Why does $\sqrt{x^2}$ seem to equal $x$ and not $|x|$ when you multiply the exponents?

I understand that $\sqrt{x^2} = |x|$ because the principal square root is positive.

But since $\sqrt x = x^{\frac{1}{2}}$ shouldn’t $\sqrt{x^2} = (x^2)^{\frac{1}{2}} = x^{\frac{2}{2}} = x$ because of the exponents multiplying together?

Also, doesn’t $(\sqrt{x})^2$ preserve the sign of $x$? But shouldn’t $(\sqrt{x})^2 = (\sqrt{x})(\sqrt{x}) = \sqrt{x^2}$?

How do I reconcile all this? What rules am I not aware of?

Edit: Since someone voted to close my question, I should probably explain the difference between my question and Proving square root of a square is the same as absolute value, regardless of how much I think the difference should be obvious to anyone who reads the questions. Cole Johnson was asking if there’s any way to prove that $\sqrt{x^2} = |x|$. I am not asking that; I already accept the equation as fact. I’m asking how to resolve some apparent contradictions that arise when considering square roots of squares, and how I should approach these types of problems. (Cameron, please read.)

Solutions Collecting From Web of "Why does $\sqrt{x^2}$ seem to equal $x$ and not $|x|$ when you multiply the exponents?"

A common source of confusion or “paradoxes” comes from not paying close attention to the (perhaps rarely exercised) restrictions or boundary conditions. These restrictions are necessary to ensure that paradoxes like you’re considering do not arise (i.e., otherwise the definitions would fail to be well-defined). For example, here’s a proper definition of rational exponents from Michael Sullivan’s College Algebra:

Definition of rational exponents

Note that we only consider real numbers here. Now, to answer your questions:

But since $\sqrt{x} = x^{\frac{1}{2}}$ shouldn’t $\sqrt{x^2} =
(x^2)^{\frac{1}{2}} = x^{\frac{2}{2}} = x$ because of the exponents
multiplying together?

The first assertion is not generally true; $\sqrt{x} = x^{\frac{1}{2}}$ only provided that $\sqrt{x}$ exists (that is, not for negative $x$). In your chained equality, the second equality is false, because the exponent in $x^{\frac{2}{2}}$ contains common factors (i.e., is not in lowest terms). These statements would, however, be true if $x$ was restricted to positive real numbers only.

Also, doesn’t $(\sqrt{x})^2$ preserve the sign of $x$? But shouldn’t
$(\sqrt{x})^2 = (\sqrt{x})(\sqrt{x}) = \sqrt{x^2}$?

All of these equalities are false for negative $x$, because in that case the expression $\sqrt{x}$ does not exist in real numbers (i.e., it’s undefined). Likewise, if you look carefully at the rule for multiplying radicals, then you’ll see the same restriction against square roots of negative numbers.

Edit: Added text from Precalculus: a right triangle approach by Ratti & McWaters. Hopefully this clarifies the rule for products of square roots (namely that only positive radicands can be generally combined or separated). Also, note the warning from the section on complex numbers that doing so in that case is illegitimate.

Product and quotient properties of square roots

Warning against combining products of complex square roots

The rule $(x^a)^b = x^{ab}$ is only true for positive values of $x$. With negative values, you need to be much more careful.

For example, $\sqrt x \sqrt x = \sqrt{x\cdot x}$ is only true for positive values of $x$, because for negative values, the left side is not defined.

First of all, $\sqrt{a}\sqrt{b} = \sqrt{ab}$ only when $a$ and $b$ are non-negative. $\sqrt{a}$ and $a^{\frac{1}{2}}$ are not the same function, as Cameron Williams points out in the comments above. Secondly, we can break your square of square root up. For positive values, $\sqrt{x^2} = x$, as one would expect. However, since $\sqrt{(-x)^2} = \sqrt{x^2}$, we notice that we get a positive answer from a negative value, getting $x$ from $-x$. We can represent this as flipping the function over the $x$-axis, or in other-words throwing a negative sign out front (For example, $y=-(x^2+5x)$ flips $y=x^2+5x$ across the $x$-axis). Thus, we get
$$\sqrt{\gamma^2}=\begin{cases}\gamma & \gamma\geq 0\\-\gamma & \gamma<0,\end{cases}$$
And this is the definition of the absolute value function

Consider the following:$$1^{\frac1n}$$Look up roots of unity.

The nth root of a number produces n different answers, creating a regular n-sided shape with a “radius” of 1 on the complex plane.

But what does this mean? Well, you could consider this…

The nth root of a positive number in general.

This removes all of the other possible answers, leaving us with only one answer. Among the answers removed was the original answer, but because real-life math usually can ignore this, we’re all good.

Now signs, as you may imagine, have been messed up. We, as a general society, have decided that it be positive. This has made math much easier for younger pupils still learning math. And as we take more roots, we pretty much lose signs. We end up with $i$’s, plusses, and minuses. It is not that you have messed up anywhere, it was just a simple decision made to make the square root function positive.

In many maths, you will find that we will say the n-th root of a number does indeed produce n amount of answers, but for lower level math, we stick with the positive answer, or the result of no positive answer.

I mean, think about it, what does $\sqrt1=-1,1$ do for us in real life? Not much, as it turns out. When we need both the positive and negative answers, it is known that that is what we are finding. But it doesn’t have much application to real life.

So the rule is that square roots are positive, and just keep a spot in the back of your mind remembering that this is not the only answer.