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Math Help - Roots, a+bi form

  1. #1
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    Roots, a+bi form

    Express the roots of the equation -6x = 2x^2 + 5 in simplest a+bi form.

    Thoughts?
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  2. #2
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    Quadratic

    2x^2+6x+5=0

    \displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}\Rightarrow \frac{-6\pm\sqrt{36-4(2)(5)}}{2(2)}=\cdots
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  3. #3
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    \displaystyle -6x = 2x^2 + 5

    \displaystyle 0 = 2x^2 + 6x + 5.


    \displaystyle x = \frac{-6 \pm \sqrt{6^2 - 4\cdot 2\cdot 5}}{2\cdot 2}.

    Simplify.
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  4. #4
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    2x^2+6x+5 = 0 \Rightarrow 2\left(x+\frac{3}{2}\right)^2+\frac{1}{2} = 0.

    Then it's just one step more to finish it up.
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  5. #5
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    Okay so I got it down to

    X= -6 +/- radical[23]/4

    but how do i proceed from there.
    and sorry about the total lack of proper mathematical typing. i have no idea how to type actual radicals, etc.
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  6. #6
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    Quote Originally Posted by MathFailure94 View Post
    Okay so I got it down to

    X= -6 +/- radical[23]/4

    but how do i proceed from there.
    and sorry about the total lack of proper mathematical typing. i have no idea how to type actual radicals, etc.
    \displaystyle\frac{-6\pm\sqrt{-4}}{4}=\frac{-6\pm 2i}{4}
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  7. #7
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    How does \displaystyle 6^2 - 4\cdot 2\cdot 5 equal \displaystyle 23?
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  8. #8
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    Quote Originally Posted by Prove It View Post
    How does \displaystyle 6^2 - 4\cdot 2\cdot 5 equal \displaystyle 23?
    36-40\equiv x \ \mbox{(mod 27)}\Rightarrow -4\equiv x \ \mbox{(mod 27)}\Rightarrow -4\equiv 23 \ \mbox{(mod 27)}

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  9. #9
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    So I'm hazarding a guess that the roots would equal 3i/2 and i/2?

    Probably completely incorrect.
    I really appreciate your help, though!
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  10. #10
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    Quote Originally Posted by dwsmith View Post
    \displaystyle\frac{-6\pm\sqrt{-4}}{4}=\frac{-6\pm 2i}{4}
    \displaystyle x=\frac{-3}{2}\pm\frac{i}{2}\Rightarrow x=\frac{-3}{2}+\frac{i}{2} \ \mbox{and} \ x=\frac{-3}{2}-\frac{i}{2}
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  11. #11
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    Quote Originally Posted by dwsmith View Post
    36-40\equiv x \ \mbox{(mod 27)}\Rightarrow -4\equiv x \ \mbox{(mod 27)}\Rightarrow -4\equiv 23 \ \mbox{(mod 27)}

    Clever! :P
    Last edited by mr fantastic; January 2nd 2011 at 08:28 PM.
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  12. #12
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    Quote Originally Posted by MathFailure94 View Post
    So I'm hazarding a guess that the roots would equal 3i/2 and i/2?

    Probably completely incorrect.
    I really appreciate your help, though!
    You have been told exactly what to do and post #6 even gives you the answer (in unsimplified form). I don't understand why (a) you would still be guessing and (b) you would be guessing wrong. You need to take a lot more care with the basic arithmetic - you have made careless and basic mistakes several times now in this thread.
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