Question

Reciprocal of sum of all the solution of equation
$\frac{-x}{(x+3)}-\frac{x}{(x-3)}=-\frac{4}{(x^2-9)}-\frac{1}{(x+3)}$

Answer 1

Solve the given equation as follows:

$\frac{-x}{(x+3)}-\frac{x}{(x-3)}=-\frac{4}{(x^2-9)}-\frac{1}{(x+3)}$

$-x(x-3)-x(x+3)=-4-(x-3)$

$-x^2+3x-x^2-3x=-4-x+3$

$-2x^2=-1-x$

$2x^2=1+x$

$2x^2-x-1=0$

Now we find the roots of above quadratic equation as shown below:

$2x^2-x-1=0$

$2x^2-2x+x-1=0$

$2x(x-1)+1(x-1)=0$

$(2x+1)=0$ and $x-1=0$

$x=-\frac{1}{2}$ and $x=1$

Therefore, $x=-\frac{1}{2}$ and $x=1$