2011 AMC 10A Problem 7

Go through the choices one at a time. For each one, either find a value of $x$ that works or explain why none can.

1. (A) $(x+7)^2 = 0$. A square is $0$ only when the number being squared is $0$. So $x + 7 = 0$ and $x = -7$ works.
2. (B) $|-3x| + 5 = 0$. Move the $5$ over: $|-3x| = -5$. An absolute value is a distance, so it is never negative. It can never equal $-5$, so this equation has no solution.
3. (C) $\sqrt{-x} - 2 = 0$ means $\sqrt{-x} = 2$, so $-x = 4$ and $x = -4$. Check: $-x = -(-4) = 4$, and $\sqrt{4} = 2$. It works. The minus sign is fine because $x$ itself is negative.
4. (D) $\sqrt{x} - 8 = 0$ means $\sqrt{x} = 8$, so $x = 64$.
5. (E) $|-3x| - 4 = 0$ means $|-3x| = 4$. Then $x = \frac{4}{3}$ works (and so does $x = -\frac{4}{3}$).

Only (B) has no solution.

The answer is $\boxed{\textbf{(B)}\ \left|-3x\right|+5=0}$.

Squares, absolute values and square roots can never come out negative. So an equation that forces one of them to equal a negative number has no solution. A minus sign inside one of them, like $\sqrt{-x}$, is not a problem by itself: it just means $x$ has to be negative or zero.