Question

若关于x的方程2mx^2+(8m+1)x+8m=0有两个不相等的实数根，则m的取值范围？

m等不等于0

Answer 1

有两个不相等的实数根
△=b^2-4ac
=(8m+1)^2-4*2m*8m>0
64m^2+16m+1-64m^2>0
16m+1>0
m>-1/16

Answer 2

m＞-(1/16)

如果原方程有两个不相等的实数根
则△＞0
△=b²-4ac
=(8m+1)²-4*2m*8m
=64m²+16m+1-64m²
=16m+1＞0
m＞-(1/16)

Answer 3

有两个不相等的实数根△>0
∵△=b²-4ac=(8m+1) ²-4×2m×8m
=64m²+16m+1-64m²
=16m+1>0
m>-1/16

Answer 4

有两个不相等的实数根
故判别式大于0
即(8m+1)^2-4*2m*8m>0
==>64m^2+16m+1-64m^2>0
==>16m+1>0
==>m>-1/16