Question

If $p=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ and $q=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ then find $p^2+q^2$.

Answer 1

$p=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{{(\sqrt{3}-\sqrt{2})}^2}{3-2}={(\sqrt{3}-\sqrt{2})}^2=3+2-2\sqrt{6}=5-2\sqrt{6}q=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}={(\sqrt{3}-\sqrt{2})}^2=3+2+2\sqrt{6}=5+2\sqrt{6}\therefore p^2+q^2={(5-2\sqrt{6})}^2+{(5+2\sqrt{6})}^2=25+24+20\sqrt{6}+25+24+20\sqrt{6}=98$