Hello,
Can someone please show me how to complete this square?
2x^2+y^2=2y+1
Thanks!
-Mark
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