4A^2+4AL-L^2=0
求A值？
我會用另外方式
4A^2+4AL+L^2-2L^2=0
(2A+L)^2=2L^2
2A+L=±(√2) L
A=(-L±√2L)/2
可是如果要帶下列公式的話要怎帶？
(-b±√(b^2-4ac))/2a
2次方公式
2014-04-13 6:49 pm
回答 (16)
2014-04-16 4:22 am
✔ 最佳答案
4A^2+4AL-L^2=0a=4, b=偶數/2=4L/2=2L, c=-L^2A=[-b±√(b^2-ac)]/a=[-2L+-√(4L^2+4*L^2)]/4=(-2L+-L√8)/4=(-2L+-2L√2)/4=(-L+-L√2)/2=(-1+-√2)L/2
2014-05-30 1:51 pm
2014-05-08 4:27 pm
2014-05-04 5:45 pm
2014-05-02 3:36 pm
2014-05-01 5:30 pm
2014-04-30 10:53 pm
2014-04-29 4:20 pm
2014-04-28 4:03 pm
2014-04-28 7:01 am
2014-04-27 10:59 pm
2014-04-25 2:53 pm
2014-04-24 1:50 pm
2014-04-17 10:26 am
2014-04-15 9:51 am
2014-04-13 6:51 pm
a=4, b=4L, c= -L^2
收錄日期: 2021-04-30 18:42:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140413000010KK02748