請詳細教我計下題 :
The equation (2k-1)x^2+(k-3)x-2=0 (where k 不等於1/2), find the value of k when one root of the equation is the negative reciprocal of the other root.
答案是3/2.
急 ! F4 Quadratic Equations
2011-10-27 9:08 pm
回答 (1)
2011-10-27 10:19 pm
✔ 最佳答案
(2k-1)x^2+(k-3)x-2=0i.e. x^2 + [(k-3)/(2k-1)]x - 2/(2k-1) = 0
Let the roots be A & -1/A
A*(-1/A) = -2/(2k-1)
-1 = -2/(2k-1)
2k-1 = 2
2k = 3
k = 3/2
參考: (X-A)(X-B) = X^2 - (A+B)X + AB
收錄日期: 2021-04-13 18:19:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111027000051KK00266