Last year in Pre-Algebra we learned about square roots. I was taught then that $\sqrt{64}=8$ and $\sqrt{100}=10$, which I understood and accepted. I was also taught that $\pm\sqrt{64} = 8,-8$ because both of those numbers squared is 64, which I also get. But this year, with a new school and teacher in a different state, our teacher is telling us that: $\sqrt{64}=8,-8$ and $\pm\sqrt{64}$ also is $8,-8$. The way to get the positive root of something is: $+\sqrt{64}=8$

And these seem to contradict each other. I was always taught that a regular square root returned a positive number and only a positive number, but now my teacher is saying a regular square root gives two numbers, and considering the square root of a number $n$ is defined as $y^2=n$ I see where he is coming from.

Upon researching this Wikipedia says:

For example, $4$ and $−4$ are square roots of $16$ because $4^{2} = (−4)^{2} = 16$

And Wolfram MathWorld says:

Note that any positive real number has two square roots, one positive and one negative. For example, the square roots of $9$ are $-3$ and $+3$

But on the other side, Wolfram Alpha, when given "The square root of 9" gives only 3.

So, which is right? Is $\sqrt{64}$ considered $8$? or is it $8,-8$?

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    $\begingroup$ It’s easy to see how you got confused! The answers below are very good, though. $\endgroup$
    – Lubin
    Commented Sep 24, 2015 at 0:40
  • 9
    $\begingroup$ It's a question of notation. You will find as you go on that not everyone agrees on notation. So: find out the notation of the current teacher and use that notation for assignments in that class. $\endgroup$
    – GEdgar
    Commented Sep 24, 2015 at 1:48
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    $\begingroup$ Based on the first paragraph it seems that both you and your teacher treat the symbol $\pm$ in an unconventional way. You: If $\sqrt{64}=8$, then $\pm\sqrt{64}=\pm 8$ is trivial and the meaning of the symbol $\pm$ is that $\pm 8 = 8, -8$ -- you don't need to think about squares or deep reasons here. Your new teacher: $X$ and $+X$ are the same object, so if $\sqrt{64}$ is multivalued ($8, -8$), then $+\sqrt{64}$ is still multivalued. $\endgroup$
    – JiK
    Commented Sep 24, 2015 at 9:46
  • 2
    $\begingroup$ The square root is a function. As such, it "returns" at most one value. Unless otherwise specified, the positive root of the equation $y^2=x$ is chosen. $\endgroup$
    – user65203
    Commented Sep 24, 2015 at 21:06
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    $\begingroup$ By definition, the square root symbol returns the positive square root, so it can only return one value, not two. $\endgroup$
    – tc216
    Commented Dec 29, 2016 at 19:21

11 Answers 11


Your new teacher is wrong. $\sqrt{\cdot}$ is the principal square root operator. That means it returns only the principal root -- the positive one. $\sqrt{64}=8$. It does NOT equal $-8$.

On the other hand, the equation $64=x^2$ DOES have $2$ solutions: $x=8$ or $x=-8$. Thus both $8$ and $-8$ are square roots of $64$.

Let's see what happens when we take the principal square root of both sides of this equation: $$\begin{align}64 &= x^2 \\ \implies \sqrt{64} &= \sqrt{x^2} \\ \implies 8 &= |x| \\ \implies x&=8 \text{ or } x=-8\end{align}$$

Thus the fact that the principal square root operation throws out the negative root isn't much of a problem as the math still works out correctly.

