Recently I was watching this interview of Andrew Wiles where a secondary school teacher asked this question:

How do you teach square roots?

He doesn't answer the question so I'd like to ask it here?

Here is another example concerning the widespread confusion about this question; in this clip from University Challenge, he asks "What are the two possible answers to the calculation 5 plus the square root of 4?"

  • 5
    $\begingroup$ Welcome to the site. I followed the link and the video has a more detailed question than what you asked her. I suggest you add the complete question to your post so that not everyone has to watch the video clip before answering. Links can break and that is another reason to have the complete details of the question here. $\endgroup$
    – Amy B
    Commented Jan 28, 2018 at 17:17
  • 4
    $\begingroup$ The question of how we teach square roots is very vague and could cover deriving square roots, square roots as exponents, square roots of negative numbers etc. If you watch the video it becomes clear that we are talking about the sign of the square roots of positive numbers. Whether the sq. root of a positive number is always positive and only pos/neg when it's the root of an equation. Or is the square root always negative. I am suggesting that you make your question clearer and not rely on a video link. $\endgroup$
    – Amy B
    Commented Jan 28, 2018 at 20:15
  • 1
    $\begingroup$ I'd disagree that one "chooses and sticks with it". Rather, discussion of square roots (already of real numbers) introduces some pervasive problems of ambiguity, which cannot truly be "conventionalized away". This "problem" deserves attention, rather than a fake (non-) solution by fiat. $\endgroup$ Commented Jan 30, 2018 at 1:26
  • 1
    $\begingroup$ @paul garrett Normally I find any (slightly-)contrarian viewpoints you hold helpful, but I'm afraid I don't see what you're advocating or warning against here. I'd be interested to learn, though of course it may instead be an answer to "how should you teach square roots?" $\endgroup$
    – pjs36
    Commented Feb 3, 2018 at 18:40
  • 3
    $\begingroup$ @pjs36, basically, I recommend not attempting to fix a convention... about whether $\sqrt{x}$ is always non-negative, or whatever. There are two square roots of non-zero real (or complex) numbers, as we know. For reals, sure, one could say "positive square root", but, really, there's no natural compulsion to take the positive one rather than the negative, whether or not some people are fond of that. I think that such details have to be made clear in given contexts, "forever", rather than pretending that "it's settled". Things like that. $\endgroup$ Commented Feb 3, 2018 at 20:20

3 Answers 3


As he says, this is a convention, and you just have to make a choice and stick with it. The usual convention (which makes the square root a function) is to take the positive branch.

The teacher seems a bit confused. He seems to think that there is a contradiction between saying $\sqrt{9} = 3$ on one hand, and on the other hand writing $x^2=9 \implies x=\pm3$ on the other.

If we stick to the convention that the square root takes only positive values, then there are two sensible ways to approach the quadratic equation.

One is to recognize that $\sqrt{x^2} = |x|$, not $x$. So $x^2 =9 \implies |x|=3 \implies x = \pm 3$.

The other way is to factor $x^2=9 \implies x^2-9=0 \implies (x-3)(x+3) = 0 \implies x = \pm3$.

In either case, the $\pm$ does not need to be part of the definition of the square root, and including it makes life more complicated because then square root is not a (single valued) function.


As mentioned in the comments, it is worth making the distinction between square roots and principle square roots. We can define $y$ to be a square root of $x$ if and only if $y^2=x$. Then every positive number $x$ has two square roots, $0$ has only $0$ as its square root, and negative numbers have no real square roots.

The relation $y^2=x$ is not a function of $x$, but if we throw out all the pairs with negative $y$ values we do obtain a function, called the "principle square roots function". This takes a nonnegative real number $x$ as input and returns its nonnegative square root as output.

Another observation: this choice is nice for many reasons, not the least of which is that it is consistent with the definition $x^a = e^{a \log(x)}$ which can be used to define any positive power function (assuming the exponential and logarithm have been defined).

This might help to clarify my remarks above the edit.

  • 9
    $\begingroup$ All of my elementary and college algebra books make a distinction between square root (i.e., solutions of $x^2 = a$) and (real) principal square root (i.e. $\sqrt{a}$). Perhaps this answer could be clarified by making use of that terminology? $\endgroup$ Commented Jan 28, 2018 at 17:09
  • 1
    $\begingroup$ @DanielR.Collins Sure. That is a good point. $\endgroup$ Commented Jan 29, 2018 at 1:16
  • 1
    $\begingroup$ @StevenGubkin: please be careful using the "±" sign: you mean by "±3" "plus or minus three", but people know this sign as meaning "more or less 3" (which includes 2.9, 3.05, ...). $\endgroup$
    – Dominique
    Commented Jan 29, 2018 at 15:58
  • 9
    $\begingroup$ @Dominique Really? I have never seen it used in this way. $\endgroup$ Commented Jan 29, 2018 at 16:58
  • 2
    $\begingroup$ I wonder, how would life be different if we didn't insist to work with functions. What if we dealt more with multiply-valued things as basic? What is gained, what is lost? $\endgroup$ Commented Feb 3, 2018 at 16:01

Original Answer

I first define that a square root of a number $a$ is a number $x$ such that $x^2 = a$. I then give some examples like: What are the square roots of 4? $2$ and $-2$ because $2^2=4$ and $(-2)^2=4$. I then state that positive real numbers have two square roots, one positive and one negative.

