a-maths!!!!

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) α²+β² = (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) α²+β² > 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