2006 AMC 10B Problem 14

For a monic quadratic $x^2 - (\text{sum})x + (\text{product})$, the constant term is the product of the roots. So from the first equation $ab = 2$, and from the second
$$q = \left(a + \frac{1}{b}\right)\left(b + \frac{1}{a}\right).$$

Expand the product term by term:
$$ab + a\cdot\frac{1}{a} + \frac{1}{b}\cdot b + \frac{1}{ab} = ab + 1 + 1 + \frac{1}{ab}.$$
With $ab = 2$ this is $2 + 2 + \frac{1}{2} = \frac{9}{2}$.

The value of $m$ (and of $p$) is never needed.

The answer is $\boxed{\textbf{(D)}\ \frac{9}{2}}$.

Vieta's formulas let you compute symmetric expressions in the roots without solving for them. The target $q$ is a product of the new roots, and that product simplifies to an expression in $ab$ alone, so the unspecified $m$ is a red herring. When a coefficient is left as a letter, expect the answer to be independent of it.

Choose a convenient $m$: with $m = 3$ the roots of $x^2 - 3x + 2$ are $1$ and $2$. The new roots are $1 + \tfrac{1}{2} = \tfrac{3}{2}$ and $2 + 1 = 3$, whose product is $\tfrac{9}{2}$.

• Trying to find a and b explicitly from x^2 - mx + 2 = 0 with m unknown, or expanding the product and dropping the cross terms.
• Using the sign convention wrongly (taking the product of the roots to be $-2$), or computing $p$ instead of $q$.