Question

设X1，X2是方程2x^2-4mx+2m^2+3m-2=0的两个实数根，当m为何值时，X1^2+

设X1，X2是方程2x^2-4mx+2m^2+3m-2=0的两个实数根，当m为何值时，X1^2+X2^2有最小值？并求出这个最小

Answer 1

已经给你改了答案
解由题知方程的
Δ≥0
即(-4m)^2-4*2(2m^2+3m-2)≥0
即16m^2-16m^2-24m+16≥0
即24m≤16
即m≤2/3
又由根与系数的关系知
x1+x2=2m，x1x2=(2m^2+3m-2)/2

故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+2-9/8
=2(m-3/4)^2+7/8 (m≤2/3)
故当m=2/3时，X1^2+X2^2有最小值2(2/3)^2-3(2/3)+2
=8/9-2+2
=8/9
综上知X1^2+X2^2有最小值8/9.
(评述一般初中不出这题)

Answer 2

解：2x²-4mx+2m²+3m-2=0
根据韦达定理得：x1+x2=2m，x1x2=（2m²+3m-2）/2
x1²+x2²
=（x1+x2）²-2x1x2
=（2m）²-2×（2m²+3m-2）/2
=4m²-2m²-3m+2
=2m²-3m+2
=2（m²-3/2m）+2
=2（m²-3/2m+9/16）-9/8+2
=2（m-3/4）²+7/8
当m=3/4时，x1²+x2²有最小值，最小值为7/8.