2021 AMC 10B Spring Problem 2

Attempt Problem 2 of the 2021 AMC 10B Spring 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 2021 AMC 10B Spring solutions, or check the answer key.

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

2.

What is the value of (3−23)2+(3+23)2? \begin{aligned} &\sqrt{\left(3-2\sqrt{3}\right)^2} \\ &{}+\sqrt{\left(3+2\sqrt{3}\right)^2}? \end{aligned}
0 0
43−6 4\sqrt{3}-6
6 6
43 4\sqrt{3}
43+6 4\sqrt{3}+6
Answer: D
Concepts:radicalabsolute value
Small hint:
Remember that u2=∣u∣\sqrt{u^2}=|u|
Big hint:
Decide the signs of 3−233-2\sqrt3 and 3+233+2\sqrt3 separately
Solution:
Because u2=∣u∣,\sqrt{u^2}=|u|, the expression equals ∣3−23∣+∣3+23∣.|3-2\sqrt3|+|3+2\sqrt3|. Since 23>3,2\sqrt3>3, this becomes (23−3)+(3+23)=43.(2\sqrt3-3)+(3+2\sqrt3)=4\sqrt3.
Thus, the correct answer is D .
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