Question

If $P=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ and $q=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$, find the value of $p^2+q^2$.

Answer 1

$p=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

$p=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

$p=\frac{5+3-2\sqrt{15}}{5-3}$

$\Rightarrow p=\frac{8-2\sqrt{15}}{2}=4-\sqrt{15}$

And $p^2={(4-\sqrt{15})}^2=16+15-8\sqrt{15}=31-8\sqrt{15}$ ......$\left(1\right)$

$q=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

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

$q=\frac{5+3+2\sqrt{15}}{5-3}$

$\Rightarrow q=\frac{8+2\sqrt{15}}{2}=4+\sqrt{15}$

And $q^2={(4+\sqrt{15})}^2=16+15+8\sqrt{15}=31+8\sqrt{15}$ ......$\left(2\right)$

$\therefore p^2+q^2=31-8\sqrt{15}+31+8\sqrt{15}=62$ from $\left(1\right)$ and $\left(2\right)$