对于实数系数或复数系数的一元四次方程 \(x^4+bx^3+cx^2+dx+e=0 \) 有以下公式解(证明过程暂时略去,先请大家看看有没有错误)。令:
\( u=1728 b^2 e - 576 b c d + 128 c^3 - 4608 c e + 1728 d^2; \)
\( v = 48 b d - 16 c^2 - 192 e; \)
\( p =\sqrt[3]{\sqrt{u^2 + 4 v^3}+u} ; \)
\( q =\sqrt{\frac{6pb^2+\sqrt[3]{4}p^2-16cp-2\sqrt[3]{2}v}{p}}; \)
\(A=\sqrt{\frac{1}{6}(6b^2-16c-\frac{2\sqrt[3]{2}v}{p}+\sqrt[3]{4}p)}; \)
\(m1=24\sqrt{6}pb^2+8\sqrt[6]{2}\sqrt{3}p^2-64\sqrt{6}cp-8\sqrt[6]{32}\sqrt{3}v ;\)
\(m2=24pqb ;\)
\(n1=2\sqrt[6]{2}\sqrt{3}bp^2+8\sqrt{6}bcp-48\sqrt{6}dp-2\sqrt[6]{32}\sqrt{3}bv ;\)
\(n2=2\sqrt[3]{4}qb^2+8cpq-2\sqrt[3]{2}qv; \)
原方程的四个根为:
\(x1=\frac{-(m1+m2)+\sqrt{(m1+m2)^2-192pq(n1+n2)}}{96pq}; \)
\(x2=\frac{-(m1+m2)-\sqrt{(m1+m2)^2-192pq(n1+n2)}}{96pq}; \)
\(x3=\frac{(m1-m2)+\sqrt{(m1-m2)^2-192pq(-n1+n2)}}{96pq}; \)
\(x4=\frac{(m1-m2)-\sqrt{(m1-m2)^2-192pq(-n1+n2)}}{96pq}; \)