Question

Simplify:

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

Answer 1

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

Rationalising the denominators,

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

$=\frac{21-7\sqrt{5}+9\sqrt{5}-15}{3^2-(\sqrt{5}{)}^2}-\frac{21+7\sqrt{5}-9\sqrt{5}-15}{3^2-(\sqrt{5}{)}^2}$

$=\frac{6+2\sqrt{5}}{9-5}-\frac{6-2\sqrt{5}}{9-5}$

$=\frac{6+2\sqrt{5}-6+2\sqrt{5}}{4}$

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

$=\sqrt{5}$