2022 AMC 10A Problem 6

Attempt Problem 6 of the 2022 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

6.

Which expression is equal to ∣a−2−(a−1)2∣\left|a-2-\sqrt{(a-1)^2}\right| for a<0?a < 0?
3−2a3 - 2a
1−a1 - a
11
a+1a + 1
33
Answer: A
Concepts:absolute valueradical
Small hint:
Replace the square root of a square by an absolute value
Big hint:
Use the sign of each expression when a is negative
Solution:
By the definition of square root, we get that (a−1)2=∣a−1∣.\sqrt{(a - 1)^2} = |a - 1|.
Since a<0,a < 0, we get that a−1<0,a - 1 < 0, which means that ∣a−1∣=1−a.|a - 1| = 1 - a.
The whole expression therefore simplifies to ∣a−2−(1−a)∣=∣2a−3∣. |a - 2 - (1 - a)| = |2a - 3|. Since a<0,a < 0, we know that 2a−3<0.2a - 3 < 0. This means that ∣2a−3∣=3−2a.|2a - 3| = 3 - 2a.
Thus, A is the correct answer.
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