# Square root of $x^3$

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

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.
• For $x\geq 0$ one has $|x|=x$, so the absulute value bars are not needed. Jan 19 '15 at 5:47
• This appears to be slightly dodging the question. Why is it not valid for negative real $x$? Which equalities there fail? Jan 19 '15 at 5:48
• 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 Jan 19 '15 at 5:59
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, $\sqrt {x^3} = x \sqrt x$