(x+1)^2+3(x+1)-4=0
(4-x)^2-5(4-x)-14=0
F4 Quadratic equations
2011-09-14 4:26 am
回答 (1)
2011-09-14 5:35 am
✔ 最佳答案
1) (x+1)^2 + 3(x+1) - 4 = 0[(x+1) + 4)][(x+1) - 1] = 0
(x+5)x = 0
x = 0 or x = -5
2) (4-x)^2 - 5(4-x) - 14 = 0
[(4-x) - 7][(4-x) + 2] = 0
(-x - 3)(-x + 6) = 0
x = -3 or x = 6
參考: Knowledge is power.
收錄日期: 2021-04-13 18:14:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110913000051KK00978