Help on Solving Equation in the complex number system

ProdigaI

New member
Joined
Jun 13, 2009
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7
Hey, I really need help on this one problem

x^3 - 64 = 0

Thanks

by the way x^3 is x to the 3rd power
 
a3b3=(ab)(a2+ab+b2)\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)

x364=(x4)(x2+4x+16)\displaystyle x^3-64 = (x-4)(x^2+4x+16)

(x4)=0\displaystyle (x-4)=0

x=4\displaystyle x=4

(x2+4x+16)=0\displaystyle (x^2+4x+16)=0

Quadratic\displaystyle Quadratic

Delta=b2ac=416=12<0[Incomplexnumbers12is12(i2)]\displaystyle Delta=b'^2-ac = 4-16= -12<0 [In complex numbers -12 is 12(i^2)]

x=2+(i)(sqrt12)/(1\displaystyle x=-2+-(i)(sqrt12)/(1

[sup:3fo67zfe]Aladdin[/sup:3fo67zfe]
 
Hello, ProdigaI!

Are you familiar with the Argand diagram?


x364=0\displaystyle x^3 - 64 \:=\: 0

We have: .\(\displaystyle x^3 \:=\:64 \quadf\Rightarrow\quad x \:=\:4\)

We have a real root: .(4,0)\displaystyle (4,0)

On the complex plane, the three cube roots are spaced equally about the origin.
So the graph looks like this:
Code:
          *   |
           \  |
            \ |
             \|
      - - - - + - - - - * - -
             /|       (4,0)
            / |
           /  |
          *   |

Using some Trig, we find that the other two points are at:
. . \(\displaystyle \left(\text{-}2,\: 2i\sqrt{3}),\;\;(\text{-}2,\: \text{-} 2i\sqrt{3})\)

Hence, the other two roots are: .-2+2i3 and -22i3\displaystyle \text{-}2+2i\sqrt{3}\:\text{ and }\:\text{-}2 -2i\sqrt{3}

 
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