Question

Given, find the value of a, if $[3–\sqrt{5}]/[3+2\sqrt{5}]=-19/11+a\sqrt{5}$

Answer 1

$[3–\sqrt{5}]/[3+2\sqrt{5}]=-19/11+a\sqrt{5}$

Let us consider LHS

$[3–\surd 5]/[3+2\sqrt{5}]$

Rationalize the denominator,

$[3–\sqrt{5}]/[3+2\sqrt{5}]=\left[\right(3–\sqrt{5}\left)\right(3–2\sqrt{5}\left)\right]/\left[\right(3+2\sqrt{5}\left)\right(3–2\sqrt{5}\left)\right]$

$=\left[3\right(3–2\sqrt{5})–\sqrt{5}(3–2\sqrt{5}\left)\right]/[32–(2\sqrt{5}\left)2\right]$

$=[9–6\sqrt{5}–3\sqrt{5}+2.5]/[9–4.5]$

$=[9–6\sqrt{5}–3\sqrt{5}+10]/[9–20]$

$=[19–9\sqrt{5}]/-11$

$=-19/11+9\sqrt{5}/11$

So when comparing with RHS

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

Hence, the value of $a=9/11$