设x1,x2是关于x的方程x2-(k+2)x+2k+1=0的两个实数根,且X1²+X2²=11

(1) x1+x2=k+2
x1x2=2k+1
(x1+x2)²=x1²+x2²+2x1x2
=11+2x1x2
(k+2)²=11+2(2k+1)
k²+4k+4=11+4k+2
k²=9
k=±3
k=3时,x²-5x+7=0
Δ=25-4*70,k=-3符合题意,故k=-3
(2) x²+x-5=0
x1+x2=-1
x1x2=-5
新方程的两根为x3,x4
x3=x1+x2=-1
x4=(x1-x2)²
=x1²+x2²-2x1x2
=11-2*(-5)
=21
x3+x4=-1+21=20
x3x4=(-1)*21=-21
新方程：x²-20x-21=0