AMC 10 · 2007 · #21

Grade 11 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Symmetric Function of Roots via Vieta

vieta-formulasquadratic-equationspolynomial-rootsfunction-evaluation vieta-formulaslogical-deductionguess-and-check ↑ Prerequisites: vieta-formulasquadratic-equations

📏 Long solution 💡 4 insights

Problem

For a quadratic, the sum of its zeros, the product of its zeros and the sum of its coefficients are all equal. Decide which description that shared value must match.

Pick an answer.

(A)

$\textrm{the\ coefficient\ of\ }x^{2}$

(B)

$\textrm{the\ coefficient\ of\ }x$

(C)

$\textrm{the\ y-intercept\ of\ the\ graph\ of\ }y=f(x)$

(D)

$\textrm{one\ of\ the\ x-intercepts\ of\ the\ graph\ of\ }y=f(x)$

(E)

$\textrm{the\ mean\ of\ the\ x-intercepts\ of\ the\ graph\ of\ }y=f(x)$

How to solve

Strategy Convert to Algebra

Every phrase in the problem — sum of zeros, product of zeros, sum of coefficients, y-intercept, x-intercept — is a sentence about a, b, c waiting to be written as a formula, so Tool #13 (Convert to Algebra) is the spine. Tool #4 (Introduce a Variable) names the zeros r and s so that Vieta's relations can be derived from the factored form rather than quoted, which also makes clear that the zeros may be complex. Tool #15 (Organize Information in More Ways) supplies the reframing that does the real damage: the sum of the coefficients is not a new object at all, it is f(1). Tool #3 (Eliminate Possibilities) is needed because "must" demands more than finding one description that fits — the other four have to fail in general, and here the strongest version of that is available: solve the hypothesis completely, get the entire family of qualifying quadratics, then test all five descriptions against the whole family at once. Tool #6 (Guess and Check) closes with one concrete member of that family as an independent numerical check.

1STEP 1

Two zeros, always

A nonzero leading coefficient means there are always two zeros.

a ≠ 0 → f(x)=a(x-r)(x-s), r,sinC with multiplicity

2STEP 2

Write the three quantities

Writing the coefficients through the zeros makes all three quantities comparable.

b=-a(r+s), c=ars → r+s=-b/a, rs=c/a, a+b+c=f(1)

3STEP 3

Use the first equality only

The first equality alone gives a relation between two coefficients.

-b/a=c/a a ≠ 0{⟹} -b=c ⟺ b+c=0

4STEP 4

Watch b and c cancel

Substituting makes them cancel, leaving the leading coefficient.

v=a+b+c=a+b+(-b)=a

5STEP 5

Check such a quadratic exists

Such quadratics really exist, so the claim is not vacuous.

a=-b/a→ b=-a², c=a²; f(x)=a(x²-ax+a), v=a, a ≠ 0

6STEP 6

Rule out (B) and (C)

Two other descriptions match only for a single leading coefficient.

b=-a²=a⇔ a=-1; f(0)=c=a²=a⇔ a=1

7STEP 7

Rule out (D) and (E)

The last two never match at all, so they are ruled out.

f(a)=a³-a³+a²=a² ≠ 0; (r+s)/2=a/2 ≠ a since a ≠ 0

8STEP 8

Test one concrete quadratic

A concrete quadratic confirms the leading coefficient, choice (A).

a=4: f(x)=4x²-16x+16=4(x-2)²; 2+2=4, 2 · 2=4, 4-16+16=4; v=4=coefficient of x² → (A)

Answer

the coefficient of x²

Test the two members that came closest to breaking the conclusion. At a=1: f(x)=x²-x+1, sum of the zeros =1, product =1, sum of the coefficients =1-1+1=1, so the hypothesis holds with v=1; here the y-intercept is also 1, so (C) happens to be right, yet the coefficient of x² is 1 as well, so (A) is still right. At a=-1: f(x)=-x²-x+1, so b=-1 and c=1; the sum of the zeros is -b/a=--1/-1=-1, the product is c/a=1/-1=-1, and the sum of the coefficients is -1-1+1=-1, so v=-1; here the coefficient of x is -1, so (B) happens to be right, and again the coefficient of x² is -1, so (A) is right too. This is the pattern the argument predicts: the distractors can coincide with the answer at isolated values of a, but only (A) survives every value. It also passes a units-style check — v=a scales with the quadratic, matching the fact that multiplying f by a constant k leaves the zeros unchanged but multiplies both a and the sum of the coefficients by k, so a description depending on the zeros alone, like (D) or (E), could never keep up.

💡Key takeaway

The sum of the coefficients is just f(1), and making the zeros' sum equal their product is exactly what cancels the middle of (1-r)(1-s) — leaving the leading coefficient behind.

Two zeros, always
Write the three quantities
Use the first equality only
Watch b and c cancel
Check such a quadratic exists
Rule out (B) and (C)
Rule out (D) and (E)
Test one concrete quadratic