Question

Find the roots of $\frac{1}{x}-\frac{1}{x-2}=3$

Answer 1

We have,

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

$\frac{x-2-x}{x(x-2)}=3$

$\frac{-2}{(x^2-2x)}=3$

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

$3x^2-6x+2=0$ ……. (1)

We know that

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Therefore,

$x=\frac{6\pm \sqrt{6^2-4\times 3\times 2}}{2\times 3}$

$x=\frac{6\pm \sqrt{36-24}}{6}$

$x=\frac{6\pm \sqrt{12}}{6}$

$x=\frac{6\pm 2\sqrt{3}}{6}$

$x=1\pm \frac{\sqrt{3}}{3}$

$x=1\pm \frac{1}{\sqrt{3}}$

Hence, the roots of this equation are $1+\frac{1}{\sqrt{3}},1-\frac{1}{\sqrt{3}}$.