Question

If $p=\frac{3-\sqrt{5}}{3+\sqrt{5}}\mathrm{and}q=\frac{3+\sqrt{5}}{3-\sqrt{5}}$, find the value of p² + q².

Answer 1

According to question,
$p=\frac{3-\sqrt{5}}{3+\sqrt{5}}\mathrm{and}q=\frac{3+\sqrt{5}}{3-\sqrt{5}}$

$$\begin{gathered} p=\frac{3-\sqrt{5}}{3+\sqrt{5}} \\ =\frac{3-\sqrt{5}}{3+\sqrt{5}}\times \frac{3-\sqrt{5}}{3-\sqrt{5}} \\ =\frac{{\left(3-\sqrt{5}\right)}^2}{{\left(3\right)}^2-{\left(\sqrt{5}\right)}^2} \\ =\frac{{\left(3\right)}^2+{\left(\sqrt{5}\right)}^2-2\left(3\right)\left(\sqrt{5}\right)}{9-5} \\ =\frac{9+5-6\sqrt{5}}{4} \\ =\frac{14-6\sqrt{5}}{4}...\left(1\right) \\ q=\frac{3+\sqrt{5}}{3-\sqrt{5}} \\ =\frac{3+\sqrt{5}}{3-\sqrt{5}}\times \frac{3+\sqrt{5}}{3+\sqrt{5}} \\ =\frac{{\left(3+\sqrt{5}\right)}^2}{{\left(3\right)}^2-{\left(\sqrt{5}\right)}^2} \\ =\frac{{\left(3\right)}^2+{\left(\sqrt{5}\right)}^2+2\left(3\right)\left(\sqrt{5}\right)}{9-5} \\ =\frac{9+5+6\sqrt{5}}{4} \\ =\frac{14+6\sqrt{5}}{4}...\left(2\right) \end{gathered}$$

Now,
$$\begin{gathered} p^2+q^2={\left(p+q\right)}^2-2pq \\ ={\left(\frac{14-6\sqrt{5}}{4}+\frac{14+6\sqrt{5}}{4}\right)}^2-2\left(\frac{14-6\sqrt{5}}{4}\right)\left(\frac{14+6\sqrt{5}}{4}\right) \\ ={\left(\frac{28}{4}\right)}^2-2\left(\frac{{\left(14\right)}^2-{\left(6\sqrt{5}\right)}^2}{16}\right) \\ ={\left(7\right)}^2-2\left(\frac{196-180}{16}\right) \\ =49-2\left(\frac{16}{16}\right) \\ =49-2 \\ =47 \end{gathered}$$

Hence, the value of p² + q² is 47.