I do not understand this question at all! Could someone please help out a person who is horrible at math? I'm desperate! $\sqrt{−x^2}$ is defined only for $x = 0$, but $\sqrt{−x}$ is defined for all non‐positive real numbers. Explain why these expressions have these domains.

• Can you tell me what the domain (the numbers you can input into) of sqrt x is? What kinds of numbers can $-x$ be? How about $-x^2= - (x^2)$ ? – Ragib Zaman Sep 15 '11 at 6:29
• It's not really true. Have a look at: en.wikipedia.org/wiki/Imaginary_number and en.wikipedia.org/wiki/… – Dan Brumleve Sep 15 '11 at 6:31
• @Dan, I'm not sure that's helpful if the student is in a class that's not doing complex numbers. – Gerry Myerson Sep 15 '11 at 6:42
• Gerry, I just think imaginary numbers are really cool and everyone should know about them regardless of what classes one is in. An answer will probably clear things up. :) – Dan Brumleve Sep 15 '11 at 6:47
• x could be any negative number to cause -x to become positive, and therefore have the square root taken from it right? Like, x=-9 there for sqrt 9 = 3? Is that what you mean? – UVic Student Sep 15 '11 at 6:59

Let’s start with $\sqrt{-x^2}$. What kind of number is $-x^2$? If $x=0$, then $-x^2 = -0^2 = -0 = 0$. That’s not a problem, because we can take the square root of $0$: $\sqrt 0 = 0$. If $x$ is anything else, however, positive or negative, then $x^2$ is positive, so $-x^2$ is negative. For instance, if $x=2$ or $x=-2$, $x^2=4$, so $-x^2$ = $-4$. The square root of a negative number isn’t defined; substitute any non-zero value for $x$, and $-x^2$ is negative, so $\sqrt{-x^2}$ isn’t defined. In other words, the only value of $x$ for which $\sqrt{-x^2}$ makes sense is $0$.
Now $\sqrt{-x}$ is a different story. What happens when you substitute a positive number for $x$? Try $x=2$, say: then $-x=-2$, and $\sqrt{-x}=\sqrt{-2}$. Oops: you’re trying to take the square root of a negative number. There was nothing special about $2$ here: any positive value of $x$ makes $-x$ negative and $\sqrt{-x}$ undefined. How about $x=0$? Then $\sqrt{-x}=\sqrt{-0}=\sqrt 0=0$, and there’s no problem. Finally, what happens if $x$ is negative, say $x=-2$? Then $-x=-(-2)=2$, which is positive, and $\sqrt{-x}=\sqrt{-(-2)}=\sqrt 2$, which is fine: $\sqrt 2$ isn’t a particularly ‘nice’ number, but it certainly exists. Something similar happens when any negative number is substituted for $x$: taking $-x$ changes the sign to a positive number, and the square root of a positive number is always meaningful. Putting the pieces together, we see that $\sqrt{-x}$ makes sense when $x$ is zero or negative, but not when $x$ is positive, so the domain of $\sqrt{-x}$ must be the set of $x$ such that $x\le 0$, i.e., the set of non-positive real numbers.