Question

If $x+\left(\frac{1}{x}\right)=6$, find $x^2+\left(\frac{1}{x^2}\right)$ and $x^4+\left(\frac{1}{x^4}\right)$

Answer 1

$x+\frac{1}{x}=6$

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

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

$\Rightarrow x^2+\frac{1}{x^2}={(x+\frac{1}{x})}^2-2=6^2-2=34$

$\Rightarrow {(x^2+\frac{1}{x^2})}^2=x^4+\frac{1}{x^4}+2x^2.\frac{1}{x^2}$

$=x^4+x^4+\frac{1}{x^4}+2$

$\Rightarrow x^2+\frac{1}{x^2}={(x^2+\frac{1}{x^2})}^2-2={34}^2-2$

$=1154.$

Hence, the answer is $1154.$