Question

Find the roots of the following equations:
(i) $x-\frac{1}{x}=3,x\ne 0$

Answer 1

(i) $x-\frac{1}{x}=3\Rightarrow x^2-3x-1=0$
On comparing this equation with $a{x}^2+bx+c=0$, we get
a = 1, b = -3 and c = -1
By using quadratic formula, we get
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$\Rightarrow$$x=\frac{-(-3)\pm \sqrt{(-3{)}^2-4\times 1\times -1}}{2\times 1}$
$\Rightarrow$ $x=\frac{3\pm \sqrt{13}}{2}$
$\therefore$ $x=\frac{3+\sqrt{13}}{2}orx=\frac{3-\sqrt{13}}{2}$