Find all values of $q$ such that the quadratic equation $qx^2 + (2q - 9)x + (q - 8) = 0$ has two positive real roots.
A quadratic with real roots has a discriminant that is either 0 or positive. Hence,
\begin{align*}
0\le (2q - 9)^2 - 4(q)(q-8) &= 4q^2- 36q + 81-4q^2+32q\\
&= -4q^2-4q+81\\
&= -4(q^2+q)+81.
\end{align*}Completing the square gives us $-4(q^2+q)-16+\allowbreak 97\le0$, or $-4(q+\tfrac 12)^2+109\le 0$. This is possible if and only if $109\le 4(q+\tfrac 12)^2\le 109+4=\allowbreak 113$, which happens if and only if $0\le (q -(-1/2))^2\le 113/4$. Since the left side is always nonnegative, it follows that $-\sqrt{113/4}\le q+ 1/2 \le \sqrt{113/4} $, which, when rearranged, gives us \[\boxed{ -\tfrac 12- \tfrac12 \sqrt{\tfrac {113}4} \le q \le -\tfrac 12 + \tfrac 12 \sqrt {\tfrac{113}4} }\]
For a quadratic equation $ax^2 + bx + c = 0$ to have two positive real roots, the discriminant must be positive, i.e., $b^2 - 4ac > 0.$
Applying this to our equation $qx^2 + (2q - 9)x + (q - 8) = 0,$ we have:
\[(2q - 9)^2 - 4q(q - 8) > 0.\]
Expanding and simplifying, we get:
\[4q^2 - 36q + 81 - 4q^2 + 32q > 0.\]
Combining like terms, we have:
\[12q + 81 > 0.\]
Subtracting $81$ from both sides, we have:
\[12q > -81.\]
Dividing both sides by $12,$ we have:
\[q > -\frac{81}{12}.\]
This simplifies to:
\[q > -\frac{27}{4}.\]
Therefore, the values of $q$ such that the quadratic equation has two positive real roots are $q \in \left(-\frac{27}{4}, \infty\right).$