2006 AMC 10B Problem 14

1. Recall a fact about quadratics like $x^2 - px + q$: if its roots are $r$ and $t$, it equals $(x - r)(x - t) = x^2 - (r + t)x + rt$. So the last number is the product of the roots.
2. Use that on the first equation $x^2 - mx + 2 = 0$: its roots $a$ and $b$ multiply to $ab = 2$.
3. Use it on the second equation: $q$ is the product of its roots,
$$q = \left(a + \frac{1}{b}\right)\left(b + \frac{1}{a}\right).$$
4. Multiply out the brackets, all four pieces:
$$a \cdot b \;+\; a \cdot \frac{1}{a} \;+\; \frac{1}{b} \cdot b \;+\; \frac{1}{b} \cdot \frac{1}{a} = ab + 1 + 1 + \frac{1}{ab}.$$
5. Put in $ab = 2$:
$$q = 2 + 1 + 1 + \frac{1}{2} = \frac{9}{2}.$$
We never needed $m$.

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

For a quadratic that starts with $x^2$, the middle number (with its sign flipped) is the sum of the roots and the last number is their product. This lets you work with the roots without solving for them. When a problem leaves a letter like $m$ unknown, that is a strong hint the answer does not depend on it.

Since $m$ does not matter, pick an easy value. With $m = 3$ the equation is $x^2 - 3x + 2 = (x-1)(x-2)$, so the roots are $1$ and $2$ (and indeed $1 \cdot 2 = 2$). The new roots are $1 + \frac12 = \frac32$ and $2 + \frac11 = 3$. Their product is $\frac32 \cdot 3 = \frac92$.

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