x² 的二次方程式 (x²)² + 2(k+1)x² + k² - 3k + 2 = 0
之公式解為
x²
= { -2(k+1) ± √[4(k+1)² - 4(k²-3k+2)] } / (2*1)
= -(k+1) ± √[(k+1)² - (k²-3k+2)]
= -(k+1) ± √(5k-1)
因為上式四根相異且恰有二個虛根,所以
5k-1 > 0 且 k+1 < √(5k-1)
5k-1 > 0
=> k > 1/5 ..... (*)
k+1 < √(5k-1)
=> (k+1)² < 5k-1
=> k² - 3k + 2 < 0
=> (k-1)(k-2) < 0
=> 1 < k < 2 ..... (**)
(*) and (**) => 1 < k < 2