Can someone please factorise 5x²- 22x + 80=0?
回答 (7)
2009-10-25 5:00 pm
✔ 最佳答案
5x^2 - 22x + 80 = 0x = [-b ± √(b^2 - 4ac)]/(2a)
a = 5
b = -22
c = 80
x = [22 ± √(484 - 1600)]/10
x = [22 ± √(-1116)]/10
x = [22 ± √(i^2 * 6^2 * 31)]/10
x = [22 ± 6i√31]/10
x = [11 ± 3i√31]/5
∴ x = [11 ± 3i√31]/5
2009-10-25 4:56 pm
No factors. Roots are complex conjugates
2009-10-25 5:11 pm
This will not factor without imaginary values so we need to use quadratic formula instead.
a = 5
b = -22
c = 80
x = [ -(-22) ± â(-22^ - 4 * 5 * 80)] / (2 * 5)
x = [22 ± â(484 - 1600) ] /10
x = (22 ± â-1116) /10
x = 2.2 ± 6iâ31
a = 5
b = -22
c = 80
x = [ -(-22) ± â(-22^ - 4 * 5 * 80)] / (2 * 5)
x = [22 ± â(484 - 1600) ] /10
x = (22 ± â-1116) /10
x = 2.2 ± 6iâ31
2009-10-25 5:06 pm
You cannot factorise as the discriminant is not a square number.
b^2-4ac = square to factorise.
5x²- 22x + 80=0 >>>> a= 5 , b= -22, c=80
=> (-22^2) -4x5x80
=> 484 - 4 x 5 x 80
= -1116
Since -1116 is not square it does not factorise.
b^2-4ac = square to factorise.
5x²- 22x + 80=0 >>>> a= 5 , b= -22, c=80
=> (-22^2) -4x5x80
=> 484 - 4 x 5 x 80
= -1116
Since -1116 is not square it does not factorise.
2009-10-25 5:03 pm
use the formula for finding complex roots
x = {-b ± â( b² - 4ac)} / 2a
x = {-b ± â( b² - 4ac)} / 2a
2009-10-25 4:56 pm
Not factorable.
2009-10-25 4:55 pm
Not factorisable without imaginaries
收錄日期: 2021-05-01 12:50:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091025085108AA3JgbI