解方程组,代入得:
4K^2X^2-8K(K+1)X+4(2K-1)^2=X^2-4
(4K^2-1)X^2-8K(2K+1)X+4(4K^2-4K+2)=0,
Δ=64K^2(2K+1)^2-32((2K+1)(2K-1)(2K^2-2K+1)
=32(2K+1)[4K^3+2K^2-4K^3+6K^2-4K+1]
=32(2K+1)[8K^2-4K+1]
=256(2K+1)[K^2-1/2K+(1/4)^2-1/16+1/8]
=256(2K+1)[(K-1/4)^2+1/16]
∵(K-1/4)^2+1/16≥1/16>0,
∴有公共点即Δ≥0,
∴2K+1≥0,
K≥-1/2.