Question

If $\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(3\sqrt{2}-2\sqrt{3}\right)}=\left(a+b\sqrt{6}\right)$, find the values of a and b.

Answer 1

$$\begin{gathered} \frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(3\sqrt{2}-2\sqrt{3}\right)} \\ =\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(3\sqrt{2}-2\sqrt{3}\right)}\times \frac{\left(3\sqrt{2}+2\sqrt{3}\right)}{\left(3\sqrt{2}+2\sqrt{3}\right)} \\ =\frac{6+2\sqrt{6}+3\sqrt{6}+6}{18-12} \\ =\frac{12+5\sqrt{6}}{6} \end{gathered}$$
$$\begin{gathered} =2+\frac{5}{6}\sqrt{6} \\ \mathrm{We}\mathrm{know}: \\ a+b\sqrt{6}=2+\frac{5}{6}\sqrt{6} \\ \therefore a=2\mathrm{and}b=\frac{5}{6} \end{gathered}$$