问题描述：
一道数学题``今天中午就要答案了`
已知X1,X2是关于方程4X^2-（3M-5）-6M^2=0的俩个实数根,且X1/X2的绝对值=3/2,求 常数M的值
问题解答：
我来补答
有两个不同的实数根
所以(3m-5)^2+96m^2>0
105m^2-30m+25>0
此不等式恒成立
x1+x2=(3m-5)/4
x1*x2=-3m^2/2
|x1/x2|=3/2
所以x1=(3/2)x2或x1=-(3/2)x2
x1=(3/2)x2
分别代入x1+x2=(3m-5)/4
x1*x2=-3m^2/2
(5/2)x2=(3m-5)/4
(3/2)x2^2=-3m^2/2
x2=(3m-5)/10
所以(3/2)[(3m-5)/10]^2=-3m^2/2
[(3m-5)/10]^2+m^2=0
则(3m-5)/10=0,且m=0
不成立
x1=-(3/2)x2
分别代入x1+x2=(3m-5)/4
x1*x2=-3m^2/2
-(1/2)x2=(3m-5)/4
(3/2)x2^2=-3m^2/2
x2=-(3m-5)/2
所以-(3/2)[(3m-5)/2]^2=-3m^2/2
[(3m-5)/2]^2=m^2
(3m-5)/2=m或(3m-5)/2=-m
m=5或m=1
