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    Question

    For what real value(s) of k does the quadratic equation

    - x² − (k + 3)x + 2k = 0, have equal real roots?
    A 2 Correct Answer Incorrect Answer
    B 1 Correct Answer Incorrect Answer
    D None of these Correct Answer Incorrect Answer

    Solution

    ATQ,

    For equal roots, discriminant D = 0. Here a = 1, b = −(k + 3), c = 2k. D = b² − 4ac = (k + 3)² − 4(1)(2k) = (k² + 6k + 9) − 8k = k² − 2k + 9 Set D = 0: k² − 2k + 9 = 0 Discriminant of this in k: Δ = (−2)² − 4(1)(9) = 4 − 36 = −32 < 0 Since Δ < 0, the equation in k has no real solution. So there is no real value of k for which the quadratic has equal real roots.

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