how to solve the following equation?
(x^2 - 2x) - 2x^2 + 4x - 3 =0
(5^2x - 125)^2 = 0
Math problem
2007-05-24 10:29 pm
回答 (2)
2007-05-24 10:46 pm
✔ 最佳答案
(x^2 - 2x) - 2x^2 + 4x - 3 =0x^2-2x+3=0
x=(2士√-8)/2
x=1士√-2
x=1士√2i
[5^(2x) - 125]^2 = 0
5^(2x)-125=0
5^(2x)=125
5^(2x)=5^3
2x=3
x=1.5
2007-05-24 14:46:33 補充：
i=√-1(虛數)
參考: me
2007-05-24 10:44 pm
1. (x^2 - 2x) - 2x^2 + 4x - 3 =0
x^2 - 2x + 3 = 0
因為 b^2 - 4ac <0 (自行驗證)
所以 x 無實數解.(但有虛數根!!!)
2. (5^2x - 125)^2 = 0
5^2x - 125 = 0
5^2x = 125
2x = log 125 / log 5
x = 3/2
x^2 - 2x + 3 = 0
因為 b^2 - 4ac <0 (自行驗證)
所以 x 無實數解.(但有虛數根!!!)
2. (5^2x - 125)^2 = 0
5^2x - 125 = 0
5^2x = 125
2x = log 125 / log 5
x = 3/2
收錄日期: 2021-05-03 04:46:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070524000051KK02351