## Completing the Square not agreeing with Quadratic

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.

### Completing the Square not agreeing with Quadratic

I'm aweful at Completing the square I don't know why but I am. I have a problem that requires me to do so and show my work. When I was done I did the same problem using the Quadratic Formula and the answers don't match. Can someone help?

Problem: 9x^2+36X+31

Here is my work Lemme know if you can see my mistake:
Completing the Square:
x^2+4x=-31/9 (Rearranged and divided by 9)
x^2+4x+4=-31/9+36/9 (Half of the coefficent, squared and added to both sides)
(x+2)^2=5/9 (simplified)
x+2 = +-sqrt(5)/3
x= -2+-sqrt(5)/3

-36+-sprt(1296-4*9*31)/2(9)
-36+-sqrt(180)/18
-36+-6sqrt(5)/18
x= -6+-sqrt(5)/3

Thanks
ShawnaRae

Posts: 2
Joined: Thu May 28, 2009 6:21 pm

You didn't start with an equation, so I'm not sure how you ended up with one...? Was the original exercise actually to find the zeroes of the quadratic by completing the square...? If so, then:

You might want to be a bit more careful when writing out the Quadratic Formula. For instance, the "2a" is supposed to be under all of "-b +/- sqrt[b^2 - 4ac]". Thus, you should have gotten "-36/18 = -2" for the first part of the solution, not "-6".

stapel_eliz

Posts: 1715
Joined: Mon Dec 08, 2008 4:22 pm

### Re: Completing the Square not agreeing with Quadratic

Nevermind.. I know what I did wrong..
For the completing the square I got:
-2 +- sqrt(5)
--------
3

But I stuffed the -2 on the same line as the rest so it looked wrong when in reality it's the same as
-6+- sqrt(5)
---------------
3

Apparently I just needed to eat lunch and come back to it.
Thanks.
ShawnaRae

Posts: 2
Joined: Thu May 28, 2009 6:21 pm

ShawnaRae wrote:Nevermind.. I know what I did wrong..
For the completing the square I got:
-2 +- sqrt(5)
--------
3

Actually, you had it right the first time:

$-2\, \pm\, \frac{\sqrt{5}}{3}$

Then the other, properly formatted, is:

$\frac{-6\, \pm\, \sqrt{5}}{3}\, =\, \frac{-6}{3}\, \pm\, \frac{\sqrt{5}}{3}$

...which simplifies to match the earlier result.

stapel_eliz

Posts: 1715
Joined: Mon Dec 08, 2008 4:22 pm