2007-06-10
17:00:42
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11 answers
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asked by
liquid_force_evolution
2
in
Science & Mathematics
➔ Mathematics
Those of you saying -5 are wrong i need 3 different roots
so -5... and i need 2 more
2007-06-10
17:08:37 ·
update #1
If you didnt take pre-calculus dont even try
2007-06-10
17:09:02 ·
update #2
nvm -5 isnt a complex root at all
2007-06-10
17:21:33 ·
update #3
-5
2007-06-10 17:02:20
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answer #1
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answered by feanor 7
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You're not going to find the 3 roots of this by algebra. You find it by using complex numbers.
The number -125 can be written as this:
5^3 * exp(i*pi) where i is the square root of 1.
You find the three cube roots by going around the complex plane to find the 3 angles that when multiplied by three will give you pi, given that the expression is periodic every 2*pi.
So the three roots are going to be:
5*exp(i*pi/3)
5*exp(i*pi)
5*exp(i*5*pi/3)
the one in the middle is equal to -5. The other two are most compactly written in how I wrote it. You can expand it out into real and complex parts, but there's no real reason to do so unless you need to work with the two components.
2007-06-10 17:15:04
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answer #2
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answered by Elisa 4
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One of the cubic roots is real and the other two are complex... take -125 and write in the vector form you get -125=125[cos(pi), sin(pi)] since pi is the argument because -125 is on the real axis... so the roots will be of the form 5[cos(pi/3+k(2pi)/3),sin(pi/3+k(2pi)/3)] where k is equal to 0,1,2
So finally you get the roots:
5[1/2,Sqrt(3)/2]=5/2+(i*5*Sqrt(3))/2; i is Sqrt(-1) and k=0
5[-1,0]= -5 (which is the real root) when k=1
5[1/2,-Sqrt(3)/2]=5/2-(i*5*Sqrt(3))/2 when k=2
2007-06-10 17:15:13
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answer #3
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answered by jsos88 2
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(125@180)^(1/3) =
Two roots are complex, and all can be represented in polar coordinates in the complex plane:
5@180, 5@60, 5@-60
In cartesian coordinates
- 5, (5/2)(1 + i√3), (5/2)(1 - i√3)
2007-06-10 17:16:25
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answer #4
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answered by Helmut 7
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To find the three answers you have to use complex numbers.
If z^3=-125, then z=-5, 5e^(pi i/3), 5e^(-pi i/3)
I think - its a while since I did this Maths at high school.
2007-06-10 17:08:45
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answer #5
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answered by Anonymous
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Your answer is -5.
Because the cube root of -1 is -1.
2007-06-10 17:06:40
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answer #6
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answered by Anonymous
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-5, 5/2 + 5isqrt(3)/2, 5/2 - 5isqrt(3)/2
Are those the 3 you wanted?
2007-06-10 17:14:58
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answer #7
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answered by hawkeye3772 4
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-5 , -5 cis 2pi/3 and -5 cis 4pi/3 where cis t = cos t + i sin t
2007-06-10 17:12:41
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answer #8
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answered by Mein Hoon Na 7
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ok o know i answered wrong so im changing it the best i know how.
5, -5, and ur mom
2007-06-10 17:19:35
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answer #9
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answered by Anonymous
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1953125
2007-06-10 17:03:13
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answer #10
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answered by Anonymous
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