Question

If both $a$ and $b$ are rational numbers then find the minimum value of $a+b$ in the following:
$\frac{3+\sqrt{5}}{3-\sqrt{5}}=a+b\sqrt{5}$

Answer 1

(i) $\frac{3+\sqrt{5}}{3-\sqrt{5}}$
$3+\sqrt{5}$ is the rationalizing factor of $3-\sqrt{5}$
$\therefore \frac{3+\sqrt{5}}{3-\sqrt{5}}=\frac{3+\sqrt{5}}{3-\sqrt{5}}\times \frac{3+\sqrt{5}}{3+\sqrt{5}}=\frac{{(3+\sqrt{5})}^2}{{\left(3\right)}^2-{\left(\sqrt{5}\right)}^2}$
$=\frac{9+5+6\sqrt{5}}{9-5}=\frac{14+6\sqrt{5}}{4}=\frac{14}{4}+\frac{6}{4}\sqrt{5}$
$=\frac{7}{2}+\frac{3}{2}\sqrt{5}=a+b\sqrt{5}$ (given)

$\therefore a=\frac{7}{2},b=\frac{3}{2}$

$a+b=\frac{7}{2}+\frac{3}{2}=5$