If $$2x + \frac{1}{{4x}} = 1{\text{,}}$$   then the value of $${x^2} + \frac{1}{{64{x^2}}}$$   is?

$$\eqalign{ & 2x + \frac{1}{{4x}} = 1 \cr & {\text{Dividing by 2 both side}} \cr & \Rightarrow x + \frac{1}{{8x}} = \frac{1}{2} \cr & {\text{Squaring both side }} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} + 2 \times x \times \frac{1}{{8x}} = \frac{1}{4} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} + \frac{1}{4} = \frac{1}{4} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} = \frac{1}{4} - \frac{1}{4} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} = 0 \cr} $$