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2021 AMC 10B Problem 2

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2021 AMC 10B Problem 2AMC 10 Step by Step
Name:                             Date: September 13, 2026
Exponents, Logarithms & Radicals · difficulty 1 of 5 · about 1 min · position P1-10

What is the value of (3−23)2+(3+23)2\sqrt{\left(3-2\sqrt{3}\right)^2}+\sqrt{\left(3+2\sqrt{3}\right)^2} ?

  1. A)

    00

  2. B)

    43−64\sqrt{3}-6

  3. C)

    66

  4. D)

    434\sqrt{3}

  5. E)

    43+64\sqrt{3} + 6

Answer

D

Key insight

√(t²) = |t|, and 2√3 ≈ 3.46 exceeds 3, so the first term is 2√3 − 3, not 3 − 2√3.

Solution

For any real tt, t2=∣t∣\sqrt{t^2} = |t|. So the expression equals
∣3−23∣+∣3+23∣.|3 - 2\sqrt{3}| + |3 + 2\sqrt{3}|.

Since 23=12>9=32\sqrt{3} = \sqrt{12} > \sqrt{9} = 3, the number 3−233 - 2\sqrt{3} is negative, so its absolute value is 23−32\sqrt{3} - 3. The second number is clearly positive, so its absolute value is 3+233 + 2\sqrt{3}.

Adding: (23−3)+(3+23)=43(2\sqrt{3} - 3) + (3 + 2\sqrt{3}) = 4\sqrt{3}.

The answer is (D) 43\boxed{\textbf{(D)}\ 4\sqrt{3}}.

Why this works

The square root symbol always denotes the nonnegative root, so t2\sqrt{t^2} is ∣t∣|t|, not tt. Whenever a problem squares something and then takes a root, check the sign of the inside; here the sign comparison 232\sqrt{3} versus 33 is the entire problem.

The trap

Cancelling the square root against the square without an absolute value, which gives (3 − 2√3) + (3 + 2√3) = 6, choice (C).

Common mistakes

  • Cancelling the square root against the square without an absolute value, which gives (3 − 2√3) + (3 + 2√3) = 6, choice (C).
  • Deciding that 3−233 - 2\sqrt{3} is positive because "3>23 > 2," without comparing 23≈3.462\sqrt{3} \approx 3.46 to 33.

Techniques

Set up the equation/formula and compute; no special trick needed