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2007-06-10 17:00:42 · 11 answers · 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

11 answers

-5

2007-06-10 17:02:20 · answer #1 · answered by feanor 7 · 0 4

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 · answer #2 · answered by Elisa 4 · 2 2

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 · answer #3 · answered by jsos88 2 · 2 1

(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 · answer #4 · answered by Helmut 7 · 1 2

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 · answer #5 · answered by Anonymous · 1 2

Your answer is -5.
Because the cube root of -1 is -1.

2007-06-10 17:06:40 · answer #6 · answered by Anonymous · 0 4

-5, 5/2 + 5isqrt(3)/2, 5/2 - 5isqrt(3)/2

Are those the 3 you wanted?

2007-06-10 17:14:58 · answer #7 · answered by hawkeye3772 4 · 1 2

-5 , -5 cis 2pi/3 and -5 cis 4pi/3 where cis t = cos t + i sin t

2007-06-10 17:12:41 · answer #8 · answered by Mein Hoon Na 7 · 2 1

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 · answer #9 · answered by Anonymous · 0 4

1953125

2007-06-10 17:03:13 · answer #10 · answered by Anonymous · 0 4

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