Question

If $\frac{x^2+1}{x^2}=27$. Find $(x-\frac{1}{x})$?

Answer 1

$x^2+\frac{1}{x^2}=27;x-\frac{1}{x}=?$
Use the formula $(a-b{)}^2=a^2+b^2-2ab$
${(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2\left(x\right)\left(\frac{1}{x}\right)$
= 27 - 2
${(x-\frac{1}{x})}^2=25$
$x-\frac{1}{x}=\pm 5$