Answer
$(\sqrt -9)^{2} = -9$
Work Step by Step
$(\sqrt -9)^{2} = \sqrt -9 \times \sqrt -9 \ne \sqrt 81$
The product rule for radicals applies to real numbers only. Square root of any negative number is a complex number. So,
$\sqrt -9 \times \sqrt -9 = i\sqrt 9 \times i\sqrt 9$
$ =3i \times 3i$
$= 9i^{2}$
$= 9(-1)$
$=-9$
