For what value of k the expression $$p + \frac{1}{4} + \sqrt p + {k^2}$$     is perfect square?

Solution (By Examveda Team)

$$\eqalign{ & p + \frac{1}{4} + \sqrt p + {k^2} \cr & = p + \sqrt p + \left( {{k^2} + \frac{1}{4}} \right) \cr & = {\left( {\sqrt p } \right)^2} + 2 \times \frac{1}{2} \times \sqrt p + \left( {{k^2} + \frac{1}{4}} \right) \cr & = {{\text{A}}^2} + {\text{2}} \times {\text{A}} \times {\text{B}} + {{\text{B}}^2} \cr & {\text{A}} = \sqrt p \cr & {{\text{B}}^2} = \left( {{k^2} + \frac{1}{4}} \right) \cr & {\text{B}} = \frac{1}{2} \cr & \therefore {k^2} + \frac{1}{4} = {\left( {\frac{1}{2}} \right)^2} \cr & \Rightarrow {k^2} + \frac{1}{4} = \frac{1}{4} \cr & \Rightarrow {k^2} = 0 \cr & \Rightarrow k = 0 \cr} $$