AL polynomial3

子路

2010-02-22 3:52 am

看圖~~~看圖~~~看圖~~~看圖~~~看圖~~~看圖

圖片參考：http://i256.photobucket.com/albums/hh182/zilu_photo/sshot-2010-02-21-19-49-04.png

It is known that
k=2010 !!!

更新1:

The answer is -105386508 there is a simpler method

更新2:

for the previous trial ans, y=f(x) is okay up to here why not sub x=g(y) into the original equation?

更新3:

[(a-b)(b-c)(c-a)]^2 =-p'(a)p'(b)p'(c) =-27p(r1)p(r2) where r1 and r2 are roots of p'(x)=0 出題出得不好, r1和r2不是整數天助你又教我一回了, 正常要消走一個term 如果我不是做過的話我也想不到上面的方法

回答 (1)

天助

2010-02-22 7:21 am

✔ 最佳答案

1. a,b,c are roots of x^3-6x^2-9x+k=0
so, (A,B,C)=(a-2, b-2, c-2) are roots of x^3-21x+k-34=0
A+B+C=0, AB+BC+CA=-21, ABC=34-k=K=-1976, B^3=21B+K

x^3-21x-K=(x-A)(x-B)(x-C) -----(*)
2. (a-b)(b-c)(c-a)=(A-B)(B-C)(C-A)
3. (A-B)(B-C)=B(A+C)-B^2-AC= -2B^2-K/B
=-(2B^3+K)/B=-(42B+3K)/B= -42(B+p)/B (p=K/14)
so, [(a-b)(b-c)(c-a)]^2=[(A-B)(B-C)(C-A)]^2
=-42^3(A+p)(B+p)(C+p)/(ABC) --- (**)
4. Putting x=-p into eq.(*), then -p^3+21p-K=-(A+p)(B+p)(C+p)
conti. eq.(**)= -42^3(p^3-21p+K)/K
=-42^3*[(K/14)^3 -1.5K+K)/K
= -27(K^2-1372) (K=-1976)
=-105386508 (so urgly!)

Note: Previous problem: a,b,c are roots of x^3-2x^2+x+k=0,
then [(a-b)(b-c)(c-a)]^2=-27k^2-4k

2010-02-21 23:48:16 補充：
您的方法很棒!只可惜r1,r2不夠漂亮.

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