Mathematics - Quadratic Equation Question with Solution | TestHub

$\because a,b,c$are in G.P.$\Rightarrow b^2=ac$$\dots \left(i\right)$
Equation$a{x}^2+2bx+c=0$
Hence,$D=4b^2-4ac=0$$\left(\because b^2=ac\right)$
$\Rightarrow$equation has equal roots$\Rightarrow$roots are$=\frac{-b}{a},\frac{-b}{a}$($\because$sum of roots$\alpha +\alpha =\frac{-2b}{a}$)
$\because$Both equations have one root in common.
$\therefore$Common root is$-\frac{b}{a}$. It will satisfy equation.
$\therefore d{\left(\frac{-b}{a}\right)}^2+2e\left(-b\right)a+f=0\Rightarrow d{b}^2-2aeb+a^{2}f=0$
$\Rightarrow d\left(ac\right)-2aeb+a^{2}f=0$$\left(\because b^2=ac\right)$
$\Rightarrow dc-2aeb+af=0$
$\Rightarrow \frac{dc}{ac}-\frac{2eb}{ac}+\frac{af}{ac}=0$(dividing by$ac$)
$\Rightarrow \frac{d}{a}-\frac{2eb}{b^2}+\frac{f}{c}=-\Rightarrow \frac{d}{a}+\frac{f}{c}=\frac{2e}{b}$
$\Rightarrow \frac{d}{a},\frac{e}{b},\frac{f}{c}$are in A.P.