Question

a、b是方程x^2-2(m+2)x+2m+2=0的两个实数根,问:m为何值时,(a-1)^2+(b-1)^2有最小值?最

Answer 1

a+b=2(m+2)=2m+4

ab=2m+2

(a-1)^2+(b-1)^2

=a^2-2a+1+b^2-2b+1

=(a+b)^2-2ab-2(a+b)+2

=(2m+4)^2-2(2m+2)-2(2m+4)+2

=4m^2+8m+6

=(2m+2)^2+2

当m=-1时有最小值2