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Math Help - How do I find the imaginary number solutions to this problem?

  1. #1
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    How do I find the imaginary number solutions to this problem?

    So I have the following equation:

    <br />
r^3 + 4r^2 - 24 = 0<br />

    I can find the real solution of 2 by using synthetic division and finding which of the multiples of 24 ends up to have no remainder.

    But, the back of the book also wants an imaginary number solution and I am lost as to how to find one. Any help would be greatly appreciated.
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  2. #2
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    (r-2)(r^2+6r+12)=r^3+4r^2-24.
    Can you find the roots of r^2+6r+12~?
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  3. #3
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    You can use the factor theorem to say that r-2 is a factor of f(r)

    Hence (r-2)(Ar^2+Br+C) = r^3+4r^2-24 where A, B and C are constants to be found. Normally. I'd equate the coefficients but you can also do long division

    edit: too slow
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  4. #4
    Super Member Quacky's Avatar
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    I wouldn't have even bothered with synthetic division. Factor theorem will work nicely.
    You've established that r=2 is a root, therefore by the factor theorem, (r-2) is a factor. So take this out (via factorization):
    <br />
(r-2)(Ar^2+Br+C)=0
    Then just complete the square or use the formula. on the quadratic for the complex solution(s).
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  5. #5
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    @Plato:

    Yeah, plug into the the quadratic equation and you get:
    x = -6+/-sqrt(36 - 4(1)(12))/2(1)
    x = -6+/-sqrt(36 - 48)/2
    x = -6+/-sqrt(-12)/2
    x = -6+/-2*sqrt(-3)/2
    x = -6+/-2i*sqrt(3)/2
    x = -3+/-i*sqrt(3)

    Thanks!
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