急~有關一元二次方程既問題!!!

2010-10-08 3:50 am

如題~
1.若α和β是二次方程5x^2-7x+c=0的根,且α:β=3:4,求c的可能值.

2.若4x^2+2kx-5=0中兩根的差是3,求k的可能值.

3.已知α和β是二次方程(x-2)(x-3)=p的根.若1/α+1/β=4,求p的可能值.

4.已知α和β是二次方程x^2+1=k(x-1)的根.若(α-β)^2=8,求k的可能值.

可唔可以列明步驟呀??十萬個thx~!!!

回答 (1)

用戶2053

2010-10-08 4:29 am

✔ 最佳答案

1)5x^2-7x+c=0So
α + β = 7/5....(1)
α β = c/5....(2)Since α : β = 3 : 4 ,Let α = 3k , β = 4k , sub into (1) :3k + 4k = 7/57k = 7/5k = 1/5So α = 3/5 , β = 4/5 , sub into (2) :(3/5)(4/5) = c/512/25 = c/5c = 12/5
2)α - β = 3α^2 - 2αβ + β^2 = 9α^2 + 2αβ + β^2 - 4αβ = 9(α + β)^2 - 4αβ = 9 .....(1)4x^2+2kx-5=0α + β = - 2k/4 = - k/2
α β = - 5/4 Sub into (1) :(- k/2)^2 - 4(- 5/4) = 9(k^2) / 4 + 5 = 9k^2 = 16k = 4 or k = - 4
3)1/α + 1/β = 4αβ (1/α + 1/β) = 4αβα + β = 4αβ ........(1)(x-2)(x-3)=px^2 - 5x + 6-p = 0So
α + β = -(-5)/1 = 5
α β = 6 - pSub into (1) :5 = 4(6 - p)5 = 24 - 4p4p = 19p = 19/4
4)(α-β)^2=8(α + β)^2 - 4αβ = 8 ......(1);x^2+1=k(x-1)x^2 - kx + k+1 = 0So
α + β = -(-k)/1 = k
αβ = k+1Sub into (1) :k^2 - 4(k+1) = 8k^2 - 4k - 12 = 0(k - 6)(k + 2) = 0k = 6 or k = - 2

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