If $0<\mathrm{a}<\mathrm{b}<\mathrm{c}$ and the roots  $\mathrm{\alpha },\mathrm{\beta }$ of the equation  ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$ are imaginary,  then

Since the roots are imaginary
$\therefore {\mathrm{b}}^2-4\mathrm{ac}<0$
The roots α and ß are given by
$\mathrm{\alpha }=\frac{-\mathrm{b}+\mathrm{i}\sqrt{4\mathrm{ac}-{\mathrm{b}}^2}}{2\mathrm{a}}\mathrm{\beta }=\frac{-\mathrm{b}-\mathrm{i}\sqrt{4\mathrm{ac}-{\mathrm{b}}^2}}{2\mathrm{a}}$
Clearly$\mathrm{\alpha }=\overline{\beta }\boxed{\therefore \left|\mathrm{\alpha }\right|=\left|\mathrm{\beta }\right|}$
And $\left|\mathrm{\alpha }\right|=\sqrt{\frac{{\mathrm{b}}^2}{4{\mathrm{a}}^2}+\frac{4\mathrm{ac}-{\mathrm{b}}^2}{4{\mathrm{a}}^2}}=\sqrt{\frac{\mathrm{c}}{\mathrm{a}}}\therefore \left|\mathrm{\alpha }\right|>1(\because \mathrm{c}>\mathrm{a})$