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(B) Given the quadratic equation $ax^2 + bx + c = 0$ with $a > 0, b < 0, c < 0$.Sum of roots ($S$.$O$.$R$.) $= \alpha + \beta = -\frac{b}{a}$. Since $

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$0 < |\alpha| < \beta$

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(B) Given the quadratic equation $ax^2 + bx + c = 0$ with $a > 0, b Sum of roots ($S$.$O$.$R$.) $= \alpha + \beta = -\frac{b}{a}$. Since $b 0$,$-\frac{b}{a} > 0$,so $\alpha + \beta > 0$.Product of roots ($P$.$O$.$R$.) $= \alpha \beta = \frac{c}{a}$. Since $c 0$,$\frac{c}{a} Since the product of the roots is negative,one root must be positive and the other negative. Let $\alpha 0$.Given $\alpha + \beta > 0$,the positive root must have a larger absolute value than the negative root. Thus,$|\beta| > |\alpha|$.Since $\beta > 0$,we have $\beta > |\alpha|$.Therefore,$0 < |\alpha| < \beta$.

If $ax^2 + bx + c = 0$ has real and distinct roots,$\alpha$ and $\beta$ $(\beta > \alpha)$. Further $a > 0, b < 0$ and $c < 0$ then :-

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