Question

Solve the following quadratic equation by factorization.
$\frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x};x\ne 0,-1,2$

Answer 1

$\frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x};x\ne 0,-1,2$
$\frac{2}{x+1}+\frac{3}{2x-4}=\frac{23}{5x}$

$\frac{2(2x-4)+3(x+1)}{(x+1)(2x-4)}=\frac{23}{5x}$

$\frac{4x-8+3x+3}{(x+1)(2x-4)}=\frac{23}{5x}$

$\frac{7x-5}{2x^2-2x-4}=\frac{23}{5x}$

$(7x-5)5x=23(2x^2-2x-4)$

$35x^2-25x=46x^2-46x-92)$

$11x^2-21x-92=0$

$11x^2-44x+23x-92=0$

$11x(x-4)+23(x-4)=0$

$(x-4)(11x+23)=0$

Therefore,

$x=4,\frac{-23}{11}$