x1，x2是关于x的方程4x*2-4mx+m+2=0的两个根，当m为何值时，y=x1*2+x2*2 取最小值 ？并求这个最小值

△=（4m)^2-16(m+2)≥0,即m≥2或m≤-1
x1 + x2 = - (-4m/4) = m
x1 x2 = (m+2)/4
y=x1^2+x2^2= (x1 + x2)^2 - 2 x1 x2
y = m^2 - (m+2)/2
= m^2 - m/2 - 1
= (m - 1/4)^2 - 17/16 (m≥2或m≤-1)
所以m =-1 时, y取最小值=1/2

y=x1*2+x2*2
= (x1 + x2)*2 - 2 x1 x2

根据韦达定理
x1 + x2 = - (-4m/4) = m
x1 x2 = (m+2)/4

所以
y = m*2 - (m+2)/2
= m*2 - m/2 - 1
= m*2 - 2 (1/4) m + (1/4)*2 - (1/4)*2 - 1
= (m - 1/4)*2 - 17/16

因此 当 m = 1/4 时, y 最小
最小值为 -17/16