how to solve X^3=1 X=?
2008-11-17 8:04 am
i heard that you can get 2 answer for X one is real number and another is imaginary number
回答 (7)
2008-11-17 8:17 am
✔ 最佳答案
For a cubic equation you get 3 answers.x³ = 1
x³ - 1 = 0
x³-1³=0
(x-1)(x²+x+1) = 0
x - 1=0
x = 1
x²+x+1 = 0
x = (-1±√(1²-4×1×1))/2
= (-1±√-3)/2
= (-1±i√3)/2
The 3 cube roots of 1:
{1, (-1+i√3)/2, (-1-i√3)/2}
Note that [(-1+i√3)/2]² = (-1-i√3)/2
and
[(-1-i√3)/2]² = (-1+i√3)/2
The two complex roots are frequently referred to as ω and ω²
2008-11-17 8:52 am
There is one real root and two imaginary roots.
x^3 = 1
x^3 - 1 = 0
(x - 1)(x^2 + x + 1) = 0
Set each factor to 0,
x - 1 = 0 ---> x = 1 (Obvious answer)
x^2 + x + 1 = 0
Using quadratic formula:
x = [-1 ± √(1 - 4)]/2
x = [-1 ± √(-3)]/2
x = (-1 ± i√3)/2
where i = √(-1)
The three solutions are:
x = 1
x = (-1 + i√3)/2
x = (-1 - i√3)/2
x^3 = 1
x^3 - 1 = 0
(x - 1)(x^2 + x + 1) = 0
Set each factor to 0,
x - 1 = 0 ---> x = 1 (Obvious answer)
x^2 + x + 1 = 0
Using quadratic formula:
x = [-1 ± √(1 - 4)]/2
x = [-1 ± √(-3)]/2
x = (-1 ± i√3)/2
where i = √(-1)
The three solutions are:
x = 1
x = (-1 + i√3)/2
x = (-1 - i√3)/2
2008-11-17 9:35 am
x^3 = 1
x = ³√1
x = 1
x = ³√1
x = 1
2016-04-10 10:59 pm
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Do you mean 3^(x^3) = (1/27)^9? 3^(x^3) = (3^(-3))^9 3^(x^3) = 3^(-27) x^3 = -27 x = -3
Do you mean 3^(x^3) = (1/27)^9? 3^(x^3) = (3^(-3))^9 3^(x^3) = 3^(-27) x^3 = -27 x = -3
2015-08-21 10:09 am
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RE:
how to solve X^3=1 X=?
i heard that you can get 2 answer for X one is real number and another is imaginary number
RE:
how to solve X^3=1 X=?
i heard that you can get 2 answer for X one is real number and another is imaginary number
參考: solve 3 1 x: https://bitly.im/zEmsv
2008-11-17 8:25 am
x^3=1
x^3 - 1 = 0
(x - 1)(x^2 + x + 1) = 0
x = 1 or x = (-1 + isqrt(3))/2 or x = (-1 - isqrt(3))/2
x^3 - 1 = 0
(x - 1)(x^2 + x + 1) = 0
x = 1 or x = (-1 + isqrt(3))/2 or x = (-1 - isqrt(3))/2
2008-11-17 8:15 am
x^3=1
x^3 - 1 = 0
(x - 1)(x^2 + x + 1) = 0
x = 1 or x = (1 + isqrt(3))/2 or x = (1 - isqrt(3))/2
x^3 - 1 = 0
(x - 1)(x^2 + x + 1) = 0
x = 1 or x = (1 + isqrt(3))/2 or x = (1 - isqrt(3))/2
2008-11-17 8:14 am
To find the third roots of one, its x= exp(i * 2pi / 3 [since third roots] * k ). k = 0,1,2 OR 1,2,3 if you perfer. For k=0 or k=3, you get same number.
I should note that it could be x= exp(i * (2pi / 3 * k +2pi*n)). Where n is an integer. This 2pi*n of course comes from the fact that the exp function is periodic
I should note that it could be x= exp(i * (2pi / 3 * k +2pi*n)). Where n is an integer. This 2pi*n of course comes from the fact that the exp function is periodic
2008-11-17 8:09 am
yeah but for the REAL set of numbers, x = 1
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