若方程有等根,求k的值
更新1:
答案係1喎
amaths
回答 (2)
2007-12-17 2:20 am
✔ 最佳答案
use this-----> D= b^2 －4ac D= 0^2 -4k(-2(k+2)+3(k+2))
=-4k(k+2)
=-4k^2 -8k =0
-4k(k+2)=0
k= -2, 0 <-----不能，因為不為零
我讀英文的，唔知有冇理解錯
2007-12-17 2:23 am
kx^2-2(k+2)+3(k+2)=0
kx^2-2k-4+3k+6=0
kx^2+(k+2)=0
since the equation has equal roots....so the discriminant of equation = 0
-4k(k+2)=0
k=0 or k=-2
I hope i can help u.....
2007-12-16 18:27:32 補充：
k=0(rejected) putx=0 into the equation,k(0)^2-2(k+2)+3(k+2) not equal to 0so 0 should be rejected
kx^2-2k-4+3k+6=0
kx^2+(k+2)=0
since the equation has equal roots....so the discriminant of equation = 0
-4k(k+2)=0
k=0 or k=-2
I hope i can help u.....
2007-12-16 18:27:32 補充：
k=0(rejected) putx=0 into the equation,k(0)^2-2(k+2)+3(k+2) not equal to 0so 0 should be rejected
參考: my amath knowledge
收錄日期: 2021-04-20 22:53:28
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https://hk.answers.yahoo.com/question/index?qid=20071216000051KK03194