Question

If, (a + 1)x^2 + 2(a+1)x + (a – 2) = 0, then, for what parameter of ‘a’ the given equation have imaginary roots?

(a) (-∞, -1)

(b) (-1, ∞)

(c) (-1, 1)

(d) (-∞, ∞)

This question was addressed to me in unit test.

Asked question is from Applications of Quadratic Equations topic in section Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer 1

The correct option is (a) (-∞, -1)

For explanation I would say: For, imaginary roots, D > 0

Where, D = b^2 – 4ac

In the equation, (a + 1)x^2 + 2(a+1)x + (a – 2) = 0

D = [2(a+1)]^2 – 4 (a + 1)(a – 2)

    = 4a^2 + 4 + 8a – 4{a^2 – 2a + a – 2}

    = 4a^2 + 4 + 8a – 4a^2 + 4 a + 8 < 0

    => 12a + 12 < 0

    => 12a < -12

    => a < -1

Therefore, a € (-∞, -1)