Question

If $\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=a+b\sqrt{6}$ find the value of a&b.

Answer 1

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

$\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=\frac{(\sqrt{2}+\sqrt{3})(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}+2\sqrt{3})}$ (Rationalization)

$=\frac{6+2\sqrt{6}+3\sqrt{6}+6}{\left(3{\sqrt{2}}^2\right)-(2\sqrt{3}{)}^2}$

$\left(\right(3{\sqrt{2}}^2)=9\times 2=18)$

$=\frac{12+5\sqrt{6}}{18-12}$ $(2\sqrt{3}{)}^2=4\times 3=12$

$=\frac{12+5\sqrt{6}}{6}$

$=2+\frac{5}{6}\sqrt{6}$

$\therefore a=2$ and

$b=\frac{5}{6}$