grove888 寫到:sorry!應是2s2+(a+3)s+2a=0
s²(s+a) + k(s+1) = 0 ...... (*)
對 (*) 微分得
2s(s+a) + s² + k'(s+1) + k = 0
當 k' = 0 時,
2s(s+a) + s² + k = 0
3s² + 2as + k = 0
(s+1)[3s² + 2as + k] = 0 ﹝假設條件:s ≠ -1﹞
(s+1)(3s² + 2as) + k(s+1) = 0
(s+1)(3s² + 2as) - s²(s+a) = 0 ﹝by (*)﹞
s[(s+1)(3s + 2a) - s(s+a)] = 0
s[3s² + 2as + 3s + 2a - s² - as] = 0
s[2s² + as + 3s + 2a] = 0
s[2s² + (a+3)s + 2a] = 0
2s² + (a+3)s + 2a = 0 ﹝假設條件:s ≠ 0﹞