180} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-8 x-8 y^2-2 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2-2 y+5 x^2-8 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-8 y^2-2 y+5 x^2-8 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2-8 x+\underline{\text{ }}\right)+\left(-8 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2-8 x+\underline{\text{ }}\right)=5 \left(x^2-\frac{8 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-\frac{8 x}{5}+\underline{\text{ }}\right)$}+\left(-8 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2-2 y+\underline{\text{ }}\right)=-8 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right): \\
5 \left(x^2-\frac{8 x}{5}+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }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}
\frac{16}{5}-5=-\frac{9}{5}: \\
5 \left(x^2-\frac{8 x}{5}+\frac{16}{25}\right)-8 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)=\fbox{$-\frac{9}{5}$} \\
\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 }-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 9:
\begin{array}{l}
-\frac{9}{5}-\frac{1}{8}=-\frac{77}{40}: \\
5 \left(x^2-\frac{8 x}{5}+\frac{16}{25}\right)-8 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=\fbox{$-\frac{77}{40}$} \\
\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{$\left(x-\frac{4}{5}\right)^2$}-8 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=-\frac{77}{40} \\
\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{} & 5 \left(x-\frac{4}{5}\right)^2-8 \fbox{$\left(y+\frac{1}{8}\right)^2$}=-\frac{77}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+6 x+5 y^2-5 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2-5 y+8 x^2+6 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
5 y^2-5 y+8 x^2+6 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(8 x^2+6 x+\underline{\text{ }}\right)+\left(5 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2+6 x+\underline{\text{ }}\right)=8 \left(x^2+\frac{3 x}{4}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+\frac{3 x}{4}+\underline{\text{ }}\right)$}+\left(5 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2-5 y+\underline{\text{ }}\right)=5 \left(y^2-y+\underline{\text{ }}\right): \\
8 \left(x^2+\frac{3 x}{4}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2-y+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }8\times \frac{9}{64}=\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{8}-5=-\frac{31}{8}: \\
8 \left(x^2+\frac{3 x}{4}+\frac{9}{64}\right)+5 \left(y^2-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 }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{5}{4}=\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{5}{4}-\frac{31}{8}=-\frac{21}{8}: \\
8 \left(x^2+\frac{3 x}{4}+\frac{9}{64}\right)+5 \left(y^2-y+\frac{1}{4}\right)=\fbox{$-\frac{21}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{4}+\frac{9}{64}=\left(x+\frac{3}{8}\right)^2: \\
8 \fbox{$\left(x+\frac{3}{8}\right)^2$}+5 \left(y^2-y+\frac{1}{4}\right)=-\frac{21}{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{3}{8}\right)^2+5 \fbox{$\left(y-\frac{1}{2}\right)^2$}=-\frac{21}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2+2 y^2+9 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+9 y+\left(9 x^2-3\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
2 y^2+9 y+9 x^2=3 \\
\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+9 y+\underline{\text{ }}\right)+9 x^2=\underline{\text{ }}+3 \\
\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): \\
\fbox{$2 \left(y^2+\frac{9 y}{2}+\underline{\text{ }}\right)$}+9 x^2=\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 }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 6:
\begin{array}{l}
3+\frac{81}{8}=\frac{105}{8}: \\
2 \left(y^2+\frac{9 y}{2}+\frac{81}{16}\right)+9 x^2=\fbox{$\frac{105}{8}$} \\
\end{array}
Step 7:
\begin{array}{l}
y^2+\frac{9 y}{2}+\frac{81}{16}=\left(y+\frac{9}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \fbox{$\left(y+\frac{9}{4}\right)^2$}+9 x^2=\frac{105}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-4 y^2+4 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+4 y+\left(-3 x^2-2\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-4 y^2+4 y-3 x^2-2 \text{from }\text{both }\text{sides}: \\
4 y^2-4 y+\left(3 x^2+2\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
4 y^2-4 y+3 x^2=-2 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(4 y^2-4 y+\underline{\text{ }}\right)+3 x^2=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2-4 y+\underline{\text{ }}\right)=4 \left(y^2-y+\underline{\text{ }}\right): \\
\fbox{$4 \left(y^2-y+\underline{\text{ }}\right)$}+3 x^2=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \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{4}{4}=1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
1-2=-1: \\
4 \left(y^2-y+\frac{1}{4}\right)+3 x^2=\fbox{$-1$} \\
\end{array}
Step 8:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \fbox{$\left(y-\frac{1}{2}\right)^2$}+3 x^2=-1 \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+4 x+5 y^2-8 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2-8 y+6 x^2+4 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
5 y^2-8 y+6 x^2+4 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+4 x+\underline{\text{ }}\right)+\left(5 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\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(5 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2-8 y+\underline{\text{ }}\right)=5 \left(y^2-\frac{8 y}{5}+\underline{\text{ }}\right): \\
