How do I simplify the square root of variables that include ranges?

Question

How would I simplify $\sqrt{x^3y^2}$ where $x>0$ and $y<0$? The answer key says $-xy\sqrt x$ but I don't understand the answer.

Answer 1

Note that $\sqrt{z^2} = |z|$. Consequently, if $z>0$ then $\sqrt{z^2} = +z$ otherwise $\sqrt{z^2} = -z$.

Answer 2

I guess you can see that it's $\sqrt{x^2}\sqrt{y^2}\sqrt{x}$.

Now $\sqrt{a^2}$ is the "positive" version of $a$, since squaring would remove the sign and then the square root would return the number back to its original magnitude, but is always nonnegative. In other words, squaring and then taking the square root effectively strips away any sign the number might carry.

The "ranges" you mention are really just telling you that $x$ is positive and $y$ is negative. So the corresponding positive versions are "$x$" (already known to be positive) and "$-y$" (positive because its opposite is the negative number $y$).

So the expression is then $x\cdot (-y)\cdot\sqrt{x}$, i.e., $-xy\sqrt{y}$.

Addendum: At the end of the day, you should take away from this that $-a$ is a positive number if it is known that $a$ is negative. Don't be fooled by the presence of the "minus sign" into thinking that it is a negative number. A minus sign only indicates a negative number if the thing it is applied to is known to be positive. So "$-2$" is negative because "$2$" is positive, but you don't know whether "$-w$" is negative or not because you don't know whether "$w$" itself is negative or not.