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Math Help - Quartic Equation

  1. #1
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    Quartic Equation

    Prove that \tan \frac{\pi}{15} is a root of the equation.

    t^4 -6 \sqrt{3}t^3 +8t^2 + 2 \sqrt{3} t -1 =0

    Give the other roots in the form \tan \frac{r \pi}{15}
    I have tired using an expression for \tan 4 \theta and got nowhere

    Thanks in Advance

    Bobak
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by bobak View Post
    I have tried using an expression for \tan 4 \theta and got nowhere
    I think maybe you're confusing π/15 radians with 15 degrees! Try tan(5θ) instead of tan(4θ). If you put the expression for tan(5θ) equal to √3, you'll get a fifth-degree equation one of whose roots is √3. Divide out by the factor t+√3 and you'll be on the right track.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    I think maybe you're confusing π/15 radians with 15 degrees! Try tan(5θ) instead of tan(4θ). If you put the expression for tan(5θ) equal to √3, you'll get a fifth-degree equation one of whose roots is √3. Divide out by the factor t+√3 and you'll be on the right track.
    I never thought of using  \tan(5 \theta) at all! So the general solution is given by \theta = \frac{(3n+1) \pi}{15} excluding the solution \theta = \frac{2 \pi}{3} ( i.e. r = 3) as this corrospends to \tan \theta = - \sqrt{3}

    Many Thanks

    Bobak
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