# Is $\sqrt{x^2}$ always $\pm x?$

I am wondering if this holds in every single case: $$\sqrt{x^2} = \pm x$$

Specifically in this case: $$\sqrt{\left(\frac{1}{4}\right)^2}$$ In this one we know that the number is positive before squaring, so after removing the square and root shouldn't we just have: $$\frac{1}{4}$$

Also in a case such as this: $$\sqrt{(\sqrt{576})-8}$$ we have $$\sqrt{\pm24-8}$$ which is either $\sqrt{16}$ or $\sqrt{-32}$ but since we care about real, principal roots only, can't we say that $\sqrt{16}$ is actually $4$ but not $-4.$

Am I mistaken somewhere in my reasoning?

Thanks!

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Did you mean $\sqrt{x^2}$ in the title and first equation? –  Daniel Fischer Sep 7 '13 at 17:56
$${\sqrt{x^2}=\color{red}{{\boldsymbol|x\boldsymbol|}}}\text{ which is always }\geqslant0$$ –  user93957 Jan 31 at 13:40

The principal square root of $x$ denoted by $\sqrt x$ is defined to be the non-negative square root of $x$. Thus $\sqrt{25}=5$, $\sqrt{3^2}=3$ and $\sqrt{(-3)^2}=\sqrt 9=3$.

Note that, $\sqrt {x^2}$ can be $x$ or $-x$ depending on whether $x$ is positive or negative or zero. For example, $\sqrt{7^2}=7$ and $\sqrt{(-7)^2}=7$. It should be clear that when $x\geq0$, $\sqrt {x^2}=x$, but when $x<0$, $\sqrt {x^2}=-x$. This can be briefly written as $\sqrt{x^2}=|x|$.

When you're talking about finding all the square roots of a positive number, then indeed there are two solutions. In other words, you're looking to solve $x^2=a$ for some positive constant $a$. Then $x$ can either equal $a$ or $-a$. This is usually written as $x=\pm a$. Note that for $a=0$, there is only one solution and for $a<0$ there are none in $\mathbb R$.

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Thank you for your clarification, would it be then ok to say then that in the second case when we have $\sqrt{16}$ that it is indeed $\pm$4, whereas in first case it can only be $\frac{1}{4}$ ? –  mzm Sep 7 '13 at 18:21
@mzm In your second case, $\sqrt{16}=4$. The expression $\sqrt x$ denotes the non-negative number which when squared gives $x$. As it turns out, for every $x\geq0$, there is exactly one such number. For $16$, it is $4$. But the solutions to $x^2=16$ are indeed $4$ and $-4$. By "the solutions to $x^2=16$", I mean the set of all $x$ that satisfies the given equation. –  Alraxite Sep 7 '13 at 18:32
@mzm Again, the reason $\sqrt {16}$ equals $4$ is that we have defined the square root function to always return the non-negative root. If you want to refer to the negative square root of $x$, then you may simply write $-\sqrt{x}$. –  Alraxite Sep 7 '13 at 18:36
Thank you, that's very clear. –  mzm Sep 7 '13 at 18:50

I assume that in your question, you meant to ask about $\sqrt{x^2}$ rather than $\sqrt{x}$. My answer is based on that assumption.

It is inaccurate to say (in the case that you are interested only in real, principal roots) that $\sqrt{x^2}=\pm x$; the correct thing to say is that $\sqrt{x^2}=\lvert x\rvert$.

This is a common confusion; I think that it usually results from the fact that if you have $\sqrt{x^2}=y$, then you get $x^2=y^2$... and this has two different solutions: $x=y$ and $x=-y$.

That is not the same thing as saying $\sqrt{(-3)^2}=\pm 3$; clearly, if you care about principal real roots only, then $\sqrt{(-3)^2}=\sqrt{9}=3$. Note, however, that $3=\lvert-3\rvert$, though; this is why you cannot simply say that $\sqrt{x^2}=x$.

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It is accurate, if you take it to mean: Is $\sqrt{x^2}$ always equal to one of $+x$ and $-x$? (And I don't see what else you could take it to mean.) –  TonyK Sep 7 '13 at 18:07
@TonyK In that sense, yes. After all $|x|=x$ or $|x|=-x$. The "common confusion" is specifically about the meaning of $\pm$ (is it an "and" or an "or"? May we pick?) –  Hagen von Eitzen Sep 7 '13 at 18:10
Yes, I see now from the OP's edit that there is some real confusion! –  TonyK Sep 7 '13 at 18:12
Yes, I meant $\sqrt{x^2}$ sorry about the confusion. –  mzm Sep 7 '13 at 18:18
@mzm: No, I meant your real confusion with $\sqrt{576} = \pm 24$. –  TonyK Sep 7 '13 at 18:21

As a matter of fact, the notation $\sqrt a$, where $a \in \mathbb R$ is non-negative, is defined to be the unique non-negative $b \in \mathbb R$ such that $b^2 = a$.

So $\sqrt {x^2}$ is either $+x$ (if $x \ge 0$) or $-x$ (if $x < 0$).

With this clarification, you can see that your equation

$$\sqrt{\sqrt{576}-8} = \sqrt{\pm24 -8}$$

is wrong (if it means anything at all). It should be simply:

$$\sqrt{\sqrt{576}-8} = \sqrt{24 -8} = 4$$

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