Can anyone show me how to complete the square on this

**Eraser147** · October 21st 2012, 07:59 PM

$x^2-8x+y^2+2y=-14$

**Prove It** · October 21st 2012, 08:06 PM

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.

**Eraser147** · October 21st 2012, 08:17 PM

It would be $(x-4)^2-2+(y+1)^2+13$ . Which is wrong. Solutions manual says it's supposed to be $(x-4)^2-16+(y+1)^2-1$

**Prove It** · October 21st 2012, 08:25 PM

Remember that this is an EQUATION.

**Eraser147** · October 21st 2012, 08:27 PM

so would it be what i had = 0 or = -14. I am so confused.

**Prove It** · October 21st 2012, 08:32 PM

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.

**Eraser147** · October 21st 2012, 08:34 PM

Can you show me? I still don't quite follow.

**Prove It** · October 21st 2012, 08:49 PM

You can show me every step you've taken, then I can guide you where you have gone wrong...

**Eraser147** · October 21st 2012, 08:57 PM

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.

**Soroban** · October 22nd 2012, 10:08 AM

Hello, Eraser147!

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

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

**Eraser147** · October 22nd 2012, 05:21 PM

Thanks Soroban, you are the best.

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