Question

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

Answer 1

Given expression is $x-\frac{1}{x}=9$.......(i)
Square on both sides, we get
$(x-\frac{1}{x}{)}^2=9^2$.........(ii)
Since, we have $(a-b{)}^2=a^2-2ab+b^2$
Substitute $a=x,b=\frac{1}{x}$ in above formula, we get

$\Rightarrow (x-\frac{1}{x}{)}^2=x^2+\frac{1}{x^2}-2(x\left)\right(\frac{1}{x})$
$\therefore (x-\frac{1}{x}{)}^2=x^2+\frac{1}{x^2}-2$......(iii)
Substitute equation (iii) in equation(ii), we get

$(x^2+\frac{1}{x^2}-2)=81$

$x^2+\frac{1}{x^2}=81+2$

$\therefore$ the value of $x^2+\frac{1}{x^2}$ is $83$.