更新1:
the book answer is 3/2, -3+3isqrt3/4, -3-3isqrt3/4 I just need help with the steps to getting there
8x^3-27=0 helppp please?
2009-01-07 9:41 am
I have been working on this problem for over 30 minutes can someone please help me solve it???
回答 (8)
2009-01-07 9:45 am
✔ 最佳答案
x = 3/28x^3 = 27
(2x)^3 = 3^3
Clear?
2009-01-07 9:48 am
8x^3 has to equal 27 so when you subtract 27 you get 0.
so x^3 has to equal 27/8
So x = 3/2
Check:
(3/2)^3 = 27/8
27/8 * 8 - 27
27 - 27 = 0
Bingo!
so x^3 has to equal 27/8
So x = 3/2
Check:
(3/2)^3 = 27/8
27/8 * 8 - 27
27 - 27 = 0
Bingo!
2009-01-07 9:46 am
3/2
2009-01-07 2:15 pm
(8x³)^1/3 = 27^1/3
2x = 3
x = 3/2
2x = 3
x = 3/2
2009-01-07 12:02 pm
8x³ = 27
x³ = 27/8
x = 3/2
x³ = 27/8
x = 3/2
2009-01-07 10:33 am
8x^3 - 27 = 0
8x^3 = 27
x^3 = 27/8
x = ³√(27/8)
x = ³√(3/2 * 3/2 * 3/2)
x = 3/2 (1.5)
8x^3 = 27
x^3 = 27/8
x = ³√(27/8)
x = ³√(3/2 * 3/2 * 3/2)
x = 3/2 (1.5)
2009-01-07 10:00 am
8x^3-27=0
8x^3 = 27
x^3 = 27/8
x = 3/2
8x^3 = 27
x^3 = 27/8
x = 3/2
2009-01-07 9:56 am
8x³ - 27 = 0
You could do:
8x³ = 27
(8x³)^⅓ = 27^⅓
2x = 3
x = 3/2
If you want all the roots (including the complex ones)
8x³ - 27 = 0
Factoring
(2x - 3)(4x² + 6x + 9) = 0
Equating each factor to 0
(2x - 3) = 0
2x = 3
x = 3/2
(4x² + 6x + 9) = 0
x = {-6 ± √[6² - 4(4)(9)]}/2(4)
.. = {-6 ± √-108}/8
.. = {-6 ± √(36)(3)(-1)}/8
.. = (-6 ± 6i√3)/8
.. = (-3 ± 3i√3)/4
.. = -¾ ± ¾ i√3
You could do:
8x³ = 27
(8x³)^⅓ = 27^⅓
2x = 3
x = 3/2
If you want all the roots (including the complex ones)
8x³ - 27 = 0
Factoring
(2x - 3)(4x² + 6x + 9) = 0
Equating each factor to 0
(2x - 3) = 0
2x = 3
x = 3/2
(4x² + 6x + 9) = 0
x = {-6 ± √[6² - 4(4)(9)]}/2(4)
.. = {-6 ± √-108}/8
.. = {-6 ± √(36)(3)(-1)}/8
.. = (-6 ± 6i√3)/8
.. = (-3 ± 3i√3)/4
.. = -¾ ± ¾ i√3
收錄日期: 2021-05-01 11:49:03
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