Factor completely 8X^2-3X+24?
回答 (8)
2008-04-08 4:13 pm
✔ 最佳答案
Let's be careful here, it is not factorable under the INTEGERS, since the discriminant < 0. We can still factor it under the complex numbers.
The solutions turn out to be,
x1 = (3/2) - (551/40)i
x2 = (3/2) + (551/40)i
Using, these we can write the factorization as
(x - (3/2)+(551/40)i)(x - (3/2) - (551/40)i)
參考: Math Geek!
2017-01-05 9:44 am
a ingredient is something which you would be able to flippantly divide a extensive decision via. 6 might nicely be divided via a million, 2, 3, and six. you're saying she basically desires 3 solutions so as that area seems wonderful. a million, 2, and 3, are all factors of 6 so those may be maximum suitable. 6 might additionally be a ingredient of 6.
2008-04-08 4:10 pm
x1 = 0.1875+1.7218721642444887i
x2 = 0.1875-1.7218721642444887i
x2 = 0.1875-1.7218721642444887i
2008-04-08 4:08 pm
It is relatively prime....... no factors
2008-04-08 4:17 pm
This equation is of form ax^2+bx+c
a = 8 b = -3 c = 24
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[3 +/-sqrt(-3^2-4(8)(24)]/(2)(8)
discriminant is b^2-4ac =-759
i^2 = -1, so √i^2 = i
No real roots: The complex roots are
x=[3 +i √(759)] / (2)(8)
x=[3 -i √(759)] / (2)(8)
x=[3+i27.54995] / 16
x=[3-i27.54995] / 16
a = 8 b = -3 c = 24
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[3 +/-sqrt(-3^2-4(8)(24)]/(2)(8)
discriminant is b^2-4ac =-759
i^2 = -1, so √i^2 = i
No real roots: The complex roots are
x=[3 +i √(759)] / (2)(8)
x=[3 -i √(759)] / (2)(8)
x=[3+i27.54995] / 16
x=[3-i27.54995] / 16
2008-04-08 4:16 pm
(4x+3) (4x-6)
2008-04-08 4:12 pm
the equation cannot be factored as if equated to zero x is a complex number.
2008-04-08 4:06 pm
8x^2 - 3x + 24
= cannot be factored
= cannot be factored
收錄日期: 2021-05-01 10:23:18
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