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Question:

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

Solution:

According to question,

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

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

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

$=\frac{(3-\sqrt{5})^{2}}{(3)^{2}-(\sqrt{5})^{2}}$

$=\frac{(3)^{2}+(\sqrt{5})^{2}-2(3)(\sqrt{5})}{9-5}$

$=\frac{9+5-6 \sqrt{5}}{4}$

$=\frac{14-6 \sqrt{5}}{4} \quad \cdots(1)$

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

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

$=\frac{(3+\sqrt{5})^{2}}{(3)^{2}-(\sqrt{5})^{2}}$

$=\frac{(3)^{2}+(\sqrt{5})^{2}+2(3)(\sqrt{5})}{9-5}$

$=\frac{9+5+6 \sqrt{5}}{4}$

$=\frac{14+6 \sqrt{5}}{4} \quad \ldots(2)$

Now,

$p^{2}+q^{2}=(p+q)^{2}-2 p q$

$=\left(\frac{14-6 \sqrt{5}}{4}+\frac{14+6 \sqrt{5}}{4}\right)^{2}-2\left(\frac{14-6 \sqrt{5}}{4}\right)\left(\frac{14+6 \sqrt{5}}{4}\right)$

$=\left(\frac{28}{4}\right)^{2}-2\left(\frac{(14)^{2}-(6 \sqrt{5})^{2}}{16}\right)$

$=(7)^{2}-2\left(\frac{196-180}{16}\right)$

$=49-2\left(\frac{16}{16}\right)$

$=49-2$

$=47$

Hence, the value of $p^{2}+q^{2}$ is 47 .