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

Small hint:

By Vieta’s formulas,

ab=2

Big hint:

q

is the product

\left(a+\tfrac1b\right)\left(b+\tfrac1a\right)

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.