If $$x = \sqrt 5 + 2{\text{,}}$$   then the value of $$\frac{{2{x^2} - 3x - 2}}{{3{x^2} - 4x - 3}}$$   is equal to?

Solution (By Examveda Team)

$$x = \sqrt 5 + 2$$
Rationalization of denominator.
$$\eqalign{ & \Rightarrow \frac{1}{x} = \frac{1}{{\sqrt 5 + 2}} \times \frac{{\sqrt 5 - 2}}{{\sqrt 5 - 2}} \cr & \Rightarrow \frac{1}{x} = \frac{{\sqrt 5 - 2}}{{5 - 4}} \cr & \Rightarrow \frac{1}{x} = \sqrt 5 - 2 \cr & \Rightarrow x - \frac{1}{x} = \sqrt 5 + 2 - \sqrt 5 + 2 \cr & \Rightarrow x - \frac{1}{x} = 4 \cr & \therefore \frac{{2{x^2} - 3x - 2}}{{3{x^2} - 4x - 3}} \cr} $$
Multiply by $$\frac{1}{x}$$ in numerator and denominator
$$\eqalign{ & = \frac{{\frac{{2{x^2}}}{x} - \frac{{3x}}{x} - \frac{2}{x}}}{{\frac{{3{x^2}}}{x} - \frac{{4x}}{x} - \frac{3}{x}}} \cr & = \frac{{2x - \frac{2}{x} - 3}}{{3x - \frac{3}{x} - 4}} \cr & = \frac{{2\left( {x - \frac{1}{x}} \right) - 3}}{{3\left( {x - \frac{1}{x}} \right) - 4}} \cr & = \frac{{2 \times 4 - 3}}{{3 \times 4 - 4}} \cr & = \frac{{8 - 3}}{{12 - 4}} \cr & = \frac{5}{8} \cr & = 0.625 \cr} $$