AMC 10 · 2018 · #5

Grade 9 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

polynomial-rootspolynomial-factoringquadratic-equationscasework systematic-enumerationcaseworkconvert-to-algebra ↑ Prerequisites: polynomial-factoringquadratic-equations

📏 Short solution 💡 2 insights

Problem

Two quadratics are given: one is fully known, x squared minus 3x plus 2; the other still has an unknown constant, x squared minus 5x plus k. For some values of that constant the two quadratics share a root, a single number making both zero. Find every such value and add them together.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Make a Systematic List

The phrase "all possible values" is a warning that a partial answer scores zero, so the candidates must be listed exhaustively. Tool #4 (Introduce a Variable) names the shared number r, which turns a vague overlap into two equations. Tool #2 (Make a Systematic List) is the engine: the fully known quadratic x² - 3x + 2 has exactly two roots, so r can only be one of two numbers, and that short list is guaranteed complete. Tool #7 (Identify Subproblems) then splits the work into one easy linear equation per candidate. Tool #3 (Eliminate Possibilities) checks each candidate actually produces a valid k before the values are added.

1STEP 1

Name the shared root

Give the shared root a name.

r² - 3r + 2 = 0 and r² - 5r + k = 0

2STEP 2

Factor the known quadratic

Factor the known quadratic.

x² - 3x + 2 = (x-1)(x-2)

3STEP 3

List the only candidates for r

There are only two candidates.

(r-1)(r-2) = 0 ⟹ r ∈ {1, 2}

4STEP 4

Case r = 1

The first candidate gives one value.

1² - 5(1) + k = 0 ⟹ 1 - 5 + k = 0 ⟹ k = 4

5STEP 5

Case r = 2

The second gives another.

2² - 5(2) + k = 0 ⟹ 4 - 10 + k = 0 ⟹ k = 6

6STEP 6

Confirm both, then add

Adding them gives 10.

x² - 5x + 4 = (x-1)(x-4), x² - 5x + 6 = (x-2)(x-3), 4 + 6 = 10 → (E)

Answer

The two working values 4 and 6 appear as choices (B) and (D), which is the trap: those are single values of k, while the question asks for their sum, 10. Choice (C) 5 would be the sum only if exactly one of the roots worked and gave k = 5, but k = 5 makes x² - 5x + 5, whose roots (5 ± √(5))/2 are irrational and therefore not 1 or 2. Choice (A) 3 matches nothing. A last sanity check on completeness: any shared root must be 1 or 2, both cases were used, and neither was double-counted, so 10 is the whole sum.

💡Key takeaway

A shared root has to be one of the roots you already know, so factor the polynomial you fully understand, test its two roots one at a time, and add up every k that survives.

Name the shared root
Factor the known quadratic
List the only candidates for r
Case r = 1
Case r = 2
Confirm both, then add