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    $\begingroup$ I am beginning to notice that not all definitions are universal. Depending on who is "talking": + the natural numbers may or may not include $0$ + 0 may or may not divide 0 + $\sqrt 9$ may or may not include $-3$ + The dihedral group $D_n$ may be of order $n$ or $2n$ $\endgroup$ Commented Sep 24, 2015 at 2:36
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    $\begingroup$ @StevenGregory, correct. And that's a bit unfortunate. The patterns and relationship are what matter, not terms and symbols. But, alas, being human, the terms and symbols can vary. (Though the square root operator is pretty standard.) $\endgroup$ Commented Sep 24, 2015 at 4:36
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    $\begingroup$ @StevenGregory Actually $\sqrt{}$ being the principal square root operator (i.e. only the positive number) is one of the more consistent notations. Almost all mathematicians will agree on this unlike, for example natural numbers including or not including 0 (Coming from set theory 0 is a natural number for me for example). The main reason that $\sqrt{}$ tends to be consistent is that you really want it to be a function and thus have a unique output. $\endgroup$
    – DRF
    Commented Sep 24, 2015 at 8:02
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    $\begingroup$ @DRF In complex analysis, we tend to use $\sqrt{\hphantom{z}\vphantom{z}}$ to denote any branch of the square root, there is no privileged branch. Only when dealing exclusively with square roots of non-negative real numbers there is a strong convention that, unless explicitly stated otherwise, $\sqrt{x}$ denotes the non-negative square root. $\endgroup$ Commented Sep 24, 2015 at 9:42
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    $\begingroup$ @DRF :-) But if contrary to expectation the universe lasts infinitely long, there still may be infinitely many exceptions. $\endgroup$ Commented Sep 24, 2015 at 10:29

Both of your professors are right. It is just an issue of notation.

The first professor defines $\sqrt{x}$ as the non-negative number that when multiplied by itself is equal to $x$, if any.
Your second professor defines $\sqrt{x}$ as the numbers that when multiplied by themselves are equal to $x$, if any.

This means that they are using the same symbol $\sqrt{x}$ to convey different concepts.
It would be better if everyone used the same words and symbols for the same concepts. But in maths as in other issues in life you will find different people using the same word or symbol for different concepts.

Since he is the professor you will have to respect his authority regarding the choice of notation. There is no significative gain between one or another notation but it is very important to choose a notation so that everyone is on the same page. And the one choosing the notation in an academic environment will be the professor. It is unfortunate that different professors of the same institution chose different notations but you will have to live with it.

  • 5
    $\begingroup$ +1 for "it's just an issue of notation", but I'd still say that any notation where $+x$ means something other than $x$, for any $x$, is rather unfortunately chosen. That said, no, OP, it's probably not worth arguing with your teacher about. Just accept that they want to see things done the way they're used to, even if there might be a different, (possibly) better way, and chalk it up as a learning experience. You now know two conventions for the sign of square roots, whereas your teacher knows (or at least insists on using) only one. That's worth something. :) $\endgroup$ Commented Sep 26, 2015 at 12:29
  • $\begingroup$ +1 for different people using different but similar definitions. I recall a math class in college where I lost points because the definition of parallel lines that I'd learned (two lines with equal slope) was different from the professor's definition (two lines that never intersect that have the same slope). $\endgroup$
    – zzzzBov
    Commented Sep 26, 2015 at 17:25
  • $\begingroup$ The problem is that when we write $\sqrt{64} in an expression we mean it to mean one specific value. We can't have a single expression change value willy-nilly or to have two values simulatiously. So the notation really needs to be just one of the two values and common sense says it should be the positive one. $\endgroup$
    – fleablood
    Commented Sep 30, 2015 at 4:45
  • $\begingroup$ I would never use the 2nd professor notation by my own choice. But no, $\sqrt{64} does not have to mean one specific value. It can mean three values, it can mean bananas or apples. You are free to give it whatever meaning you wish as long as you are coherent and consistent. And it is quite possible to do correct, coherent and consistent math using the 2nd professor notation. $\endgroup$ Commented Sep 30, 2015 at 11:25

$\sqrt{\cdot}:[0,\infty)\to [0,\infty)$ is a function that to each $x\ge0$ assigns a $y\ge 0$ such that $y^2=x$. A very different thing is the set of solutions of the equation $x^2=9$, for example. In fact the only reason we have a canonical square root function in $\mathbb{R}$ is because $\mathbb{R}$ is often considered to have a total order $<$ that lets you pick a solution of the equation $x^2=9$. If you were doing only algebra (i.e. no order relation), $\sqrt{\cdot}$ might not be definable.

