Question

Find the roots of the equation $\frac{x^2-1}{x}=3$$,x\ne 0$.

Answer 1

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

Given equation is

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

$\Rightarrow$ $x^2-1=3x$

$\Rightarrow$$x^2-3x-1=0$

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

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

From the given equation we have,

$a=1,b=-3,c=-1$

Step 2: Determine the roots of the equation

$\therefore$The roots of the equation are,

$x=\frac{-\left(-3\right)\pm \sqrt{{\left(-3\right)}^2-4\times 1\times \left(-1\right)}}{2\times 1}$

$=\frac{3\pm \sqrt{9+4}}{2}$

$=\frac{3\pm \sqrt{13}}{2}$

$=\frac{3+\sqrt{13}}{2},\frac{3-\sqrt{13}}{2}$

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