It is known that the zeros of the function y = x ^ 2 + 2mx + 2m + 3 (M belongs to R) are x1, x2, Find the minimum value of X1 ^ 2 + x2 ^ 2

x^2+2mx+2m+3=0
x1+x2=-2m
x1x2=2m+3
x1²+x2²
=(x1+x2)²-2x1x2
=4m²-2(2m+3)
=4m²-4m-6
=4m²-4m+1-7
=(2m-1)²-7
So X1 & sup2; + x2 & sup2; > = - 7
The minimum value is - 7

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