Question

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

Answer 1

$P=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{3+2-2\sqrt{6}}{1}=5-2\sqrt{6}$

$q=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{3+2+2\sqrt{6}}{1}=5+2\sqrt{6}$

$P^2+q^2=25+24-20\sqrt{6}+25+24+20\sqrt{6}$

$=49\times 2$

$=98$