解方程:(1)2x^4+3x^3-x^2+2x+3=0 (2)2x^4-7x^3+9x^2-7x+2=0

budusheng
2011-02-03 · 超过10用户采纳过TA的回答
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(2)可以分解因式:2x^4-7x^3+9x^2-7x+2=(-2 + x) (-1 + 2 x) (1 - x + x^2)。根为:2,1/2,1/2-Sqrt[3]/2,1/2+Sqrt[3]/2。
(1)无法像(2)有一个较好的分解,有四次方程的求根公式,但超级复杂,用mathematica、matlab等可以求得(1)有两实根两虚根,很复杂……大约为: -1.82727,-0.8082,0.567737 - 0.832696 i,0.567737 + 0.832696 i。精确值:
-(3/8) + 1/2 Sqrt[43/48 + 1/6 (1429/2 - (3 Sqrt[152949])/2)^(1/3) + 1/6 (1/2 (1429 + 3 Sqrt[152949]))^(1/3)] - 1/2 Sqrt[43/24 - 1/6 (1429/2 - (3 Sqrt[152949])/2)^(1/3) - 1/6 (1/2 (1429 + 3 Sqrt[152949]))^(1/3) - 115/(32 Sqrt[43/48 + 1/6 (1429/2 - (3 Sqrt[152949])/2)^(1/3) + 1/6 (1/2 (1429 + 3 Sqrt[152949]))^(1/3)])]
-(3/8) + 1/2 Sqrt[43/48 + 1/6 (1429/2 - (3 Sqrt[152949])/2)^(1/3) + 1/6 (1/2 (1429 + 3 Sqrt[152949]))^(1/3)] + 1/2 Sqrt[43/24 - 1/6 (1429/2 - (3 Sqrt[152949])/2)^(1/3) - 1/6 (1/2 (1429 + 3 Sqrt[152949]))^(1/3) - 115/(32 Sqrt[43/48 + 1/6 (1429/2 - (3 Sqrt[152949])/2)^(1/3) + 1/6 (1/2 (1429 + 3 Sqrt[152949]))^(1/3)])]
-(3/8) - 1/2 Sqrt[43/48 + 1/6 (1429/2 - (3 Sqrt[152949])/2)^(1/3) + 1/6 (1/2 (1429 + 3 Sqrt[152949]))^(1/3)] - 1/2 Sqrt[43/24 - 1/6 (1429/2 - (3 Sqrt[152949])/2)^(1/3) - 1/6 (1/2 (1429 + 3 Sqrt[152949]))^(1/3) + 115/(32 Sqrt[43/48 + 1/6 (1429/2 - (3 Sqrt[152949])/2)^(1/3) + 1/6 (1/2 (1429 + 3 Sqrt[152949]))^(1/3)])]
-(3/8) - 1/2 Sqrt[43/48 + 1/6 (1429/2 - (3 Sqrt[152949])/2)^(1/3) + 1/6 (1/2 (1429 + 3 Sqrt[152949]))^(1/3)] + 1/2 Sqrt[43/24 - 1/6 (1429/2 - (3 Sqrt[152949])/2)^(1/3) - 1/6 (1/2 (1429 + 3 Sqrt[152949]))^(1/3) + 115/(32 Sqrt[43/48 + 1/6 (1429/2 - (3 Sqrt[152949])/2)^(1/3) + 1/6 (1/2 (1429 + 3 Sqrt[152949]))^(1/3)])]
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