If $$\frac{{\sqrt 7 - 2}}{{\sqrt 7 + 2}} = a\sqrt 7 + b{\text{,}}$$     then the value of a is?

Solution(By Examveda Team)

$$\eqalign{ & \frac{{\sqrt 7 - 2}}{{\sqrt 7 + 2}} = a\sqrt 7 + b \cr & {\text{L}}{\text{.H}}{\text{.S }}\frac{{\sqrt 7 - 2}}{{\sqrt 7 + 2}} \times \frac{{\sqrt 7 - 2}}{{\sqrt 7 - 2}} \cr & {\text{ }}\left( {{\text{Rationalisation}}} \right) \cr & = \frac{{{{\left( {\sqrt 7 - 2} \right)}^2}}}{{{{\left( {\sqrt 7 } \right)}^2} - \left( 4 \right)}} \cr & = \frac{{7 + 4 - 4\sqrt 7 }}{{7 - 4}} \cr & = \frac{{11 - 4\sqrt 7 }}{3} \cr & = \frac{{11}}{3} - \frac{4}{3}\sqrt 7 \cr & = - \frac{4}{3}\sqrt 7 + \frac{{11}}{3} \cr & = a\sqrt 7 + b{\text{ }}\left( {{\text{ R}}{\text{.H}}{\text{.S}}{\text{.}}} \right) \cr} $$
(Compare the coefficients of $$\sqrt 7 $$ and constant term)
$$\eqalign{ & {\text{a}} = - \frac{4}{3} \cr & b = \frac{{11}}{3} \cr} $$