Question

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

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

Answer 1

The given equation is

$$\begin{gathered} x-\frac{1}{x}=3,x\ne 0 \\ \Rightarrow \frac{x^2-1}{x}=3 \\ \Rightarrow x^2-1=3x \\ \Rightarrow x^2-3x-1=0 \end{gathered}$$

This equation is of the form $a{x}^2+bx+c=0$, where a = 1, b = −3 and c = −1.

∴ Discriminant, D = $b^2-4ac={\left(-3\right)}^2-4\times 1\times \left(-1\right)=9+4=13>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{13}$

$$\begin{gathered} \therefore \alpha =\frac{-b+\sqrt{D}}{2a}=\frac{-\left(-3\right)+\sqrt{13}}{2\times 1}=\frac{3+\sqrt{13}}{2} \\ \beta =\frac{-b-\sqrt{D}}{2a}=\frac{-\left(-3\right)-\sqrt{13}}{2\times 1}=\frac{3-\sqrt{13}}{2} \end{gathered}$$

Hence, $\frac{3+\sqrt{13}}{2}$ and $\frac{3-\sqrt{13}}{2}$ are the roots of the given equation.