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
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![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](/uploads/image/z/14339376-0-6.jpg?t=%E5%B7%B2%E7%9F%A5x1%2Cx2%2C%E6%98%AF%E6%96%B9%E7%A8%8Bx%26%23178%3B-%EF%BC%88k%2B1%EF%BC%89x%2Bk%26%23178%3B%2F4%2B1%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E5%AE%9E%E6%A0%B9%2C%E6%B1%82S%3Dx1%26%23178%3B%2Bx2%26%23178%3B%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC)
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
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
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