Question

If $a=3-2\sqrt{2}$, find the value of $a^2-\frac{1}{a^2}$.

Answer 1

Given: $a=3-2\sqrt{2}$
To find: $a^2-\frac{1}{a^2}$

We know that $(a-b)(a+b)=a^2-b^2$

$\therefore a^2-\frac{1}{a^2}=(a-\frac{1}{a})(a+\frac{1}{a})$
$\therefore \frac{1}{a}=\frac{1}{3-2\sqrt{2}}\times \frac{3+2\sqrt{2}}{3+2\sqrt{2}}$

$=\frac{3+2\sqrt{2}}{3^2-(2\sqrt{2}{)}^2}$

$=3+2\sqrt{2}$
$\therefore a-\frac{1}{a}=3-2\sqrt{2}-3-2\sqrt{2}$

$=-4\sqrt{2}$

$a+\frac{1}{a}=3-2\sqrt{2}+3+2\sqrt{2}$

$=6$

Therefore, $(a-\frac{1}{a})(a+\frac{1}{a})$ is

$(-4\sqrt{2})\left(6\right)=-24\sqrt{2}$