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2022 AMC 10A Problem 6

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2022 AMC 10A Problem 6AMC 10 Step by Step

Name: Date: September 13, 2026

Absolute Value & Inequalities · difficulty 2 of 5 · about 2 min · position P1-10

Which expression is equal to $\left|a-2-\sqrt{(a-1)^2}\right|$ for $a<0?$

A)
$3-2a$
B)
$1-a$
C)
$1$
D)
$a+1$
E)
$3$

Answer

Key insight

For a < 0, sqrt((a-1)^2) = |a-1| = 1 - a, so the inside becomes 2a - 3, which is negative; its absolute value is 3 - 2a.

Solution

The square root of a square is an absolute value: $\sqrt{(a-1)^2} = |a - 1|$ . Because $a < 0$ , the quantity $a - 1$ is negative, so $|a - 1| = 1 - a$ .

The expression inside the outer bars is then
$a - 2 - (1 - a) = 2a - 3.$
Since $a < 0$ , we have $2a - 3 < -3 < 0$ , so the outer absolute value flips the sign:
$|2a - 3| = 3 - 2a.$

The answer is $\boxed{\textbf{(A)}\ 3-2a}$ .

Why this works

Every absolute value (including the disguised one, $\sqrt{u^2}$ ) is removed by the same routine: decide the sign of the inside using the given range, then drop the bars with the appropriate sign. Doing this twice, inside first, is the whole problem.

Alternative approach

Test $a = -1$ : the expression is $|-1 - 2 - \sqrt{4}| = |-5| = 5$ . The choices give (A) $5$ , (B) $2$ , (C) $1$ , (D) $0$ , (E) $3$ , so only (A) survives.

The trap

Simplifying sqrt((a-1)^2) to a - 1 without the absolute value, which collapses the expression to |-1| = 1, choice (C).

Common mistakes

Simplifying sqrt((a-1)^2) to a - 1 without the absolute value, which collapses the expression to |-1| = 1, choice (C).
Getting $2a - 3$ correctly but forgetting the outer bars flip its sign, and looking for $2a - 3$ among the choices.

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer