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由 Raceleader 於 星期日 四月 06, 2003 4:41 pm
It is also true for -1<=a<=2-sqrt(8)
Re: [數學]Find the possible values of a.
由 --- 於 星期日 四月 06, 2003 4:12 pm
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
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
[數學]Find the possible values of a.
由 E.T 於 星期日 四月 06, 2003 3:25 pm
Given that (a+1)x2 + ax + 1 > 0 for all values IN THE REGION x<1 .
Find the possible values of a.
Find the possible values of a.