设a,B是方程4x^2-4mx+m+2=0的两个实根,f(m)=a^2+B^2,求实数m的取值范围和f(m)得最小值

Question

设a,B是方程4x^2-4mx+m+2=0的两个实根,f(m)=a^2+B^2,求实数m的取值范围和f(m)得最小值

Answer 1

a,B是方程4x^2-4mx+m+2=0的两个实根,
所以
a+b=m
ab=(m+2)/4
Δ=16m²-16(m+2)>=0
m²-m-2>=0
(m+1)(m-2)>=0
m>=2或m<=-1
f(m)=a²+b²=(a+b)²-2ab
=m²-(m+2)/2
=m²-1/2m-1
=(m-1/4)²-17/16
所以m=-1时,取最小值=1-1/2=1/2