SOLUTION: {{{w^3=1}}}, Find the exact values of {{{(1+2w+3w^2)}}} & {{{(1+3w+2w^2)}}}, given that these are roots of the eq'n {{{x^2 + 3x +3=0}. I did the Q, however I am just stuck on say

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson  -> Lesson -> SOLUTION: {{{w^3=1}}}, Find the exact values of {{{(1+2w+3w^2)}}} & {{{(1+3w+2w^2)}}}, given that these are roots of the eq'n {{{x^2 + 3x +3=0}. I did the Q, however I am just stuck on say      Log On


   



Question 1152853: w%5E3=1, Find the exact values of %281%2B2w%2B3w%5E2%29 & %281%2B3w%2B2w%5E2%29, given that these are roots of the eq'n or %281%2B3w%2B2w%5E2%29?
Thank you so much

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
No complex imaginary number can be said to be "larger" or "smaller" than any
other complex numbers.  Only real numbers can be said to be larger or smaller
than other real numbers.  

We should do all the possibilities.

The problem from the very beginning:

Since 
%281%2B2w%2B3w%5E2%29 

and

%281%2B3w%2B2w%5E2%29

are roots of the equation 

x%5E2+%2B+3x+%2B+3=0

their sum is equal to the coefficient of x with the opposite sign:

So:

%281%2B2w%2B3w%5E2%29%2B%281%2B3w%2B2w%5E2%29=-3

or

2%2B5w%2B5w%5E2=-3
5w%5E2%2B5w%2B5=0
w%5E2%2Bw%2B1=0
w=%28-1+%2B-+i%2Asqrt%283%29%29%2F2

[We can check that their product will be the constant term 3, but
we don't need that.

(1) When we substitute 

%28-1+%2B+i%2Asqrt%283%29%29%2F2

into 

%281%2B2w%2B3w%5E2%29 

we get

%28-3+-+i%2Asqrt%283%29%29%2F2

-----

(2)  When we substitute 

%28-1+-+i%2Asqrt%283%29%29%2F2

into 

%281%2B2w%2B3w%5E2%29 

we get

%28-3+%2B+i%2Asqrt%283%29%29%2F2

------

(3) When we substitute 

%28-1+%2B+i%2Asqrt%283%29%29%2F2

into 

%281%2B3w%2B2w%5E2%29 

we get the value

%28-3+%2B+i%2Asqrt%283%29%29%2F2

------

(4) When we substitute 

%28-1+-+i%2Asqrt%283%29%29%2F2

into 

%281%2B3w%2B2w%5E2%29 

we get the value

%28-3+-+i%2Asqrt%283%29%29%2F2

---

So, the exact values of both  %281%2B2w%2B3w%5E2%29 and %281%2B3w%2B2w%5E2%29, 
given that

%281%2B2w%2B3w%5E2%29 

and

%281%2B3w%2B2w%5E2%29

are roots of the equation



Note that for %281%2B2w%2B3w%5E2%29, the sign of the imaginary part is
OPPOSITE the sign of the imaginary part of what is substituted, and
that for %281%2B3w%2B2w%5E2%29, the sign of the imaginary part is the 
SAME AS the sign of the imaginary part of what is substituted.

Edwin