Question

If $x-\frac{1}{x}=5,$ then find $x^2-\frac{1}{x^2}$.

Answer 1

Given $x-\frac{1}{x}=5$

Squaring on both sides

${(x-\frac{1}{x})}^2=(5{)}^2$

$\Rightarrow x^2+\frac{1}{x^2}-2x\cdot \frac{1}{x}=25$

$\Rightarrow x^2+\frac{1}{x^2}=25+2$ .......... $(a-b{)}^2=a^2-2ab+b^2$

$\Rightarrow x^2+\frac{1}{x^2}=27$

Since $x^2+\frac{1}{x^2}=27$, then

$x^2+\frac{1}{x^2}+2=27+2$

and as $(a+b{)}^2=a^2+2ab+b^2$

So, ${(x+\frac{1}{x})}^2=29$

$x+\frac{1}{x}=\sqrt{29}$

Then

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

$\Rightarrow \left(\sqrt{29}\right)\left(5\right)=x^2-\frac{1}{x^2}$ ......... $a^2-b^2=(a-b)(a+b)$

$\therefore$ So, the answer is $5\sqrt{29}$.