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I understand the concept behind the expression $\sqrt{x^2} = |x|$.

So, then why is the square root of $x^3$ NOT equal to $|x|\sqrt{x}$? Specifically, I can write $\sqrt{x^3}$ as $\sqrt{x^2\times x}$. Can I not now write this as $|x|\times \sqrt{x}$?

user163862
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2 Answers2

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You can: Like you say, (for $x \geq 0$,) $$\sqrt{x^3} = \sqrt{x^2 \cdot x} = \sqrt{x^2} \sqrt{x} = |x| \sqrt{x} = x \sqrt{x},$$ so the two functions are the same.

Now, if $\sqrt{\cdot}$ is the usual real square root function, which is a map $[0, \infty) \to \mathbb{R}$, then each of these expressions is only valid for $x \in [0, \infty)$: For values $x < 0$, they would involve functions evaluated at points not in their domain and so are not defined.

Travis Willse
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  • My precalc teacher indicated that square root (2X^2) = abs(x)root 2; however, the square root (2x^3) = xsqrt(2x). Unclear why the x does not have absolute value bars as it did in sqrt (2x^2) – user163862 Jan 19 '15 at 05:41
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    For $x\geq 0$ one has $|x|=x$, so the absulute value bars are not needed. – Janko Bracic Jan 19 '15 at 05:47
  • This appears to be slightly dodging the question. Why is it not valid for negative real $x$? Which equalities there fail? – Mario Carneiro Jan 19 '15 at 05:48
  • YES, this is my question precisely! For REAL values of x, it would see the sqrt (x^3) would be abs(x) * sqrt(x). – user163862 Jan 19 '15 at 05:54
  • If you allow square roots of negative numbers, you are right, but in pre-calc we usually are not talking about complex numbers, so $\sqrt{x}$ is left undefined for $x<0$. @user163862 – Thomas Andrews Jan 19 '15 at 05:59
  • It's not that the equalities fail exactly, it's that the none of the expressions make sense if we regard the square root as the usual real function $\sqrt{\cdot}: [0, \infty) \to \mathbb{R}, as they would involve evaluating a function at a point outside its domain. – Travis Willse Jan 19 '15 at 06:44
  • One can still ask for what values the involved identities hold. It's true for all $x$ that $x^3 = x^2 \cdot x$ and $\sqrt{x^2} = |x|$, but it's only true for $x \geq 0$ that $|x| = x$. Similarly, the identity $\sqrt{a b} = \sqrt{a} \sqrt{b}$ only holds for $a, b \geq 0$. Is this what you were looking for? – Travis Willse Jan 19 '15 at 06:49
  • @MarioCarneiro It certainly answers the question as stated, which asks why two expressions are not equal, whereas in fact they are equal as functions on the largest subset of $\mathbb{R}$ on which each makes sense. Why the expressions are not defined on a larger interval is a separate question, no? Anyway, since OP is interested in that point, I've added a short explanation of that too. – Travis Willse Jan 19 '15 at 07:44
  • Thanks for that. And sorry for the strange choice of words, of course it fails because of domain restrictions and not because it is defined but incorrect in that region (although $\sqrt{ab}\ne\sqrt a\sqrt b$ was the first thing I thought when I saw the question, it's moot if you are using real square root). – Mario Carneiro Jan 19 '15 at 10:38
  • You're welcome, I'm glad you found it useful. – Travis Willse Jan 19 '15 at 14:49
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Definitely, $\sqrt {x^3} = |x| \sqrt x$. But here x must be positive. If $x$ is negative then $x^3$ is also negative. That would imply $\sqrt {x^3}$ doesn't exists. So, $x \ge 0$, for this case $|x|=x$. So it is not require to denote $|x|$. Hence, $\sqrt {x^3} = x \sqrt x$

curious_mind
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