# Soft question about the square root [duplicate]

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I got to thinking about the square root the other day, and there's this thing that bugs me in the back of my mind. As far as I know, $\sqrt{4}$ is unambiguously $2$, and nothing else, as the square root of a number is defined as the positive root of that number. Yet, when solving algebraic equations, people (myself included) seem to follow this logic:

Solve: $x^2 = 9$

Solution: $\sqrt{x^2} = \sqrt{9} \Rightarrow x = \pm3$

All of a sudden, people love the minus sign! But this is obviously notationally incorrect, even though $x$ really is $\pm 3$. For myself, I made a deal with myself a long time ago: square roots of numbers are always positive, square roots of unknowns always have 2 roots (at least in $\mathbb C$ (counted with multiplicity)).

This itch really needs to be scratched, driving me crazy! :) Thanks in advance.

## marked as duplicate by 6005, Parcly Taxel, JonMark Perry, Dylan, user91500Oct 9 '16 at 7:30

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• I am not able to see a question in your post. What is your question? – MJD Jun 19 '14 at 20:16

## 8 Answers

Consider the function $f(x) = \sqrt{x}$. This is a function that gives the positive square root of a number. As you have noted a positive real number has two square roots; one is positive and the other is negative. It is by convention the let $\sqrt{x}$ denote the positive square root. If I want the negative square root I explicitly write $g(x) = -\sqrt{x}$. Note $f(x)$ and $g(x)$ are respectively the top and bottom half of the parabola given by $y^2 = x$. Note we cannot achieve this parabola by a single function of $x$ (as I would teach my college algebra students this parabola fails the vertical line test). Thus is summary $\sqrt{x}$ can only denote the positive square root because we want $\sqrt{x}$ to be a well-defined function. Otherwise $\sqrt{x}$ is a multivalued function. For those you you who have been exposed to complex analysis you know that $\log(z)$ is a multivalued function and we often pick a branch to work with. Something similar is going on here.

Now consider solving equations. When solving equations we typically want all solutions. Thus we will say $x = \pm \sqrt{9} = \pm 3$ when solving $x^2 = 9$. Here we are showing that there are two solutions. If you want to leave $\sqrt{x}$ as a multivalued function it would not be necessary to write it this way with the $\pm$. In most cases people like a "principle" (positive) square root so that $\sqrt{x}$ is really a function. This makes it necessary to explictly denote $-\sqrt{x}$ when we want it.

• THIS!!! Never thought about thinking thinking about it this way. If I reflect back on my thin understanding of complex analysis, I want to remember that for for $z= re^i\theta$, $log(z)$ is not uniquely defined, and so we pick the principal branch because $log(z)$ cant possibly be welldefined, as it fails to be injective, it repeats for all $2i\pi n$ where n is an integer. Now I want to think about $f(x) = \sqrt{x}$ like we're choosing the a branch of "square root", is this pretty close? Feel free to use mature mathematical language, I'm more knowledgeable than the question lets on. – JuliusL33t Jun 19 '14 at 20:51
• PS: the positivity of the square root is just convention, defining $\sqrt{x}$ as positive is just convention, math would work just as fine with the negative definition, it would just be very cumbersome! It this true? – JuliusL33t Jun 19 '14 at 20:55
• Right, $\log(re^{i\theta}) = \log(r) + i\theta'$ for $\theta' = \theta + 2i\pi n$ for any integer $n$. Where on the right side we use the regular real logarithm. Then we make it well-defined by picking a branch. For example pick the unique $\theta'$ so that $0 \leq \theta' < 2\pi$. Or we could just as well pick $-\pi < \theta' \leq \pi$ or any other half open interval of length $2\pi$. – John Machacek Jun 19 '14 at 20:59
• Yes, you could by convention let $\sqrt{x}$ denote the negative square root. Then you would explicitly right $+\sqrt{x}$ for the positive square root. Like you say math would still work, but this convention would be odd having to put in the $+$ sign. But I think you got it. The main thing is $\sqrt{x}$ can mean one of two things, we have to pick which one we want it to mean. – John Machacek Jun 19 '14 at 21:03
• Awesome! Thanks! :) – JuliusL33t Jun 19 '14 at 21:04

How about this? \eqalign{\rm x^2=9&\iff\rm x^2-9=0 \\&\rm\iff x^2-3^2=0\\&\rm\iff (x-3)(x+3)=0\\&\rm\iff x=3\color{grey}{\text{ or }}x=-3\\&\rm\iff x=\pm3.} You got a negative solution because if you square it, you will get a positive number, whose square root is also a positive number. It's all because $\color{white}{\overline{\color{black}{\rm\sqrt{(x)^2}=\sqrt{(-x)^2}}}}.$

If $x^2 = 9$ then $\sqrt{x^2} = \sqrt{9}$. What you seem to be missing is that $\sqrt{x^2}$ is not $x$ but $|x|$ so the equation simplifies to $|x| = 3$. The only two values of $x$ which have absolute value $3$ are $x = 3$ and $x = -3$, i.e. $x = \pm 3$.

I'm not quite sure, but I suppose that $\sqrt{x^2}=|x|$, so when you solve for |x|, the solution really is unique, but for x there are two solutions.

Square roots of real numbers are always positive. However, when solving an equation, you're not actually "taking the square root of both sides of the equation". When you say $x^2 = 9 \implies x = \pm 3$, the steps behind it are: \begin{align} x^2 &= 9 \\ x^2 - 9 &= 0 \\ (x-3)(x+3) &= 0 \\ \implies x = 3 \quad \mbox{or} \quad x &= -3 \end{align} Do not have doubts, $\sqrt{9} = 3$, always. In general, $\sqrt{x^2} = |x|$.

Your mistake is assuming that the inverse operation of the function $x^2$ is $\sqrt{x}$, when, in fact, it is $\pm \sqrt{x}$.

e.g. if we're solving $x^2=9$, the operation that undoes the $x^2$ is $\pm \sqrt{...}$, so we get: $$\pm \sqrt{x^2}=\pm9 \iff \pm x= \pm 9,$$ the only solutions to which are $x\in \{-3,3\}.$

Here's one way to think about it. Square root is really something to the $\frac 12$ power. So $\sqrt x = x^{\frac 12}$.

Exponents are all about multiplication. So if $x>1$ then we can say:$$x^a > x^b \implies a>b$$ With that in mind, $$x^0 < x^{\frac 12} < x^1$$ So $$1 < x^{\frac 12} < x$$

The problem is that $\sqrt{x^2}$ is not equal to $x$. Rather, $\sqrt{x^2} = |x|$. So the solution to the problem is: $$x^2 = 9$$ $$\sqrt{x^2} = \sqrt{9}$$ $$|x| = 3$$ $$x = \pm 3$$

This error (of thinking that $\sqrt{x^2} = x$) comes up a lot, in a lot of contexts. See my answer to this question.