Question

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

Answer 1

$\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}$

$=\frac{(7+3\sqrt{5})(3-\sqrt{5})-(7-3\sqrt{5})(3+\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}$

$=\frac{(6+2\sqrt{5})-(6-2\sqrt{5})}{3^2-(\sqrt{5}{)}^2}$

$=\frac{4\sqrt{5}}{4}$

$=\sqrt{5}$

Now,

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

$\Rightarrow a=0,b=1$