Question

若关于x的方程x^2-ax+1=0在x∈(1/2,3)上有实数根，则实数a的取值范围？

Answer 1

x^2-ax+1=0在x∈(1/2,3)上有实数根,则
x1+x2=a,x1x2=1,得
a=x1+1/(x1)=[(x1)^2+1]/(x1)
因x∈(1/2,3),故
a∈(5/2,10/3)

Answer 2

a大雨等于2小于10/3

Answer 3

2<=a<=23 a∈(2,3)

追问

2<=a<=23还是a∈(2,3) ?

追答

2<=a<=3