Question

Find the values of $a$ and $b$ if
$\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=a+\sqrt{5}b$

Answer 1

$\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=a+\sqrt{5}b$

$\frac{7+3\sqrt{5}}{3+\sqrt{5}}\times \frac{3-\sqrt{5}}{3-\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}\times \frac{3+\sqrt{5}}{3+\sqrt{5}}=a+\sqrt{5}b$

$\frac{\left(7+3\sqrt{5}\right)\left(3-\sqrt{5}\right)}{9-5}-\frac{\left(7-3\sqrt{5}\right)\left(3+\sqrt{5}\right)}{9-5}=a+\sqrt{5}b$

$\frac{21-7\sqrt{5}+9\sqrt{5}-15-21-7\sqrt{5}+9\sqrt{5}+15}{4}=a+\sqrt{5}b$

$\frac{-14\sqrt{5}+18\sqrt{5}}{4}=a+\sqrt{5}b$

$\frac{4\sqrt{5}}{4}=a+\sqrt{5}b$

$\sqrt{5}=a+\sqrt{5}b$

Comparing $a=0,b=1$