Re: [中學] 解方程式...
※ 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
--
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