Given that x2+2mx+m2-2m+3=0 has 2 roots α and β (with α＜β).
(1) Find α+β and αβ in terms of m.
(2) Find(α-β)2 in terms of m.
(3) If β-α=2, find m.
Maths question
2007-11-11 9:34 pm
回答 (2)
2007-11-11 9:43 pm
✔ 最佳答案
(1) α+β = -2m and αβ = m2-2m+3(2) (α-β)2 = (α+β)^2 - 4αβ
= 4m^2 - 4(m2-2m+3)
= 8m - 12
(3) If β-α=2, (α-β)2 = 4
8m - 12 = 4
8m = 16
m = 2
2007-11-11 9:54 pm
所有一元二次方程都係咁既(要背):
x^2 - (sum of roots)x + (product of roots) = 0
或者背呢樣野:
sum of roots = -b/a
product of roots = c/a
(1) α+β = -2m
αβ = m^2-2m+3
(2) (α+β)^2 = α^2 + 2αβ + β^2
(-2m)^2 = α^2 + β^2 + 2(m^2-2m+3)
α^2 + β^2 = 4m^2 -2m^2 + 4m -6
α^2 + β^2 = 2m^2 + 4m -6
(α-β)^2 = α^2 - 2αβ + β^2
(α-β)^2 = α^2 + β^2 - 2αβ
(α-β)^2 = 2m^2 + 4m -6 - 2(m^2-2m+3)
(α-β)^2 = 8m-12
(3) β-α=2
-(α-β) = 2
(α-β)^2=2^2
8m-12 = 4
m = 2
x^2 - (sum of roots)x + (product of roots) = 0
或者背呢樣野:
sum of roots = -b/a
product of roots = c/a
(1) α+β = -2m
αβ = m^2-2m+3
(2) (α+β)^2 = α^2 + 2αβ + β^2
(-2m)^2 = α^2 + β^2 + 2(m^2-2m+3)
α^2 + β^2 = 4m^2 -2m^2 + 4m -6
α^2 + β^2 = 2m^2 + 4m -6
(α-β)^2 = α^2 - 2αβ + β^2
(α-β)^2 = α^2 + β^2 - 2αβ
(α-β)^2 = 2m^2 + 4m -6 - 2(m^2-2m+3)
(α-β)^2 = 8m-12
(3) β-α=2
-(α-β) = 2
(α-β)^2=2^2
8m-12 = 4
m = 2
收錄日期: 2021-04-13 18:12:00
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https://hk.answers.yahoo.com/question/index?qid=20071111000051KK01866