(由於冇開方根ge符號,所以要用文字表達)
開方根 x^2 - 4x - 8 = 2
F.4 maths-----含有平方根號ge方程
2006-11-18 10:36 pm
回答 (3)
2006-11-18 10:42 pm
✔ 最佳答案
√(x^2 - 4x - 8 )= 2 x^2 - 4x - 8 = 4
x^2 - 4x - 12 = 0
(x-6)(x+2)=0
x=6 or x=-2
2006-11-18 10:47 pm
√(4x^2-4x-8)=2
4x^2-4x-8=4
4x^2-4x-12=0
x={-(-4)士√[(-4)^2-4(4)(-12)]}/2(1)
x=(4士√208)/2
x=2士2√13
4x^2-4x-8=4
4x^2-4x-12=0
x={-(-4)士√[(-4)^2-4(4)(-12)]}/2(1)
x=(4士√208)/2
x=2士2√13
2006-11-18 10:45 pm
both sqare
x^2-4x-8=4
x^2-4x-10=0
再用 x= - b 正負開方根 b^2 - 4ac 除以 2a .......
2006-11-18 14:48:49 補充：
ar!!!!! 計錯了!!!! sorry both sqare x^2-4x-8=4 x^2-4x-12=0x=6 or -2
x^2-4x-8=4
x^2-4x-10=0
再用 x= - b 正負開方根 b^2 - 4ac 除以 2a .......
2006-11-18 14:48:49 補充：
ar!!!!! 計錯了!!!! sorry both sqare x^2-4x-8=4 x^2-4x-12=0x=6 or -2
收錄日期: 2021-04-23 00:08:11
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061118000051KK02488