For what valsue (s) of k the expression $$p+\frac{1}{4}\sqrt{p}+k^{2}$$ is a perfect square ?
$$p+\frac{1}{4}\sqrt{p}+k^{2}$$
= $$(\sqrt{p})^{2} + 2 * \sqrt{p} * \frac{1}{8} + (\frac{1}{8})^{2} - (\frac{1}{8})^{2} + k^{2}$$
=> $$k^{2} = (\frac{1}{8})^{2}$$
=> $$k = \pm\frac{1}{8}$$
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