AMC 10 · 2020 · #5
Grade 8 arithmeticPick an answer.
AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #7 (Subproblems) is the spine: |y| = 2 splits into two cases, y = 2 and y = -2, each its own small equation. Tool #1 (Diagram) makes the two cases visible — rewrite x² - 12x + 34 = (x - 6)² - 2 and picture an upward parabola with vertex (6, -2). The horizontal line y = 2 cuts the parabola at two points symmetric around x = 6; the line y = -2 just touches the vertex at one point. Tool #5 (Pattern) — the symmetry around x = 6 means each pair of intersection x-values averages to 6, so we don't have to solve, just count and sum.
Complete the square: x² - 12x + 34 = (x - 6)² - 2.
Adding and subtracting 36 turns the messy quadratic into a clean shifted square.
6.EE.A.3Draw A DiagramAbsolute value 2 splits into two cases: (x - 6)² - 2 = 2 or (x - 6)² - 2 = -2.
An absolute value equals 2 exactly when the inside is +2 or -2.
6.EE.A.4Identify SubproblemsCase 1: (x - 6)² = 4 gives x = 4 or 8, symmetric about x = 6, so their sum is 12.
Square equals 4 means the inside is ± 2 — Grade 8 square-root reasoning.
8.EE.A.2Identify SubproblemsCase 2: (x - 6)² = 0 has the single root x = 6 — the parabola's vertex.
Square equals 0 has exactly one root — Grade 8 square-root reasoning.
8.EE.A.2Identify SubproblemsAdd the distinct roots: 4 + 8 + 6 = 18, choice (C).
Case 1's two roots sum to 12 by symmetry; add the lone Case 2 root 6 to get 18.
4.OA.A.3Look For A PatternThis AMC 10 problem only needs Grade 8 "square roots" you already know — rewrite as (x - 6)² - 2, the y = 2 case gives a symmetric pair summing to 12 and the y = -2 case gives just x = 6, for a total of 18.