# Square root theory

The problem is: $\sqrt{x^3y}$. It's not clear to me why when I take out the $x^2$ that I must put an absolute value around it, since when x remains inside the square root there is an automatic domain restriction of x being non-negative. I would think that this domain restriction eliminates the need for absolute value bars.

I can see that if there is no x remaining in the square root that I must include the absolute value bars, but unclear why if an x remains inside.

Can someone help me see this?

• What if $y <0$ ? – Improve Dec 1 '16 at 23:21
• Because $\sqrt{x^2}=|x|$ and $\sqrt{x^3 y}=\sqrt{x^2 \cdot x y}=\sqrt{x^2} \sqrt{x y}=|x| \sqrt{x y}$. – dxiv Dec 1 '16 at 23:22
• Yes, I thought this might be the case. Does that mean that if there had been no y remaining inside the square root, then it would have eliminated the need for the absolute value? – user163862 Dec 1 '16 at 23:22
• In this case, yes. However check that you understand why you need the absolute value if you are considering $\sqrt{x^4y}$. – Improve Dec 1 '16 at 23:25
• what if the question is $\sqrt(x^3)$ Doesn't the x remaining inside the square root negate the need for the absolute value? – user163862 Dec 1 '16 at 23:28