已知抛物线y＝x²-2x-3 1.求出y＞0与y＜0时x的取值范围

来源：学生作业学帮网编辑：学帮网时间：2024/05/31 09:24:47

x²-2x-31=0
x-2x+1=32
(x-1)²=32
x-1=±4根号2
x=±4根号2+1
抛物线a＞0,开口向上,
所以y＞0,x＞4根号2+1或x＜-4根号2+1
y＜0,-4根号2+1＜x＜4根号2+1