Question

Find the value of $a$ in the following:
$\frac{3-\sqrt{5}}{3+2\sqrt{5}}=a\sqrt{5}-\frac{19}{11}$.

Answer 1

$LHS=\frac{3-\sqrt{5}}{3+2\sqrt{5}}=\frac{3-\sqrt{5}}{3+2\sqrt{5}}\times \frac{3-2\sqrt{5}}{3-2\sqrt{5}}$

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

$=\frac{9-6\sqrt{5}-3\sqrt{5}+10}{9-20}=\frac{19-9\sqrt{5}}{-11}$

Now, $\frac{19-9\sqrt{5}}{-11}=a\sqrt{5}-\frac{19}{11}$

$\Rightarrow \frac{-19}{11}+\frac{9}{11}\sqrt{5}=a\sqrt{5}-\frac{19}{11}$

$\Rightarrow \frac{9}{11}\sqrt{5}-\frac{19}{11}=a\sqrt{5}-\frac{19}{11}$

Hence, $a=\frac{9}{11}$.