# Finding values of quadratic equation

• May 6th 2011, 03:24 AM
yorkey
Finding values of quadratic equation
Got a math sum I'm working on thats troubling me.

\$\displaystyle {x}^{2} + 4x - 8\$ can be written in the form \$\displaystyle {(x + p)}^{2} + q \$

Find the values of \$\displaystyle p\$ and \$\displaystyle q\$
• May 6th 2011, 03:27 AM
mr fantastic
Quote:

Originally Posted by yorkey
Got a math sum I'm working on thats troubling me.

\$\displaystyle {x}^{2} + 4x - 8\$ can be written in the form \$\displaystyle {(x + p)}^{2} + q \$

Find the values of \$\displaystyle p\$ and \$\displaystyle q\$

You have two options, both require that you have the necessary knowledge and understanding.

Option 1: Complete the square for x^2 + 4x - 8 and compare with (x + p)^2 + q.

Option 2: Expand (x + p)^2 + q and equate the coefficients of the various powers of x with the corresponding coefficients in x^2 + 4x - 8.
• May 6th 2011, 03:41 AM
yorkey
Ok great! Thanks

p = 2
q = -8

Does that work?
• May 6th 2011, 03:59 AM
DrSteve
\$\displaystyle {(x + 2)}^{2} - 8= x^2+2x+2x+4-8= x^2+4x-4 \$.

So p looks correct, but q does not. You almost got it though.
• May 6th 2011, 04:07 AM
yorkey
Oh I'm a retard. My reasoning went something like this:

(x + 2)^2 is expanded to x^2 + 4, which it obviously is NOT. So q is -12, right?
• May 6th 2011, 05:19 AM
masters
Quote:

Originally Posted by yorkey
Oh I'm a retard. My reasoning went something like this:

(x + 2)^2 is expanded to x^2 + 4, which it obviously is NOT. So q is -12, right?

Hi yorkey,

Yes, q = -12

Complete the square on \$\displaystyle x^2+4x+4\$

\$\displaystyle (x^2+4x\: {\color{red}+4})-8 \: {\color{red}-4}\$

\$\displaystyle (x \: {\color{red}+2})^2\: {\color{red}-12}\$
• May 6th 2011, 05:28 AM
yorkey
Y'all are legends, thanks.

/this case is closed