Question

Given that $\frac{3-\sqrt{5}}{3+2\sqrt{5}}=a\sqrt{5}-\frac{19}{11}$, then $a=\frac{9}{m}$. Value of $m$ is

Answer 1

Given, $\frac{3-\sqrt{5}}{3+2\sqrt{5}}=a\sqrt{5}-\frac{19}{11}$
Multiply and divide it by $3-2\sqrt{5}$
$\therefore \frac{(3-\sqrt{5})(3-2\sqrt{5})}{(3+2\sqrt{5})(3-2\sqrt{5})}$ = $a\sqrt{5}-\frac{19}{11}$
$\frac{9-3\sqrt{5}-6\sqrt{5}+10}{-11}=a\sqrt{5}-\frac{19}{11}$
$\frac{-9\sqrt{5}+19}{11}=a\sqrt{5}-\frac{19}{11}$
$\frac{-9\sqrt{5}+19}{11}$ + $\frac{19}{11}$ = $a\sqrt{5}$
$\therefore a=\frac{9}{11}$

Thus the value of $m$ is $11$.