Question

If $x=3-2\sqrt{2}$, find $x^2+\frac{1}{x^2}$.

Answer 1

We have, $x=3-2\sqrt{2}$

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

$=\frac{3+2\sqrt{2}}{(3{)}^2-(2\sqrt{2}{)}^2}=\frac{3+2\sqrt{2}}{9-8}=3+2\sqrt{2}$

Thus, $x^2+\frac{1}{x^2}=(3-2\sqrt{2})+(3+2\sqrt{2}{)}^2$

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

$=9+8-12\sqrt{2}+9+8+12\sqrt{2}=34$