If a 0, b 0, c 0,then both the roots of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

Have negative real parts

Vedclass · Answer

(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 $ax^2 + bx + c = 0$:

Similar Questions

If $\frac{1 + \sqrt{3}i}{2}$ is a root of the equation $x^4 - x^3 + x - 1 = 0$,then its real roots are:

If $\frac{1+\sqrt{3}i}{2}$ is a root of the equation $x^4-x^2+x-1=0$,then its real roots are:

If $a > 0$,then the value of $\sqrt{a + \sqrt{a + \sqrt{a + \dots \infty}}}$ is:

If $x$ is a solution of the equation $\sqrt{2x + 1} - \sqrt{2x - 1} = 1$ for $x \ge \frac{1}{2}$,then $\sqrt{4x^2 - 1}$ is equal to:

If $p$ and $q$ are odd integers, then the roots of the equation $2px^{2} + (2p + q)x + q = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module