AMC 10 · 2018 · #21

Grade 11 algebra
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
polynomial-rootssign-analysisexponentslogical-deduction easier-related-problempattern-recognitionlogical-deduction ↑ Prerequisites: polynomial-rootsexponents
📏 Medium solution 💡 3 insights
Problem
Five expressions are listed. Each one, set equal to zero, has exactly one real solution. The task is to decide whose real solution sits farthest to the right on the number line. Four of the five are high-degree polynomials built from two odd powers plus one; the fifth is linear. None of the four can be solved in closed form, so the roots have to be ranked without ever being found.

Pick an answer.

(A)
$x^{19}+2018x^{11}+1$
(B)
$x^{17}+2018x^{11}+1$
(C)
$x^{19}+2018x^{13}+1$
(D)
$x^{17}+2018x^{13}+1$
(E)
2019x+2018
How to solve
Strategy Eliminate Possibilities

Solving a degree-19 equation is off the table, so the route has to avoid roots entirely. Tool #16 (Change Focus) supplies the pivot that makes the whole problem collapse: stop asking where each curve crosses zero and ask which curve is lower. For two rising functions on the same window, the lower graph is the one still short of zero when the higher one has already arrived, so it must catch up farther to the right — lowest graph, greatest root. To use that pivot the roots must first be confined to one common window, which is Tool #9 (Solve an Easier Related Problem): instead of solving, plug in the two easiest inputs, x=-1 and x=0, and let the sign change trap all four big roots in (-1,0). Tool #5 (Look for a Pattern) then supplies the one fact that ranks the graphs, namely that on (-1,0) a higher odd power is a larger number. With that, Tool #3 (Eliminate Possibilities) finishes the job as a knockout: three choices fall to the height comparison, and the linear choice falls to a single sign test.

1STEP 1

Trap the four big roots in (-1,0)

All four roots sit in one narrow interval.

A(-1)=B(-1)=C(-1)=D(-1)=-2018 ＜ 0, A(0)=B(0)=C(0)=D(0)=1 ＞ 0
2STEP 2

Lower graph means later crossing

The lower graph crosses later.

g(x) ＜ f(x) on (-1,0) ⟹ g(r_f) ＜ 0 ⟹ r_g ＞ r_f
3STEP 3

Bigger odd exponent, bigger value

On the negatives, a bigger exponent gives a bigger value.

-1 ＜ x ＜ 0 ⟹ x¹¹ ＜ x¹³ ＜ x¹⁷ ＜ x¹⁹ ＜ 0
4STEP 4

Knock out (A), (C) and (D)

Differencing knocks out three choices at once.

A-B=x¹⁷(x²-1) ＞ 0, D-B=2018x¹¹(x²-1) ＞ 0, C-B=(A-B)+(D-B) ＞ 0
5STEP 5

Test the linear choice (E)

Checking the linear one too, the seventeenth-power expression wins.

B (-2018/2019) ＜ 1-2018(1-11/2019) ＜ -1996 ＜ 0 ⟹ r_B ＞ -2018/2019 (B)
Answer
x¹⁷+2018x¹¹+1
Pin the roots down numerically and check the ranking survives. Write a root of (B) as -t with 0 ＜ t ＜ 1; the equation becomes t¹⁷+2018t¹¹=1. Since t ＜ 1 forces t¹⁷ ＜ t¹¹, the two terms are trapped between 2018t¹¹ ＜ 1 and 2019t¹¹ ＞ 1, so 1/2019 ＜ t¹¹ ＜ 1/2018 and t≈ 2018⁻¹/11≈ 0.5007. The choices carrying x¹³ instead obey 2018t¹³≈ 1, giving t≈ 2018⁻¹/13≈ 0.5569. Refined numerically, the five roots are (B) ≈-0.5006709, (A) ≈-0.5006711, (D) ≈-0.5568955, (C) ≈-0.5568969, and (E) =-0.9995047 exactly. That is precisely the order the elimination produced, with (B) on top. It also explains the shape of the answer: the coefficient 2018 sits on the low-exponent slot, so the 11-versus-13 decision moves the root by about 0.056 while the 17-versus-19 decision only moves it in the seventh decimal place — and (E), pinned near -1, was never in contention.
💡Key takeaway

You do not have to find a root to compare roots: when two graphs both rise, the one running lower has to travel farther right before it reaches zero.

  • Trap the four big roots in (-1,0)
  • Lower graph means later crossing
  • Bigger odd exponent, bigger value
  • Knock out (A), (C) and (D)
  • Test the linear choice (E)

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