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B Why is k not smaller than zero?

  1. Jul 4, 2016 #1
    Question:

    The equation kx2+kx+3-k = 0, where k is a constant, has no real roots

    a) Show that 5k2-12k < 0
    b) Find the set of possible values of k

    Solution attempted:
    a) For no real roots, b2-4ac < 0

    a = k
    b = k
    c = 3 - k

    k2 - 4k(3 - k) < 0
    k2 -12k + 4k2 < 0
    5k2 - 12k < 0
    Hence shown

    b) 5k2 - 12k < 0
    k(5k-12) < 0
    So k < 0 or (5k-12) < 0, k< 12/5, k<2.4

    So k < 2.4

    But the answer says 0 < k < 2.4

    Why?
     
    Last edited: Jul 4, 2016
  2. jcsd
  3. Jul 4, 2016 #2
    Your ##b## should be squared.
     
  4. Jul 4, 2016 #3
    Done.
    I'm self-studying maths, and it doesn't say why in the book.
     
  5. Jul 4, 2016 #4

    mfb

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    Staff: Mentor

    If k<0, then k(5k-12) < 0 cannot be true because both factors are negative.
     
  6. Jul 4, 2016 #5
    Because if k<0, (5k-12) would be positive :wink:
    got it
     
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