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Math Help - Factors which include imaginary numbers

  1. #1
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    Factors which include imaginary numbers

    I am sure this is a very basic procedure, but I am stuck. Can someone please take me through how you determine the solution of the following? Thanks, Frostking

    (1 - x) (3 - x) = 1

    also

    (2 - x) (- 2 - x) = -5

    I need these to determine Eigenvalues and I am lost.
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  2. #2
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    Quote Originally Posted by Frostking View Post
    I am sure this is a very basic procedure, but I am stuck. Can someone please take me through how you determine the solution of the following? Thanks, Frostking

    (1 - x) (3 - x) = 1

    also

    (2 - x) (- 2 - x) = -5

    I need these to determine Eigenvalues and I am lost.
    3 - 1x - 3x + x^2 = 1
    x^2 -4x + 3 = 1
    x^2 - 4x + 3 -1 = 0
    x^2 - 4x + 2 = 0

    Next step
    x^2 - 4x + 4 = -2+4
    x^2 - 4x + 4 = 2
    (x-2)^2 = 2
    Square root both sides of the equation
    x-2 = +- 1.414
    x = 2 +- 1.414
    x = 3.414, or x = 0.586
    Those are your roots/x-coordinates.
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    Factoring with imaginaries

    Thanks very much for showing me that I did not have to use i on that one. Could you please tell me how to if I do, say on this one:

    I have ( 1 - lambda) (- 3 - lambda) + 5 = 0

    So I get -3 + 2 lambda + lambda^2 + 5 = 0

    lambda^2 + 2 lambda + 2 = 0

    I know from the book that the answer is -1 + i and - 1 - i but I do not know how to get this. Do I just plug values into the quad formula? Thanks for your time. Frostking
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    Quote Originally Posted by Frostking View Post
    Thanks very much for showing me that I did not have to use i on that one. Could you please tell me how to if I do, say on this one:

    I have ( 1 - lambda) (- 3 - lambda) + 5 = 0

    So I get -3 + 2 lambda + lambda^2 + 5 = 0

    lambda^2 + 2 lambda + 2 = 0

    I know from the book that the answer is -1 + i and - 1 - i but I do not know how to get this. Do I just plug values into the quad formula? Thanks for your time. Frostking
    Hi Frostking,

    Yes, you just plug them into the quadratic formula.

    \lambda^2+2\lambda+2=0

    \lambda=\frac{-2\pm\sqrt{2^2-(4)(1)(2)}}{2(1)}

    \lambda=\frac{-2\pm\sqrt{-4}}{2}

    \lambda=\frac{-2\pm 2i}{2}

    \lambda=-1\pm i
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