x^2-2mx+m+2=0
△=4m^2-4(m+2)≥0
m^2-m-2≥0
(m-2)(m+1) ≥0
m≥2,m≤-1
x1^2+x2^2=(x1+x2)^2-2x1x2
=(2m)^2-2*2
=4m^2-4
因m≥2,m≤-1
所以4m^2-4≥0
即x1^2+x2^2的最小值为0
x^2-2mx+m+2=0
△=4m^2-4(m+2)≥0
m^2-m-2≥0
(m-2)(m+1) ≥0
m≥2,m≤-1
x1^2+x2^2=(x1+x2)^2-2x1x2
=(2m)^2-2*2
=4m^2-4
因m≥2,m≤-1
所以4m^2-4≥0
即x1^2+x2^2的最小值为0