△ = 4(k + 1)² - 4(k² - 1) ≥ 0
解得:k ≥ -1
根據(jù)韋達(dá)定理
x1 + x2 = -2(k + 1)
x1 * x2 = k² - 1
x1² + x2²
= (x1 + x2)² - 2x1 * x2
= 4(k + 1)² - 2(k² - 1)
= 4k² + 8k + 4 - 2k² + 2
= 2k² + 8k + 6
= 2(k + 2)² - 2
當(dāng) k = - 1 時,取最小值 0