Question

$21.Ifp=\frac{3-\sqrt{5}}{3+\sqrt{5}}andq=\frac{3+\sqrt{5}}{3-\sqrt{5}},Findthevalueof{p}^2+q^2.$

Answer 1

Dear Student,
$$\begin{gathered} p=\frac{3-\sqrt{5}}{3+\sqrt{5}} \\ q=\frac{3+\sqrt{5}}{3-\sqrt{5}} \\ {\left(p+q\right)}^2=p^2+q^2+2pq \\ {p}^2+q^2=\frac{{\left(p+q\right)}^2}{2pq}=\frac{\left[\frac{3-\sqrt{5}}{3+\sqrt{5}}+\frac{3+\sqrt{5}}{3-\sqrt{5}}\right]}{2\left(\frac{3+\sqrt{5}}{3-\sqrt{5}}\right)\left(\frac{3-\sqrt{5}}{3+\sqrt{5}}\right)}=\frac{1}{2}\left[\frac{{\left(3-\sqrt{5}\right)}^2+{\left(3+\sqrt{5}\right)}^2}{-2}\right] \\ {p}^2+q^2=\frac{1}{2}\left[\frac{28}{-2}\right]=-7 \\ Regards \end{gathered}$$