2002 AMC 10B Problems/Problem 10

Problem

Suppose that $a$ and $b$ are nonzero real numbers, and that the equation $x^2+ax+b=0$ has positive 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

From Vieta's Formulas, $ab=b$ and $a+b=-a$ . Since $b\ne 0$ , we have $a=1$ , and hence $b=-2$ . Our answer is $\boxed{(1,-2)\Rightarrow\text{(C)}}$ .

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2002 AMC 10B Problems/Problem 10

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