已知抛物线y=x的平方+(m-1）x+2m与x轴的两个交点在y轴同侧,他们的距离等于5,求m的值

设两根分别为x1,x2,由韦达定理得
x1+x2=1-m
x1x2=2m 交点在x轴同侧,x1,x2同号,x1x2>0 2m>0 m>0
(x1-x2)²=(x1+x2)²-4x1x2=(1-m)²-4(2m)=m²-10m+1
|x1-x2|=√(m²-10m+1)=5
m²-10m+1=25
m²-10m-24=0
(m-12)(m+2)=0
m=12或m=-2(