Question

设α,β是方程4x^2-4mx+m+2=0的两个实根,f(m)=α^2+β^2

1.求实数m的取值范围 2.求f（m）的最小值

Answer 1

(-4m)^2-4(4)(m+2)>0
m^2-m-2>0
(m-2)(m+1)>0
m>2 or m<-1

α+β=m
αβ=(m+2)/4
f(m)= α^2+β^2
= (α+β)^2-2αβ
= m^2-(m+2)/2
=m^2-m/2 -1
=(m-1/4)^2 - 17/16
min f(m) at m= -1 as m>2 or m<-1
min f(m) = 25/16-17/16 = 1/2