2002 AMC 12B Problems/Problem 6

Problem

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

$\mathrm{(A)}\ (-2,1) \qquad\mathrm{(B)}\ (-1,2) \qquad\mathrm{(C)}\ (1,-2) \qquad\mathrm{(D)}\ (2,-1) \qquad\mathrm{(E)}\ (4,4)$

Solution

Since $(x-a)(x-b) = x^2 - (a+b)x + ab = x^2 + ax + b = 0$ , it follows by comparing coefficients that $-a - b = a$ and that $ab = b$ . Since $b$ is nonzero, $a = 1$ , and $-1 - b = 1 \Longrightarrow b = -2$ . Thus $(a,b) = (1,-2) \Rightarrow \mathrm{(C)}$ . Another method is to use Vieta's formulas. The sum of the solutions to this polynomial is equal to the opposite of the $x$ coefficient, since the leading coefficient is 1; in other words, $a + b = -a$ and the product of the solutions is equal to the constant term (i.e, $a*b = b$ ). Since $b$ is nonzero, it follows that $a = 1$ and therefore (from the first equation), $b = -2a = -2$ . Hence, $(a,b) = (1,-2) \Rightarrow \mathrm{ (C)}$

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2002 AMC 12B Problems/Problem 6

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