更新1:
i can get to u^2-5u+4=0 where u=x^2
x^4-5x^2+4=0 solve by substitution?
回答 (4)
2008-05-30 3:49 pm
✔ 最佳答案
Then factor:(u - 4)(u -1) = 0
u = 4 or u = 1
x² = 4 or x² = 1
x = ±2 or x = ±1
2008-05-30 4:09 pm
(x² - 4)(x² - 1) = 0
x² = 4 , x² = 1
x = ± 2 , x = ± 1
x² = 4 , x² = 1
x = ± 2 , x = ± 1
2008-05-30 3:54 pm
x^4 - 5x^2 + 4 = 0
(x^2 - 4)(x^2 - 1) = 0
(x + 2)(x - 2)(x + 1)(x - 1) = 0
x + 2 = 0
x = -2
x - 2 = 0
x = 2
x + 1 = 0
x = -1
x - 1 = 0
x = 1
ⴠx = ±2 , ±1
(x^2 - 4)(x^2 - 1) = 0
(x + 2)(x - 2)(x + 1)(x - 1) = 0
x + 2 = 0
x = -2
x - 2 = 0
x = 2
x + 1 = 0
x = -1
x - 1 = 0
x = 1
ⴠx = ±2 , ±1
2008-05-30 3:51 pm
now that u = x^2 ....
u^2 - 5u +4 = 0
(u-4) (u-1) = 0 [u=4][u=1]
if u=x^2 = 4 .....u= -+ 2
if u=x^2 = 1 ... u= +- 1
u^2 - 5u +4 = 0
(u-4) (u-1) = 0 [u=4][u=1]
if u=x^2 = 4 .....u= -+ 2
if u=x^2 = 1 ... u= +- 1
收錄日期: 2021-05-01 10:33:21
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080530074459AAfoHuk