If $$a, b, c$$  are in G.P., then the equations $$a{x^2} + 2bx + c = 0$$     and $$d{x^2} + 2ex + f = 0$$     have a common root if $$\frac{d}{a},\frac{e}{b},\frac{f}{c}$$   are in-

Solution :
$$\eqalign{ & a,b,c{\text{ are in G}}{\text{.P}}{\text{.}} \cr & \,\,\,\,\,\,\,\,\,\,\,{b^2} = ac\,\,\,\,\,\,\,.....\left( 1 \right) \cr & \,\,\,\,\,\,\,\,\,\,a{x^2} + 2bx + c = 0 \cr & {\text{and }}d{x^2} + 2ex + f = 0{\text{ have a common root }} \cr & {\text{Let it be }}\alpha {\text{, then }}a{\alpha ^2} + 2b\alpha + c = 0 \cr & \,\,\,\,\,\,\,d{\alpha ^2} + 2e\alpha + f = 0 \cr & \Rightarrow \,\,\frac{{{\alpha ^2}}}{{2\left( {bf - ec} \right)}} = \frac{\alpha }{{cd - af}} = \frac{1}{{2\left( {ae - bd} \right)}} \cr & \Rightarrow \,\,{\alpha ^2} = \frac{{bf - ce}}{{ae - bd}};\alpha = \frac{{cd - af}}{{2\left( {ae - bd} \right)}} \cr & {\text{Substituting the value of }}\alpha {\text{, we get}} \cr & \,\,\,\,\,\,\,\frac{{{{\left( {cd - af} \right)}^2}}}{{4{{\left( {ae - bd} \right)}^2}}} = \frac{{bf - ce}}{{ae - bd}} \cr & \Rightarrow {\left( {cd - af} \right)^2} = 4\left( {ae - bd} \right)\left( {bf - ce} \right) \cr & {\text{Dividing both sides by }}{a^2}{c^2}{\text{ we get}} \cr & {\left( {\frac{d}{a} - \frac{f}{c}} \right)^2} = 4\left( {\frac{e}{c} - \frac{{bd}}{{ac}}} \right)\left( {\frac{{bf}}{{ac}} - \frac{e}{a}} \right) \cr & {\left( {\frac{d}{a} - \frac{f}{c}} \right)^2} = 4\left( {\frac{e}{c} - \frac{d}{b}} \right)\left( {\frac{f}{c} - \frac{e}{a}} \right)\,\,\,\,\,\left[ {{\text{Using eq}}{\text{.}}\left( 1 \right)} \right] \cr & \Rightarrow \,\frac{{{d^2}}}{{{a^2}}} + \frac{{{f^2}}}{{{c^2}}} - \frac{{2df}}{{ac}} = \frac{{4ef}}{{cb}} - \frac{{4{e^2}}}{{ac}} - \frac{{4df}}{{{b^2}}} + \frac{{4de}}{{ab}} \cr & \Rightarrow \,\frac{{{d^2}}}{{{a^2}}} + \frac{{{f^2}}}{{{c^2}}} + \frac{{4{e^2}}}{{{b^2}}} + 2\frac{d}{a}.\frac{f}{c} - 4\frac{e}{b}.\frac{f}{c} - 4\frac{d}{a}.\frac{e}{b} = 0\,\,\,\,\,\,\left[ {{\text{Using eq}}{\text{.}}\left( 1 \right)} \right] \cr & \Rightarrow {\left( {\frac{d}{a} + \frac{f}{c} - 2\frac{e}{b}} \right)^2} = 0 \cr & \Rightarrow \frac{d}{a} + \frac{f}{c} = \frac{{2e}}{b} \cr & \Rightarrow \frac{d}{a},\frac{e}{b},\frac{f}{c}{\text{ are in A}}{\text{.P}}{\text{.}} \cr} $$