2006 AMC 10B Problem 14

Attempt Problem 14 of the 2006 AMC 10B 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 2006 AMC 10B solutions, or check the answer key.

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14.

Let $a$ and $b$ be the roots of the equation $x^2-mx+2=0.$ Suppose that $a+\tfrac1b$ and $b+\tfrac1a$ are the roots of the equation $x^2-px+q=0.$ What is $q?$

$\dfrac{5}{2}$

$\dfrac{7}{2}$

$4$

$\dfrac{9}{2}$

$8$

Answer: D

Concepts:Vieta’s Formulas symmetry (algebra)

Difficulty rating: 1480

Solution:

Since $a$ and $b$ are roots of $x^2-mx+2,$ we have $ab=2.$

The value $q$ is the product of the new roots: $\begin{aligned} q&=\left(a+\tfrac1b\right)\left(b+\tfrac1a\right)\\ &=ab+1+1+\tfrac{1}{ab}\\ &=2+2+\tfrac12=\tfrac92. \end{aligned}$

Thus, the correct answer is D.

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