2006-11-08 17:14:30 · 10 answers · asked by latoya W 1 in Science & Mathematics ➔ Mathematics
(x^3+8)=(x+2)(x^2-2x+4)=0
x+2=0
x=-2
x^2-2x+4=0
x=[2+/-rt(4-16)]/2
=2+/-2irt3/2
=1+/-irt3
so the real zero=-2
and the imaginary rootsare
1+irt3 and 1-irt3
2006-11-08 17:35:51 · answer #1 · answered by raj 7 · 0⤊ 0⤋
x=-2
2006-11-09 01:16:23 · answer #2 · answered by Mech_Eng 3 · 0⤊ 0⤋
There are 3, 1 real and a pair of complex conjugates.
x^3+8 = 0 ........ then factor
(x+2)(x² - 2x + 4) = 0 .... and set each factor equal to 0
so x+2 = 0, x = -2 and
x² - 2x + 4 = 0
x² - 2x = -4
x² - 2x + 1 = -3
(x - 1)² = -3
x - 1 = ± â-3
x = 1 ± i â3 ....... the 2 complex roots
2006-11-09 01:23:52 · answer #3 · answered by Philo 7 · 0⤊ 0⤋
It's a very simple question.
Let f(x) = 0, i.e. x^3 = -8
Just take cube root of -8 (with a calculator if you want), you will get -2. No factoristaion or using of complex number is required.
[(-2)*(-2)*(-2)=-8]
2006-11-09 01:31:52 · answer #4 · answered by orhhai 2 · 0⤊ 0⤋
-2
2006-11-09 02:48:14 · answer #5 · answered by Anonymous · 0⤊ 0⤋
f(x) = 0
x^3 + 8 = 0
x^3 = -8
x=-2
2006-11-09 01:39:48 · answer #6 · answered by Anonymous · 0⤊ 0⤋
x^3 + 8 =0
x^3= -8
x= 2,2,-2
2006-11-09 01:38:54 · answer #7 · answered by ibrahim s 1 · 0⤊ 0⤋
x^3 = -8
Then you can put -8 in euler notation
exp(j * Theta) = Cos(Theta) + i * Sin(Theta)
and then take the 1/3 root or
exp(j * Theta/3) which will give you the answer.
2006-11-09 01:24:24 · answer #8 · answered by Julio Cesar C 2 · 0⤊ 0⤋
this is fairly simple
x^3 = -8
so x = -2 , -2w, -2w^2 where w is cube root of 1
also knows as cis 120 degrees
2006-11-09 02:05:06 · answer #9 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
sum of cubic
(x^3 + b^3) = (x + a)(x^2 -ax + a^2)
2006-11-09 01:22:18 · answer #10 · answered by The Clueless Philospher 2 · 0⤊ 0⤋