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Homework Help: Complex numbers help needed

  1. Nov 18, 2005 #1
    Complex numbers ... help needed!!

    In our exercises we are told to solve for x (element of a complex number)
    1. x^2 - 6x + 25=0
    The answer is x=3+4i or x=3-4i
    Can anyone tell me how these answers were derived??
     
    Last edited: Nov 18, 2005
  2. jcsd
  3. Nov 18, 2005 #2

    NateTG

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    Are you familiar with the quadratic formula?
     
  4. Nov 18, 2005 #3
    Yes, I may be able to get the answer using the quad. formula for this one but what about if I had:

    x^4+3x^2-4=0

    or

    x^2+ix+6=0
     
  5. Nov 18, 2005 #4
    can't you just use the quadratic equation?

    edit:

    whoops i was too slow.
     
  6. Nov 18, 2005 #5
    Quad. form. can only be used when the x-term is of degree 2 ... any idea on how to solve for x in the above eqns?
     
  7. Nov 18, 2005 #6
    is it possible to write it into polar complex form and solve the equation?
     
  8. Nov 18, 2005 #7
    We haven't covered this topic yet so I'm not quite sure how you would use it
     
  9. Nov 18, 2005 #8
    You could always complete the square, and set that equal to -16.

    x^2 - 6x + 9 + (25-9) = 0

    x^2 - 6x + 9 = -16

    can you take it from there?

    quadratic formula does basically the same thing, and you can skip half a dozen steps if you remember it..
     
    Last edited: Nov 18, 2005
  10. Nov 18, 2005 #9

    NateTG

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    Those are both quite doable with the quadratic formula:
    [tex]x^4+3x^2-4=\left(x^2\right)^2+3 \left(x^2\right) + 3[/tex]
    so
    [tex]x^2= \frac{-3 \pm \sqrt{3^2-4(1)(3)}}{2}[/tex]
    and
    [tex]x^2+ix+6[/tex]
    is no problem:
    [tex]x=\frac{-i \pm \sqrt{i^2-4(1)(6)}}{2}[/tex]
     
  11. Nov 18, 2005 #10
    Thanks that does work ... any ideas on the other ones?
     
  12. Nov 18, 2005 #11
    Thanks for the help. I appreciate it. :smile:
     
  13. Nov 18, 2005 #12
    x^4 + 3x^2 - 4 = 0 factors into two "pieces". One has a complex solution, the other does not.

    Note: there are four roots to the equation. You should have two complex, and two real solutions.
     
    Last edited: Nov 18, 2005
  14. Nov 18, 2005 #13
    In this case the quadratic formula works, but for other high polynomial funtions you're probably better off with synthetic devision, factoring, and the like. For instance:

    (1)^4 + 3(1)^2 - 4 = 0
    So:
    x^4 + 3x^2 - 4 = (x-1)(x^3 + x^2 + 4x + 4) = 0
    (-1)^3 + (-1)^2 + 4(-1) + 4 = 0
    So:
    (x-1)(x+1)(x^2 + 4) = 0

    Real roots are 1 and -1
    Complex roots are +2i and -2i
     
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