利用二次公式解下列二次方程,請詳細列明公式。(答案以根式表示)
1)x(2x+1)-1/3=0
2)(x-1/2)(x-2)=3/2x+4
F.4 maths(方程)
2009-09-30 6:53 pm
回答 (3)
2009-10-01 3:46 am
✔ 最佳答案
1) x(2x + 1) – 1/3 = 02x^2 + x – 1/3 = 0
x = [-1 +/- √(1 + 8/3)] / 4
x = [-1 +/- √(11/3)] / 4
x = -1/4 +/- (√33) / 12
2) (x – 1/2)(x – 2) = 3x/2 + 4
x^2 – 2.5x + 1 = 1.5x + 4
x^2 – 4x – 3 = 0
x = [4 +/- √(16 + 12)] / 2
x = 2 +/- √7
2009-10-01 1:35 am
001,use the quadratic formula
2009-09-30 7:18 pm
1)
x (2x + 1) - 1/3 = 0
2x^2 + x - 1/3 = 0
6x^2 + 3x - 1 = 0
(3x - 1) (2x + 1) = 0
x = 1/3 or x = -1/2
2)
(x - 1/2) (x - 2) = 3/2 x + 4
x^2 - 2x - 1/2x + 1 = 3/2 x + 4
x^2 - 5/2 x + 1 = 3/2 x + 4
x^2 - 4x -3 = 0
(x - 1) (x - 3) = 0
x = 1 or x = 3
x (2x + 1) - 1/3 = 0
2x^2 + x - 1/3 = 0
6x^2 + 3x - 1 = 0
(3x - 1) (2x + 1) = 0
x = 1/3 or x = -1/2
2)
(x - 1/2) (x - 2) = 3/2 x + 4
x^2 - 2x - 1/2x + 1 = 3/2 x + 4
x^2 - 5/2 x + 1 = 3/2 x + 4
x^2 - 4x -3 = 0
(x - 1) (x - 3) = 0
x = 1 or x = 3
收錄日期: 2021-04-23 23:19:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090930000051KK00342