Question

Find the values of 'a' and 'b' in $\frac{\sqrt{5}+\sqrt{3}}{2\sqrt{5}-3\sqrt{3}}=a-b\sqrt{15}$

Answer 1

$\frac{\sqrt{5}+\sqrt{3}}{2\sqrt{5}-3\sqrt{3}}$

Rationalizing the denominator, we get

$\frac{\sqrt{5}+\sqrt{3}}{2\sqrt{5}-3\sqrt{3}}\times \frac{2\sqrt{5}+3\sqrt{3}}{2\sqrt{5}+3\sqrt{3}}$

$=\frac{2\times 5+3\sqrt{15}+2\sqrt{15}+3\times 3}{{\left(2\sqrt{5}\right)}^2-{\left(3\sqrt{3}\right)}^2}$

$=\frac{10+3\sqrt{15}+2\sqrt{15}+9}{4\times 5-9\times 3}$

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

$=\frac{-19-5\sqrt{15}}{7}$

$=-\frac{19}{7}-\frac{5}{7}\sqrt{15}$

Comparing with $a-b\sqrt{15}$ we get

$a=-\frac{19}{7},b=\frac{5}{7}$

$\therefore a=-\frac{19}{7},b=\frac{5}{7}$