3x^2+10x+10=0
點計........
比答案我
3x^2+10x+10=0點計...10分.急.......!!!
2006-11-08 6:00 am
回答 (4)
2006-11-09 1:56 am
✔ 最佳答案
3x^2+10x+10=03x2+10x+10-10＝--10
3+10x=－５
3+10x－３＝－５－３
１０ｘ＝－８
ｘ＝０．８
2006-11-08 6:14 am
no real solutions
參考: .................................................................................................
2006-11-08 6:13 am
做二次方程式有兩個方法，最麻煩不過一定計到答案的方法是用Quadratic Formula
即是: let ax^2 + bx + c= 0,
x=[-b +or- (b^2 - 4ac)^(1/2)] / (2a)
in your case, 3x^2+10x+10=0
x = {-10 +or- [10^2 - 4(3) (10)]^(1/2)} / [2 (3)]
in which the discriminant ( b^2 - 4ac) = -20 < 0,
therefore, it has no real solutions.
2006-11-09 23:48:36 補充：
what the hell? 你唔識唔好亂咁做
即是: let ax^2 + bx + c= 0,
x=[-b +or- (b^2 - 4ac)^(1/2)] / (2a)
in your case, 3x^2+10x+10=0
x = {-10 +or- [10^2 - 4(3) (10)]^(1/2)} / [2 (3)]
in which the discriminant ( b^2 - 4ac) = -20 < 0,
therefore, it has no real solutions.
2006-11-09 23:48:36 補充：
what the hell? 你唔識唔好亂咁做
參考: by myself
2006-11-08 6:03 am
no solution
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