Why is the range of radical functions all non-negative real numbers?

Question

The way I see it, the square root of a number is +/-. For example, the square root of 9 is +3 or -3 since squaring either number gives 9. So why is the range of radical functions restricted to positive numbers or 0, only?

E.g. $y = \sqrt{x-2}$

• Domain is clearly = 2 or larger.

• But for range-- say x was 3, we'd then get $ y = \sqrt{3-2} = \sqrt1 = \pm 1.$ Every possible number we can put for x results in either 0 or a +/- number. Thus, if anyone could explain why our range is only non-negative numbers (as opposed to all real numbers), I would greatly appreciate it. Surprisingly I haven't found any online source that explains it.

Answer 1

The definition of the square root function is that for a non-negative real number $x$, $\sqrt x$ is the unique non-negative number $y$ with $y^2=x$.

This $\sqrt 9=3$. There cannot be more values because a function by definition only "returns" one value per input.

However, the solutions of the equation $x^2=9$ are $y=\pm\sqrt 9=\pm3$, which is a shorthand notation for "$y=3$ or $y=-3$".

Answer 2

The key word is function. If you're indeed talking about the square root function, then it should return exactly one value when applied to a given number.