# Solving Special Quartic Equation

• Jan 24th 2014, 03:34 AM
fireboy25
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...

• Jan 24th 2014, 03:53 AM
romsek
Re: Solving Special Quartic Equation
4 roots, 2 per image. Have fun.

Attachment 30072

Attachment 30073
• Feb 18th 2014, 04:14 AM
fireboy25
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.
• Feb 18th 2014, 07:28 AM
romsek
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.
• Feb 18th 2014, 07:47 AM
romsek
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\}$$
• Feb 18th 2014, 08:23 AM
Hartlw
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.