Solve the following quadratic equation by factorization:
$\frac{1}{x}\ –\ \frac{1}{x\ -\ 2}\ =\ 3$

Given quadratic equation is $\frac{1}{x}\ –\ \frac{1}{x\ -\ 2}\ =\ 3$.

We have to solve the given quadratic equation by factorization. 

$\frac{1}{x}\ –\ \frac{1}{x\ -\ 2}\ =\ 3$ can be written as,

$\frac{(x-2)-(x)}{x(x-2)}=3$ 

$x-2-x=3(x^2-2x)$     (On cross multiplication)

$-2=3x^2-6x$

$3x^2-6x+2=0$

$3x^2-3x-3x-2=0$

$3x^2-(3+\sqrt3)x-(3-\sqrt3)x+((\sqrt3)^2-(\sqrt1)^2)=0$

$(\sqrt3)^2x^2-\sqrt3(\sqrt3+1)x-\sqrt3(\sqrt3-1)x+(\sqrt3+1)(\sqrt3-1)=0$

$\sqrt{3}x(\sqrt{3}x-(\sqrt{3}+1))-(\sqrt3-1)(\sqrt{3}x-(\sqrt{3}+1))=0$

$(\sqrt{3}x-(\sqrt{3}+1))(\sqrt{3}x-(\sqrt{3}-1))=0$

$(\sqrt{3}x-(\sqrt{3}+1))=0$ or $(\sqrt{3}x-(\sqrt{3}-1))=0$

$\sqrt{3}x=\sqrt{3}+1$ or $\sqrt{3}x=\sqrt{3}-1$

$x=\frac{\sqrt{3}+1}{\sqrt3}$ or $x=\frac{\sqrt{3}-1}{\sqrt3}$

The roots of the given quadratic equation are $x=\frac{\sqrt{3} +1}{\sqrt{3}} \ or\ x=\frac{\sqrt{3} -1}{\sqrt{3}}$.