If a, b and c(a 0, c 0) are in GP, then consider the following in...

Given that $a, b, c$ are in GP $\therefore \quad b^2=a c$ Now, $a x^2+b x+c=0$$\begin{aligned} \mathrm{D} & =b^2-4 a c \\ & =a c-4 a c=-3 a c \lt 0 \end{aligned}$1. $\therefore$ Roots are imaginary (correct) Let, $\quad a=2, b=4, c=8\begin{aligned} \therefore \quad x^2+2 x+4 & =0 \\ \Rightarrow \quad x & =\frac{-2 \pm \sqrt{4-16}}{2}=-1 \pm \sqrt{3} \end{aligned}$ 2. Ratio of roots $=\frac{-1+\sqrt{3}}{-1-\sqrt{3}}=\frac{\omega}{\omega^2}=\frac{1}{\omega}$ (correct) 3.\begin{aligned} \text{ Product of roots } & =\frac{(-1+\sqrt{3})(-1-\sqrt{3})}{2} \times 4 \\ & =\omega \cdot \omega^2 \cdot 4=4=\frac{b^2}{a^2} \end{aligned}$$ (correct) Hence, statements 1,2 and 3 are correct.