For what value (s) of a is $$x + \frac{1}{4}\sqrt x + {a^2}$$    a perfect square?

A. $$ \pm \frac{1}{{18}}$$

B. $$\frac{1}{8}$$

C. $$ - \frac{1}{5}$$

D. $$\frac{1}{4}$$

Answer: Option B

Solution(By Examveda Team)

$$\eqalign{ & x + \frac{1}{4}\sqrt x + {a^2} \cr & = {\left( {\sqrt x } \right)^2} + 2 \times \frac{1}{8} \times \sqrt x + {a^2} \cr & \left[ {\left( {{{\text{A}}^2} + {\text{2AB}} + {{\text{B}}^2}} \right) = {{\left( {{\text{A}} + {\text{B}}} \right)}^2}} \right] \cr & {\text{Here, A}} = \sqrt x {\text{ and }} \cr & {\text{B}} = a \cr & {\text{B}} = \frac{1}{8} \cr & \therefore a = \frac{1}{8} \cr} $$