AMC 10 · 2011 · #7
Grade 8 arithmeticWhich of the following equations does NOT have a solution?
Pick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Five equations are listed. Find the one equation for which no real number $x$ can ever make it true.
Givens: (A) $(x+7)^2=0$; (B) $|-3x|+5=0$; (C) $\sqrt{-x}-2=0$; (D) $\sqrt{x}-8=0$; (E) $|-3x|-4=0$
Unknowns: Which single equation has no solution — no value of $x$ that satisfies it
Understand
Restated: Five equations are listed. Find the one equation for which no real number $x$ can ever make it true.
Givens: (A) $(x+7)^2=0$; (B) $|-3x|+5=0$; (C) $\sqrt{-x}-2=0$; (D) $\sqrt{x}-8=0$; (E) $|-3x|-4=0$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #14 Extreme Principle
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.
Execute — Answer: B
6.EE.B.5 Step 1 Decide what makes an equation solvable
- An equation has a solution when at least one real number $x$ makes the left side equal the right side.
- So the plan is simple: try to solve each choice.
- The moment we find a real $x$ that works, that choice is safe and gets eliminated.
- The choice we can never satisfy is the answer.
💡 Solving an equation is really asking "is there a number that fits?" — if none can fit, there is no solution.
8.EE.A.2 Step 2 Solve the square and square-root choices
- Choice (A): $(x+7)^2=0$ needs the inside to be $0$, so $x=-7$ works.
- Choice (D): $\sqrt{x}=8$ gives $x=64$.
- Choice (C): $\sqrt{-x}=2$ means $-x=4$, so $x=-4$.
- Each of these produces a real $x$, so (A), (C), and (D) all have solutions and are eliminated.
💡 A square root or a square set equal to a non-negative number can always be undone to find $x$.
6.NS.C.7 Step 3 Solve the other absolute-value choice
- Choice (E): $|-3x|-4=0$ means $|-3x|=4$.
- Since $4$ is positive, this is reachable: $-3x=4$ or $-3x=-4$, giving $x=-\tfrac{4}{3}$ or $x=\tfrac{4}{3}$.
- So (E) has solutions and is eliminated too.
💡 An absolute value can equal any non-negative number, so setting it equal to a positive number always has answers.
6.NS.C.7 Step 4 Test the last choice against the smallest possible value
- Only (B) is left: $|-3x|+5=0$, which rearranges to $|-3x|=-5$.
- But an absolute value is never negative — its smallest possible value is $0$ — so $|-3x|+5$ is always at least $5$ and can never reach $0$.
- No $x$ works.
- Since (A), (C), (D), and (E) each had a solution, the equation with no solution is (B).
💡 If the smallest a quantity can ever be is already above $0$, it can never equal $0$.
6.EE.B.5 An equation has a solution when at least one real number $x$ makes the left side 8.EE.A.2 Choice (A): $(x+7)^2=0$ needs the inside to be $0$, so $x=-7$ works. Choice (D): 6.NS.C.7 Choice (E): $|-3x|-4=0$ means $|-3x|=4$. Since $4$ is positive, this is reachabl 6.NS.C.7 Only (B) is left: $|-3x|+5=0$, which rearranges to $|-3x|=-5$. But an absolute v Review
Reasonableness: The four eliminated choices each yielded a concrete $x$ ($-7$, $64$, $-4$, and $\pm\tfrac43$), so they genuinely have solutions. Choice (B) fails for a structural reason, not an arithmetic slip: adding $5$ to something that is already $\ge 0$ 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.
Alternative: Skip the algebra and just read the required value of each non-negative expression. (A) square $=0$: fine. (C),(D) square root $=2$ and $=8$: fine. (E) absolute value $=4$: fine. (B) absolute value $=-5$: impossible, because absolute values are never negative. Only (B) demands a negative output from something that can never be negative, so (B) is the answer.
CCSS standards used (min grade 8)
6.EE.B.5Understand solving an equation or inequality as a process of finding values (Framing "no solution" as: no real $x$ can make the two sides equal, which sets up the elimination strategy.)8.EE.A.2Use square root and cube root symbols to represent solutions (Solving the square and square-root choices (A) $x=-7$, (C) $x=-4$, and (D) $x=64$ to confirm they have solutions.)6.NS.C.7Understand ordering and absolute value of rational numbers (Solving (E) $|-3x|=4$, and recognizing that $|-3x|\ge 0$ makes $|-3x|+5=0$ impossible.)
⭐ An absolute value can never be negative, so anything that forces it below zero has no solution.
⭐ An absolute value can never be negative, so anything that forces it below zero has no solution.
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