Answer
The solution set is incomplete because the negative square root was not considered.
Work Step by Step
By the Square Root Principle, $x^2=c$ ($c$ is a positive constant) implies $x=\pm\sqrt{c}.$ In the same way, $x^2=16$ implies
\begin{align*}
x&=\pm\sqrt{16}
\\&=
\pm4
.\end{align*}
Thus, the solution set should have been $\{-4,4\}$. Hence, the solution set, $\{4\}$, of the student is incomplete because the negative square root was not considered.
