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Math Help - Solving Special Quartic Equation

  1. #1
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    Solving Special Quartic Equation

    Hi guys!..somebody gave me this problem they really need to solve this for their research.

    PROBLEM: How to find the roots of a quartic equation, ax^4+bx^3+cx2+dx+e=0, where b=0. This is a special case thats why I really need your help. Thank you...

    Please I really need your help Masters...
    Last edited by fireboy25; January 24th 2014 at 03:39 AM.
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  2. #2
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    Re: Solving Special Quartic Equation

    4 roots, 2 per image. Have fun.

    Solving Special Quartic Equation-clipboard01.jpg

    Solving Special Quartic Equation-clipboard01.jpg
    Thanks from fireboy25
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  3. #3
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    Re: Solving Special Quartic Equation

    ..Sir may i ask if how did you input it there? or in what site did you edited that? because i want to try in making a quartic formula where b and c are equal to 0. I really need your help sir.
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  4. #4
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    Re: Solving Special Quartic Equation

    I just dumped your equation into Mathematica and had it solved with the Solve function.

    Go to Wolframalpha.com and there is probably a general solution to the roots of the quartic polynomial in terms of a,e (the 5 coefficients). Take those and zero out the coefficients you want and maybe they'll simplify a bit (maybe not).

    Ok wolframalpha.com choked on the quartic with b and c zeroed out but Mathematica solved it. The reply to this post has the 4 roots.
    Last edited by romsek; February 18th 2014 at 07:49 AM.
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  5. #5
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    Re: Solving Special Quartic Equation

    $$\left\{\left\{x\to \frac{1}{2} \sqrt{\frac{4 e \sqrt[3]{\frac{2}{3}}}{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}+\frac{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}{\sqrt[3]{2} 3^{2/3} a}}-\frac{1}{2} \sqrt{-\frac{4 e \sqrt[3]{\frac{2}{3}}}{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}-\frac{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}{\sqrt[3]{2} 3^{2/3} a}-\frac{2 d}{a \sqrt{\frac{4 e \sqrt[3]{\frac{2}{3}}}{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}+\frac{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}{\sqrt[3]{2} 3^{2/3} a}}}}\right\},\left\{x\to \frac{1}{2} \sqrt{-\frac{4 e \sqrt[3]{\frac{2}{3}}}{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}-\frac{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}{\sqrt[3]{2} 3^{2/3} a}-\frac{2 d}{a \sqrt{\frac{4 e \sqrt[3]{\frac{2}{3}}}{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}+\frac{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}{\sqrt[3]{2} 3^{2/3} a}}}}+\frac{1}{2} \sqrt{\frac{4 e \sqrt[3]{\frac{2}{3}}}{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}+\frac{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}{\sqrt[3]{2} 3^{2/3} a}}\right\},\left\{x\to -\frac{1}{2} \sqrt{-\frac{4 e \sqrt[3]{\frac{2}{3}}}{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}+\frac{2 d}{a \sqrt{\frac{4 e \sqrt[3]{\frac{2}{3}}}{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}+\frac{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}{\sqrt[3]{2} 3^{2/3} a}}}-\frac{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}{\sqrt[3]{2} 3^{2/3} a}}-\frac{1}{2} \sqrt{\frac{4 e \sqrt[3]{\frac{2}{3}}}{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}+\frac{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}{\sqrt[3]{2} 3^{2/3} a}}\right\},\left\{x\to \frac{1}{2} \sqrt{-\frac{4 e \sqrt[3]{\frac{2}{3}}}{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}+\frac{2 d}{a \sqrt{\frac{4 e \sqrt[3]{\frac{2}{3}}}{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}+\frac{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}{\sqrt[3]{2} 3^{2/3} a}}}-\frac{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}{\sqrt[3]{2} 3^{2/3} a}}-\frac{1}{2} \sqrt{\frac{4 e \sqrt[3]{\frac{2}{3}}}{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}+\frac{\sqrt[3]{9 a d^2+\sqrt{3} \sqrt{27 a^2 d^4-256 a^3 e^3}}}{\sqrt[3]{2} 3^{2/3} a}}\right\}\right\}$$
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  6. #6
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    Re: Solving Special Quartic Equation

    If you have a math library handy, general closed solutions for cubics and quartics are given in "Dickson, New First Course in the Theory of Equations." But you have to wade through some algebra.

    It's also available at AbeBooks Official Site - New & Used Books, Textbooks, & Rare Books for a few $ if you really need it.
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