二次方程 x^2+(3k-2)x+1=0有等根,求k的值
答案:4/3 or 0
數學問題1條(須列式)
2008-08-16 1:33 am
回答 (1)
2008-08-16 1:44 am
✔ 最佳答案
(3k-2)^2-4(1)(1) = 09k^2-12k 4-4 = 0
9k^2-12k = 0
3k(3k-4) = 0
3k = 0 or 3k-4 = 0
k = 0 or k = 4/3
2008-08-15 17:44:24 補充：
9k^2-12k+4-4 = 0
參考: Me
收錄日期: 2021-04-13 15:57:24
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https://hk.answers.yahoo.com/question/index?qid=20080815000051KK02230