Question

Find sum of solution of the equation
$\frac{1}{(x-4)}+\frac{1}{(x+4)}=\frac{x^2}{(x^2-16)}$ is

Answer 1

Solve the given equation as follows:

$\frac{1}{(x-4)}+\frac{1}{(x+4)}=\frac{x^2}{(x^2-16)}$

$\frac{(x+4+x-4)}{(x^2-16)}=\frac{x^2}{(x^2-16)}$

$2x=x^2$

$x^2-2x=0$

$x(x-2)=0$

$x=0$ and $x-2=0$

$x=0$ and $x=2$

Therefore, $x=0$ and $x=2$ .