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Real find both roots of the equation

  1. Oct 14, 2005 #1

    UnD

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    x^2 +6x +k=0 has one root (a) where Im (a) =2, If k is real find both roots of the equation and k

    So i got b+ 2i is the root

    (b+2i)^2 +6(x+2i) +k=0
    and after expanding it out, i have no clue what to do. Please help. THanks
     
  2. jcsd
  3. Oct 14, 2005 #2
    Try using the quadratic formula and thinking about what comes underneath the square root in relation to Im(a) = 2
     
  4. Oct 14, 2005 #3

    Tide

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    You know that if the roots are complex then the two roots are complex conjugates of each other. The sum of the roots is -6 (negative ratio of linear coefficient to quadratic coefficient) so you should be able figure out what what the real part has to be. Once you have a root you can find k.
     
  5. Oct 15, 2005 #4

    HallsofIvy

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    I wish you had shown us what you got by expanding it! Clearly that "6(x+ 2i)" should be "6(b+ 2i)" but I don't know whether that's a typo or you actually left the x in your calculation.
    Expand it out and set it equal to 0. For a complex number to be equal to 0, both real and imaginary parts must be 0. That gives you two (simple) equations for the two (real) unknown numbers, b and k.
     
  6. Nov 1, 2006 #5
    sorry for bumping this topic

    but could anyone please explain in detail how this question is done?
     
  7. Nov 1, 2006 #6

    HallsofIvy

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    First try doing it yourself! You said "after expanding it out, i have no clue what to do." and I asked you to show what you got after expanding. You should get a complex number depending on b and k. As I said before, set real and imaginary parts equal to 0 and you get two equations for b and k. Solve those equations.
     
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