$ax^2+2bx+c=0$ and $px^2+2qx+r=0$ has a common root. If terms $\frac ap, \frac bq, \frac cr$ lie on consecutive arithmetic series, Prove that terms p, q, r lie on consecutive geometric series.
My Try
Since they have a common root I derived the expression,
$$\frac{rb-cq}{qa-bp}=\frac{(ra-cp)^2}{4(bp-qa)^2}$$
How can I continue? any hint is appreciated. Thank you!
Since the two polynomial have a common root, we can write $ax^2+2bx+c=a(x-\alpha)(x-\beta)=0$ and $px^2+2qx+r=p(x-\alpha)(x-\gamma)=0$. In the following, thanks to the hypotesis of arithmetic progression we will suppose $p,q,r,\gamma,\alpha\neq 0$. Also $a\neq 0$ to have a polynomial of degree $2$.
Thanks to the hypotesis we have the following informations (the first four equations are the Vieta's formula, and the last one is the arithmetic progression): \begin{equation} \begin{cases} \alpha\beta = \frac{c}{a}\\ \alpha+\beta = -\frac{2b}{a}\\ \alpha\gamma = \frac{r}{p}\\ \alpha + \gamma = -\frac{2q}{p}\\ \frac{b}{q}-\frac{a}{p} = \frac{c}{r} - \frac{b}{q} \end{cases} \quad\longrightarrow \quad \begin{cases} c = a\alpha\beta\\ b = -\frac{a(\alpha+\beta)}{2}\\ r = p\alpha \gamma\\ q= - \frac{p(\alpha+\gamma)}{2}\\ \frac{a(\alpha+\beta)}{p(\alpha+\gamma)}- \frac{a}{p} = \frac{a\beta}{p\gamma}- \frac{a(\alpha+\beta)}{p(\alpha+\gamma)} \end{cases} \end{equation}
Now continue to compute the last equation: we can simplify the factor $\frac{a}{p}$ and: \begin{gather} \frac{\alpha+\beta}{\alpha+\gamma} - 1 = \frac{\beta}{\gamma} - \frac{\alpha+\beta}{\alpha+\gamma}\\ \frac{\gamma(\alpha+\beta)-\gamma(\alpha+\gamma)}{\gamma(\alpha+\gamma)} = \frac{\beta(\alpha+\gamma)-\gamma(\alpha+\beta)}{\gamma(\alpha+\gamma)}\\ \alpha\gamma+\beta\gamma-\alpha\gamma-\gamma^2 = \alpha\beta+\beta\gamma-\alpha\gamma-\beta\gamma\\ \gamma^2-\alpha\gamma-\beta\gamma+\alpha\beta =0\\ (\gamma-\alpha)(\gamma-\beta)=0 \end{gather} So we have only two possible cases: $\gamma=\alpha$ or $\gamma=\beta$.
Now $p,q,r$ are in geometric series if and only if $$ \frac{p}{q}=\frac{q}{r} \ \Longleftrightarrow \ q^2 = pr $$ and sobstituting $r$ and $q$ (as they appear in the system above) we obtain: \begin{gather} q^2 = pr\\ p^2\frac{(\alpha+\gamma)^2}{4} = p^2\alpha\gamma\\ (\alpha+\gamma)^2 = 4\alpha\gamma\\ (\gamma-\alpha)^2 = 0 \end{gather} So if we require that the two starting polynomial have exactly one root in common (or $\alpha=\gamma=\beta$), we have the statement because the condition of arithmetic progression forces to $\gamma=\alpha$.
If not, it's possible that $\gamma=\beta$ and in this case $p,q,r$ are not in geometric progression (in general). Take for example $\beta=\gamma=1$, $\alpha=2$, $a=p=1$.
