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If $\sqrt{x^2}$ can be simplified as follows:

$\sqrt{x^2} = (x^2)^\frac{1}{2} = x^{\frac{2}{1}\times\frac{1}{2}} =x^\frac{2}{2} = x^1 = x$

Then why would $\sqrt{x^2} = \pm x$?

logsofrhythms
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    By definition, for a real positive number $a$, the symbol $\sqrt{a}$ means the positive quadratic root of $a$. Then $\sqrt{x^2}=|x|$. On the other hand, if you are looking for \textit{all} the real numbers $y$ such that raised to the square give you $x^2$ then you must consider $x$ and $-x$. – Ángel Mario Gallegos Dec 23 '14 at 03:30
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    It is not that $\sqrt{x^2} = \pm x$; if $x\ge 0$, then $\sqrt{x^2} = x$, and if $x\le 0$, then $\sqrt{x^2} = - x$. – peterwhy Dec 23 '14 at 05:20

2 Answers2

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Expanding on Mario G's comment, by definition, $\sqrt{a}$ where $a$ is any non-negative number is defined to be positive. So to the quantity $\sqrt{x^2}$ positive, we assign it the value $|x|$ because we don't know from the outset whether $x$ is positive or negative.

As for the plus-minus sign, that really doesn't occur when taking square roots, but rather when you solve equations. Consider the equation $a^2=x^2$. Now solving for $a$ will give you two solutions, $x$ and $-x$. But solving this equation is not the same thing as taking a square root on both sides, but something a little different. That's why $\sqrt{x^2} \neq\pm{x}$.

sayantankhan
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  • I don't think I agree that when I write a radical symbol on both sides of an equation I'm not taking a square root. – Kevin Carlson Dec 23 '14 at 03:42
  • But my point is, taking a square root on both sides is not the same as finding the roots of the equation. When you take a square root, you're essentially losing some information, which you don't want to lose if you're looking for the roots. – sayantankhan Dec 23 '14 at 03:51
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Because $\sqrt{x^2}$ can't be simplified "as follows." To be more precise: the notation $\sqrt{x^2}$ naturally designates a number $y$ such that $y^2=x^2$. Among real numbers there are two such $y$, namely, $y=-x$ and $y=x$. Your mistake is in extending the rule $(x^a)^b=x^{ab}$, valid for $a$ and $b$ integers, to allow $a$ and $b$ to be fractions.

Kevin Carlson
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  • That is an excellent point about the scope of $a$ and $b$ being restricted to integers! I should pay more attention next time to such qualifications. – logsofrhythms Dec 23 '14 at 03:38
  • Yes, but it's just the kind of thing that school courses often don't focus on, so it's understandable. – Kevin Carlson Dec 23 '14 at 03:38
  • $\sqrt[2]{}$ is understood by convention to denote the principal square root, i.e. the nonnegative root. – epimorphic Dec 26 '14 at 15:58
  • When solving an equation, I write the expression $\sqrt{x^2}=\pm x$. That is the situation the OP was in, or the question wouldn't have come up. Anyway, this is only a convention. It's much more important to understand the multivalued nature of the square root than to remember the arbitrary standard single valued shadow of it. Indeed, I know of math teachers who have been asked this question in interviews, with the goal that they first say $\pm x$ and specialize to $|x|$ in the special cases when we need to pick a single value. – Kevin Carlson Dec 26 '14 at 18:03