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Solving polynominals equations (relationship of roots)

Quadratic equation - $alpha$ and $beta$ RootsTechnique to simplify algebraic calculations on roots of polynomialInterval of Polynomial Root FindingFind $alpha^3 + beta^3$ which are roots of a quadratic equation.sum and product of roots of polynomials: finding equations for rootsSolving two Cubic Equation on their Roots.Finding an equation with related rootsFind the roots of $acx^2-b(c+a)x+(c+a)^2=0$If $3x^2-6x+p=0$ has roots $alpha$ and $beta$, then find a quadratic with roots $(alpha+beta)/alpha$ and $(alpha+beta)/beta$Find the roots of $3x^3-4x-8$

2

0

$begingroup$

The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$fracalpha+betaomega+fracalpha+omegabeta+fracbeta+omegaalpha$$

So far I have found:
$$alpha+beta+omega=frac-ba = 4 \
alphabeta+betaomega+alphaomega=fracca = 1 \
alpha×beta×omega=frac-da = -6$$
And evaluated the above fractions creating
$$fracalpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2alphabetaomega$$

I don't know how to continue evaluating the question.

Note:

The answer I have been given is $-dfrac113$

polynomials roots

share|cite|improve this question

edited 34 mins ago

Lee David Chung Lin

4,51851342

asked 1 hour ago

Alex Alex

186

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For latex, you use instead of /.
  $endgroup$
  – BadAtGeometry
  1 hour ago
  

  
  
  
  

add a comment | 

2

0

$begingroup$

The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$fracalpha+betaomega+fracalpha+omegabeta+fracbeta+omegaalpha$$

So far I have found:
$$alpha+beta+omega=frac-ba = 4 \
alphabeta+betaomega+alphaomega=fracca = 1 \
alpha×beta×omega=frac-da = -6$$
And evaluated the above fractions creating
$$fracalpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2alphabetaomega$$

I don't know how to continue evaluating the question.

Note:

The answer I have been given is $-dfrac113$

polynomials roots

share|cite|improve this question

edited 34 mins ago

Lee David Chung Lin

4,51851342

asked 1 hour ago

Alex Alex

186

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For latex, you use instead of /.
  $endgroup$
  – BadAtGeometry
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$fracalpha+betaomega+fracalpha+omegabeta+fracbeta+omegaalpha$$

So far I have found:
$$alpha+beta+omega=frac-ba = 4 \
alphabeta+betaomega+alphaomega=fracca = 1 \
alpha×beta×omega=frac-da = -6$$
And evaluated the above fractions creating
$$fracalpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2alphabetaomega$$

I don't know how to continue evaluating the question.

Note:

The answer I have been given is $-dfrac113$

polynomials roots

share|cite|improve this question

edited 34 mins ago

Lee David Chung Lin

4,51851342

asked 1 hour ago

Alex Alex

186

$endgroup$

share|cite|improve this question

edited 34 mins ago

Lee David Chung Lin

4,51851342

asked 1 hour ago

Alex Alex

186

share|cite|improve this question

edited 34 mins ago

Lee David Chung Lin

4,51851342

asked 1 hour ago

Alex Alex

186

edited 34 mins ago

Lee David Chung Lin

4,51851342

edited 34 mins ago

Lee David Chung Lin

4,51851342

asked 1 hour ago

Alex Alex

186

asked 1 hour ago

Alex Alex

186

4 Answers
4

active

oldest

votes

1

$begingroup$

$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta$$

$$= fracalpha + beta + omega - omegaomega + fracbeta + omega + alpha - alphaalpha + fracalpha + omega + beta - betabeta$$

$$ = (alpha + beta + omega) left(frac1alpha + frac1beta + frac1omegaright) - 3$$

$$ = (alpha + beta + omega) left(fracbetaomegaalphabetaomega + fracalphaomegaalphabetaomega + fracalphabetaalphabetaomegaright) - 3$$

$$ = fracalpha + beta + omegaalphabetaomega(betaomega + alphaomega + alphabeta) - 3$$

I think you should be able to take it from there.

