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(D) Given the equation $ax^2 + bx + c = 0$ with $a 0, c > 0$.Multiply by $-1$ to get $-ax^2 - bx - c = 0$,where $-a > 0, -b < 0, -c < 0$.Let

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$\alpha < 0 < |\alpha| < \beta$

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(D) Given the equation $ax^2 + bx + c = 0$ with $a 0, c > 0$.Multiply by $-1$ to get $-ax^2 - bx - c = 0$,where $-a > 0, -b Let $f(x) = ax^2 + bx + c$. Since $a Since $f(0) = c > 0$,the $y$-intercept is positive.Since the product of roots $\alpha \beta = \frac{c}{a} Since $\alpha The sum of roots $\alpha + \beta = -\frac{b}{a}$. Since $a 0$,$-\frac{b}{a} > 0$,so $\alpha + \beta > 0$.This implies $|\beta| > |\alpha|$.Thus,$\alpha 0$ means the positive root $\beta$ has a larger absolute value than the negative root $\alpha$,i.e.,$|\beta| > |\alpha|$.Given $\alpha |\alpha|$,the correct relation is $\alpha < 0 < |\alpha| < \beta$.

Let $a, b, c \in \mathbb{R}$ and $\alpha, \beta$ be the real roots of the equation $ax^2 + bx + c = 0$. If $a < 0, b > 0, c > 0$ and $\alpha < \beta$,then:

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