Question

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

Answer 1

Step 1: Estimate the value of $a,b$ and $c$

Given equation is,

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

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

$\Rightarrow$ $3\left(x^2-2x\right)=-2$

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

According to the Sridharacharya formula, for a given equation $a{x}^2+bx+c$, the roots of this equation is given by

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

From the given equation we have,

$a=3,b=-6,c=2$

Step 2: Determine the roots of the equation

$\therefore$The roots of the equation are,

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

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

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

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

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

$=1+\frac{1}{\sqrt{3}},1-\frac{1}{\sqrt{3}}$

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