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Math Help - Real and complex solutions

  1. #1
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    Real and complex solutions

    Can someone help me with this one?

    Find the real and complex solution of the equation x[squared]-8=0.

    Possible answers:
    1. -2, 2, 2cis(pi/3)
    2. 2, 2cis(2pi/3), 2cis(4pi/3)
    3. -2, 2cis(pi/3), 2cis(2pi/3)
    4. 2cis(pi/3), 2cis(2pi/3), 2cis(4pi/3)
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mike1 View Post
    Can someone help me with this one?

    Find the real and complex solution of the equation x[squared]-8=0.

    Possible answers:
    1. -2, 2, 2cis(pi/3)
    2. 2, 2cis(2pi/3), 2cis(4pi/3)
    3. -2, 2cis(pi/3), 2cis(2pi/3)
    4. 2cis(pi/3), 2cis(2pi/3), 2cis(4pi/3)
    Are you sure that you have typed the question correctly. the only solutions
    of x^2-8=0, are x=\pm \sqrt{8}.

    From the list of solutions I would guess you are asked for the solutions of:

    x^3-8=0,

    in which case the answer will be 2, which you can check by cubing each of
    the solutions in the lis to check that they come to 8.

    RonL
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  3. #3
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    Quote Originally Posted by mike1 View Post
    Can someone help me with this one?

    Find the real and complex solution of the equation x[squared]-8=0.

    Possible answers:
    1. -2, 2, 2cis(pi/3)
    2. 2, 2cis(2pi/3), 2cis(4pi/3)
    3. -2, 2cis(pi/3), 2cis(2pi/3)
    4. 2cis(pi/3), 2cis(2pi/3), 2cis(4pi/3)
    If x[squared] -8 = 0
    is x^2 -8 = 0
    then,
    (x+sqrt(8))(x-sqrt(8)) = 0
    x = +,-sqrt(8)
    x = +,-2sqrt(2) ------------answer.

    Another way,
    x^2 -8 = 0
    x^2 = 8
    x = +,-sqrt(8)
    .....

    The answers are not in the multiple choice.
    The answers are all real numbers. No complex.
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  4. #4
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    Now I'm really confused. Do you really think my worksheet is wrong?
    Last edited by CaptainBlack; January 14th 2007 at 05:33 AM.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by mike1 View Post
    Now I'm really confused. Do you really think my worksheet is wrong?

    The equation given to solve is a quadratic, the solution sets contain
    three solutions, which is a charateristic in the complex domain of the
    roots of a cubic.

    So yes there is a mistake somewhere.

    RonL
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