# Thread: Need help with an equation

1. ## Need help with an equation

Can anyone finish this equation?

(10)^2=(x/2)^2+((x+4)/2)^2

2. I know that the next step looks like this, but I don't understand where the 2x that is underlined came from??

Would appreciate it if anyone could answer that question

100=x^2/4+x^2/4+2x+4

3. Originally Posted by Yaymath
Can anyone finish this equation?

(10)^2=(x/2)^2+((x+4)/2)^2

$10^2=\left(\frac{x}{2}\right)^2+\left(\frac{x+4}{2 }\right)^2$

$100=\frac{x^2}{4}+\frac{x^2+8x+16}{4}$

$400=x^2+x^2+8x+16$

$2x^2+8x-384=0$

$x^2+4x-192=0$

$(x-12)(x+16)=0$

$x=12 \ \ or \ \ x=-16$

4. I understand everything except the last part

(did you do the additive inverse of 400 or 16?)

(how did this become to (x - 12)(x+16)=0

The rest is clear for me, it's just there i'm confused

5. Originally Posted by masters

$10^2=\left(\frac{x}{2}\right)^2+\left(\frac{x+4}{2 }\right)^2$

$100=\frac{x^2}{4}+\frac{x^2+8x+16}{4}$

$400=x^2+x^2+8x+16$

$2x^2+8x-384=0$

$x^2+4x-192=0$

I factored the above equation. You could have used the quadratic formula if you wanted to get the values for x.

$(x-12)(x+16)=0$

I used the zero product property here. The zero product property says that if ab=0, then a=0 or b=0.

Either x - 12= 0 or x + 16 = 0

Then I solved each equation to get:

$x=12 \ \ or \ \ x=-16$
..

6. Thanx alot you are indeed the master

Bye and thnx again