Answer
$\sqrt{-9}$ is not a real number because there is no real number that when squared gives $-9$.
Work Step by Step
Note that:
$\sqrt{9} = 3$ because $3^2=9$
$\sqrt{16}=4$ because $4^2=16$
The given radical is NOT A REAL NUMBER because there is no real number that when squared gives $-9$.