△ ≥ 0
(k - 1)² - 4(2k - 1) ≥ 0
k² - 2k + 1 - 8k + 4 ≥ 0
k² - 10k + 5 ≥ 0
k ≤ 5 - 2√5 或 k ≥ 5 + 2√5
设方程两根为 x1、x2 ,则
x1 + x2 = k - 1
x1 * x2 = 2k - 1
因为两根都大于2
所以 x1 - 2 > 0 ,x2 - 2 > 0
所以 x1 - 2 + x2 - 2 > 0 且 (x1 - 2)(x2 - 2) > 0
x1 - 2 + x2 - 2 > 0
x1 + x2 - 4 > 0
k - 1 - 4 > 0
k > 5
(x1 - 2)(x2 - 2) > 0
x1 * x2 - 2(x1 + x2) + 4 > 0
2k - 1 - 2(k - 1) + 4 > 0
5 > 0 恒成立
综上：k ≥ 5 + 2√5