For what value(s) of k will the expression $ p + \frac{1}{9} \sqrt{p} + k^2$ be a perfecr square ?
Answer & explanation
Correct answer: option 4
$ 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 $k = ±\frac{1}{18}$