I found a couple of interesting properties. First stating the obvious: $ p_1 + p_2 = - \frac {b}{a}$ $ p_1p_2 = \frac {c}{a}$ $ q_1 + q_2 = - \frac {b}{c}$ $ q_1q_2 = \frac {a}{c}$ Now the properties. $ p_1p_2q_1q_2 = 1$ $ q_1 + q_2 = \frac {p_1 + p_2}{p_1p_2}$ $ p_1 + p_2 = \frac {q_1 + q_2}{q_1q_2}$ $ q_1 - p_1 + q_2 - p_2 = \frac {p_1 + p_2}{p_1p_2} - \frac {q_1 + q_2}{q_1q_2} = p_1p_2(q_1 + q_2) - q_1q_2(p_1 + p_2) = \frac {b(a - c)}{ac}$ $ ap_1 = cq_1$ $ ap_2 = cq_2$ Unfortunatley, no results . Maybe this will be useful to someone else..

Let $ f,g$ be the two quadratics respectively; then $ g(x) = x^2f\left(\frac 1x\right)$. If $ f(0) = c = 0$ then $ g$ does not have two distinct roots, so the roots of $ g$ are $ \frac 1{p_1}, \frac 1{p_2}$. If $ p_1,\, \frac{1}{p_2},\, p_2,\, \frac{1}{p_1}$ is an arithmetic progression then $ p_1 + \frac{1}{p_1} = p_2 + \frac{1}{p_2}$; but $ x + \frac{1}{x}$ is a monotonically increasing function for $ x \ge 1$ and hence we have either $ p_1 = p_2$ or $ q_1 = p_2$, both of which make the progression constant and contradicting distinct roots. If $ p_1,\, \frac {1}{p_1},\, p_2,\, \frac{1}{p_2}$ is an arithmetic progression, then $ p_1$ and $ p_2$ have opposite signs (if they have same sign then consider the distance of their absolute value from $ 1$), so $ p_1 - \frac 1{p_1} = p_2 - \frac{1}{p_2}$. Now $ x - \frac 1x$ is strictly increasing, and as $ p_1 \neq p_2$ we must have $ p_1 = -\frac{1}{p_2} \Longrightarrow p_1p_2 = \frac{-c}{a} = 1$ as desired.