delta=(8m)^2-16(m-4)=16(4m^2-m+4)>0,方程必有两个不等实根.
x1+x2=-2m,
4x1^2+8mx1+m-4=0
x2^2+2mx2+m/4-1=0
上两式相减：4x1^2-x2^2+8mx1-2mx2+3m/4-3=0
x2=-2m-x1,代入：
4x1^2-x2^2+8mx1-2m(-2m-x1)+3m/4-3=0
4x1^2-x2^2+10mx1+4m^2+3m/4-3=0
对比4x1^2-x2^2+10mx1+1/4=0,得：4m^2+3m/4-3=1/4
16m^2+3m-13=0
(16m -13)(m+1)=0
m=13/16 or m=-1
这里X1>X2条件没有用到.