F.4 maths

Method 1 :
Sum of roots = α + β = - (-2)/1 = 2 , i.e. β = 2 - α
α² + 2β
= α² + 2(2 - α)
= α² - 2α + 4
= (α² - 2α - 6) + 10
= 0 + 10 ... (α² - 2α - 6 = 0 since α is a root of x² - 2x - 6 = 0)
= 10

Method 2 :
x² - 2x - 6 = 0
α + β = 2
α β = - 6

(α² + 2β) - (β² + 2α)
= (α² - β²) - 2(α - β)
= (α - β)(α + β) - 2(α - β)
= (α - β) (α + β - 2)
= (α - β) (2 - 2)
= 0
∴ α² + 2β = β² + 2α

Hence α² + 2β
= ( (α² + 2β) + (β² + 2α) ) / 2
= ( (α + β)² - 2αβ + 2(α + β) ) / 2
= ( 2² - 2(- 6) + 2(2 ) / 2
= 10


2014-10-22 07:01:37 補充：
多謝Mr.cat的鼔勵 ~
你的Method 3 叫 Method 1.5 更貼切, 個人認為基本上它是Method 1 的另一種表達而已。

2014-10-22 07:04:12 補充：
Method 3 :
x² - 2x - 6 = 0
α = 1 ± √7 , β = 1干√7
∴ α² + 2β = (1 ± 2√7 + 7) + (2 干 2√7) = 10

2014-10-24 01:41:34 補充：
You are 貓sir.

Method 3:
We usually reduce the degree of the term, so you may consider:

α² - 2α - 6 = 0
α² = 2α + 6 [From degree 2 to degree 1]

Therefore,
α² + 2β
= 2α + 6 + 2β
= 2(α + β) + 6
= 2(2) + 6
= 10

2014-10-21 23:23:33 補充：
不要氣餒，上次輸了：
https://hk.knowledge.yahoo.com/question/question?qid=7014081900173

今次一定勝！

╭∧---∧╮
│ .✪‿✪ │
╰/) ⋈ (\\╯

2014-10-23 00:52:18 補充：
I concur.

You are so welcome, but are you sure I am a Mr.?

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