Question

If $\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=a+b\sqrt{15}$, find $a$ and $b$.

Answer 1

Consider the given equation

$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=a+b\sqrt{15}$

$LHS$ is,

$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

$\Rightarrow \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

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

$\Rightarrow \frac{5+3+2\sqrt{15}}{2}$

$\Rightarrow \frac{8+2\sqrt{15}}{2}$

$\Rightarrow 4+\sqrt{15}$

$RHS$ is,

$\Rightarrow a+b\sqrt{15}$

On comparing with the $RHS$, we get,

$a=4,b=1$