Question

Find the value of a & b : $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=a+b\sqrt{6}$

Answer 1

$\begin{array}{c}Wehave\\ \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=a+b\sqrt{6}\\ Now\\ \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\ =\frac{{\left(\sqrt{3}+\sqrt{2}\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}\\ =\frac{3+2+2\sqrt{3\times 2}}{3-2}\\ =\frac{5+2\sqrt{6}}{1}\\ =5+2\sqrt{6}\\ Now,\\ 5+2\sqrt{6}=a+b\sqrt{6}\\ Oncomparingbothsidesweget\\ a=5\\ and\\ b=2\end{array}$