Question

If $x-\frac{1}{x}=9$, find the value of $x^2+\frac{1}{x^2}$

Answer 1

We have,$x-\frac{1}{x}=9$
On squaring both sides, we get
${(x-\frac{1}{x})}^2=9^2$
$\Rightarrow x^2-2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=81$
$\Rightarrow x^2-2+{\left(\frac{1}{x}\right)}^2=81$
$\Rightarrow x^2+\frac{1}{x^2}=81+2$
[On transposing -2 on RHS]
$\Rightarrow x^2+\frac{1}{x^2}=83$