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?

  • 1
    $\begingroup$ I would say this can answer your question. See also this. $\endgroup$ – user8101 Mar 11 '11 at 9:49
  • 4
    $\begingroup$ 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 $x=-3$. $\endgroup$ – 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 $-1$.


The positive value is taken because it's more valuable in general, since the domains of elementary functions which aren't defined on the entire real line are often the positive or non-negative numbers. For instance, taking the positive value lets you say $\sqrt{\sqrt{x}} = \sqrt[4]{x}$ or $\ln{\sqrt{x}} = \frac{1}{2} \ln{x}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.