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f(x) = a(x-2)(x-b) = ax^2 - a(2+b)x  + 2ab
f(f(x)) = a^3[x^2 - (2+b)x + (2ab - 2)/a][x^2 - (2+b)x + (2ab - b)/a] = a^3*I*II.
Since 5 is the only root of f(f(x)), it must be 2 or 4 multiple root.
So, at least one of the follwoing is true.
I = (x -5)^2
or
II = (x-5)^2.
for either case, we get b = 8. The rest is easy.

所有跟帖： 

• 谢谢您给我打气, -jinjing- ♀ (119 bytes) () 07/15/2011 postreply 11:13:27

• 回复：谢谢您给我打气, -wxcfan123- ♂ (234 bytes) () 07/15/2011 postreply 14:13:18

• If you solve a from I, should prove II does not have real root, -wxcfan123- ♂ (0 bytes) () 07/15/2011 postreply 14:41:19

• Wrong way is always as wrong reason. -jinjing- ♀ (152 bytes) () 07/15/2011 postreply 15:21:12

• Math is not riddle -15少- ♂ (655 bytes) () 07/15/2011 postreply 15:20:39

• 回复：Math is not riddle -jinjing- ♀ (273 bytes) () 07/15/2011 postreply 16:03:10

• Why make stupid calcul when -15少- ♂ (815 bytes) () 07/15/2011 postreply 17:33:47

• You totally misunderstanding me, f(x)=a(X-5)^2+b,why b=2? you te -jinjing- ♀ (37 bytes) () 07/15/2011 postreply 18:04:50

• you tell me,please,not 8. -jinjing- ♀ (0 bytes) () 07/15/2011 postreply 18:07:08

• Ms. Kang has explained, her explanation is not as clear as you d -15少- ♂ (461 bytes) () 07/16/2011 postreply 06:30:03

• Why f(8)=0?Cs we use a Thm, It is unknow for most people here. -jinjing- ♀ (493 bytes) () 07/16/2011 postreply 08:19:30

• You like calcul -15少- ♂ (329 bytes) () 07/16/2011 postreply 09:10:20

• 回复：You totally misunderstanding me, f(x)=a(X-5)^2+b,why b=2? you -15少- ♂ (1365 bytes) () 07/16/2011 postreply 06:09:01