唔識做功課24

As a and β are the roots of the equation x^2－2x－6＝0, so
a＋β＝2 ⋯⋯⋯⋯⋯ (i) (sum of roots)
a^2－2a－6＝0 ⋯⋯ (a is one of the root)
a^2＝2a＋6 ⋯⋯⋯ (ii)
Therefore,
a^2＋2β
＝2a＋6＋2β ⋯⋯⋯ (from (ii))
＝6＋2(a＋β)
＝6＋2*2 ⋯⋯⋯⋯⋯ (from (i))
＝10

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

sum of roots = -b/a

借問一下, 2(a+β) 之後果步點變做 2[-(-2)/1] ?

你答啦～

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2014-08-20 01:20:07 補充：
大家加油呀～

請 HK~ 作答～

﹝｡◕‿◕｡◕‿◠｡﹞

x² - 2x - 6 = 0 with roots a and β,

{ a + β = -(-2)/1 = 2
( aβ = (-6)/1 = -6

x^2-2x-6=0
⇒a^2-2a-6=0 ⇒ a^2=2a+6

∴a^2+2β
=2a+6+2β
=2(a+β)+6
=2[-(-2)/1]+6
=2(2)+6
=4+6=10

2014-08-20 00:28:03 補充：
無所謂啦 0.0

大家都係幫發問者 =]

而家點數已經唔重要了 =]