Let $a$ and $b$ be the roots of the equation $x^2-mx+2=0$. Suppose that $a + \frac{1}{b}$ and $b + \frac{1}{a}$ are the roots of the equation $x^2-px+q=0$. What is $q$?
By Vieta's formulas, $ab = 2.$  Then
\[q = \left( a + \frac{1}{b} \right) \left( b + \frac{1}{a} \right) = ab + 1 + 1 + \frac{1}{ab} = 2 + 1 + 1 + \frac{1}{2} = \boxed{\frac{9}{2}}.\]
\frac{9}{2}
Level 2
Intermediate Algebra
Problem: Let $a$ and $b$ be the roots of the equation $x^2-mx+2=0$. Suppose that $a + \frac{1}{b}$ and $b + \frac{1}{a}$ are the roots of the equation $x^2-px+q=0$. What is $q$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of \( q \), we first need to determine the roots \( a \) and \( b \) of the equation \( x^2 - mx + 2 = 0 \). By Vieta's formulas, we know that \( a + b = m \) and \( ab = 2 \).

Next, we calculate \( a + \frac{1}{b} \) and \( b + \frac{1}{a} \). We can express these as follows:

1. \( a + \frac{1}{b} = a + \frac{1}{\frac{2}{a}} = a + \frac{a}{2} = \frac{3a}{2} \)
2. \( b + \frac{1}{a} = b + \frac{1}{\frac{2}{b}} = b + \frac{b}{2} = \frac{3b}{2} \)

Now, we have the new roots \( \frac{3a}{2} \) and \( \frac{3b}{2} \). We can find the sum and product of these new roots:

- The sum of the new roots is:
  \[
  \frac{3a}{2} + \frac{3b}{2} = \frac{3(a + b)}{2} = \frac{3m}{2}
  \]

- The product of the new roots is:
  \[
  \left(\frac{3a}{2}\right) \left(\frac{3b}{2}\right) = \frac{9ab}{4} = \frac{9 \cdot 2}{4} = \frac{18}{4} = \frac{9}{2}
  \]

From the quadratic equation \( x^2 - px + q = 0 \), we identify that \( q \) is equal to the product of the roots. Therefore, we conclude that \( q = \frac{9}{2} \).<|END_OF_LATENT|> The final answer is $\frac{9}{2}$.