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    $\begingroup$ This is excellent. The important thing is that in modern mathematics, we don’t use multi-valued functions. The square root function has to have a single value, and we choose the nonnegative one. $\endgroup$
    – Lubin
    Commented Sep 24, 2015 at 0:39
  • $\begingroup$ I'll go out on a limb and disagree with @Lubin here. This question is about mere set-valued (inverse) functions which are a much simpler beast than the premodern notion of multi-valued functions. I've added an answer sketching the justification of the set-valued approach. $\endgroup$ Commented Sep 24, 2015 at 2:26
  • $\begingroup$ So is $\sqrt{-1}$ definable? $\endgroup$
    – Henry
    Commented Sep 24, 2015 at 10:05
  • $\begingroup$ Are you implying the square root operator cannot properly be used on anything other than non-negative reals (e.g. negative reals, complex numbers)? That's a new one on me. $\endgroup$
    – abligh
    Commented Sep 24, 2015 at 17:05
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    $\begingroup$ For @ASCIIAdvocate, I’ll say that you pays your money and makes your choice. I’ll take the simpler route here and insist on all functions being single-valued. $\endgroup$
    – Lubin
    Commented Sep 24, 2015 at 18:16

I had exactly the same problem when I was younger. Eventually, I was taught that if you solve an $\color{blue}{\mathrm{equation}}$ containing an unknown variable say $x$; such as: $$x^2=81$$ Then the equation has solutions given by $x=9$ and $x=-9$.

But if you are just given the $\color{red}{\mathrm{expression}}$: $$\sqrt{81}$$ then the expression can only reduce to $9$ (Not $-9$).

So the number of solutions really simplifies to whether the radical in question belongs to an $\color{blue}{\mathrm{equation}}$ or an $\color{red}{\mathrm{expression}}$; where the latter will only take the principal root.

  • 2
    $\begingroup$ A useful distinction, neatly put $\endgroup$
    – Silverfish
    Commented Sep 24, 2015 at 20:18
  • $\begingroup$ The last paragraph is badly worded. If the radical is originally part of an equation, then only the principle root is taken. If the radical is introduced while solving an equation, then both roots are taken. $\endgroup$ Commented Jun 14, 2020 at 14:40
  • $\begingroup$ But you are correct- if the radical is part of an expression, then only the principle root is taken. $\endgroup$ Commented Jun 14, 2020 at 14:47
  • $\begingroup$ Might I add, when you introduce a square root (or other even root radical) to an equation to solve it, you should always add the $\pm$ symbol to avoid abuse of notation and ambiguity. $\endgroup$ Commented Jun 18, 2020 at 2:18

In many cases we simply write the positive number for the square root. However, if you are writing things properly the square root should have a plus and a negative sign in front of it. The best example that I can think of is the equation for the solution of a quadratic equations. Between the -b and the square root we have plus and minus since the result of a quadratic equation has two solutions.


My daughter's high school algebra teacher is using the same notation, $\sqrt{16}=\pm4$, and she is extremely confused, so I know what you're going through. While I agree a teacher can use whatever notation they wish and define it to mean whatever they want, I am bothered by the fact (and you seem to be, too) that if you define $\sqrt{16}=\pm4$, then that would imply $4=-4$, unless your teacher and I have an entirely different definition of $=$. So, while I won't go so far as to say your teacher's notation is wrong, I will say that it is very poorly chosen. The fact that you are bothered by this use of notation and that you questioned it shows that you are carefully thinking about the ideas as well as the notation used to represent them. Keep up the good work.


That convention, where $\sqrt{u}$ is the set of $y$ with $y^2 = u$, is used sometimes in order to make (more) steps in an algebraic derivation reversible. This runs into the complication that one needs to either allow complex square roots when $u < 0$, or to also make sure the logic can correctly handle the case where the set is empty, but in principle it can be set up in a consistent and well-defined way. In the same way, inverse functions can be handled as set-valued.

