Question

x1x2为2x2-4mx+2m2+3m-2=0的两实根，当m为何值时，（x1）2+（x2）2有最小值，求出最小值

Answer 1

有实根，判别式大于等于0
16m²-16m²-24m+16>=0
m<=2/3

韦达定理
x1+x2=2m
x1x2=(2m²+3m-2)/2
所以x1²+x2²=(x1+x2)²-2x1x2
=4m²-2m²-3m+2
=2m²-3m+2
=2(m-3/4)²+7/8
m<=2/3
所以定义域在对称轴左边，函数值随m增大而减小
所以m=2/3
x1²+x2²最小=8/9

Answer 2

韦达定理得:
x1+x2=4m/2=2m
x1x2=(2m2+3m-2)/2

x1^2+x2^2=(x1+x2)^2-2x1x2
=4m^2-(2m^2+3m-2)
=2m^2-3m+2
=2(m-3/4)^2+7/8

判别式=16m^2-8(2m2+3m-2)>=0
-24m+16>=0
m<=2/3

所以,当m=2/3时,有最小值是:2*(2/3-3/4)^2+7/8=8/9