AMC 10 · 2022 · #6
Grade 8 algebraProblem
Which expression is equal to a−2−(a−1)2 for a<0?
Pick an answer.
AMC 10 2022 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: Simplify the expression $\left|a-2-\sqrt{(a-1)^2}\right|$ assuming $a$ is a negative number. The answer must be one of five algebraic expressions in $a$.
Givens: Expression: $\left|a-2-\sqrt{(a-1)^2}\right|$; Condition: $a < 0$; Answer choices: (A) $3-2a$, (B) $1-a$, (C) $1$, (D) $a+1$, (E) $3$
Unknowns: Which of the five expressions equals the original for every negative $a$
Understand
Restated: Simplify the expression $\left|a-2-\sqrt{(a-1)^2}\right|$ assuming $a$ is a negative number. The answer must be one of five algebraic expressions in $a$.
Givens: Expression: $\left|a-2-\sqrt{(a-1)^2}\right|$; Condition: $a < 0$; Answer choices: (A) $3-2a$, (B) $1-a$, (C) $1$, (D) $a+1$, (E) $3$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #9 Solve an Easier Related Problem, #6 Guess and Check
This is a multiple-choice problem where every answer choice is a specific function of $a$. Tool #3 (Eliminate Possibilities) is the AMC headline move — pick one concrete negative number, evaluate the original expression, then test each choice. Tool #9 (Easier Related Problem) is what we lean on to pick the test value: instead of carrying $a$ symbolically, set $a = -1$ (smallest absolute value that stays negative). Tool #6 (Guess and Check) is the verification step. The hard symbolic property $\sqrt{(a-1)^2}=|a-1|$ never needs to be stated abstractly — a single numerical evaluation makes it obvious. If two choices happened to tie at $a=-1$, we would simply repeat with $a=-2$ to break the tie.
Execute — Answer: A
6.NS.C.5 Step 1 - Pick the simplest negative number to test: $a = -1$.
- (It satisfies $a < 0$ and keeps the arithmetic tiny.)
💡 Replacing the unknown $a$ with a single concrete negative number turns a symbolic problem into a counting check.
8.EE.A.2 Step 2 - Compute the inside square root first.
- With $a=-1$: $a-1 = -2$, so $(a-1)^2 = 4$, and the principal square root is the positive root, $\sqrt{4} = 2$.
💡 $\sqrt{x^2}$ always gives back the positive size, never the negative — the square already erased the minus sign.
6.NS.C.7 Step 3 - Plug into the outer absolute value: $a - 2 - 2 = -1 - 2 - 2 = -5$.
- Take its absolute value.
💡 Absolute value strips the sign — the distance from $0$ on the number line.
6.EE.A.2 Step 4 - Now evaluate each answer choice at $a=-1$ and keep the ones that equal $5$.
- (A) $3-2(-1)=5$.
- (B) $1-(-1)=2$.
- (C) $1$.
- (D) $(-1)+1=0$.
- (E) $3$.
- Only (A) matches.
💡 Substitute the same number into every choice; the right one returns the same value as the original.
6.EE.A.4 Step 5 - To make sure we did not get lucky at $a=-1$, repeat with $a=-2$.
- Inner: $\sqrt{(-3)^2}=3$.
- Outer: $|-2-2-3| = |-7| = 7$.
- Choice (A): $3-2(-2)=7$.
- Still matches; no other choice does.
💡 One match could be coincidence; two matches in a row across different inputs confirm the algebraic identity.
6.NS.C.5 Pick the simplest negative number to test: $a = -1$. (It satisfies $a < 0$ and k 8.EE.A.2 Compute the inside square root first. With $a=-1$: $a-1 = -2$, so $(a-1)^2 = 4$, 6.NS.C.7 Plug into the outer absolute value: $a - 2 - 2 = -1 - 2 - 2 = -5$. Take its abso 6.EE.A.2 Now evaluate each answer choice at $a=-1$ and keep the ones that equal $5$. (A) 6.EE.A.4 To make sure we did not get lucky at $a=-1$, repeat with $a=-2$. Inner: $\sqrt{( Review
Reasonableness: Two independent test values $a=-1$ and $a=-2$ both produced exactly the value of $3-2a$. Since the four other choices each disagreed at $a=-1$, only (A) can be the correct identity. As a sanity check on the structure: when $a$ is negative, $a-2-\sqrt{(a-1)^2} = a-2-(1-a) = 2a-3$, which is negative, so taking its absolute value flips the sign to $3-2a$ — exactly choice (A).
Alternative: Tool #13 (Convert to Algebra). Use $\sqrt{x^2}=|x|$. Since $a<0$ means $a-1<0$, $|a-1|=1-a$. The expression becomes $|a-2-(1-a)|=|2a-3|$. Since $a<0$ implies $2a-3<0$, $|2a-3|=3-2a$. The plug-in path is faster and skips the two sign-of-absolute-value case splits, which is the main place students slip.
CCSS standards used (min grade 8)
6.NS.C.5Understand that positive and negative numbers describe quantities (Choosing $a=-1$ as a legal value satisfying the condition $a<0$ and computing with negative numbers.)8.EE.A.2Use square root and cube root symbols to represent solutions (Evaluating $\sqrt{(-2)^2}=2$ — the principal square root returns the non-negative size.)6.NS.C.7Understand ordering and absolute value of rational numbers (Evaluating $|-5|=5$ and $|-7|=7$ — absolute value as distance from zero.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Substituting $a=-1$ into each of the five answer expressions and computing the result.)6.EE.A.4Identify when two expressions are equivalent (Confirming that $3-2a$ matches the original expression at a second test value $a=-2$, ruling out coincidence.)
⭐ This AMC 10 problem only needs Grade 8 square-root reasoning you already know — plug in $a=-1$, follow the order of operations, and test each choice; the one that lands on the same number is (A) $3-2a$.
⭐ This AMC 10 problem only needs Grade 8 square-root reasoning you already know — plug in $a=-1$, follow the order of operations, and test each choice; the one that lands on the same number is (A) $3-2a$.