Multiple choice

If $\alpha$ and $\beta(\alpha<\beta)$ be two different real roots of the equation $ax^2+bx+c=0,$ then

  1. $\displaystyle\alpha>-\frac { b }{ 2a } $
  2. $\displaystyle\beta <-\frac { b }{ 2a } $
  3. $\displaystyle\alpha<-\frac { b }{ 2a } <\beta $
  4. $\displaystyle\beta <-\frac { b }{ 2a } <\alpha $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a quadratic ax^2+bx+c=0 with roots alpha and beta, the vertex of the parabola is at x = -b/(2a). If alpha < beta, the vertex lies between the roots.

AI explanation

For any quadratic equation ax^2 + bx + c = 0 with two distinct real roots, the axis of symmetry is the vertical line passing through the vertex at x = -b / 2a. Because the parabola opens either upward or downward, the two roots must lie strictly on opposite sides of this axis of symmetry. Therefore, the smaller root alpha must be less than -b / 2a, and the larger root beta must be greater than -b / 2a, proving alpha < -b / 2a < beta.

Keep practicing — related questions