1、已知x1,x2是方程x^2+mx+m-1=0的两个实数根,且x1^2+x2^2=17,求m的值

1题x1+x2=-m
x1x2=-1
则x1^2+x2^2
=(x1+x2)^2-2x1x2
=m^2+2=17
m=±√15
判别式大于等于0
m^2-4(m-1)>=0
(m-2)^2>=0
恒成立
所以m=±√152题有两个实数根
x-3x-a-a+2=0
x-3x-a-a+9/4-1/4=0
x-3x+9/4-a-a-1/4=0
(x-3x+9/4)-(a+a+1/4)=0
(x-3/2)-(a+1/2)=0
(x-3/2)=(a+1/2)
x-3/2=±(a+1/2)
x=3/2±(a+1/2)
x1=2+a
x2=1-a望采纳哦～