Question

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

Answer 1

△=（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