Given two real numbers $1<p<q$ so that $\frac{1}{p} + \frac{1}{q} = 1$ and $pq = \frac{9}{2}$, what is $q$?
Solving $pq = \frac{9}{2}$ for $p$ we see that $p = \frac{9}{2q}$. Plugging this into $\frac{1}{p} + \frac{1}{q} = 1$ we then get \[ \frac{2q}{9} + \frac{1}{q} = 1 \Rightarrow 2q^2 - 9q +9 = 0 .\] Applying the quadratic equation we then see that  \[ q = \frac{9 \pm \sqrt{81-72}}{4} = \frac{9 \pm 3}{4} .\] Now, the smaller root corresponds to $p$ and the larger to $q$, so we see that $\boxed{q = 3}$.
3
Level 4
Algebra
Problem: Given two real numbers $1<p<q$ so that $\frac{1}{p} + \frac{1}{q} = 1$ and $pq = \frac{9}{2}$, what is $q$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for $q$, we start with the equations given: $\frac{1}{p} + \frac{1}{q} = 1$ and $pq = \frac{9}{2}$. 

From the first equation, we can express $q$ in terms of $p$:
1. Rearranging gives $\frac{1}{q} = 1 - \frac{1}{p}$.
2. This simplifies to $\frac{1}{q} = \frac{p-1}{p}$.
3. Taking the reciprocal, we find $q = \frac{p}{p-1}$.

Next, we substitute this expression for $q$ into the second equation:
1. We have $p \cdot q = p \cdot \frac{p}{p-1} = \frac{p^2}{p-1}$.
2. Setting this equal to $\frac{9}{2}$ gives the equation $\frac{p^2}{p-1} = \frac{9}{2}$.

To eliminate the fraction, we cross-multiply:
1. This results in $2p^2 = 9(p - 1)$.
2. Expanding the right side yields $2p^2 = 9p - 9$.

Rearranging this into standard quadratic form:
1. We get $2p^2 - 9p + 9 = 0$.

Next, we apply the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 2$, $b = -9$, and $c = 9$:
1. Calculating the discriminant: $b^2 - 4ac = (-9)^2 - 4 \cdot 2 \cdot 9 = 81 - 72 = 9$.
2. Thus, the square root of the discriminant is $\sqrt{9} = 3$.

Now substituting back into the quadratic formula:
1. We find $p = \frac{9 \pm 3}{4}$.
2. This gives two potential solutions: $p = \frac{12}{4} = 3$ and $p = \frac{6}{4} = \frac{3}{2}$.

Since we know $1 < p < q$, we analyze both cases:
1. If $p = 3$, then substituting into $q = \frac{p}{p-1}$ gives $q = \frac{3}{3-1} = \frac{3}{2}$, which contradicts $p < q$.
2. If $p = \frac{3}{2}$, then substituting gives $q = \frac{\frac{3}{2}}{\frac{3}{2}-1} = \frac{\frac{3}{2}}{\frac{1}{2}} = 3$.

This satisfies the condition $p < q$. Therefore, the value of $q$ is confirmed to be $3$.<|END_OF_LATENT|> The final answer is $3$.