Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \frac{6}{3\sqrt{2}-2\sqrt{3}}=a\sqrt{2}+b\sqrt{3} \\ \Rightarrow \frac{6}{3\sqrt{2}-2\sqrt{3}}\times \frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}=\frac{6\times \left(3\sqrt{2}+2\sqrt{3}\right)}{{\left(3\sqrt{2}\right)}^2-{\left(2\sqrt{3}\right)}^2} \\ =\frac{18\sqrt{2}+12\sqrt{3}}{18-12} \\ =\frac{6\times \left(3\sqrt{2}+2\sqrt{3}\right)}{6} \end{gathered}$$
$$\begin{gathered} 3\sqrt{2}+2\sqrt{3}=a\sqrt{2}+b\sqrt{3} \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ a=3\mathrm{and}b=2 \end{gathered}$$