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2002 AMC 12B Problem 6

Problem 6 of 25EasierAlgebra

Suppose that aa and bb are nonzero real numbers, and that the equation x2+ax+b=0x^2+ax+b=0 has solutions aa and b.b. Then the pair (a,b)(a,b) is

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Solution

Since aa and bb are the roots, x2+ax+b=(x−a)(x−b)=x2−(a+b)x+ab. \begin{gathered} x^2+ax+b \\ {}=(x-a)(x-b) \\ {}=x^2-(a+b)x+ab. \end{gathered} Matching coefficients gives a+b=−aa+b=-a and ab=b.ab=b. As b≠0,b\neq0, the second equation gives a=1,a=1, and then a+b=−aa+b=-a gives b=−2.b=-2. So (a,b)=(1,−2).(a,b)=(1,-2). Thus, the correct answer is C.
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Tagged: Vieta’s Formulas · quadratic

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