# F.4 maths

If α and β are the roots of the equation of x^2 - 2x -6 = 0, find the values of α^2 + 2β

### 2 個解答

• 6 年前
最愛解答

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.

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• 6 年前

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?q...

今次一定勝！

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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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