AMC 10 · 2020 · #5

Grade 8 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

absolute-valuepolynomial-rootsvieta-formulascasework caseworkidentify-subproblems ↑ Prerequisites: absolute-valuepolynomial-roots

📏 Medium solution 💡 3 insights

Problem

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

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Find the sum of every real number $x$ that makes $|x^2 - 12x + 34| = 2$ true.

Givens: The expression inside the absolute-value bars is $x^2 - 12x + 34$; Its absolute value equals $2$; Answer choices: (A) $12$, (B) $15$, (C) $18$, (D) $21$, (E) $25$

Unknowns: All real $x$ satisfying the equation, then their sum

Understand

Restated: Find the sum of every real number $x$ that makes $|x^2 - 12x + 34| = 2$ true.

Givens: The expression inside the absolute-value bars is $x^2 - 12x + 34$; Its absolute value equals $2$; Answer choices: (A) $12$, (B) $15$, (C) $18$, (D) $21$, (E) $25$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #5 Look for a Pattern

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 $x^2 - 12x + 34 = (x - 6)^2 - 2$ 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.

Execute — Answer: C

#1 Draw a Diagram 6.EE.A.3 Step 1

Complete the square to rewrite the inside expression with a vertex form.
$x^2 - 12x + 34 = (x^2 - 12x + 36) - 2 = (x - 6)^2 - 2$.

$$x^2 - 12x + 34 = (x - 6)^2 - 2$$

💡 Adding and subtracting $36$ turns the messy quadratic into a clean shifted square.

#7 Identify Subproblems 6.EE.A.4 Step 2

Split into two subproblems via the definition of absolute value.
The equation becomes $(x - 6)^2 - 2 = 2$ OR $(x - 6)^2 - 2 = -2$.

$$(x - 6)^2 - 2 = 2 \quad \text{or} \quad (x - 6)^2 - 2 = -2$$

💡 An absolute value equals $2$ exactly when the inside is $+2$ or $-2$.

#7 Identify Subproblems 8.EE.A.2 Step 3

Case 1: $(x - 6)^2 = 4$.
Taking square roots gives $x - 6 = 2$ or $x - 6 = -2$, so $x = 8$ or $x = 4$.
These two solutions are symmetric around $x = 6$, so their sum is $4 + 8 = 12$.

$$(x - 6)^2 = 4 \;\Rightarrow\; x = 4\ \text{or}\ 8, \quad \text{sum} = 12$$

💡 Square equals $4$ means the inside is $\pm 2$ — Grade 8 square-root reasoning.

#7 Identify Subproblems 8.EE.A.2 Step 4

Case 2: $(x - 6)^2 = 0$.
The only solution is $x = 6$.
(This is the parabola's vertex touching the horizontal line $y = -2$.)

$$(x - 6)^2 = 0 \;\Rightarrow\; x = 6$$

💡 Square equals $0$ has exactly one root — Grade 8 square-root reasoning.

#5 Look for a Pattern 4.OA.A.3 Step 5

Add all distinct solutions.
Total sum $= 4 + 8 + 6 = 18$, which is choice (C).

$$4 + 8 + 6 = 18 \;\Rightarrow\; \textbf{(C)}$$

💡 Case 1's two roots sum to $12$ by symmetry; add the lone Case 2 root $6$ to get $18$.

[1] #1 6.EE.A.3 Complete the square to rewrite the inside expression with a vertex form. $x^2 -

[2] #7 6.EE.A.4 Split into two subproblems via the definition of absolute value. The equation be

[3] #7 8.EE.A.2 Case 1: $(x - 6)^2 = 4$. Taking square roots gives $x - 6 = 2$ or $x - 6 = -2$,

[4] #7 8.EE.A.2 Case 2: $(x - 6)^2 = 0$. The only solution is $x = 6$. (This is the parabola's v

[5] #5 4.OA.A.3 Add all distinct solutions. Total sum $= 4 + 8 + 6 = 18$, which is choice (C).

Review

Reasonableness: Check each root in the original equation. $x = 4$: $16 - 48 + 34 = 2$, $|2| = 2\ \checkmark$. $x = 8$: $64 - 96 + 34 = 2$, $|2| = 2\ \checkmark$. $x = 6$: $36 - 72 + 34 = -2$, $|-2| = 2\ \checkmark$. All three valid, and $4 + 8 + 6 = 18$ matches (C).

Alternative: Tool #1 (Diagram) standalone — plot the parabola $y = (x - 6)^2 - 2$ (vertex $(6, -2)$, opens up) and the two horizontal lines $y = 2$ and $y = -2$. The intersection picture immediately shows two roots from $y = 2$ symmetric about $x = 6$ (summing to $12$) and one root from $y = -2$ at the vertex ($x = 6$), grand total $18$.

CCSS standards used (min grade 8)

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Adding the three roots $4 + 8 + 6$ at the end.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Rewriting $x^2 - 12x + 34$ as $(x - 6)^2 - 2$ by adding and subtracting $36$.)
6.EE.A.4 Identify when two expressions are equivalent (Splitting $|y| = 2$ into the two equivalent cases $y = 2$ and $y = -2$.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Solving $(x - 6)^2 = 4$ and $(x - 6)^2 = 0$ by taking square roots.)

⭐ This 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 $18$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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