share|cite|improve this answer

answered 1 hour ago

user1952500user1952500

1,5251016

$endgroup$

add a comment | 

2

$begingroup$

Hint: We can write $$frac4-ww+frac4-betabeta+frac4-alphaalpha$$ and this is $$4left(fracalphabeta+alpha w+wbetaalpha beta wright)-3$$ and this is $$-frac23left(1-beta w-alpha w+alpha w+beta wright)$$
This simplifies to $$-frac23-3=-frac113$$

share|cite|improve this answer

answered 1 hour ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79.7k42867

$endgroup$

add a comment | 

0

$begingroup$

That follows from your results, since we get: $dfrac4-omegaomega+dfrac4-betabeta+dfrac4-alphaalpha=dfrac4(omegabeta+omega alpha+betaalpha)-3omegabetaalphaomega beta alpha=dfrac4+18-6=-dfrac113$.

share|cite|improve this answer

answered 1 hour ago

Chris CusterChris Custer

14.7k3827

$endgroup$

add a comment | 

0

$begingroup$

Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
alpha =-1, beta =2,omega=3.$$
Hence:
$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta=\
frac-1+ 23 + frac2 + 3-1 + frac-1 + 32=\
frac13-5+1=\
-frac113.$$

share|cite|improve this answer

answered 11 mins ago

farruhotafarruhota

22.5k2942

$endgroup$

add a comment | 

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1

$begingroup$

$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta$$

$$= fracalpha + beta + omega - omegaomega + fracbeta + omega + alpha - alphaalpha + fracalpha + omega + beta - betabeta$$

$$ = (alpha + beta + omega) left(frac1alpha + frac1beta + frac1omegaright) - 3$$

$$ = (alpha + beta + omega) left(fracbetaomegaalphabetaomega + fracalphaomegaalphabetaomega + fracalphabetaalphabetaomegaright) - 3$$

$$ = fracalpha + beta + omegaalphabetaomega(betaomega + alphaomega + alphabeta) - 3$$

I think you should be able to take it from there.

share|cite|improve this answer

answered 1 hour ago

user1952500user1952500

1,5251016

$endgroup$

add a comment | 

1

$begingroup$

$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta$$

$$= fracalpha + beta + omega - omegaomega + fracbeta + omega + alpha - alphaalpha + fracalpha + omega + beta - betabeta$$

$$ = (alpha + beta + omega) left(frac1alpha + frac1beta + frac1omegaright) - 3$$

$$ = (alpha + beta + omega) left(fracbetaomegaalphabetaomega + fracalphaomegaalphabetaomega + fracalphabetaalphabetaomegaright) - 3$$

$$ = fracalpha + beta + omegaalphabetaomega(betaomega + alphaomega + alphabeta) - 3$$

I think you should be able to take it from there.

share|cite|improve this answer

answered 1 hour ago

user1952500user1952500

1,5251016

$endgroup$

add a comment | 

$begingroup$

$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta$$

$$= fracalpha + beta + omega - omegaomega + fracbeta + omega + alpha - alphaalpha + fracalpha + omega + beta - betabeta$$

$$ = (alpha + beta + omega) left(frac1alpha + frac1beta + frac1omegaright) - 3$$

$$ = (alpha + beta + omega) left(fracbetaomegaalphabetaomega + fracalphaomegaalphabetaomega + fracalphabetaalphabetaomegaright) - 3$$

$$ = fracalpha + beta + omegaalphabetaomega(betaomega + alphaomega + alphabeta) - 3$$

I think you should be able to take it from there.

