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Math Help - Finding complex roots of polynomials

  1. #1
    Rae
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    Finding complex roots of polynomials

    Find all solutions of the equation:

    x^4 + 2x^2 - 8 = 0


    I know the roots are either complex or irrational because none of the rational possibilities work out. Is there a way to find the other roots without using a graphing calculator? Thanks for any help.
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  2. #2
    Master Of Puppets
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    start by saying a = x^2 then solve a^2 + 2a - 8 = 0

    what do you get?
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  3. #3
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    Let u = x^2

    u^2 + 2u - 8 = (u-2)(u+4) = 0

    u = x^2 = 2 \: \implies x = \pm \sqrt{2}

    OR

    u = x^2 = -4 \: \implies x = \pm 2i \: \text{  where  } i^2 = -1
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    Also, by Descartes Rule of Signs, we know there are 1 positive and 1 negative solution; hence, we have a pair of complex numbers.

    If you use the formula \displaystyle \frac{P}{Q}, you can use synthetic division to find your 2 real solutions and then use \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} to find the complex pair.
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  5. #5
    Rae
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    Quote Originally Posted by dwsmith View Post
    Also, by Descartes Rule of Signs, we know there are 1 positive and 1 negative solution; hence, we have a pair of complex numbers.

    If you use the formula \displaystyle \frac{P}{Q}, you can use synthetic division to find your 2 real solutions and then use \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} to find the complex pair.
    Doesn't the formula P/Q only find the rational real roots and not the irrational?
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  6. #6
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    You equation doesn't contain irrational values.
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  7. #7
    Rae
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    The equation has 2 complex roots and 2 irrational roots. I tried using P/Q and none of the rational possibilties were roots.
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  8. #8
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    Why not tackle it head on?

    Factor it: (x^2+4)(x^2-2).

    From that the roots pop right out: x=\pm 2i~\&~\pm\sqrt2~.
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  9. #9
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    Quote Originally Posted by Rae View Post
    The equation has 2 complex roots and 2 irrational roots. I tried using P/Q and none of the rational possibilties were roots.
    You are correct. I really didn't look at the question. haha
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  10. #10
    Rae
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    Thanks, I understand how to find the solutions now.
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