2
a(i),2x +2x-1=0
2 2
(ii),2x +2ax-a =0
3 2
b,4x +5x -x =0
答案要surd form,請列明steps
更新1:
a(i) => 2x既2次 a(ii) => 2x既2次, a既2次 b => 4x既3次 , 5x既2次
數學qraduatic equation~~
2008-10-05 8:41 pm
唔該可唔可以幫我解答以下3題數?
2
a(i),2x +2x-1=0
2 2
(ii),2x +2ax-a =0
3 2
b,4x +5x -x =0
答案要surd form,請列明steps
2
a(i),2x +2x-1=0
2 2
(ii),2x +2ax-a =0
3 2
b,4x +5x -x =0
答案要surd form,請列明steps
回答 (1)
2008-10-05 10:04 pm
✔ 最佳答案
a(i) 2x^2 +2x - 1 = 0
x = {-2√[ 2^2 - 4(2)(-1)]} /[2(2)]
x = (-1 √3)/2
(a)(ii)
2x^2 + 2ax-a^2 = 0
x = {-2a √[(2a)^2 - 4(2)(-a^2)] }/[2(2)]
x = (-aa√3)/2
= (-1 √3)a/2
(b)
4x^3 +5x^2 - x = 0
x(4x^2 +5x - 1) = 0
x = 0 or 4x^2 +5x - 1 = 0
x= 0 or x = {-5√[5^2 - 4(4)(-1)]}/ [2(4)]
x = 0 or x = (-5√41)/8
收錄日期: 2021-04-18 17:24:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081005000051KK00903