Question

已知抛物线y=5x^2+(m-1)+m与x轴两交点在y轴同侧,它们的距离的平方等于49/25,求m值是多少?

Answer 1

y=5x^2+(m-1)x+m

令两交点的横坐标为x1,x2,x2>x1

x1+x2=-(m-1),x1*x2=m

∵两交点在y轴同侧

∴x1*x2>0,m>0

(x2-x1)^2=49/25

(x2+x1)^2-4x1*x2=49/25

[-(m-1)]^2-4m=49/25

m=3±√249/5

∵m>0

∴m=3+√249/5