# Completing the Square

1. Mar 2, 2006

### thomasrules

Ok so can you please tell me where I go wrong here. I want to put
y=3x^2+2x-1 into y=a(x-h)^2+K

ax^2-2ah+ah^2+k=3x^2-2x+ah^2-1

ax^2=3x^2 -----> a=3
-2ahx=-2x
-2(3)hx=-2x ----->h=1/3
K=----->-1

Therefore:

y=3(x-1/3)^2-1

But I think it's suppost to be: y=3(x+1/3)^2-4/3

Is that right??? So what is wrong....

2. Mar 2, 2006

### topsquark

There's nothing wrong with your method...except that I find it very confusing!

I know you are trying to compare terms to find h and k here, but how did the 2x get a negative sign?

I'd recommend looking at it like this:
$$y=3x^2+2x-1$$
$$y=3x^2+2x+D-D-1$$

Now, a perfect square in the form $$a(x-h)^2$$ looks like:
$$ax^2-2ahx+ah^2=3x^2+2x+D$$
We see that -2ah=2 and ah^2=D.
We of course know what a is, right? I leave it to you to finish finding h and D.

Then you've got:
$$y=(3x^x+2x+D)-D-1$$
$$y=a(x-h)^2-D-1$$
with whatever D value you have.

-Dan

3. Mar 2, 2006

### mathman

In going from the first line to the second, it looks quite screwed up.

4. Mar 2, 2006

### thomasrules

Yea thanks Dan for your help I now clearly understand it..

Btw ya that was a mistake with the -2 , what should have bee +2

5. Mar 3, 2006

### HallsofIvy

This is where you went wrong: you want
$$ax^2- 2ah+ ah^2+ k= 3x^2- 3x-1$$
not "ah2- 1".

No, ah2+ k= -1. Since a= 3 and h= 1/3, ah2= 1/3 and k= -1- 1/3= -4/3.