Question

If $x^2+\left(\frac{1}{x^2}\right)=27,$ find the value of $x+\left(\frac{1}{x}\right).$

Answer 1

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

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

$\Rightarrow (x{)}^2+(\frac{1}{x}{)}^2+2\times x\times \left(\frac{1}{x}\right)=29$

Using formula, we have
$a^2+b^2+2ab=(a+b{)}^2$

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

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