判别式 = (k+1)^2-4(k+2) = k^2-2k-7=(k-1)^2-8=(k-1+2√2)(k-1-2√2) ≥ 0
k ≤ 1-2√2,或k≥1+2√2
韦达定理：
x1+x2=k+1,x1x2=k+2
x1^2+x^2^2=(x1+x2)^2-2x1x2=(k+1)^2-2(k+2)=k^2-3=6
k^2=9
k=-3,或k=3
k=3＜1+2√2不合题意,舍去
∴k=-3