Question

If $x^2+\frac{1}{x^2}=27$, find the value of $x-\frac{1}{x}$

Answer 1

$x-\frac{1}{x}$
We have,
${(x-\frac{1}{x})}^2=x^2-2\times x\times \frac{1}{x}+\frac{1}{x^2}$
$\Rightarrow {(x-\frac{1}{x})}^2=x^2-2+\frac{1}{x^2}$
$\Rightarrow {(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2$
$\Rightarrow {(x-\frac{1}{x})}^2=27-2[\because x^2+\frac{1}{x^2}=27$ given]
$\Rightarrow {(x-\frac{1}{x})}^2=25\Rightarrow {(x-\frac{1}{x})}^2$
$\Rightarrow x-\frac{1}{x}=\pm 5$
[Taking square root of both sides]