Question

已知方程2(m+1)x^2+4mx+3m-2=0是关于x的一元二次方程,那么m的取值范围是

Answer 1

一元二次方程满足两个条件：1、x^2的系数不为0，2、△>=0
首先考虑第一个，得m≠-1
然后考虑△=b^2-4ac=16m^2-4*2(m+1)*(3m-2)
=16m^2-(8m-8)(3m-2)
=16m^2-24m^2+40m-16
=-8m^2+40m-16
=-8(m^2+5m-2)>=0
所以
m^2+5m-2<=0
m^2+5m+(5/2)^2<=2+(5/2)^2
(m+5/2)^2<=33/4
-sqrt33/4<=m+5/2<=sqrt33/4
(-sqrt33-5)/4<=m<=(sqrt33-5)/4

所以m的取值范围是(-sqrt33-5)/4<=m<=(sqrt33-5)/4，同时m≠-1

Answer 2

方程是一元二次方程，二次项系数≠0
2(m+1)≠0
m≠-1

Answer 3

m不等于-1