Question

If $\frac{2}{\sqrt{3}+\sqrt{5}}+\frac{5}{\sqrt{3}-\sqrt{5}}=a\sqrt{3}+b\sqrt{5}$, find $a$ & $b$

Answer 1

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

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

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

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

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

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

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

On comparing both sides we get

$=a=\frac{-7}{2}andb=\frac{-3}{2}$