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Math Help - Find all the real numbers m so that..

  1. #1
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    Find all the real numbers m so that..

    Can any one help me with this:


    Find all the REAL numbers m so that the equation

    z^3 + (3+i)z^2 -3z - (m+i) = 0

    has at least one REAL root.



    NOTE: i = (-1)^1/2
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  2. #2
    MHF Contributor red_dog's Avatar
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    The equation can be written as
    z^3+3z^2-3z-m+i(z^2-1)=0
    If z is a real root then
    \left\{\begin{array}{ll}z^3+3z^2-3z-m=0\\z^2-1=0\end{array}\right.
    From the second equation we have z=\pm 1.
    Plugging z in the first equation we get m=1, \ m=5
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by red_dog View Post
    The equation can be written as
    z^3+3z^2-3z-m+i(z^2-1)=0
    If z is a real root then
    \left\{\begin{array}{ll}z^3+3z^2-3z-m=0\\z^2-1=0\end{array}\right.
    From the second equation we have z=\pm 1.
    Plugging z in the first equation we get m=1, \ m=5
    that was a lot easier than i thought it would be. you don't want to know the things than were going through my head!
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  4. #4
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    10x
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by N1CKYYY View Post
    10x
    What is this supposed to mean?

    -Dan
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  6. #6
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    Quote Originally Posted by topsquark View Post
    What is this supposed to mean?

    -Dan
    inappropriate personal remark removed
    Last edited by CaptainBlack; January 20th 2008 at 08:51 AM.
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  7. #7
    Bar0n janvdl's Avatar
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    Quote Originally Posted by N1CKYYY View Post
    inappropriate personal remark removed
    Why don't you learn some manners and show respect to our senior members?
    Last edited by CaptainBlack; January 20th 2008 at 08:52 AM.
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  8. #8
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    I think this thread has run its course.
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