更新1:
（完整長解）~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
數學【多項式】問題?
回答 (2)
2010-04-05 10:02 pm
2010-04-05 9:58 pm
令f(x) = 15x^2 - (b+3)x - 1
f(-b/3) = 0
15(-b/3)^2 - (b+3)(-b/3) - 1 = 0
15(b^2/9) + (b^2+3b)/3 - 1 = 0
5b^2 + b^2+3b - 3 = 0
6b^2 + 3b - 3 = 0
2b^2 + b - 1 = 0
(2b-1)(b+1) = 0
b=1/2 或 b= -1
f(-b/3) = 0
15(-b/3)^2 - (b+3)(-b/3) - 1 = 0
15(b^2/9) + (b^2+3b)/3 - 1 = 0
5b^2 + b^2+3b - 3 = 0
6b^2 + 3b - 3 = 0
2b^2 + b - 1 = 0
(2b-1)(b+1) = 0
b=1/2 或 b= -1
參考: Knowledge is power.
收錄日期: 2021-04-23 20:40:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100405000051KK00506