Question

find the value of $a$ and $b.$
$\frac{2-\sqrt{5}}{2+3\sqrt{5}}=a\sqrt{5}+b$

Answer 1

Given $\frac{2-\sqrt{5}}{2+3\sqrt{5}}=a\sqrt{5}+b$,

$\Rightarrow \frac{2-\sqrt{5}}{2+3\sqrt{5}}=\frac{(2-\sqrt{5})(2-3\sqrt{5})}{2^2-{\left(3\sqrt{5}\right)}^2}$

$=\frac{4+(3\times 5)-6\sqrt{5}-2\sqrt{5})}{4-(9\times 5)}$

$=\frac{4+15-8\sqrt{5}}{4-45}$

$=\frac{19-8\sqrt{5}}{-41}$

$=\frac{-19}{41}+\frac{8}{41}\sqrt{5}$

$\therefore a=\frac{8}{41}\&b=\frac{-19}{41}$