If ax^2 + bx + c = 0 has real and distinct ro | vedclass

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Question

(B) Given the quadratic equation $ax^2 + bx + c = 0$ with $a > 0, b

Vedclass · Accepted Answer

$0 < |\alpha| < \beta$

Vedclass · Answer

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

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

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