if 3x^2-(2k+3)X+4=0 has a double real root,then
-(2k+3)^2=48
(2K+3)^2=48
which one is right?pls explain
math 二次方式問題
2008-05-04 5:11 am
回答 (2)
2008-05-06 3:22 am
✔ 最佳答案
3x^2-(2k+3)X+4=0has a double real root
so,[-(2k+3]^2-16(3)>0(判別式)
answer=(2k+3)^2>48
點解唔連埋個負號,係因為個負號2次方左,負負得正,所以無左!
2008-05-05 19:28:30 補充：
3x^2-(2k+3)X+4=0
has a double real root
so,[-(2k+3]^2-16(3)=0(判別式)
answer=(2k+3)^2=48
點解唔連埋個負號,係因為個負號2次方左,負負得正,所以無左!
2008-05-04 5:44 am
二個相等實根=delta=0
b^2-4ac=0
(2k+3)^2-4(3)(4)=0
(2k+3)^2=48
b^2-4ac=0
(2k+3)^2-4(3)(4)=0
(2k+3)^2=48
收錄日期: 2021-04-13 15:30:50
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