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Imaginary numbers entertwined with quadratics

  1. Apr 6, 2006 #1
    Hi, I'm not sure if this is calculas based or algebra based so here's the question.

    (A) 2i, -2i

    For this question i don't know what is being asked so i guess the pairs could be x...
    I know this:
    ax^2+bx+c, a(x-h)^2+k and a(x-s)(x-t)
    So the problem is how can i use the things that i know to answer the question. If you can please use analogies and don't give me the answer i just need to know what to do.
  2. jcsd
  3. Apr 6, 2006 #2


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    Think about the process you go through to solve a quadratic and apply it in reverse. Take note of the complex numbers though.


    Note to mentors/admin suggest moving this thread to pre calc forum.
  4. Apr 6, 2006 #3
    Im trying this.... I just used (a+bi)(c+di)
    -from previous question 2i, -2i I represented a = 0.... and so forth
  5. Apr 6, 2006 #4
    Although i don't know what 2i represents.... Is it x b c?
  6. Apr 6, 2006 #5


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    From what I can see the solution is pretty simple

    [tex](x - (2i))(x - (-2i)) = 0[/tex]

    or am I missing something?

    Remember that [itex]i^2 = -1 [/itex]

    [EDIT] I've edited my post to change my stupid mistake
    Last edited: Apr 7, 2006
  7. Apr 6, 2006 #6
    I doubt that hootenay... I'm thinking of x = -b -+ Square root of b^2-4ac over 2a equation and deriving it so that something will occur... im confused...
  8. Apr 6, 2006 #7


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    Like hoot implied(I think slightly incorrectly), it's always true that if r1 and r2 are the two roots to a quadratic then the characteristic quadratic equation is (x-r1)(x-r2)=0. Then you just expand that to get it into the standard form.

    The same argument can be extended to higher degree polynomial equations.
  9. Apr 6, 2006 #8
    My davey jones's locker i think ive got It!!
    Correct me if i'm right
    am i correct?
    now if the pairs were 1+3i, 1-3i I would have sub the same way like
    and solve from that correct?
  10. Apr 6, 2006 #9


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    Are you correct?

    It is very easy to check your work. What are the roots of

    [tex] x^2 + 4 =0 [/tex] ?
  11. Apr 7, 2006 #10


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    You made a small mistake when going from the first line to the second line. Your second line should reads:
    0 = (x - 1 - 3i) (x - 1 + 3i).
    Now just expand all the terms out, simplify it, and it's all done.
    Can you go from here? :)
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