This began as a comment to Foool's answer, but grew.
The ratios given lead to a set of solutions $r = q + k$, $p = q + 3k$, where the term ratio of the geometric progression is $\left(\frac{3}{2}\right)^\frac{1}{k}$. (Foool's answer is therefore the special case $q=k=1$). Note that $k$ is integral and non-zero, but may be negative; $q$ can be any integer.
Therefore you have $px^2 + 2qx - 2r = (q+3k)x^2 + 2qx - 2(q+k)$
Jumping straight to the quadratic equation, $\begin{eqnarray}x & = & \frac{-q \pm \sqrt{q^2 + 2(q+3k)(q+k)} }{q+3k} \\ & = & \frac{-q \pm \sqrt{3q^2 + 8qk + 6k^2} }{q+3k}\end{eqnarray}$
So it looks quite complicated.
We can derive a Sturm sequence:
$\begin{eqnarray} P_0 & = & (q+3k)x^2 + 2qx - 2(q+k) \\ P_1 & = & (q+3k)x + q \\ P_2 & = & -qx + 2(q+k) \\ P_3 & = & \frac{q^2+8qk+6k^2}{q} \end{eqnarray}$ where I've scaled P_1 by a positive scalar as a computational optimisation. If we consider sign changes in
$\begin{eqnarray} P_0(0) & = & - 2(q+k) \\ P_1(0) & = & q \\ P_2(0) & = & 2(q+k) \\ P_3(0) & = & \frac{q^2+8qk+6k^2}{q} \end{eqnarray}$ we have one between $P_0(0)$ and $P_2(0)$, and a second if $ \frac{(q+k)(q^2+8qk+6k^2)}{q} < 0$
If we consider sign changes in
$\begin{eqnarray} P_0(1) & = & q+5k \\ P_1(1) & = & 2q+3k \\ P_2(1) & = & q+2k \\ P_3(1) & = & \frac{q^2+8qk+6k^2}{q} \end{eqnarray}$
And again we see that it's quite complicated. With some patience one can draw a graph showing the regions where the number of sign changes is the same, and hence derive the number of roots, but I think there may be some criterion missing from the question.