2002 AMC 12B Problem 6 — Solution | LIVE by Po-Shen Loh

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2002 AMC 12B Problem 6

Below is the professionally curated solution for Problem 6 of the 2002 AMC 12B, 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.

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 quadratic

Difficulty rating: 1190

6.

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

$(-2,1)$

$(-1,2)$

$(1,-2)$

$(2,-1)$

$(4,4)$

Solution:

Since $a$ and $b$ are the roots, $x^2+ax+b=(x-a)(x-b)=x^2-(a+b)x+ab.$ 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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