AMC 10 · 2002 · #10

Grade 8 algebra
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
quadratic-equationsvieta-formulas convert-to-algebra ↑ Prerequisites: quadratic-equationsvieta-formulas
📏 Medium solution 💡 2 insights
Problem
The equation x² + ax + b = 0 is built from two numbers a and b, and its two solutions turn out to be those very same numbers a and b. Both a and b are nonzero. Find the pair (a,b) that makes this happen.

Pick an answer.

(A)
$\ (-2,1)$
(B)
$\ (-1,2)$
(C)
$\ (1,-2)$
(D)
$\ (2,-1)$
(E)
$\ (4,4)$

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Convert to Algebra

The whole sentence 'the roots are a and b' is a word fact, so Tool #13 (Convert to Algebra) turns it into symbols: rebuild the quadratic in factored form and match it term-by-term to x²+ax+b, which produces two equations linking a and b. Tool #3 (Eliminate Possibilities) then uses the 'nonzero' rule to throw away a stray branch when one equation could split two ways. Tool #6 (Guess and Check) confirms the winning pair by plugging it back and reading off the roots.

1STEP 1

Rebuild the quadratic from its roots

Roots a and b rebuild the quadratic directly: x²+ax+b = (x-a)(x-b) = x² - (a+b)x + ab — one quadratic, written two ways.

x²+ax+b=(x-a)(x-b)=x²-(a+b)x+ab
2STEP 2

Match the like parts

The two forms must agree term by term: the x-coefficients give a = -(a+b), the constants give b = ab.

a=-(a+b) and b=ab
3STEP 3

Apply the nonzero rule to the constant equation

Rearrange b = ab and factor: b(a-1)=0. Since b is promised nonzero, that branch dies and a = 1 is forced.

b=ab→ ab-b=0→ b(a-1)=0→ a=1 (since b ≠ 0)
4STEP 4

Solve for b and read off the pair

Put a=1 into a = -(a+b): 1 = -1 - b, so b = -2 and the pair is (1,-2) — choice (C).

1=-(1+b)→ 1=-1-b→ b=-2→ (a,b)=(1,-2) (C)
Answer
(1,-2)
Check the pair by rebuilding the equation: with a=1 and b=-2 it reads x²+x-2=0, which factors as (x-1)(x+2)=0, so its roots are 1 and -2 — exactly a and b, and both are nonzero as required. It is worth noting the algebra also permits a=b=-1/2, but that pair is not offered among the choices, so among the five options only (1,-2) survives.
💡Key takeaway

If a quadratic's own roots are its coefficients, rebuild it as (x-a)(x-b), match the pieces to x²+ax+b, and let the 'nonzero' rule pick the real answer.

  • Rebuild the quadratic from its roots
  • Match the like parts
  • Apply the nonzero rule to the constant equation
  • Solve for b and read off the pair

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