唔識做功課24
2014-08-20 7:56 am
If a and β are the roots of the equation x^2-2x-6=0 , find the values of a^2+2β .
回答 (6)
2014-08-20 4:37 pm
✔ 最佳答案
As a and β are the roots of the equation x^2－2x－6＝0, soa＋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
2014-08-20 9:23 am
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
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
2014-08-20 8:51 am
sum of roots = -b/a
2014-08-20 8:49 am
借問一下, 2(a+β) 之後果步點變做 2[-(-2)/1] ?
2014-08-20 8:24 am
你答啦～
我差ｄ見你唔到就答左～
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
我差ｄ見你唔到就答左～
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
2014-08-20 8:22 am
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 =]
⇒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 =]
收錄日期: 2021-04-15 16:17:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140819000051KK00173