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(D) Given the quadratic equation $ax^2 + bx + c = 0$ with $a  0, c > 0$.Multiplying by $-1$,we get $-ax^2 - bx - c = 0$,where the coefficients

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

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(D) Given the quadratic equation $ax^2 + bx + c = 0$ with $a 0, c > 0$.Multiplying by $-1$,we get $-ax^2 - bx - c = 0$,where the coefficients are now $A = -a > 0, B = -b The product of the roots is $\alpha \beta = \frac{C}{A} = \frac{-c}{-a} = \frac{c}{a}$. Since $c > 0$ and $a The sum of the roots is $\alpha + \beta = -\frac{B}{A} = -\frac{-b}{-a} = -\frac{b}{a}$. Since $b > 0$ and $a 0$.Since the sum of the roots is positive and the product is negative,the root with the larger absolute value must be positive.Given $\alpha Since the sum $\alpha + \beta > 0$,we have $|\beta| > |\alpha|$,which means $\beta > |\alpha|$ is false,but rather the positive root $\beta$ has a larger magnitude than the negative root $\alpha$,i.e.,$|\beta| > |\alpha|$.Thus,$\alpha < 0 < |\alpha| < \beta$.

Let $a, b, c \in R$ 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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