Question

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

Answer 1

$x=1+\sqrt{2}$
$\Rightarrow \frac{1}{x}=\frac{1}{1+\sqrt{2}}$
$=\frac{1}{1+\sqrt{2}}\times \frac{1-\sqrt{2}}{1-\sqrt{2}}$
$=\frac{1-\sqrt{2}}{1-2}=\frac{1-\sqrt{2}}{-1}=-(1-\sqrt{2})$
$\therefore x-\frac{1}{x}=(1+\sqrt{2})-\{-(1-\sqrt{2}\left)\right\}$
$=1+\sqrt{2}+1-\sqrt{2}=2$
Hence, ${(x-\frac{1}{x})}^2=2^2=4$.