So what does $\sqrt{a}$ mean? There are two choices, right?

Define that the symbol $\sqrt{a}$ always denotes the positive square root of $a$.

Then I given a couple of examples: The square roots of 4 are $\sqrt{4}=2$ and $-\sqrt{4}=-2$. The square roots of 2 are $\sqrt{2}$ and $-\sqrt{2}$.

$n$-th roots are defined in a similar way, as solutions of $x^n=a$. Example: $-2$ is cube root of $-8$ because $(-2)^3=-8$ To entice students, I like to mention that there are actually three cube roots of $-8$: $-2$ and two complex roots (which they will see in their future)

So what does $\sqrt[3]{a}$ mean? There are three choices?

Answer: The symbol $\sqrt[3]{a}$ always denotes the real $3$-rd root of $a$.

Example: $\sqrt[3]{-8} = -2$

The symbol $\sqrt[n]{a}$ denotes the real $n$-th root of $a$, unless there are two real $n$-th roots, in which case pick the positive $n$-th root. Alternatively, break it up according to whether $n$ is odd or even.

Added in response to a question in the comments:

The formulas $(\sqrt{x})^2 = x$ and $\sqrt{x^2} = |x|$ and often confuse students. Here's how I approach it.

Question to students: $(\sqrt{x})^2$ = ?

By definition, $\sqrt{x}$ is the non-negative number such that $(\sqrt{x})^2 = x$.

So $(\sqrt{x})^2 = x$.

Example: $(\sqrt{5})^2 = 5$

Question to students: $\sqrt{x^2}$ = ?

Some students will say $\sqrt{x^2}$ = x$.

Let's test this formula.

Example: $\sqrt{2^2} = \sqrt{4} = 2$. That works.

Example: $\sqrt{(-2)^2}$ = ?. Since $(-2)^2 = 4$, we have $\sqrt{(-2)^2} = \sqrt{4} = 2$. But the suggested formula $\sqrt{x^2} = x$ says that $\sqrt{(-2)^2} = -2$. The formula doesn't work.

Some student will probably now suggest $\sqrt{x^2} = |x|$. If not, you should suggest it.

Now I explain why it works.

By definition, $\sqrt{x^2}$ is the non-negative number such that $(\sqrt{x^2})^2 = x^2$.

So either $\sqrt{x^2} = x$ (because $x^2 = x^2$) or $\sqrt{x^2} = -x$ (because $(-x)^2 = x^2$). How to choose? By definition, pick the non-negative one. If $x \geq 0$, then $\sqrt{x^2} = x$. If $x < 0$, then $\sqrt{x^2} = -x$.

Let's write this as a piecewise function:

$$ \sqrt{x^2} = \left\{ \begin{array}{ll} x & \text{if } x \geq 0\\ -x & \text{if } x < 0\\ \end{array} \right. $$

Does this function seems familiar? This exactly the formula for $|x|$. Therefore $$ \sqrt{x^2} = |x|. $$

NOTE: I am assuming here I have introduced the piecewise formula for the absolute value $|x|$.

  • $\begingroup$ Thank you for your answer. Do you use the approach that taking a square root "undoes" squaring? $\endgroup$
    – user 85795
    Commented Feb 17, 2018 at 14:12
  • 1
    $\begingroup$ @skullpatrol I added something to my answer to explain how I approach the ``undoing'' $\endgroup$
    – LucasSilva
    Commented Feb 17, 2018 at 20:26

The answer in the video was a bit unsatisfying for you, and I understand why. You expect a firm answer.

IRL, when I get this question, I need to quickly determine the level the student is at in her studies. When square roots (or nth roots for that matter) are first discussed, we are only concerned about the positive value. After all, my square with an area of 4 square units has a side that’s a positive number.

But then the students move on the the square root as part of the quadratic equation solving, and we must be mindful of both positive and negative roots.

At some point, the 4th root of 16 has 4 solutions, as the +/-2i is also a root.

It’s for the adult math teacher to understand what level the class is at, and whether that level contains the negative or imaginary root as a solution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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