F.4 Quadratic equation .URGENT!!! 20 points will be given.

2008-09-14 3:20 am
If the equation 2x²+(m+1)x+(m+1) = 0 has one double real root, find the value of m.


P.S.
(Delta = b²-4ac)
using this formula
更新1:

仲有一條 if the equation kx²-2kx+(k+6) = 0 has no real roots, find the range of values k.

更新2:

第一條既答案係 m < 25/12

回答 (2)

2008-09-14 3:30 am
✔ 最佳答案
2x+(m+1)x+(m+1)=0 has double root ,
i.e. b^2-4ac=0
(m+1)^2-4(2)(m+1(=0
m^2+2m+1-8m-8=0
m^2-6m-7=0
(m-7)(m+1)=0
m=7 or m=-1//



2008-09-14 14:07:05 補充:
kx²-2kx+(k+6) = 0 has no real roots
i.e. b^2-4ac<0
4k^2-4k(k+6)<0
-24k<0
k>0//
2008-09-14 3:36 am
1.
(1 + m)^2 - 8(1 +m) = 0
(1 + m)(1 + m -8) = 0
(1 +m)(m -7) = 0
m = -1 and 7.
2.
4k^2 - 4k(k+6) < 0
k^2 - k^2 -6k < 0
6k> 0
k> 0


收錄日期: 2021-04-25 22:38:05
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