急!!用求根公式解一元二次方程(20點)
回答 (3)
2012-10-22 4:27 am
參考: My Maths World
2012-10-22 4:17 am
Question and Answer :
(7-4√(3)) x^2+(2-√(3)) x = 2
(7-4√(3)) x^2+(2-√(3)) x - 2 = 0
x ={ -(2-√(3)) ± √ [(2-√(3)) ^2 - 4 (7-4√(3)) (-2)] } / 2(7-4√(3))
x = [ √3 + 2 ± √ 63-36√(3) ] / 14-8√(3)
x = 3.73 or -7.46 ( cor. to 2 d. p. )
(7-4√(3)) x^2+(2-√(3)) x = 2
(7-4√(3)) x^2+(2-√(3)) x - 2 = 0
x ={ -(2-√(3)) ± √ [(2-√(3)) ^2 - 4 (7-4√(3)) (-2)] } / 2(7-4√(3))
x = [ √3 + 2 ± √ 63-36√(3) ] / 14-8√(3)
x = 3.73 or -7.46 ( cor. to 2 d. p. )
2012-10-22 4:16 am
禁機-.-
x=3.732050808 or-7.464101615
x=3.732050808 or-7.464101615
收錄日期: 2021-04-13 19:04:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121021000051KK00479