I will give my answer with which some will take issue (i.e.@lookagain).
I worked with and taught foundation of mathematics for both in mathematics & philosophy departments.
So my take: \(\displaystyle \pm 2\) is a symbol, a short hand notation, for "positive two or negative two" \(\displaystyle 2 \vee -2\).
So the short cut \(\displaystyle \sqrt{16}=\pm 4\). But in my advanced courses I would mark that incorrect, zero credit.
\(\displaystyle \sqrt{9}\) is a number,
one number not two numbers. The square roots of \(\displaystyle 9\) are \(\displaystyle \pm 3\) .
Note how the verb is plural because there are
two of them.
But \(\displaystyle \large\sqrt 9=3\) and \(\displaystyle \large\sqrt 9\ne -3\).
To an answer for your question.
1) I taught that the radical for even roots applied only to non-negative real numbers and gives one number.
Thus \(\displaystyle \sqrt 9=3\) and \(\displaystyle \sqrt[3]{-27}=-3\)
2) There are two square roots of \(\displaystyle -49\) which are \(\displaystyle \pm 7\bf{i}\) but neither is \(\displaystyle \sqrt {-49}\) because that would be a notation error.
3) There are in the complex field there are \(\displaystyle n\text{ nth roots of any number}\)
So what are the cube roots of \(\displaystyle -27~?\). Well
SEE HERE. There are three.
4) There are six sixth roots of \(\displaystyle -64\)
See Here Note that \(\displaystyle 2\bf{i}\) is one of the roots as well as \(\displaystyle -2\bf{i}\).
But it would be a mistake to use the notation \(\displaystyle \sqrt[6]{-64}=\pm 2\bf{i}\).
Do you see why?