F. 4 Maths

2013-10-02 2:59 am
Given that α and β are the roots of the equation x2 – 3x – 5 = 0. find the values of each of the following expressions.(a) (α + 2β) (2α + β)(b) (α + β)2 (c) (α –α/β) (β – β/α)
更新1:

αβ = c/a=-5?

回答 (2)

2013-10-02 5:39 am
✔ 最佳答案
兩根的和 = α + β = - (-3/1) = 3
兩根的積 = αβ = -5/1 = -5

(a) (α + 2β)(2α + β)
= 2α^2 + αβ + 4αβ + 2β^2
= 2α^2 + 2β^2 + 5αβ
= 2(α^2 + β^2) + 5αβ
= 2(α^2 + β^2 + 2αβ - 2αβ) + 5αβ
= 2[(α^2 + 2αβ + β^2 ) - 2αβ] + 5αβ
= 2[(α + β)^2 - 2αβ] + 5αβ
= 2[3^2 - 2(-5)] + 5(-5)
= 2[9 - (-10)] + (-25)
= 2(9 + 10) - 25
= 2(19) - 25
= 13

(b) (α + β)^2
= 3^2
= 9

(c) (α - α/β)(β - β/α)
= α(β - β/α) - α/β(β - β/α)
= αβ - β - α + 1
= αβ - (β + α) + 1
= -5 - 3 + 1
= -7
2013-10-02 3:40 am
假設你並非懶,而是被嚇怕了,幫你小小。

由於 α 和 β 為根,有

α + β = -(b/a) = 3
αβ = 5

(a) (α + 2β) (2α + β) 乘開來
= 2α^2 + 4αβ + αβ + 2β^2
= 2(α^2 + β^2) + 5αβ
= 2(α^2 + β^2 + 2αβ - 2αβ) + 5αβ
= 2[(α^2 + 2αβ + β^2 ) - 2αβ] + 5αβ
= 2[(α + β)^2 - 2αβ] + 5αβ
把 α + β = 3 和 αβ = 5 代入便可以了

(b)直接代入 α + β = 3

(c) 乘開之後會等於 αβ - α - β + 1 ,自己再做啦!


2013-10-01 21:05:03 補充:
對不起,我錯了,αβ = c/a=-5 才是正確的。


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