第一題. k=3 or k=-4

f(k/2) = 2(k/2)^2 + (k/2) - 4

2(k^2/4) + k/2 -4 = 2
k^2/2 + k/2 =6
k^2 +k =12
k^2+k-12=0
k=3 or k=-4

第2題

P(3) = (3)^3 + a(3)^2 +b(3) +15 =66

27+9a+3b+15 =66
3(3a+b)=24
3a+b=8 ------- (i)

P(-3) = (-3)^3 + a(-3)^2 +b(-3) +15 =0
-27+9a-3b+15=0
3(3a-b)=12
3a-b=4 -------- (ii)

From (i)
b=8-3a ------- (iii)

Sub (iii) into (ii)

3a-(8-3a)=4
6a=12
a=2

Sub a=2 into (iii)

b=8-3(2)
b=2

[ 本帖最後由 cat. 於 2006-12-27 01:26 PM 編輯 ]