Question

If $x+\frac{1}{x}=9$ and $x^2+\frac{1}{x^2}=5$. Find the value of $x-\frac{1}{x}$.

Answer 1

It is given that $x^2+\frac{1}{x^2}=5$.

Consider ${(x-\frac{1}{x})}^2$ and solve as follows:

${(x-\frac{1}{x})}^2={\left(x\right)}^2+{\left(\frac{1}{x}\right)}^2-(2\times x\times \frac{1}{x})(\because (a+b{)}^2=a^2+b^2+2ab)=x^2+\frac{1}{x^2}-2=5-2(Given{x}^2+\frac{1}{x^2}=5)$

$=3\Rightarrow {(x-\frac{1}{x})}^2=3\Rightarrow x-\frac{1}{x}=\sqrt{3}$

Hence, $(x-\frac{1}{x})=\sqrt{3}$.