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...
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.
$$\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\}$$
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.