1989

JEE Main 1989 Mathematics question on Complex Numbers and Quadratic Equations

Let a,b,ca, b, c be real numbers, a≠0a \neq 0. If α\alpha is a root of a2x2+bx+c=0a^2x^2 + bx + c = 0, β\beta is the root of a2x2−bx−c=0a^2x^2 - bx - c = 0 and 0<α<β0 < \alpha < \beta, then the equation a2x2+2bx+2c=0a^2x^2 + 2bx + 2c = 0 has a root γ\gamma that always satisfies
  1. A.
    γ=α+β2\gamma = \dfrac{\alpha + \beta}{2}
  2. B.
    γ=α+β2\gamma = \alpha + \dfrac{\beta}{2}
  3. C.
    γ=α\gamma = \alpha
  4. D.
    α<γ<β\alpha < \gamma < \beta
Show answer

Correct option: D

[1989]

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