Find the values of 'a' and 'b':

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

Given,

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

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

$\Rightarrow \dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} \times \dfrac{2 + \sqrt{3}}{2 + \sqrt{3}} \\[1em] \Rightarrow \dfrac{(2 + \sqrt{3})^2}{2^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{2^2 + (\sqrt{3})^2 + 2 \times 2 \times \sqrt{3}}{2^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{4 + 3 + 4\sqrt{3}}{4 - 3} \\[1em] \Rightarrow 7 + 4\sqrt{3}.$

Comparing $7 + 4\sqrt{3} \text{ with } a + b\sqrt{3}$, we get :

a = 7 and b = 4.

Hence, a = 7 and b = 4.