If $\dfrac{3 + \sqrt{2}}{3 - \sqrt{2}} = a + b\sqrt{2}$, find the values of 'a' and 'b'.

Given,

Equation : $\dfrac{3 + \sqrt{2}}{3 - \sqrt{2}} = a + b\sqrt{2}$

Rationalizing L.H.S. of the above equation :

$\Rightarrow \dfrac{3 + \sqrt{2}}{3 - \sqrt{2}} \times \dfrac{3 + \sqrt{2}}{3 + \sqrt{2}} \\[1em] \Rightarrow \dfrac{(3 + \sqrt{2})^2}{(3)^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{(3)^2 + (\sqrt{2})^2 + 2 \times 3 \times \sqrt{2}}{9 - 2} \\[1em] \Rightarrow \dfrac{9 + 2 + 6\sqrt{2}}{7} \\[1em] \Rightarrow \dfrac{11 + 6\sqrt{2}}{7} \\[1em] \Rightarrow \dfrac{11}{7} + \dfrac{6\sqrt{2}}{7}$

Comparing $\dfrac{11}{7} + \dfrac{6}{7}\sqrt{2} \text{ with } a + b\sqrt{2}$, we get :

$a = \dfrac{11}{7} \text{ and } b = \dfrac{6}{7}.$

Hence, $a = \dfrac{11}{7} \text{ and } b = \dfrac{6}{7}$.