AMC 10 · 2011 · #7

Grade 8 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
absolute-valuesign-analysissigned-square-root logical-deduction ↑ Prerequisites: absolute-value
📏 Medium solution 💡 1 insight
Problem
Five equations in the single unknown x are listed below, and exactly one of them has no solution — find the equation that no real number x can ever make true.

Pick an answer.

(A)
$(x+7)^2=0$
(B)
|-3x|+5=0
(C)
$\sqrt{-x}-2=0$
(D)
$\sqrt{x}-8=0$
(E)
|-3x|-4=0

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Eliminate Possibilities

There are only five choices, so Tool #3 (Eliminate Possibilities) fits: try to solve each equation, and any equation that produces a real x is eliminated. The one left standing is the answer. Tool #14 (Extreme Principle) sharpens the check — instead of guessing, we compare each equation against the smallest value its left side can reach. A square, a square root, and an absolute value all bottom out at 0, so if an equation forces one of them to equal a negative number, it is instantly impossible.

1STEP 1

Decide what makes an equation solvable

An equation is solvable when some real x makes both sides equal, so solve each choice and eliminate every one that yields such an x.

solution exists ⇔ some real x makes both sides equal
2STEP 2

Solve the square and square-root choices

(x+7)2=0(x+7)^2=0 gives x=−7x=-7, x=8\sqrt{x}=8 gives x=64x=64, −x=2\sqrt{-x}=2 gives x=−4x=-4 — so (A), (C), (D) are eliminated.

(x+7)²=0→ x=-7; √(x)=8→ x=64; √(-x)=2→ x=-4
3STEP 3

Solve the other absolute-value choice

(E) asks for ∣−3x∣=4|-3x|=4, and 4 is positive, so x=43x=\frac{4}{3} or x=−43x=-\frac{4}{3} — (E) is eliminated too.

|-3x|=4→ -3x=± 4→ x=∓4/3
4STEP 4

Test the last choice against the smallest possible value

Only (B) is left, and it demands ∣−3x∣=−5|-3x|=-5; an absolute value is never negative, so (B) has no solution.

|-3x| ≥ 0→ |-3x|+5 ≥ 5 ＞ 0 → (B)
Answer
|-3x|+5=0
The four eliminated choices each yielded a concrete x (-7, 64, -4, and ±4/3), so they genuinely have solutions. Choice (B) fails for a structural reason, not an arithmetic slip: adding 5 to something that is already ≥ 0 can never land on 0. The contrast with (E) is the key — |-3x|=4 (positive) is fine, but |-3x|=-5 (negative) is impossible. This confirms (B) is the unique no-solution equation.
💡Key takeaway

An absolute value can never be negative, so anything that forces it below zero has no solution.

  • Decide what makes an equation solvable
  • Solve the square and square-root choices
  • Solve the other absolute-value choice
  • Test the last choice against the smallest possible value

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