Question

If $x=\sqrt{3}+1$, find the values of ${(x-\frac{2}{x})}^2$.

Answer 1

$x=\sqrt{3}+1$

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

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

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

$=4$