Question

Solve the given quadratic equation by factorization method
$\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0$

Answer 1

We have
$\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0$

$\Rightarrow 2x(2x+3)+(x-3)+3x+9=0\Rightarrow 4x^2+6x+x-3+3x+9=0\Rightarrow 4x^2+10x+6=0$

$\Rightarrow 2x^2+5x+3=0$

$\Rightarrow 2x^2+2x+3x+3=0\Rightarrow 2x(x+1)+3(x+1)=0\Rightarrow (2x+3)(x+1)=0$

$\Rightarrow x=-1$ and $x=\frac{-3}{2}$