Question

If $a+\frac{1}{a}=6$ and $a\ne 0$; find
$a^2-\frac{1}{a^2}$

Answer 1

Given:$a+\frac{1}{a}=6$

$\Rightarrow {(a+\frac{1}{a})}^2=36$

$\Rightarrow a^2+\frac{1}{a^2}+2\times a\times \frac{1}{a}=36$

$\Rightarrow a^2+\frac{1}{a^2}=36-2=34$ .......$\left(1\right)$

${(a-\frac{1}{a})}^2=a^2+\frac{1}{a^2}-2\times a\times \frac{1}{a}$

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

$=34-2=32$ from $\left(1\right)$

$\therefore {(a-\frac{1}{a})}^2=32$

$\therefore a-\frac{1}{a}=\sqrt{32}=4\sqrt{2}$

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

$=6\times 4\sqrt{2}=24\sqrt{2}$

$\therefore a^2-\frac{1}{a^2}=24\sqrt{2}$