Arguably this is the correct way, since there is no algebraically natural way to privilege one square root over the other, or even to be able to tell the twins apart without additional information. The reason for canonizing the positive square root of positive real numbers is that this case comes up most often and it is an easy convention to remember.


Wolfram Alpha recognizes the notion of a principal square root. Wikipedia also explicitly notes that the operator $\sqrt{}$ is the "principal square root function" though it is often only referred to as the "square root function" hence the confusion.


I come down on the side of a single number result. I look at this question in terms of how functions are formally defined: as a relation between a domain (input) and codomain (output). A one-directional mapping between sets.

If the square root function is defined informally as the inverse of the square (this is how Euclid treated it), let's consider the domain and codomain of the square function. Both would be the real numbers.

So then, the inverse function (the square root) would also have a codomain of the real numbers. Meaning that the square root of 9 must be 3. Simply 3. Because "3 and -3" is not a real number, it is a set of real numbers. (And the positive can be preferred over the negative due to the basis of the square root in geometry. Squares in Euclidean geometry do not have sides with a negative measure.)

In order to conclude that the square root of 9 is 3, -3 we must believe that the codomain of the square root function is the set of pairs of real numbers that are negatives of each other. But the inverse of such a square root function would then be a function of pairs of reals to reals - different than what the square function is normally taught to mean. QED


The answers above have been great. Here is a negative example to highlight the meaning:

$$-4 = \sqrt {x}$$ gives no solutions. The answer is NOT $16$.

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    $\begingroup$ So $x=\neg16$? :^) $\endgroup$ Commented Dec 19, 2019 at 1:05
  • $\begingroup$ Isn't this just an alternate formulation of the original question? $\endgroup$
    – wberry
    Commented Dec 21, 2022 at 17:32

Ultimately, it's a matter of notation. There is no right or wrong answer.

Over time, for better or for worse, mathematicians have adopted the radical sign to mean the principal (nonnegative) square root. In reality, true square roots are either positive or negative.

The first thing to understand is that when you are discussing true square roots, $ \sqrt{25} \ne 5 $. In mathematics, having equality means having one and only one distinct solution. The true $ \sqrt{25} $ has two distinct solutions, so there is nothing equal to it.

Consider the following,

$$ x^2 = 100 \Longleftrightarrow x = \sqrt{100} \Longleftrightarrow x = 10 \lor x = -10 $$

The $ x = 10 \lor x = -10 $ at the end is key here. Both $ x = 10 $ as well as $ x = -10 $ satisfy the first equation, so the assertion is true. However, it is untrue to say (in any context) that $ x = 10 \land x = -10$, since it implies that $ 10 = -10 $, which is obviously false.

Simultaneously, $ x = 10 \lor x = -10 $, but $ x \ne 10 $ (and $ x \ne -10 $). That's not a mistake. When combined, either $ x = 10 $ or $ x = -10 $ are true, but neither are true alone. Both statements directly contradict each other, therefore neither contract each other.

In the equation $ \sqrt{25} = \pm{5} $, plus-or-minus is being used to represent a value with two solutions, i.e. $ \pm{5} = 5 \lor \pm{5} = -5 $. $ \pm{5} $ is not a number, but rather, a value with two solutions; $ \pm{5} \in \{-5, 5\} $, nothing more.

Using plus-or-minus on a true square root does nothing, since the square root is already positive or negative; $ \pm{(\pm{n})} \in \{-(\pm{n}), (\pm{n})\} = \{\pm{n}, \pm{n}\} = \{\pm{n}\} \therefore \pm{(\pm{n})} = \pm{n} $. By the same token, negating a square root or making it positive have no effect.

In the quadratic formula,

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, ax^2 + bx + c = 0 $$

The $ \pm $ can be substituted with either $ + $ or $ - $ and it still holds true. The quadratic formula works regardless of whether you believe $ \sqrt{} $ means principal or true square root. Neat, huh?


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