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匿名 發問於 科學及數學數學 · 6 年前

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

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  • 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!

    ╭∧---∧╮

    │ .✪‿✪ │

    ╰/) ⋈ (\\╯

    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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