Question

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$\frac{1}{x}-\frac{1}{x-2}=3,x\ne 0,2$ [CBSE 2010]

Answer 1

The given equation is

$$\begin{gathered} \frac{1}{x}-\frac{1}{x-2}=3,x\ne 0,2 \\ \Rightarrow \frac{x-2-x}{x\left(x-2\right)}=3 \\ \Rightarrow \frac{-2}{x^2-2x}=3 \\ \Rightarrow -2=3x^2-6x \end{gathered}$$

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

This equation is of the form $a{x}^2+bx+c=0$, where a = 3, b = −6 and c = 2.

∴ Discriminant, D = $b^2-4ac={\left(-6\right)}^2-4\times 3\times 2=36-24=12>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{12}=2\sqrt{3}$

$$\begin{gathered} \therefore \alpha =\frac{-b+\sqrt{D}}{2a}=\frac{-\left(-6\right)+2\sqrt{3}}{2\times 3}=\frac{6+2\sqrt{3}}{6}=\frac{3+\sqrt{3}}{3} \\ \beta =\frac{-b-\sqrt{D}}{2a}=\frac{-\left(-6\right)-2\sqrt{3}}{2\times 3}=\frac{6-2\sqrt{3}}{6}=\frac{3-\sqrt{3}}{3} \end{gathered}$$

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