If $p, q, r$ are in G.P. and the equations: $$px^2 + 2qx + r = 0$$ $$dx^2 + 2ex + f = 0$$ Have a common root, then show that $$\frac{d}{p}, \frac{e}{q}, \frac{f}r$$ are in A.P.
Well I tried taking the sum and products of the root, but that didn't get me anywhere. I was told that there is an extremely short way of doing this problem. Can anyone help me?
Given that $p, q, r$ are in G.P., let $q=pR, r=pR^2$. $$\begin{align} px^2+2qx+r&=0\\ px^2+2pRx+rR^2&=0\\ x^2+2Rx+R^2&=0\\ (x-R)^2&=0\\ x&=-R \end{align}$$
Let $\dfrac dp=\lambda, \dfrac eq=\mu, \dfrac fr=\gamma$.
$$\begin{align} dx^2+2ex+f&=0\\ \lambda px^2+2\mu qx+\gamma r &=0\\ \lambda px^2+2\mu pRx+\gamma pR^2&=0\\ \lambda x^2+2\mu Rx+\gamma R^2&=0\\ \end{align}$$
We are told that both equations have the same root, hence $x=-R$ is a root.
$$\begin{align} \lambda R^2-2\mu R^2+\gamma R^2&=0\\ \lambda -2\mu +\gamma &=0\\ \gamma -\mu&=\mu-\lambda\\ \end{align}$$
i.e. $\lambda, \mu, \gamma$ are in A.P.