Question

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

Answer 1

It is given that $\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=a+b\sqrt{15}$

Simplify the term on the L.H.S. by rationalizing,

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

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

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

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

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

$=4+\sqrt{15}$

Now,

$4+\sqrt{15}$$=a+b\sqrt{15}$

$\Rightarrow a=4$ and $b=1$