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(B) The roots of the quadratic equation $ax^2 + bx + c = 0$ are given by the quadratic formula:$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$Case $1$: If th

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Have negative real parts

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(B) The roots of the quadratic equation $ax^2 + bx + c = 0$ are given by the quadratic formula:$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$Case $1$: If the discriminant $D = b^2 - 4ac \ge 0$,the roots are real.Since $a, b, c > 0$,we have $\sqrt{b^2 - 4ac} Thus,both roots $x_1 = \frac{-b + \sqrt{D}}{2a}$ and $x_2 = \frac{-b - \sqrt{D}}{2a}$ are negative because the numerator is negative and $a > 0$.Case $2$: If $D = b^2 - 4ac $x = \frac{-b}{2a} \pm i \frac{\sqrt{4ac - b^2}}{2a}$The real part of these roots is $	ext{Re}(x) = -\frac{b}{2a}$.Since $a > 0$ and $b > 0$,the real part $-\frac{b}{2a}$ is always negative.Conclusion: In all cases,the roots have negative real parts.

If $a > 0, b > 0, c > 0$,then both the roots of the equation $ax^2 + bx + c = 0$:

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