由 Raceleader 於 星期日 四月 06, 2003 3:45 pm
(a+1)x2+ax+1>0
2 cases:
1. (a+1)x2+ax+1>0 for all real x:
2. (a+1)k2+ak+1=0, where k≧1
Case 1:
Discriminant=a2-4(a+1)=(a-2)2-8
(a-2)2-8<0
2-√8<a<2+√8
But a+1>0
a>-1
So, 2-√8<a<2+√8
Case 2:
k=1, a=-1, it is true for all x<1
when k>1 and a>-1:
k={-a-√[a2-4a-4]}/2(a+1)
if k>1, then a>-1
Combine both cases: -1≦a<2+√8