三次方的方程怎么求解我在做自动控制原理，画根轨迹中 自动控制的根轨迹画法，渐近线的k怎么知道它取值多少？如图所示...
\u4e09\u6b21\u65b9\u7684\u65b9\u7a0b\u600e\u4e48\u6c42\u89e3\u7528\u5361\u4e39\u516c\u5f0f
\u786e\u5b9a\u4e00\u822c\u7684\u4e09\u6b21\u65b9\u7a0b\u7684\u6839\u7684\u516c\u5f0f.
\u5982\u679c\u7528\u73b0\u5728\u7684\u6570\u5b66\u8bed\u8a00\u548c\u7b26\u53f7,\u5361\u4e39\u516c\u5f0f\u7684\u7ed3\u8bba\u53ef\u4ee5\u501f\u52a9\u4e8e\u4e0b\u9762\u8fd9\u6837\u4e00\u79cd\u6700\u57fa\u672c\u7684\u8bbe\u60f3\u5f97\u51fa\u3002
\u5047\u5982\u7ed9\u6211\u4eec\u4e00\u4e2a\u4e00\u822c\u7684\u4e09\u6b21\u65b9\u7a0b\uff1a
ax3+3bx2+3cx+d=0 \uff081\uff09
\u5982\u679c\u4ee4
x=y-b/a
\u6211\u4eec\u5c31\u628a\u65b9\u7a0b\uff081\uff09\u63a8\u5bfc\u6210
y3+3py+2q=0 \uff082\uff09
\u5176\u4e2d p=c/a-b2/a2\uff0c2q=2b3/a3-3bc/a2+d/a \u3002
\u501f\u52a9\u4e8e\u7b49\u5f0f
y=u-p/u
\u5f15\u5165\u65b0\u53d8\u91cfu \u3002\u628a\u8fd9\u4e2a\u8868\u8fbe\u5f0f\u5e26\u5165\uff082\uff09\uff0c\u5f97\u5230\uff1a
\uff08u3\uff092+2qu3-p3=0 \uff083\uff09
\u7531\u6b64\u5f97
u3=-q\u00b1\u221a\uff08q2+p3\uff09\uff0c
\u4e8e\u662f
y=3\u221a\uff08-q\u00b1\u221a\uff08q2+p3\uff09\uff09-p/3\u221a\uff08-q\u00b1\u221a\uff08q2+p3\uff09\uff09 \u3002
=3\u221a\uff08-q+\u221a\uff08q2+p3\uff09\uff09+3\u221a\uff08-q-\u221a\uff08q2+p3\uff09\uff09 \u3002
\uff08\u6700\u540e\u8fd9\u4e2a\u7b49\u5f0f\u91cc\u7684\u4e24\u4e2a\u7acb\u65b9\u6839\u7684\u79ef\u7b49\u4e8e-p \u3002\uff09
\u4f60\u968f\u4fbf\u4ee3\u51e0\u4e2a\u6570\u8bd5\u8bd5 \u6bd4\u5982 0 1 2 3 4 \u4f60\u5c31\u4f1a\u53d1\u73b0 \u5176\u5b9e\u5c31\u51e0\u6761\u800c\u5df2 \u591a\u4f59\u7684\u6e10\u8fdb\u7ebf\u90fd\u662f\u91cd\u5408\u7684
解三次方程：ax^3+bx^2+cx+d=0先计算判别式
t=b/a/3
u=c/a/3-t^2
v=t^3+(d-ct)/a/2
w=u^3+v^2
如果w<0就有3个实数根
x0=2pcosq-t，x1,2=2pcos(q±120°)-t
其中：p=√-u，q=arccos(-v/u)/3
否则
有一个实数根：x0=p-q-t
两个复数根：x2,3=(q-p)/2-t±(p+q)√3i/2
其中：p=(√w-v)^(1/3)，q=(√w+v)^(1/3)
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绛旓細瑙ｄ竴鍏冧笁娆℃柟绋嬬殑涓绉嶅父瑙佹柟娉曟槸浣跨敤 Cardano 鍏紡銆侰ardano 鍏紡缁欏嚭浜嗗疄鏁版牴鐨勮〃杈惧紡锛屽畠鍖呭惈涓涓疄鏍瑰拰涓や釜鍏辫江澶嶆牴鐨勫拰銆侰ardano 鍏紡濡備笅锛歺 = [(q + (q^2 + r^3)^0.5)^(1/3) + (q - (q^2 + r^3)^0.5)^(1/3)] - (b / (3a))鍏朵腑锛宷 = (3ac - b^2) / (9a...
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