# Square roots — positive and negative

It is perhaps a bit embarrassing that while doing higher-level math, I have forgot some more fundamental concepts. I would like to ask whether the square root of a number includes both the positive and the negative square roots.

I know that for an equation $x^2 = 9$, the solution is $x = \pm 3$. But if simply given $\sqrt{9}$, does one assume that to mean only the positive root? And when simply talking about the square root of a number in general, would one be referring to both roots or just the positive one, when neither is specified?

• I would say this can answer your question. See also this. – user8101 Mar 11 '11 at 9:49
• These are often confused because students believe that $\sqrt{x^2}=x$, but actually $\sqrt{x^2}=|x|$. So: $\sqrt{x^2}=\sqrt{9}$ implies $|x|=3$, and so there are two possibilites: $x=3$ or $x=-3$. – rschwieb May 21 '13 at 21:06

If you want your square-root function $\sqrt x$ to be a function, then it needs to have the properties of a function, in particular that for each element of the domain the function gives a single value from the codomain. If you take a function to be a set of ordered pairs, then each of the initial values of the pairs must appear exactly once.

So to be a function, square-root needs to be single valued; the multi-valued version is really a relation, at which point you might get into issues of principal values.

For convenience, the square root of non-negative real numbers is usually taken to be the non-negative real value, but there is nothing other than practicality to stop you from taking some other pattern. Such arbitrary choices can raise significant issues when considering, for example, cube-root functions defined on the real and complex numbers.

For positive real $x$, $\sqrt x$ denotes the positive square root of $x$, by definition. Wikipedia agrees with me on this.

Consider the similar equation $x^4 = 9$. This one has four solutions. The most obvious solution is $\sqrt 3$, which is roughly 1.732. Another solution is $-\sqrt 3 \approx - 1.732$. The same as the previous solution, just multiplied by $-1$.

There are two other solutions: $x = \sqrt{-3}$ and $x = -\sqrt{-3}$, which are roughly 1.732, the former multiplied by $i$ and the latter multiplied by $-i$. These are all the same distance away from 0.

If you just need one solution, you might as well take the solution that is a positive real number. If you need the other solutions, just multiply the solution you have by units other than 1.

So to solve $x^2 = 9$, the solution $x = 3$ might be enough, but if you need the other solution you just multiply the previous solution by the unit $-1$.