1. ## Completing the square

Hello,

Can someone please show me how to complete this square?

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

Thanks!
-Mark

2. You are sure it isn't $\displaystyle 2xy$ 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 ?

3. Originally Posted by Bacterius
You are sure it isn't $\displaystyle 2xy$ 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?

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

4. I think there should be a way to factorize $\displaystyle y$ using a square identity, leaving alone a constant based on $\displaystyle x$ (since you are completing the square) :

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

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

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

I would say $\displaystyle x = 1$, therefore :

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

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

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

Woah ! A square identity !

$\displaystyle (y - 1)^2 = 0$

And I successfully completed the square.

Try with $\displaystyle x = 2$ :

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

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

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

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

And then the magic operates :

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

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

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

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

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

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

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

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

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

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

Finally :

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

I think that is what the book wants

5. Great stuff, Thank you!