Let α,βare the roots of the quadratic equation x^2 -2 (k+1)x+2(k-1)=0 ,
where k is real.
(a) prove that αandβare real and unequal.
(b)(i)Express (α-5)(β-5) in terms of k.
(ii) Express α^2 +β^2 in terms of k.
(c) if α<5<β and α^2 +β^2 >16 ,find the least integral value of k.
=_= B PART 唔係用 sum of root product of root果d去計嗎?
計極都計吾到個岩個ans>_<
如果有人識計 可以教下我嗎 ?! thx
a maths inequalities
2008-09-30 6:05 am
回答 (2)
2008-09-30 6:11 am
✔ 最佳答案
With reference to my previous answer:http://hk.knowledge.yahoo.com/question/question?qid=7007082003911
a) △ = { - 2 ( k + 1 )}^2 - 4 ( 1 ){ 2 ( k - 1 ) }
= 4 ( k + 1 )^2 - 8 ( k - 1 )
= 4 ( k^2 + 2k + 1 ) - 8k + 8
= 4k^2 + 8k + 4 - 8k + 8
= 4k^2 + 12 > 0
所以α,β是兩個不的實數。
bi) α+β= 2 ( k + 1 ), αβ= 2 ( k - 1 )
(α-5)(β-5) = αβ- 5 (α+β) + 25
= 2 ( k - 1 ) - 5 { 2 ( k + 1 ) } + 25
= 2k - 2 - 10k - 10 + 25
= -8k + 13
ii) α+β = (α+β)^2 - 2αβ
={ 2 ( k + 1 ) } ^2 - 2 { 2 ( k - 1 ) }
= 4 ( k + 1 )^2 - 4 ( k - 1 )
= 4 ( k^2 + 2k + 1 ) - 4k + 4
= 4k^2 + 8k + 4 - 4k + 4
= 4k^2 + 4k + 8
c) α< 5 < β
α- 5 < 0 , β - 5 > 0
(α-5)(β-5) < 0
-8k + 13 < 0
13 < 8k
8k > 13
k > 13 / 8
ii) α+β > 16
4k^2 + 4k + 8 > 16
4k^2 + 4k - 8 > 0
k^2 + k - 2 > 0
( k + 2 )( k - 1 ) > 0
k < - 2 或 k > 1
所以,
k > 13 / 8
k的最小整數值 = 2
參考: My Maths Knowledge
2008-10-01 2:48 am
(a)
Delta= [2(k+1)]^2-4(x)[2(k-1)]
=4k^2+12
Because k^2 must larger than 0, so delta>0
(b)α+β=2(k+1)
αβ =2(k-1)
(i)(α-5)(β-5)
=αβ-5α-5β+25
=αβ+25
=2(k-1)+25
=2k+23
(ii)α^2 +β^2 +2αβ-2αβ
=(α+β)^2-2(αβ)...
Delta= [2(k+1)]^2-4(x)[2(k-1)]
=4k^2+12
Because k^2 must larger than 0, so delta>0
(b)α+β=2(k+1)
αβ =2(k-1)
(i)(α-5)(β-5)
=αβ-5α-5β+25
=αβ+25
=2(k-1)+25
=2k+23
(ii)α^2 +β^2 +2αβ-2αβ
=(α+β)^2-2(αβ)...
收錄日期: 2021-04-25 13:14:44
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080929000051KK02259