2002 AMC 10B Problem 10 — Solution | LIVE by Po-Shen Loh

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2002 AMC 10B Problem 10

Below is the professionally curated solution for Problem 10 of the 2002 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:Vieta’s Formulas system of equations

Difficulty rating: 1280

10.

Suppose that $a$ and $b$ are nonzero real numbers, and that the equation $x^2 + ax + b = 0$ has solutions $a$ and $b.$ What is the pair $(a, b)?$

$(-2, 1)$

$(-1, 2)$

$(1, -2)$

$(2, -1)$

$(4, 4)$

Solution:

Since the roots are $a$ and $b,$ Vieta's formulas give $a + b = -a$ and $ab = b.$

From $ab = b$ with $b \ne 0,$ we get $a = 1.$ Then $a + b = -a$ gives $1 + b = -1,$ so $b = -2.$

Thus $(a, b) = (1, -2),$ and the correct answer is C.

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