Is the cube root of -8 an imaginary number?
2007-11-30 19:22:14 · 7 answers · asked by sillyboys_trucksare4girls 2 in Science & Mathematics ➔ Mathematics
No. It is -2
2007-11-30 19:25:07 · answer #1 · answered by chasrmck 6 · 0⤊ 0⤋
It's not an imaginary number. It will be an imaginary number if it is a square root of a negative number. It is possible to cube root a negative number.
THE CUBE ROOT OF -8: -2
because
(-2) (-2) (-2) =-8
2007-12-01 03:37:46 · answer #2 · answered by gm_2422 2 · 0⤊ 0⤋
There are 3 cube roots of -8, which are:
-2
1 + iâ3
1 - iâ3
So, one root is real, the other two are imaginary, or complex.
2007-12-01 03:28:52 · answer #3 · answered by Scythian1950 7 · 2⤊ 0⤋
no it is not
cube root of 08 is equal to -2 because (-2)^3 = -2 * (-2) * (-2) = -8.
even roots (2, 4, 6, 8,...) of negative numbers are complex numbers
2007-12-01 03:26:06 · answer #4 · answered by Anonymous · 0⤊ 0⤋
In real domain, it is not. But in complex domain, it has three solutions, x = -2, 1屉3 i, which can be obtained by solving x^3+8 = (x+2)(x^2-2x+4) = 0
2007-12-01 03:28:54 · answer #5 · answered by sahsjing 7 · 0⤊ 0⤋
imaginary number is "i"
i = sqrt. -1
i^2=-1
cubic root -8
(break down to)
cubic root of 8 times cubic root of -1
cubic root=2i
so yes it is imaginary
2007-12-01 03:33:19 · answer #6 · answered by i.heart.u 5 · 0⤊ 2⤋
22222222222222 its 2
2007-12-01 03:26:10 · answer #7 · answered by Anonymous · 0⤊ 0⤋