根與係數關係變化(四)

提供兩種方法



一 : 做出以A、B、C、D為根的四次方程式

令 f(x) = x^4+2x^3+x+1、g(x) = x^4*f(1/x) = x^4+x^3+2x+1

a+b+c+d = -2 , a^2+b^2+c^2+d^2 = 4 , x^3 = -2x^2 - 1 - 1/x

-> a^3+b^3+c^3+d^3= -8 - 4 + 1 = -11

-> A+B+C+D = -7


A^2 = a^4(a+1)^2 = a^6+2a^5+a^4 = -3a^3-a^2-a-1

->A^2+B^2+C^2+D^2 = 33-4+2-4 = 27

->AB+BC+AC+AD+CD+BD=(49-27)/2 = 11


(a+1)(b+1)(c+1)(d+1) = f(-1) = -1、abcd=1

-> ABCD = -1


1/A = 1/(a^2(a+1)) = 1/a^2 - 1/a + 1/(a+1)

- f'(-1)/ f(-1) = 1/(a+1) + 1/(b+1) + 1/(c+1) + 1/(d+1) = 3

-> 1/A + 1/B + 1/C + 1/D = 1+1+3 = 5

-> ABC+ACD+ABD+BCD = -5

-> (x-A)(x-B)(x-C)(x-D) = x^4+7x^3+11x^2+5x-1 = h(x)

(A+B+C)(A+B+D)(A+C+D)(B+C+D) = h(-7) = 503


二 :

令 f(x) = (x-a)(x-b)(x-c)(x-d) = x^4+2x^3+x+1

g(x) = x^3+x^2+7 = (x-α)(x-β)(x-γ)

-> f(x) = g(x)(x+1)-(x^2+6x+6)

x^2+6x+6 = (x+(3+√3))(x+(3-√3))

由一,A+B+C+D=-7

(A+B+C)(A+B+D)(A+C+D)(B+C+D)= g(a)g(b)g(c)g(d)

= f(α)f(β)f(γ) = -g(-3+√3)g(-3-√3) = (24√3 + 33)(24√3 - 33)

= 503

阿 眼幹太嚴重

看到w1^2+6w1+6太高興居然忘了是相乘

出醜了真抱歉

砍掉好了呵呵

兩個錯誤:
1. f(x) = g(x)(-x-1)+(x^2+6x+6) 應為 f(x) = g(x)(-x-1) - (x^2+6x+6)
這沒影響, 因(-1)^4 = 1
2. g(a)g(b)g(c)g(d)=f(w1)f(w2)f(w3) 對
=(w1^2+6w1+6)+(w2^2+6w2+6)+(w3^2+6w3+6) 錯!(不是相加應相乘)

2009-02-20 22:59:47 補充：
喔喔!第一個錯也造成差一個負號!