Question

设X1和X2试方程2x平方-4mx+2m平方+3m-2=0的两个实数根.当m为何值时,X1平方+x2平方有最小值?

Answer 1

X1平方+x2平方≤ 2X1*X2
X1*X2=(2m平方+3m-2)/2
所以X1平方+x2平方≤2*(2m平方+3m-2)/2
=2m平方+3m-2
对称轴=(-b)/2a
=(-3)/2*2
=-3/4
所以M=3/4是有最小值 最小值7/8

Answer 2

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

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

m取3/4时，最小值7/8

Answer 3

2x^2-4mx+2m^2+3m-2=0
△=2m^2-2m^2-3m+2>=0,
m<=2/3
x1^2+x2^2
=(x1+x2)^2-2x1x2
=(2m)^2-2*((2m^2+3m-2)/2)
=4m^2-2m^2-3m+2
=2m^2-3m+2
=2(m-3/4)^2+7/8
m=2/3
2(m-3/4)^2+7/8 =2*(2/3-3/4)^2+7/8=1/72+7/8=64/72=8/9