6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2-\frac{8 y}{5}+\underline{\text{ }}\right)$}=\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{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}-1=-\frac{1}{3}: \\
6 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+5 \left(y^2-\frac{8 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{1}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-8}{5}}{2}\right)^2=\frac{16}{25} \text{on }\text{the }\text{left }\text{and }5\times \frac{16}{25}=\frac{16}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{16}{5}-\frac{1}{3}=\frac{43}{15}: \\
6 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)+5 \left(y^2-\frac{8 y}{5}+\frac{16}{25}\right)=\fbox{$\frac{43}{15}$} \\
\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{$\left(x+\frac{1}{3}\right)^2$}+5 \left(y^2-\frac{8 y}{5}+\frac{16}{25}\right)=\frac{43}{15} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{8 y}{5}+\frac{16}{25}=\left(y-\frac{4}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x+\frac{1}{3}\right)^2+5 \fbox{$\left(y-\frac{4}{5}\right)^2$}=\frac{43}{15} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2+9 x+8 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 x^2+9 x+(8 y+2)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 y+2 \text{from }\text{both }\text{sides}: \\
5 x^2+9 x=-8 y-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(5 x^2+9 x+\underline{\text{ }}\right)=(-8 y-2)+\underline{\text{ }} \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2+9 x+\underline{\text{ }}\right)=5 \left(x^2+\frac{9 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+\frac{9 x}{5}+\underline{\text{ }}\right)$}=(-8 y-2)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\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}
(-8 y-2)+\frac{81}{20}=\frac{41}{20}-8 y: \\
5 \left(x^2+\frac{9 x}{5}+\frac{81}{100}\right)=\fbox{$\frac{41}{20}-8 y$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2+\frac{9 x}{5}+\frac{81}{100}=\left(x+\frac{9}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$\left(x+\frac{9}{10}\right)^2$}=\frac{41}{20}-8 y \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2-2 x-4 y^2+6 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+6 y+7 x^2-2 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-4 y^2+6 y+7 x^2-2 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(7 x^2-2 x+\underline{\text{ }}\right)+\left(-4 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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(-4 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\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): \\
7 \left(x^2-\frac{2 x}{7}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }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}
\frac{1}{7}-5=-\frac{34}{7}: \\
7 \left(x^2-\frac{2 x}{7}+\frac{1}{49}\right)-4 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{34}{7}$} \\
\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}
-\frac{34}{7}-\frac{9}{4}=-\frac{199}{28}: \\
7 \left(x^2-\frac{2 x}{7}+\frac{1}{49}\right)-4 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$-\frac{199}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{2 x}{7}+\frac{1}{49}=\left(x-\frac{1}{7}\right)^2: \\
7 \fbox{$\left(x-\frac{1}{7}\right)^2$}-4 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=-\frac{199}{28} \\
\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{} & 7 \left(x-\frac{1}{7}\right)^2-4 \fbox{$\left(y-\frac{3}{4}\right)^2$}=-\frac{199}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2+7 x-6 y^2+9 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+9 y-5 x^2+7 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
-6 y^2+9 y-5 x^2+7 x=4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2+7 x+\underline{\text{ }}\right)+\left(-6 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2+7 x+\underline{\text{ }}\right)=-5 \left(x^2-\frac{7 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2-\frac{7 x}{5}+\underline{\text{ }}\right)$}+\left(-6 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2+9 y+\underline{\text{ }}\right)=-6 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right): \\
-5 \left(x^2-\frac{7 x}{5}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)$}=\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 }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{5}}{2}\right)^2=\frac{49}{100} \text{on }\text{the }\text{left }\text{and }-5\times \frac{49}{100}=-\frac{49}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
4-\frac{49}{20}=\frac{31}{20}: \\
-5 \left(x^2-\frac{7 x}{5}+\frac{49}{100}\right)-6 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{31}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }-6\times \frac{9}{16}=-\frac{27}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{31}{20}-\frac{27}{8}=-\frac{73}{40}: \\
-5 \left(x^2-\frac{7 x}{5}+\frac{49}{100}\right)-6 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$-\frac{73}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{7 x}{5}+\frac{49}{100}=\left(x-\frac{7}{10}\right)^2: \\
-5 \fbox{$\left(x-\frac{7}{10}\right)^2$}-6 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=-\frac{73}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{2}+\frac{9}{16}=\left(y-\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 \left(x-\frac{7}{10}\right)^2-6 \fbox{$\left(y-\frac{3}{4}\right)^2$}=-\frac{73}{40} \\
\end{array}
| khanacademy | amps |