Question

Solve-
$\frac{\sqrt{5}+\sqrt{3}}{2\sqrt{5}-3\sqrt{3}}=a-b\sqrt{15}$

Answer 1

We have,

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

On rationalize and we get,

$\frac{\sqrt{5}+\sqrt{3}}{2\sqrt{5}-3\sqrt{3}}\times \frac{2\sqrt{5}+3\sqrt{3}}{2\sqrt{5}+3\sqrt{3}}=a-b\sqrt{15}$

$\Rightarrow \frac{10+3\sqrt{15}+2\sqrt{15}+9}{{\left(2\sqrt{5}\right)}^2-{\left(3\sqrt{3}\right)}^2}=a-b\sqrt{15}$

$\Rightarrow \frac{19+5\sqrt{15}}{20-27}=a-b\sqrt{15}$

$\Rightarrow \frac{19+5\sqrt{15}}{-7}=a-b\sqrt{15}$

$\Rightarrow \frac{19}{-7}+\frac{5\sqrt{15}}{-7}=a-b\sqrt{15}$

Comparing that, and we get,

$a=-\frac{19}{7}$ and

$-b=-\frac{5}{7}$

$\Rightarrow b=\frac{5}{7}$

Hence, this is the answer.