一元四次方程的公式解
本帖最后由 TSC999 于 2021-1-7 07:02 编辑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：
\( u=1728 b^2 e - 576 b c d + 128 c^3 - 4608 c e + 1728 d^2; \)
\( v =48 b d - 16 c^2 - 192 e; \)
\(p =\sqrt{\sqrt{u^2 + 4 v^3}+u} ; \)
\( q =\sqrt{\frac{6pb^2+\sqrt{4}p^2-16cp-2\sqrt{2}v}{p}};\)
\(A=\sqrt{\frac{1}{6}(6b^2-16c-\frac{2\sqrt{2}v}{p}+\sqrt{4}p)};\)
\(m1=24\sqrt{6}pb^2+8\sqrt{2}\sqrt{3}p^2-64\sqrt{6}cp-8\sqrt{32}\sqrt{3}v ;\)
\(m2=24pqb ;\)
\(n1=2\sqrt{2}\sqrt{3}bp^2+8\sqrt{6}bcp-48\sqrt{6}dp-2\sqrt{32}\sqrt{3}bv ;\)
\(n2=2\sqrt{4}qb^2+8cpq-2\sqrt{2}qv; \)
原方程的四个根为：
\(x1=\frac{-(m1+m2)+\sqrt{(m1+m2)^2-192pq(n1+n2)}}{96pq}; \)
\(x2=\frac{-(m1+m2)-\sqrt{(m1+m2)^2-192pq(n1+n2)}}{96pq}; \)
\(x3=\frac{(m1-m2)+\sqrt{(m1-m2)^2-192pq(-n1+n2)}}{96pq}; \)
\(x4=\frac{(m1-m2)-\sqrt{(m1-m2)^2-192pq(-n1+n2)}}{96pq}; \)
本帖最后由 TSC999 于 2021-1-7 07:01 编辑
(* 一元四次方程 x^4+bx^3+cx^2+dx+e=0 的公式解 *)
Clear["Global`*"];
b = 11; c = 46; d = 96; e = 120;(* 结果正确，两对虚 *)
b = 11; c = -46; d = 9.6; e = 120;(* 结果正确，二实二虚 *)
b = -11; c = -4.6; d = 0.9766; e = -23;(* 结果正确，二实二虚 *)
b = -1.231; c = -14.6; d = 20.9766; e = -2.7563;(* 结果正确，四个实根 *)
b = 11; c = 46; d = 96; e = 10;(* 结果正确，两实两虚 *)
b = -1.1; c = 4.6; d = 96; e = 10;(* 结果正确，两实两虚 *)
b = -1.1; c = 1 - I 4.6; d = Sqrt; e = 10;(* 结果正确，两对虚 *)
b = -1.1 I; c = 1 - I 4.6; d = Sqrt + 2 I; e = 1;(* 结果正确，两对虚 *)
b = 5 + 1.1 I; c = Sqrt - I 4.6; d = Sqrt + 2 I; e =
1 - I Sqrt;(* 结果正确，两对虚 *)
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 = Power + u, (3)^-1];q = Sqrt[(
6 p b^2 + Power p^2 - 16 c p - 2 Power v)/p];
A = Sqrt[1/
6 (6 b^2 - 16 c - (2 Power v)/p + Power p)];
m1 = 24 Sqrt p b^2 + 8 Power Sqrt p^2 -
64 Sqrt c p - 8 Power Sqrt v;m2 = 24 p q b;
n1 = 2 Power Sqrt b p^2 + 8 Sqrt b c p -
48 Sqrt d p - 2 Power Sqrt b v;n2 =
Power q p^2 + 8 c q p - 2 Power v q;
x1 = (-(m1 + m2) + Sqrt[(m1 + m2)^2 - 192 p q (n1 + n2)])/(
96 p q); x2 = (-(m1 + m2) - Sqrt[(m1 + m2)^2 - 192 p q (n1 + n2)])/(
96 p q);
x3 = ((m1 - m2) + Sqrt[(m1 - m2)^2 - 192 p q (-n1 + n2)])/(
96 p q); x4 = ((m1 - m2) - Sqrt[(m1 - m2)^2 - 192 p q (-n1 + n2)])/(
96 p q);
Print["x1 = ", N, "，x2 = ", N, "，x3 = ", N,
"，x4 = ", N];
NSolve
程序的最后用 NSolve 求数字解，目的是与上述公式解进行对照，两者应当相同，不然就是程序中有错了。
何苦何必呢？找本数学手册，上面就有公式
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