# Completing the square

• Nov 23rd 2009, 05:34 PM
l flipboi l
Completing the square
Hello,

Can someone please show me how to complete this square?

2x^2+y^2=2y+1

Thanks!
-Mark
• Nov 23rd 2009, 05:46 PM
Bacterius
You are sure it isn't $2xy$ in the right-hand side ? If yes, then it is easy, if no, then there is a little more thinking to do. Have you tried using square identities to factor out both unknowns together ?
• Nov 23rd 2009, 06:13 PM
l flipboi l
Quote:

Originally Posted by Bacterius
You are sure it isn't $2xy$ in the right-hand side ? If yes, then it is easy, if no, then there is a little more thinking to do. Have you tried using square identities to factor out both unknowns together ?

Yes, the right side is what shows in the book.

I was thinking, is this the right way to start it?

$2x^2 - 1 = 2y - y^2$
• Nov 23rd 2009, 09:07 PM
Bacterius
I think there should be a way to factorize $y$ using a square identity, leaving alone a constant based on $x$ (since you are completing the square) :

$2x^2 + y^2 = 2y + 1$

$2x^2 - 2y + y^2 = 1$

I know what follows is not very mathematic but perhaps it could be a good idea to try with some numbers substituting $x$ and $y$ and investigate what happens, if a pattern doesn't show up ... but you have to find working numbers (Sweating) ...
Here is what I would do :

I would say $x = 1$, therefore :

$2 - 2y + y^2 = 1$

$y^2 - 2y + 2 = 1$

$y^2 - 2y + 1 = 0$

Woah ! A square identity !

$(y - 1)^2 = 0$

And I successfully completed the square.

Try with $x = 2$ :

$8 - 2y + y^2 = 1$

$y^2 - 2y + 8 = 1$

Erm ... we have to do some working there :

$y^2 - 2y + 1 = -6$ (substract 7 from both sides)

And then the magic operates :

$(y - 1)^2 = -6$

Thus, it seems that regardless of what value $x$ takes, we always have $(y - 1)^2 = c$, where $c$ is dependent on $x$.

Basically, we know that $c = 1 - (2x^2 - 1)$ (since we always take off the left-hand side just the needed amount to leave $1$ on the left-hand side)

Which simplifies to : $c = 2 - 2x^2$, and finally, $c = 2(1 - x^2)$.

Now let's put it all together, and detail the steps more mathematically (what was shown up there was brainwork) :

$2x^2 + y^2 = 2y + 1$

$y^2 - 2y = 1 - 2x^2$

But we need a little 1 on the left-hand side to successfully complete the square :

$y^2 - 2y + 1 = 1 - 2x^2 + 1$

$y^2 - 2y + 1 = 2 - 2x^2$

Finally :

$(y - 1)^2 = 2 - 2x^2$

I think that is what the book wants (Nod)
• Nov 23rd 2009, 09:46 PM
l flipboi l
Great stuff, Thank you!