# Can anyone show me how to complete the square on this

• Oct 21st 2012, 08:59 PM
Eraser147
Can anyone show me how to complete the square on this
$x^2-8x+y^2+2y=-14$
• Oct 21st 2012, 09:06 PM
Prove It
Re: Can anyone show me how to complete the square on this
Quote:

Originally Posted by Eraser147
$x^2-8x+y^2+2y=-14$

Treat the x terms and the y terms separately. Complete the square on each.
• Oct 21st 2012, 09:17 PM
Eraser147
Re: Can anyone show me how to complete the square on this
It would be $(x-4)^2-2+(y+1)^2+13$. Which is wrong. Solutions manual says it's supposed to be $(x-4)^2-16+(y+1)^2-1$
• Oct 21st 2012, 09:25 PM
Prove It
Re: Can anyone show me how to complete the square on this
Remember that this is an EQUATION.
• Oct 21st 2012, 09:27 PM
Eraser147
Re: Can anyone show me how to complete the square on this
so would it be what i had = 0 or = -14. I am so confused.
• Oct 21st 2012, 09:32 PM
Prove It
Re: Can anyone show me how to complete the square on this
Start with your initial equation, then whatever you add to complete the square on the x terms and the y terms, add to the other side as well, to keep the equation balanced.
• Oct 21st 2012, 09:34 PM
Eraser147
Re: Can anyone show me how to complete the square on this
Can you show me? I still don't quite follow.
• Oct 21st 2012, 09:49 PM
Prove It
Re: Can anyone show me how to complete the square on this
You can show me every step you've taken, then I can guide you where you have gone wrong...
• Oct 21st 2012, 09:57 PM
Eraser147
Re: Can anyone show me how to complete the square on this
okay this is what I did, I separated the x and y and completed the square for both x and y terms so I resulted in this $(x-4)^2-2+(y+1)^2+13$. I took $x^2-8x+14$ and compeleted the square first and I did the same by adding 14 on the y terms as well.
• Oct 22nd 2012, 11:08 AM
Soroban
Re: Can anyone show me how to complete the square on this
Hello, Eraser147!

Quote:

$\text{Complete the square: }\:x^2-8x+y^2+2y\:=\:\text{-}14$

We have: . $x^2 - 8x \qquad + y^2 + 2y \qquad \;=\;\text{-}14$

Then:. . . . $x^2 - 8x {\color{red}\:+\:16} + y^2 + 2y {\color{green}\:+\:1} \;=\;\text{-}14 {\color{red}\:+\:16} {\color{green}\:+\:1}$

Therefore: . . . . . . $(x-4)^2 + (y+1)^2 \;=\;3$

• Oct 22nd 2012, 06:21 PM
Eraser147
Re: Can anyone show me how to complete the square on this
Thanks Soroban, you are the best.