# complete the square

• Dec 11th 2008, 04:56 PM
TJ #99
complete the square
someone please show me how this works?

im supposed to be putting a standard formed quadratic equation into a vertex equation like this one.

4/3x^2 - 40x + 294
• Dec 11th 2008, 05:00 PM
euclid2
Quote:

Originally Posted by TJ #99
someone please show me how this works?

im supposed to be putting a standard formed quadratic equation into a vertex equation like this one.

4/3x^2 - 40x + 294

Do you know how to complete the square?
• Dec 11th 2008, 05:04 PM
TJ #99
to tell u the truth i know nothing thats why im askin
• Dec 11th 2008, 05:08 PM
euclid2
Quote:

Originally Posted by TJ #99
someone please show me how this works?

im supposed to be putting a standard formed quadratic equation into a vertex equation like this one.

4/3x^2 - 40x + 294

Check out this link, I will get you started
Completing the Square
$\displaystyle \frac{4}{3}x^2 - 40x + 294$
$\displaystyle (\frac{4}{3}x^2 - 40x) + 294$
[tex] (\frac{4}{3}x^2 - 40x + 400 - 400) + 294{/math]
• Dec 11th 2008, 05:15 PM
TJ #99
thanks i will give it a shot
• Dec 11th 2008, 05:32 PM
euclid2
Quote:

Originally Posted by TJ #99
thanks i will give it a shot

did you get it?
• Dec 11th 2008, 08:23 PM
Prove It
Quote:

Originally Posted by euclid2
Check out this link, I will get you started
Completing the Square
$\displaystyle \frac{4}{3}x^2 - 40x + 294$
$\displaystyle (\frac{4}{3}x^2 - 40x) + 294$
[tex] (\frac{4}{3}x^2 - 40x + 400 - 400) + 294{/math]

You can only complete the square when your a value is 1.

So...

$\displaystyle \frac{4}{3}x^2 - 40x + 294 = \frac{4}{3}(x^2 - 30x + \frac{441}{2})$

Go from there...