Question

已知x1 x2为方程x^2-2mx+6+m=0的两实根,则（x1-1）^2+(x2-1)^2的最小值为——

Answer 1

x1+x2=2m.x1x2=6+m

（x1-1）^2+(x2-1)^2

=(x1+x2)^2-2x1x2-2(x1+x2)+2

=4mm-2(6+m)-4m+2

=4mm-6m-10

=4(m-3/4)^2-9/4-10

所以最小值是当m=3/4时,值为（-49/4）