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This question already has an answer here:

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.

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marked as duplicate by 6005, Parcly Taxel, JonMark Perry, Dylan, user91500 Oct 9 '16 at 7:30

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

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  • $\begingroup$ 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. $\endgroup$ – JuliusL33t Jun 19 '14 at 20:51
  • $\begingroup$ 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? $\endgroup$ – JuliusL33t Jun 19 '14 at 20:55
  • $\begingroup$ 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 $2\pi$. $\endgroup$ – John Machacek Jun 19 '14 at 20:59
  • $\begingroup$ 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. $\endgroup$ – John Machacek Jun 19 '14 at 21:03
  • $\begingroup$ Awesome! Thanks! :) $\endgroup$ – JuliusL33t Jun 19 '14 at 21:04
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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 $\color{white}{\overline{\color{black}{\rm\sqrt{(x)^2}=\sqrt{(-x)^2}}}}.$

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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. $x = \pm 3$.

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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.

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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, $\sqrt{x^2} = |x|$.

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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 $x\in \{-3,3\}.$

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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 $$1 < x^{\frac 12} < x$$

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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.

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