If $d,e,f$ are in GP  and the two quadratic equations $a{x}^2+2bx+c=0$ and $d{x}^2+2ex+f=0$ have a common root, then prove that $\frac{d}{a},\frac{e}{b},\frac{f}{c}$ are in HP.

Given equations

$$\begin{gathered} a{x}^2+2bx+c=0 \\ d{x}^2+2ex+f=0 \end{gathered}$$

$$\begin{gathered} x=\frac{-2b\pm \sqrt{4b^2-4ac}}{2a} x= \frac{-2e\pm \sqrt{4e^2-4df}}{2d} \\ x= \frac{-b\pm \sqrt{b^2-ac}}{a} x=\frac{-e\pm \sqrt{e^2-df}}{d} \end{gathered}$$

Now as $d,e,f$ are in GP hence $e^2=df$

$$\begin{gathered} \frac{-b+\sqrt{b^2-ac}}{a} = \frac{-e}{d} \\ -bd+d\sqrt{b^2-ac} = -ae \\ d\sqrt{b^2-ac} =(bd-ae) \end{gathered}$$

squaring on both sides;

$$\begin{gathered} d^2(b^2-ac) = b^2{d}^2-2abde+a^2{e}^2 \\ 2abde= a^2{e}^2+ae{d}^2 \to \left(1\right) \end{gathered}$$

as $\frac{d}{a},\frac{e}{b}.\frac{f}{c}$ are in H.P.

Hence ; $2\frac{b}{e}=\frac{a}{d}+\frac{c}{f}$

Therefore dividing Eqn (1) with $ad{e}^2$

$$\begin{gathered} \frac{2abde}{ad{e}^2}= \frac{a^2{e}^2}{ad{e}^{2 }}+\frac{ac{d}^2}{ad{e}^2} \\ \frac{2b}{e}=\frac{a}{d}+\frac{cd}{e^2}\to \left(2\right) \end{gathered}$$

Now $e^2=df$ putting in Eqn (2)

$$\begin{gathered} \frac{2b}{e}=\frac{a}{d}+\frac{cd}{df} \\ \frac{2b}{e}=\frac{a}{d}+\frac{c}{f} \end{gathered}$$

Hence Proved.