设x1,x2是方程x^2-2kx+1=k^2的两个实数根,求x1^2+x2^2的最小值

方程为x^2-2kx+1-k^2=0
由伟达定理
x1+x2=2k
x1x2=1-k^2
x1^2+x2^2=(x1+x2)^2-2x1x2
=(2k)^2-2+2k^2
=6k^2-2
所以k=0时原式值最小，为-2。

x1^2+x2^2
=(x1+x2)^2-2x1*x2
=(2k)^2-2(1-k^2)
=4k^2-2+2k^2
=6k^2-2
△=4k^2-4(1-k^2)≥0
k≥√2 /2或k≤-√2 /2
当k=√2 /2时,x1^2+x2^2=6k^2-2取到最小值为1

x1+x2=-(b/a)=-(-2k)=2k
x1x2=c/a=1-k^2
x1^2+x2^2
=(x1+x2)^2-2x1x2
=(2k)^2-2(1-k^2)
=4k^2-2+2k^2
=6k^2-2
△=4k^2-4(1-k^2)≥0
所以k^2>=1/2
即x1^2+x2^2=6k^2-2的最小值是1

x^2-2kx+1=k^2
x^2-2kx+1-k^2=0
x1+x2=-(b/a)=-(-2k)=2k
x1x2=c/a=1-k^2
x1^2+x2^2
=(x1+x2)^2-2x1x2
=(2k)^2-2(1-k^2)
=4k^2-2+2k^2
=6k^2-2
6k^2-2=0
3k^2-1=0
k^2=1/3
k1=(根号3)/3,k2=(-根号3)/3
所以最小值为(-根号3)/3

x^2-2kx+1=k^2x^2-2kx+1-k^2=0x1+x2=2kx1x2=1-k^2x1^2+x2^2=(x1+x2)^2-2x1x2=4k^2-2(1-k^2)=4k^2-2+2k^2=6k^2-2判别式=4k^2-4(1-k^2)=4k^2-4+4k^2=8k^2-4>=0k^2>=1/2k=根号2/2x1^2+x2^2的最小值1