2022 AMC 10A Problem 6

1. Try a number first. Take $a = -1$. Then $(a - 1)^2 = (-2)^2 = 4$ and $\sqrt4 = 2$. Notice the square root gave $2$, not $a - 1 = -2$. A square root is never negative.
2. What the square root really does. $\sqrt{(a-1)^2}$ is the size of $a - 1$ with its sign thrown away, that is, $|a - 1|$. Since $a < 0$, the number $a - 1$ is negative, so its size is $-(a - 1) = 1 - a$.
3. Simplify the inside. The expression inside the outer bars becomes
$$a - 2 - (1 - a) = 2a - 3.$$
4. Remove the outer bars. With $a$ negative, $2a - 3$ is less than $-3$, so it is negative. Taking the absolute value flips its sign:
$$|2a - 3| = 3 - 2a.$$
5. Check with $a = -1$. The original is $|-1 - 2 - 2| = |-5| = 5$, and $3 - 2(-1) = 5$.

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

The square root of something squared is an absolute value in disguise. To remove any absolute value, first decide whether the inside is positive or negative using what you are told (here, $a < 0$), then keep or flip the sign. Do the inner one first, then the outer one.

Plug $a = -1$ into every choice. The expression equals $5$, and the choices give (A) $5$, (B) $2$, (C) $1$, (D) $0$, (E) $3$. Only (A) matches.