Question

If $a^4+\frac{1}{a^4}=322$, find the value of $a^2+\frac{1}{a^2}.$

Answer 1

Given : $a^4+\frac{1}{a^4}=322---\left(1\right)$

To find: $a^2+\frac{1}{a^2}$

In eq $\left(1\right)$ add $2$ both sides

$a^4+\frac{1}{a^4}+2=322+2$

$\Rightarrow (a^2{)}^2+\frac{1}{(a^2{)}^2}+2(a^2).\frac{1}{a^2}=324$ $[a^2+b^2+2ab=(a+b{)}^2]$

$\Rightarrow {(a^2+\frac{1}{a^2})}^2=324$

$\Rightarrow a^2+\frac{1}{a^2}=\pm 18$