Question 694174: find all 12 solutions to x^12=1
Already know 1 and -1 are solutions, but I'm not sure how to factor the equation.
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Hint:
x^12=1
x^12 - 1 = 0
(x^4)^3 - 1^3 = 0
(x^4 - 1)((x^4)^2 + x^4*1 + 1^2) = 0
(x^4 - 1)(x^8 + x^4 + 1) = 0
(x^2 - 1)(x^2 + 1)(x^8 + x^4 + 1) = 0
(x - 1)(x + 1)(x^2 + 1)(x^8 + x^4 + 1) = 0
Note:
To solve x^8 + x^4 + 1 = 0, let z = x^4 to get z^2 + z + 1 = 0. Solve for z and use this to find solutions to x.
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Complex_Numbers/694174 (2012-12-13 21:23:13): find all 12 solutions to x^12=1
Already know 1 and -1 are solutions, but I'm not sure how to factor the equation.
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x^12-1 = 0
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((x^4)^3-1) = 1
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(x^4-1)((x^4)^2 + x^4 + 1) = 0
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(x^2+1)(x^2-1)((x^4)^2+x^4+1) = 0
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(x+i)(x-i)(x-1)(x+1)(factors of the quadratic) = 0
At this point you have zeroes at -i,i,1,-1
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Working with the quadratic:
x^4 = [-1 +- sqrt(1-4*1)]/2
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x^4 = [(-1/2) +- (i/2)sqrt(3)]
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That gives you an additional 8 roots
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Comment: Normally you would use DeMoivre's Method to
find the 12 roots.
x = cis[(90+360n)/12] where n = 0,1,2,3,4,....,10,11
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Cheers,
Stan H.
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