All friends are nice except cma.

f(x)=ax^2+bx+c ,if we have 3 indepent conditions,we can decide  a,b,c and get anser.5 is ff's repeat root,2 is f's root,3 conditionss,so we got answer. But cma did not say yes, so I thoght again.3-2=1,our answer should have a parametric variablee.following is my best for cma:

f(x)=a(x-2)(x-p),p is not a real number,(15less is right,he likes saying white horse is not horse)

ff(x)=a^3((x-2)(x-P)-2/a)(x-2)(x-p)-p/a),ff(5)=a^3(3(5-p)-2/a)(3(5-p)-p/a))=0 we get

p=15a/(3a+1) or p=5a-2/3. f(0)=30a^2/(3a+1) or -2a(5a-2/3).a can take any number except p being real number.In fact a can't be real number and some special complex singular points.

 

所有跟帖： 

• ...some special complex singular points is empt set. -jinjing- ♀ (0 bytes) () 07/21/2011 postreply 07:39:43

• That means a can take any number except real number. -jinjing- ♀ (0 bytes) () 07/24/2011 postreply 17:27:52

• For cma's question -yma16- ♂ (85 bytes) () 07/23/2011 postreply 10:19:59

• I misread the question. I think KangMM has the right answer. -yma16- ♂ (0 bytes) () 07/23/2011 postreply 12:03:39

• MM has the right answer for 5 is repeated root, Mine is for real -jinjing- ♀ (0 bytes) () 07/25/2011 postreply 10:04:51

• You are too fast too be right as I did,...f(f(X) has 5 and -1 tw -jinjing- ♀ (0 bytes) () 07/23/2011 postreply 17:10:00