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determine the value of k for which the fuction f(x)=4x^2-3x+2kx+1 has 2 zeros.

  • Math -

    To have 2 zeros, or two real solutions, the discriminant must be positive
    for 4x^2 - 3x + 2kx + 1
    a = 4
    b = 2k-3
    c = 1

    b^2 - 4ac > 0
    (2k-3)^2 - 4(4)(1) > 0
    4k^2 - 12k -7 > 0
    (2k + 1)(2k - 7) > 0
    k < -1/2 OR k > 7/2

    You might want to put that into the interval notation you have been taught.

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