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Nature of roots of quadratic equations

  1. Dec 10, 2011 #1
    1. The problem statement, all variables and given/known data
    The equation kx2 - 3x + (k+2) = 0 has two distinct real roots. Find the set of possible values of k.


    2. Relevant equations
    Since the equation has two distinct real roots, b2 - 4ac > 0

    3. The attempt at a solution
    b2-4ac>0
    9-4(k+2)(k)>0
    9-4(k2+2k) >0
    9-4k2-8k>0
    = -4k2-8k+9>0
    Multiply both sides by -1,
    4k2+8k-9>0
    (4k-3)(k+3)>0

    -3<k<3/4

    However the answer is -2.46<k<0.458
    I'm lost, help please! :confused:
     
  2. jcsd
  3. Dec 10, 2011 #2

    SammyS

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    Your factoring is incorrect.

    (4k-3)(k+3) = 4k2 + 9k - 9 .

    Also, multiplying by -1 will change > to < .

    Solve 4k2+8k-9 = 0 by using the quadratic formula --- or by completing the square.
     
  4. Dec 10, 2011 #3
    ((-8 ± √208)/8) < 0

    -2.80 < k < 0.80

    I'm getting closer to the "answer" (-2.46<k<0.458) am I wrong, or is the "answer" wrong?
     
  5. Dec 10, 2011 #4

    eumyang

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    I'm getting the same roots (-2.80 and 0.80).

    When I played around with the coefficients of the original quadratic, I found that if you made the coefficient of the x2 term 2k:
    2kx2 - 3x + (k+2) = 0
    You will get the original answer that you stated: -2.4577 < k < 0.4577. So it looks like either you copied the problem incorrectly or the book has a typo somewhere.
     
  6. Dec 10, 2011 #5

    SammyS

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    The text book's answer is consistent with -8k2 - 16k +9 > 0 . equivalent to -4k2 - 8k + 9/2 >0

    It's hard to see how that's from a simple Typo -- unless the coefficient of x is should have been 3/√2 in the initial equation.
     
  7. Dec 12, 2011 #6
    I guess everyone makes mistakes.. it gets extremely annoying when textbooks provide you with the wrong answers or have a typo.
     
  8. Dec 13, 2011 #7

    epenguin

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    They are put there to help you to build self-confidence. :biggrin:
     
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