For what value of k the expression $$p + \frac{1}{4} + \sqrt p + {k^2}$$ is perfect square?
A. 0
B. $$ \pm \frac{1}{4}$$
C. $$ \pm \frac{1}{8}$$
D. $$ \pm \frac{1}{2}$$
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & p + \frac{1}{4} + \sqrt p + {k^2} \cr & = p + \sqrt p + \left( {{k^2} + \frac{1}{4}} \right) \cr & = {\left( {\sqrt p } \right)^2} + 2 \times \frac{1}{2} \times \sqrt p + \left( {{k^2} + \frac{1}{4}} \right) \cr & = {{\text{A}}^2} + {\text{2}} \times {\text{A}} \times {\text{B}} + {{\text{B}}^2} \cr & {\text{A}} = \sqrt p \cr & {{\text{B}}^2} = \left( {{k^2} + \frac{1}{4}} \right) \cr & {\text{B}} = \frac{1}{2} \cr & \therefore {k^2} + \frac{1}{4} = {\left( {\frac{1}{2}} \right)^2} \cr & \Rightarrow {k^2} + \frac{1}{4} = \frac{1}{4} \cr & \Rightarrow {k^2} = 0 \cr & \Rightarrow k = 0 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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