看板 Math 關於我們 聯絡資訊
※ 引述《Ahome (繼續挑戰)》之銘言: : 請問一題解方程式的難題: : 4 3 2 : x + 10x + 14x -10x + 1 = 0 : thx... 顯然 x = 0 不是解 兩邊同除以 x^2 得 x^2 + 10x + 14 - 10/x + 1/x^2 = 0 令 t = x - 1/x 則 t^2 = x^2 - 2 + 1/x^2 原式可寫為 t^2 + 10t + 16 = 0 這可以簡單解得 t = -2, -8 (過程略) 再分別解 x - 1/x = -2 => x^2 + 2x - 1 = 0 => x = -1±√2 x - 1/x = -8 => x^2 + 8x - 1 = 0 => x = -4±√17 -- 這種係數對稱的一元四次方程式 (ax^4 + bx^3 + cx^2 ± bx + a = 0) 都可以用一樣的方法來做 一次項可以跟三次項不同號 (像這題) 如果同號則改令 t = x + 1/x 即可 -- 'Oh, Harry, don't you see?' Hermione breathed. 'If she could have done one thing to make absolutely sure that every single person in this school will read your interview, it was banning it!' ---'Harry Potter and the order of the phoenix', P513 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 122.118.121.182
Ahome :鬼神啊~~好強啊~~我都不知有這種解法耶... 05/22 20:52
Ahome :不過想再請問...除了這種還有其他解法嗎?thx... 05/22 20:53
Vulpix :有啊...根式解也可以用啊...可是很難看 05/22 21:44
WalterbyJeff:好強!! 05/23 23:02
bn51401 :強大= =+ (筆記) 05/23 23:05