If 4x = √5 + 2, then the value of $$\left( {x - \frac{1}{{16x}}} \right)$$  is

Solution (By Examveda Team)

$$\eqalign{ & {\text{Given,}}\,4x = \sqrt 5 + 2 \cr & \Rightarrow 16x = 4\left( {\sqrt 5 + 2} \right) \cr & \Rightarrow 16x = 4\sqrt 5 + 8 \cr & \therefore \frac{1}{{16x}} = \frac{1}{{4\sqrt 5 + 8}} \cr & \Rightarrow \frac{1}{{16x}} = \frac{{4\sqrt 5 - 8}}{{\left( {4\sqrt 5 + 8} \right)\left( {4\sqrt 5 - 8} \right)}} \cr & \left[ {{\text{Rationalising the denominator}}} \right] \cr & \Rightarrow \frac{1}{{16x}} = \frac{{4\sqrt 5 - 8}}{{80 - 64}} \cr & \Rightarrow \frac{1}{{16x}} = \frac{{4\sqrt 5 - 8}}{{16}} \cr & \Rightarrow \frac{1}{{16x}} = \frac{{4\left( {\sqrt 5 - 2} \right)}}{{16}} \cr & \Rightarrow \frac{1}{{16x}} = \frac{{\sqrt 5 - 2}}{4} \cr & \therefore \left( {x - \frac{1}{{16x}}} \right) \cr & = \frac{{\sqrt 5 + 2}}{4} - \frac{{\sqrt 5 - 2}}{4} \cr & = \frac{{\sqrt 5 + 2 - \sqrt 5 + 2}}{4} \cr & = \frac{4}{4} \cr & = 1 \cr} $$