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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 $|x|\times \sqrt{x}$?

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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.

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  • $\begingroup$ My precalc teacher indicated that square root (2*X^2) = abs(x)*root 2; however, the square root (2*x^3) = x*sqrt(2*x). Unclear why the x does not have absolute value bars as it did in sqrt (2*x^2) $\endgroup$
    – user163862
    Jan 19 '15 at 5:41
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    $\begingroup$ For $x\geq 0$ one has $|x|=x$, so the absulute value bars are not needed. $\endgroup$ Jan 19 '15 at 5:47
  • $\begingroup$ This appears to be slightly dodging the question. Why is it not valid for negative real $x$? Which equalities there fail? $\endgroup$ Jan 19 '15 at 5:48
  • $\begingroup$ YES, this is my question precisely! For REAL values of x, it would see the sqrt (x^3) would be abs(x) * sqrt(x). $\endgroup$
    – user163862
    Jan 19 '15 at 5:54
  • $\begingroup$ 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 $\endgroup$ Jan 19 '15 at 5:59
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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, $\sqrt {x^3} = x \sqrt x$

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