For what value(s) of k will the expression $ p + \frac{1}{9} \sqrt{p} + k^2$ be a perfecr square ?

$k = ±\frac{1}{8}$ $k = ±\frac{1}{9}$ $k = ±\frac{1}{21}$ $k = ±\frac{1}{18}$

$k = ±\frac{1}{18}$

$ p + \frac{1}{9} \sqrt{p} + k^2$ ( a + b )2 = a2 + b2 + 2ab For $ p + \frac{1}{9} \sqrt{p} + k^2$ a perfect square = (\(\sqrt {p}\) + \(\frac{1}{18}\))2 So, the value of k $k = ±\frac{1}{18}$