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Help with finding all real solutions

  1. Sep 24, 2009 #1

    akp

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    1. The problem statement, all variables and given/known data
    x^4 - 8x^2 + 2 = 0


    2. Relevant equations
    quadratic formula?


    3. The attempt at a solution
    i would've tried using the quadratic formula, but im not sure if this would work with that seeing as how it's not ax^2 + bx + c
     
  2. jcsd
  3. Sep 24, 2009 #2
    You will need to use the factor/remainder theorem.
     
  4. Sep 24, 2009 #3

    symbolipoint

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    The example equation is in quadratic form, or something like it. First, you may use a substitution, something like, t=x^2. This can give you like, this:

    t^2 - 8t + 2 = 0

    Now, you can solve THAT one and you may obtain two solutions, but those solutions are for t. Now, solve each of those solutions for x (remember to first replace t with x^2 )
     
  5. Sep 24, 2009 #4
    The polynomial fortunately is quadratic, but in x2.

    If u = x2 then you have x4 -8x2 + 2 = u2 - 8u + 2 = 0.

    You can use the quadratic formula to solve for x2 keeping in mind that any negative roots from the formula must be discarded since x2 can never be negative.

    --Elucidus
     
  6. Sep 24, 2009 #5

    lurflurf

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    We have this field now called complex numbers that has negitive squares.
    i^2=-1 for example
     
  7. Sep 25, 2009 #6
    The thread title is "help with finding all real solutions" so I figured that complex solutions weren't needed.

    --Elucidus
     
  8. Sep 25, 2009 #7

    lurflurf

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    Good point. I like to keep track of the complex roots to assist in tracking the real roots. The fundamental theorem of a algebra impies each complex root means one less real root to find.
     
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