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Imaginary roots

  1. Oct 27, 2014 #1
    1. The problem statement, all variables and given/known data
    If both roots of the equation ax^2 + x + c - a = 0 are imaginary and c > -1, then:

    Ans: 3a < 2+4c

    2. Relevant equations
    Discriminant < 0 for img roots
    Vieta

    3. The attempt at a solution
    1-4(a)(c-a)<0
    4ac > 4a^2 + 1
    Minimum value of 4a^2 + 1 is 1 so
    4ac>1

    I can't think of anything beyond this and I have no idea how to incorporate c>-1 into my equations.
     
  2. jcsd
  3. Oct 27, 2014 #2

    HallsofIvy

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    You mean "both roots are complex" not "imaginary", don't you? It is impossible for that equation to have imaginary roots.
     
  4. Oct 27, 2014 #3

    PeroK

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    Okay, so far. What sort of inequality is that in relation to a?

    ... although, I don't see how you get that answer and c > 1 would make more sense. Are you sure you've got the question right?
     
    Last edited: Oct 27, 2014
  5. Oct 27, 2014 #4
    Both roots are complex is the same as both roots are imaginary, I think.

    The question is correct, at least should be since it's a question from a test I took yesterday.
    I don't have any idea what sort of inequality that is in relation to a.
     
  6. Oct 27, 2014 #5

    PeroK

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    A complex number has real and imaginary parts. If both roots were imaginary, they would be ##\pm bi## with no real part.

    It's a quadratic inequality in a. Never miss a quadratic!

    Here's how I would prove the question is wrong:

    Let c = 0 (which is > -1), so we have:

    ##ax^2 + x - a = 0##

    The discriminant is ##4a^2+1## which is +ve, so this equation never has complex roots. And the condition 3a < 2 (=2 + 4c) is not relevant.

    Ergo the question is wrong.
     
  7. Oct 27, 2014 #6
    Wait, why is the condition 3a<2 not relevant?
     
  8. Oct 27, 2014 #7

    PeroK

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    The answer says that the equation has complex roots when 3a < 2 + 4c. In this case, when 3a < 2 (as c = 0).

    But, with c = 0 the equation never has complex roots. Whether 3a < 2 or not has nothing to do with it.

    As an exercise, try c = 1. You should be able to work out when the equation has complex roots and show that it not when 3a < 6
     
  9. Oct 28, 2014 #8
    The equation always has complex roots for c=1.
    That was neat. Thank you!
     
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