Question

If $x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ and $y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$, find the value of (x² + y²).
or
Simplify: $\frac{(7+3\sqrt{5})}{(3+\sqrt{5})}-\frac{(7-3\sqrt{5})}{(3-\sqrt{5})}$

Answer 1

$$\begin{gathered} \text{Given}, \\ x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}},y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} \\ \mathrm{Therefore}, \\ {x}^2={\left(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\right)}^2=\frac{5+3+2\sqrt{15}}{5+3-2\sqrt{15}}=\frac{8+2\sqrt{15}}{8-2\sqrt{15}} \\ \mathrm{And} \\ {y}^2={\left(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)}^2=\frac{5+3-2\sqrt{15}}{5+3+2\sqrt{15}}=\frac{8-2\sqrt{15}}{8+2\sqrt{15}} \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \\ {x}^2+y^2=\frac{8+2\sqrt{15}}{8-2\sqrt{15}}+\frac{8-2\sqrt{15}}{8+2\sqrt{15}} \\ =\frac{{\left(8+2\sqrt{15}\right)}^2+{\left(8-2\sqrt{15}\right)}^2}{\left(8-2\sqrt{15}\right)\left(8+2\sqrt{15}\right)} \\ =\frac{{\left(8\right)}^2+{\left(2\sqrt{15}\right)}^2+2\times 8\times 2\sqrt{15}+{\left(8\right)}^2+{\left(2\sqrt{15}\right)}^2-2\times 8\times 2\sqrt{15}}{{\left(8\right)}^2-{\left(2\sqrt{15}\right)}^2} \\ =\frac{64+60+64+60}{64-60} \\ =\frac{248}{4} \\ =62 \end{gathered}$$

OR

$$\begin{gathered} \frac{(7+3\sqrt{5})}{(3+\sqrt{5})}-\frac{(7-3\sqrt{5})}{(3-\sqrt{5})} \\ =\frac{(7+3\sqrt{5})(3-\sqrt{5})-(7-3\sqrt{5})(3+\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})} \\ =\frac{21-7\sqrt{5}+9\sqrt{5}-15-21-7\sqrt{5}+9\sqrt{5}+15}{{\left(3\right)}^2-{\left(\sqrt{5}\right)}^2} \\ =\frac{4\sqrt{5}}{9-5} \\ =\frac{4\sqrt{5}}{4}=\sqrt{5} \end{gathered}$$