share|cite|improve this answer

answered 1 hour ago

user1952500user1952500

1,5251016

$endgroup$

share|cite|improve this answer

answered 1 hour ago

user1952500user1952500

1,5251016

answered 1 hour ago

user1952500user1952500

1,5251016

answered 1 hour ago

user1952500user1952500

1,5251016

2

$begingroup$

Hint: We can write $$frac4-ww+frac4-betabeta+frac4-alphaalpha$$ and this is $$4left(fracalphabeta+alpha w+wbetaalpha beta wright)-3$$ and this is $$-frac23left(1-beta w-alpha w+alpha w+beta wright)$$
This simplifies to $$-frac23-3=-frac113$$

share|cite|improve this answer

answered 1 hour ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79.7k42867

$endgroup$

add a comment | 

2

$begingroup$

Hint: We can write $$frac4-ww+frac4-betabeta+frac4-alphaalpha$$ and this is $$4left(fracalphabeta+alpha w+wbetaalpha beta wright)-3$$ and this is $$-frac23left(1-beta w-alpha w+alpha w+beta wright)$$
This simplifies to $$-frac23-3=-frac113$$

share|cite|improve this answer

answered 1 hour ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79.7k42867

$endgroup$

add a comment | 

$begingroup$

Hint: We can write $$frac4-ww+frac4-betabeta+frac4-alphaalpha$$ and this is $$4left(fracalphabeta+alpha w+wbetaalpha beta wright)-3$$ and this is $$-frac23left(1-beta w-alpha w+alpha w+beta wright)$$
This simplifies to $$-frac23-3=-frac113$$

share|cite|improve this answer

answered 1 hour ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79.7k42867

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79.7k42867

answered 1 hour ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79.7k42867

answered 1 hour ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79.7k42867

0

$begingroup$

That follows from your results, since we get: $dfrac4-omegaomega+dfrac4-betabeta+dfrac4-alphaalpha=dfrac4(omegabeta+omega alpha+betaalpha)-3omegabetaalphaomega beta alpha=dfrac4+18-6=-dfrac113$.

share|cite|improve this answer

answered 1 hour ago

Chris CusterChris Custer

14.7k3827

$endgroup$

add a comment | 

0

$begingroup$

That follows from your results, since we get: $dfrac4-omegaomega+dfrac4-betabeta+dfrac4-alphaalpha=dfrac4(omegabeta+omega alpha+betaalpha)-3omegabetaalphaomega beta alpha=dfrac4+18-6=-dfrac113$.

share|cite|improve this answer

answered 1 hour ago

Chris CusterChris Custer

14.7k3827

$endgroup$

add a comment | 

$begingroup$

That follows from your results, since we get: $dfrac4-omegaomega+dfrac4-betabeta+dfrac4-alphaalpha=dfrac4(omegabeta+omega alpha+betaalpha)-3omegabetaalphaomega beta alpha=dfrac4+18-6=-dfrac113$.

share|cite|improve this answer

answered 1 hour ago

Chris CusterChris Custer

14.7k3827

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Chris CusterChris Custer

14.7k3827

answered 1 hour ago

Chris CusterChris Custer

14.7k3827

answered 1 hour ago

Chris CusterChris Custer

14.7k3827

0

$begingroup$

Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
alpha =-1, beta =2,omega=3.$$
Hence:
$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta=\
frac-1+ 23 + frac2 + 3-1 + frac-1 + 32=\
frac13-5+1=\
-frac113.$$

share|cite|improve this answer

answered 11 mins ago

farruhotafarruhota

22.5k2942

$endgroup$

add a comment | 

0

$begingroup$

Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
alpha =-1, beta =2,omega=3.$$
Hence:
$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta=\
frac-1+ 23 + frac2 + 3-1 + frac-1 + 32=\
frac13-5+1=\
-frac113.$$

share|cite|improve this answer

answered 11 mins ago

farruhotafarruhota

22.5k2942

$endgroup$

add a comment | 

$begingroup$

Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
alpha =-1, beta =2,omega=3.$$
Hence:
$$fracalpha + betaomega + fracbeta + omegaalpha + fracalpha + omegabeta=\
frac-1+ 23 + frac2 + 3-1 + frac-1 + 32=\
frac13-5+1=\
-frac113.$$

share|cite|improve this answer

answered 11 mins ago

farruhotafarruhota

22.5k2942

$endgroup$

share|cite|improve this answer

answered 11 mins ago

farruhotafarruhota

22.5k2942

answered 11 mins ago

farruhotafarruhota

22.5k2942

answered 11 mins ago

farruhotafarruhota

22.5k2942