AMC 10 · 2021 · #2

Grade 8 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

absolute-valuesigned-square-rootsign-analysisexponents identify-subproblemscasework ↑ Prerequisites: absolute-valueexponents

📏 Short solution 💡 2 insights

Problem

Evaluate √((3-2√(3))²) + √((3+2√(3))²), using the fact that √(a²) equals |a| for any real a.

Pick an answer.

(A)

(B)

$~4\sqrt{3}-6$

(C)

(D)

$~4\sqrt{3}$

(E)

$~4\sqrt{3}+6$

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

Tool #9 (Easier Problem) — replace the messy expression √((□)²) with the cleaner rule |□|. That turns the problem into a sum of two absolute values, which is concrete. Tool #7 (Subproblems) splits the work into "sign of 3 - 2√(3)" and "sign of 3 + 2√(3)" — each handled separately, then added. Tool #3 (Eliminate) gives a fast sanity check: 2√(3) ≈ 3.46, so 3 - 2√(3) ≈ -0.46 and 3 + 2√(3) ≈ 6.46; their absolute values sum to ≈ 6.92 ≈ 4√(3). Only (D) matches.

1STEP 1

Rewrite each root-of-a-square as an absolute value: √(a²) = |a|, since a principal root is never negative.

√((3-2√(3))²) + √((3+2√(3))²) = |3-2√(3)| + |3+2√(3)|

2STEP 2

Compare by squaring: 3² = 9 but (2√(3))² = 12, and both are positive, so 2√(3) ＞ 3.

3² = 9 ＜ 12 = (2√(3))² → 2√(3) ＞ 3

3STEP 3

Peel the bars: 3-2√(3) is negative so |3-2√(3)| = 2√(3) - 3, while 3+2√(3) stays positive so |3+2√(3)| = 3+2√(3).

|3-2√(3)| = 2√(3) - 3, |3+2√(3)| = 3 + 2√(3)

4STEP 4

Add the pieces: the -3 and +3 cancel, leaving 2√(3) twice, which is 4√(3) — choice (D).

(2√(3) - 3) + (3 + 2√(3)) = 4√(3) → (D)

5STEP 5

Decimal check: 2√(3) ≈ 3.46, so the pieces ≈ 0.46 and 6.46, summing to ≈ 6.92, matching 4√(3) ≈ 6.93 — only (D) fits.

0.46 + 6.46 ≈ 6.92 ≈ 4√(3)

Answer

~4√(3)

Both terms inside the square roots are real, so each square root is well-defined and non-negative. The sum 4√(3) ≈ 6.93 is positive and lies between (C) 6 and (E) 4√(3) + 6 ≈ 12.93, exactly where the rough estimate places it. The -3 and +3 cancellation is the key structural fingerprint of choice (D).

💡Key takeaway

This AMC 10 problem only needs Grade 8 "√(a²) = |a|" — once you peel each square root into an absolute value and notice 2√(3) ＞ 3, the -3 and +3 cancel, leaving 4√(3)!

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