Question

Solve an equation: $\left(x^2+\frac{1}{x^2}\right)-3\left(x-\frac{1}{x}\right)-2=0$.

Answer 1

Given equation is

$\left(x^2+\frac{1}{x^2}\right)-3\left(x-\frac{1}{x}\right)-2=0$ ..... $\left(i\right)$

Now, $x^2+\frac{1}{x^2}={(x-\frac{1}{x})}^2+2$

Let $(x-\frac{1}{x})=t$

$\Rightarrow t^2+2-3t-2=0$ ....... From $\left(i\right)$

$\Rightarrow t(t-3)=0$

$\Rightarrow t=0$ or $t=3$

Now,

$t=0\Rightarrow x=\frac{1}{x}\Rightarrow x=\pm 1$

&

$t=3\Rightarrow (x-\frac{1}{x})=3$

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

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