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    By using a substitution to remove the term in  x^2 , or otherwise, find the single real solution to the equation
     \displaystyle 16x^3+96x^2+180x+99=0 .
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    how far have you got ?
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    (Original post by the bear)
    how far have you got ?
    Used substitution  x=y-2 to get it down to  16y^3-12y-5=0 .
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    this should help:

    *
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    (Original post by Ano123)
    Used substitution  x=y-2 to get it down to  16y^3-12y-5=0 .
    Remember cos 3x = 4 cos^3x - 3cosx
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    (Original post by Mathemagicien)
    Remember cos 3x = 4 cos^3x - 3cosx
    I'm not sure if that helps here.
    I'd get  \cos 3\alpha = 5/4 if I just did it like that.
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    (Original post by Ano123)
    I'm not sure if that helps here.
    I'd get  \cos 3\alpha = 5/4 if I just did it like that.
    So use cos x = \dfrac{e^{ix} + e^{-ix}}{2}

    so \dfrac{e^{3i \alpha }+e^{-3i \alpha } }{2} = 5/4

    which is a quadratic in e^{3i \alpha }

    which gives you e^{3i \alpha }=2,1/2

    take logs, we get \alpha = \pm \dfrac{log 2}{3i}

    and you can find cos \alpha (and thus y and thus x) from that

    so
    y = cos \alpha = \dfrac{e^{i \alpha} + e^{-i \alpha}}{2} = \dfrac{e^{ \pm i \dfrac{log 2}{3i} } + e^{- \pm i \dfrac{log 2}{3i} }}{2} = \dfrac{e^{ \dfrac{log 2}{3} } + e^{- \dfrac{log 2}{3} }}{2} = \dfrac{2^{ \dfrac{1}{3} }+2^{- \dfrac{1}{3} }}{2}= 2^{-2/3}+2^{-4/3}

    (which satisfies our equation)
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    (Original post by Mathemagicien)
    So e^{3i \alpha }+e^{-3i \alpha } = 5/2

    which is a quadratic in e^{3i \alpha }

    which gives you e^{3i \alpha }=2,1/2

    take logs, we get \alpha = +- \dfrac{log 2}{3i} = -+ \dfrac{i log 2}{3}

    and you can find cos \alpha from that

    I get cos \alpha = 2^{-2/3}+2^{-4/3}
    Not very nice looking solutions though are they? What have you done to get  \cos \alpha from there?
    Oh, I see.
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    (Original post by Ano123)
    Not very nice looking solutions though are they? What have you done to get  \cos \alpha from there?
    Compared to this...

    (Original post by the bear)
    this should help:
    *
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    (Original post by Mathemagicien)
    So use cos x = \dfrac{e^{ix} + e^{-ix}}{2}

    so \dfrac{e^{3i \alpha }+e^{-3i \alpha } }{2} = 5/4

    which is a quadratic in e^{3i \alpha }

    which gives you e^{3i \alpha }=2,1/2

    take logs, we get \alpha = +- \dfrac{log 2}{3i}

    and you can find cos \alpha (and thus y and thus x) from that

    I get cos \alpha = 2^{-2/3}+2^{-4/3} (which satisfies our equation)
    What does the  \mathbf{+ -} mean?
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    (Original post by Ano123)
    What does the  \mathbf{+ -} mean?
    Plus or minus (don't know the latex for it)
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    (Original post by Mathemagicien)
    Plus or minus (don't know the latex for it)
    \pm
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    (Original post by Mathemagicien)
    Plus or minus (don't know the latex for it)
    Thought so. \pm - code for it.
    I'll tell you I have got the solution to this equation but I used  y=\cosh \alpha . It's slightly easier doing this but I actually like your method a lot. Thanks!
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    (Original post by Ano123)
    Thought so. \pm - code for it.
    I'll tell you I have got the solution to this equation but I used  y=\sinh \alpha . It's slightly easier doing this but I actually like your method a lot. Thanks!
    You could also use y=cosh t, cosh 3t = 4 cosh^3 t - 3 cosh t, its equivalent
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    (Original post by Mathemagicien)
    You could also use y=cosh t, cosh 3t = 4 cosh^3 t - 3 cosh t, its equivalent
    That's what I meant I used cosh not sinh.
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    Exams are over. Chill the **** out.
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    (Original post by Randall13)
    Exams are over. Chill the **** out.
    Why?
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    (Original post by Randall13)
    Exams are over. Chill the **** out.
    this!
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    (Original post by Randall13)
    Exams are over. Chill the **** out.
    I don't do this for exams. Exams are over and I'm bored so may as well do some more maths, what could I be doing that's more interesting than that?
 
 
 
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