已知x1,x2,是方程x²-（k+1）x+k²/4+1=0的两个实根,求S=x1²+x2²的最小值

方程有二根
所以(k+1)²-4（k²/4+1)≥0
即k²+2k+1-k²-4≥0
得k≥3/2
x1+x2=k+1
x1x2=k²/4+1
S=x1²+x2²
=(x1+x2)²-2x1x2
=(k+1)²-2(k²/4+1)
=k²+2k+1-k²/2-2
=k²/2+2k-1
=(1/2)(k²+4k)-1
=(1/2)(k+2)²-3
当k=3/2时,有最小值为(1/2)(3/2+2)²-3=25/8