如題..如果要解
0=45x^2-x^4-500 用配方法可以解到ma? 有其更快既方法..要列出詳細步驟
一元四次方程的問題€
2007-08-09 8:12 pm
回答 (2)
2007-08-09 8:20 pm
✔ 最佳答案
0=45x^2-x^4-500x^4-45x^2+500=0
(x^2-20)(x^2-25)=0
x^2=20 or x^2=25
x=+-2開方5 or x=+-5
唔明再問la
參考: own
2007-08-09 8:21 pm
45x^2-x^4-500 = 0
- x^4 + 45x^2 - 500 = 0
x^4 - 45x^2 + 500 = 0
(x^2)^2 - 45x^2 + 500 = 0
(x^2 - 20) (x^2 - 25) = 0
(x^2 - 20) [ x^2 - (5^2) ] = 0 ------------------> A^2 - B^2 = (A + B) (A - B)
(x^2 - 20) [ (x+5) (x-5) ] = 0
(x+5) (x-5) (x^2 - 20) = 0
- x^4 + 45x^2 - 500 = 0
x^4 - 45x^2 + 500 = 0
(x^2)^2 - 45x^2 + 500 = 0
(x^2 - 20) (x^2 - 25) = 0
(x^2 - 20) [ x^2 - (5^2) ] = 0 ------------------> A^2 - B^2 = (A + B) (A - B)
(x^2 - 20) [ (x+5) (x-5) ] = 0
(x+5) (x-5) (x^2 - 20) = 0
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https://hk.answers.yahoo.com/question/index?qid=20070809000051KK01656