费斐数学6-2

zcx

2、If for all x, x^2+ax+b = (x+c)^2, then a =?
（1）b=0；
（2）c=-3；

key b

为什么条件1不可以呢？b=0 =>c=0 =>a=0?

lwei1016

1b=0,x^2+ax=(x+c)^2-->x^2+ax=x^2+2cx+c^2

只能算出a=c=0.偶觉得条件1可以.请牛牛指点!

[此贴子已经被作者于2004-6-17 11:55:54编辑过]

juniya

这道题目以前讨论过，搜一下了

lucre

earlier discussion is linked here http://forum.chasedream.com/dispbbs.asp?BoardID=22&ID=46316

one simple way is to transform x^2+ax+b = (x+c)^2 to (a-2c)+b-C^2=0,

hence a=2c;b=c^2

so if b=9 a=3/2 or -3/2, thus b=9 is ruled out