Question

If $x^2+\frac{1}{x^2}=79$, find the value of $x+\frac{1}{x}$.

Answer 1

Given $x^2+\frac{1}{x^2}=79$
As we know that ${(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2$
$\Rightarrow {(x+\frac{1}{x})}^2=79+2$
$\Rightarrow {(x+\frac{1}{x})}^2=81$
$\Rightarrow x+\frac{1}{x}=\sqrt{(9{)}^2}$
$\therefore x+\frac{1}{x}=\pm 9$