#ed_op#br#ed_cl#你指的根是 1/(m+n) 及 1/(m-n) 吧...#ed_op#br#ed_cl#找不到簡單的方程，只有下面這：#ed_op#br#ed_cl#a(x-n)(x-m) = ax#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-a(m+n)x+a(mn) = ax#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+bx+c#ed_op#br#ed_cl#=> m+n = -b/a, mn = c/a, m-n = (√(b#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-4ac))/a#ed_op#br#ed_cl#k(x-1/(m+n))(x-1/(m-n)) = k(x + a/b)(x - a/√(b#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-4ac))#ed_op#br#ed_cl#let k = b√(b#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-4ac)#ed_op#br#ed_cl#M = b(√(b#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-4ac))x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl# + a(b-√(b#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-4ac))x - a#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#tangpakchiu 寫到:#ed_op#div#ed_cl#若a,b,c,m,n為實數,且m,n為方程ax^2+bx+c=0的根,#ed_op#/div#ed_cl##ed_op#div#ed_cl#求二次方程M使得1/m+n,1/m-n為該方程的根。#ed_op#/div#ed_cl#
+ / -檢視主題
Re: [問題]方程的根
由 G@ry 於 星期二 三月 27, 2007 7:18 am
[問題]方程的根
由 tangpakchiu 於 星期一 三月 26, 2007 9:39 pm
#ed_op#DIV#ed_cl#若a,b,c,m,n為實數,且m,n為方程ax^2+bx+c=0的根,#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#求二次方程M使得1/m+n,1/m-n為該方程的根。#ed_op#/DIV#ed_cl#