Originally Posted by
masters Not so bad...
$\displaystyle 10^2=\left(\frac{x}{2}\right)^2+\left(\frac{x+4}{2 }\right)^2$
$\displaystyle 100=\frac{x^2}{4}+\frac{x^2+8x+16}{4}$
$\displaystyle 400=x^2+x^2+8x+16$
$\displaystyle 2x^2+8x-384=0$
$\displaystyle x^2+4x-192=0$
I factored the above equation. You could have used the quadratic formula if you wanted to get the values for x.
$\displaystyle (x-12)(x+16)=0$
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:
$\displaystyle x=12 \ \ or \ \ x=-16$