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2020 AMC 10A Problem 5

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2020 AMC 10A Problem 5AMC 10 Step by Step

Name: Date: September 13, 2026

Absolute Value & Inequalities · difficulty 2 of 5 · about 3 min · position P1-10

What is the sum of all real numbers $x$ for which $|x^2-12x+34|=2?$

A)
$12$
B)
$15$
C)
$18$
D)
$21$
E)
$25$

Answer

Key insight

Split into x^2-12x+34 = 2 (roots 4 and 8) and x^2-12x+34 = -2 (double root 6); sum the distinct roots 4 + 8 + 6.

Solution

The absolute value equals $2$ when the inside is $2$ or $-2$ .

Case 1: $x^2 - 12x + 34 = 2$ , i.e. $x^2 - 12x + 32 = 0$ , which factors as $(x-4)(x-8) = 0$ . Roots $4$ and $8$ .

Case 2: $x^2 - 12x + 34 = -2$ , i.e. $x^2 - 12x + 36 = 0$ , which is $(x-6)^2 = 0$ . The single root is $6$ .

The real numbers satisfying the equation are $4$ , $6$ , $8$ , and their sum is $18$ .

The answer is $\boxed{\textbf{(C)}\ 18}$ .

Why this works

An absolute-value equation is two ordinary equations. The quadratic $x^2 - 12x + 34 = (x-6)^2 - 2$ has minimum value $-2$ , so the second case touches the parabola at its vertex and produces only one number, not two. Always check whether a case yields a repeated root before adding root sums.

The trap

Applying Vieta to both quadratics and adding 12 + 12 = 24, which double-counts the repeated root 6 (and is not even a choice), or ignoring the negative case and answering 12.

Common mistakes

Applying Vieta to both quadratics and adding 12 + 12 = 24, which double-counts the repeated root 6 (and is not even a choice), or ignoring the negative case and answering 12.
Solving $x^2 - 12x + 34 = 2$ only and stopping at $12$ , choice (A).

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta · Split into exhaustive cases and handle each