2002 AMC 12B Problem 6

Attempt Problem 6 of the 2002 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12B solutions, or check the answer key.

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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

$(-2,1)$

$(-1,2)$

$(1,-2)$

$(2,-1)$

$(4,4)$

Answer: C

Concepts:Vieta’s Formulas quadratic

Difficulty rating: 1190

Solution:

Since $a$ and $b$ are the roots, $\begin{gathered} x^2+ax+b \\ {}=(x-a)(x-b) \\ {}=x^2-(a+b)x+ab. \end{gathered}$ Matching coefficients gives $a+b=-a$ and $ab=b.$

As $b\neq0,$ the second equation gives $a=1,$ and then $a+b=-a$ gives $b=-2.$ So $(a,b)=(1,-2).$

Thus, the correct answer is C.

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