2011 AMC 10A Problem 7

Test each equation.

(A) $(x+7)^2 = 0$ has the solution $x = -7$.

(B) $|-3x| + 5 = 0$ requires $|-3x| = -5$. An absolute value is never negative, so there is no solution.

(C) $\sqrt{-x} - 2 = 0$ means $\sqrt{-x} = 2$, so $-x = 4$ and $x = -4$. This is valid because $-x = 4 \ge 0$.

(D) $\sqrt{x} - 8 = 0$ gives $x = 64$.

(E) $|-3x| - 4 = 0$ gives $|3x| = 4$, so $x = \pm\frac{4}{3}$.

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

Squares, absolute values, and principal square roots all output nonnegative numbers. An equation that forces one of them to equal a negative number has no real solution. A minus sign inside such an expression (as in $\sqrt{-x}$) does not create a problem by itself; what matters is whether the required output is possible.

• Rejecting sqrt(-x) - 2 = 0 because of the minus sign under the radical, forgetting that x = -4 makes -x positive.
• Picking (E) by mixing up $|{-3x}| - 4 = 0$ with $|{-3x}| + 4 = 0$; only the sign of the constant matters.