Question

若不等式ax²+bx+c＞0的解集为{x|-3＜x＜4}，求1/c·x²-bx+a＞0的解集

是（1/c·x²）-bx+a>0

Answer 1

ax^2+bx+c＞0的解集为{x|-3＜x＜4}
=> a<0
x1,x2 are roots of equation : ax^2+bx+c=0
x1+x2= -b/a = -3+4
-b/a = 1
b=-a (1)

x1.x2 = c/a = (-3)(4)
c/a =-12
c= -12a >0

1/(cx^2-bx+a) ＞0
cx^2-bx+a >0
x^2 -(b/c)x + (a/c) >0
x^2- (1/12)x - 1/12 >0
12x^2-x-1 >0
(4x+1)(3x-1)>0
x>1/3 or x<-1/4

追问

题目打的有点问题，你再看一下