If $\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}} = p + q\sqrt{5}$ , find the values of p and q where p and q are rational numbers.

Since, it is given that

$\dfrac{(7 + 3\sqrt{5})}{3 + \sqrt{5}}$ - $\dfrac{(7 - 3\sqrt{5})} {(3 - \sqrt{5})}$ = p + q$\sqrt{5}$

On solving,

$\dfrac{(7 + 3\sqrt{5})(3 - \sqrt{5}) - (7 - 3\sqrt{5})(3 + \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} \\[1.5em] \Rightarrow \dfrac{(21 - 7\sqrt{5} + 9\sqrt{5} - 15) - (21 + 7\sqrt{5} - 9\sqrt{5} - 15)}{3^2 - {(\sqrt{5})}^2} \\[1.5em] \Rightarrow \dfrac{(21 - 7\sqrt{5} + 9\sqrt{5} - 15) - (21 + 7\sqrt{5} - 9\sqrt{5} - 15)}{9-5} \\[1.5em] \Rightarrow \dfrac{21 - 7\sqrt{5} + 9\sqrt{5} - 15 - 21 - 7\sqrt{5} + 9\sqrt{5} + 15}{9-5} \\[1.5em] \Rightarrow \dfrac{18{\sqrt{5}} - 14{\sqrt{5}}}{4} \\[1.5em] \Rightarrow \dfrac{4 × {\sqrt{5}}}{4} = \sqrt{5} \\[1.5em] \therefore\sqrt{5} = p + q{\sqrt{5}} \\[1.5em] \Rightarrow 0 + 1 × \sqrt{5} = p + q{\sqrt{5}}$

Hence,value of p = 0 and q = 1.