If M belongs to R X1 and X2 are two zeros of the square of the function f (x) = x-2mx + 1-m, then the minimum value of the square of X1 plus the square of X2 is——————

x1+x2=m,
x1x2=-m,
x1^2+x2^2=(x1+x2)^2-2x1x2=m^2-2(-m)=m^2+2m,
When m = - 1, m ^ 2 + 2m is the minimum,
That is, (m ^ 2 + 2m) min = 1-2 = - 1

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