Solving polynominals equations (relationship of roots)Quadratic equation - $alpha$ and $beta$ RootsTechnique...

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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 $3x^3-4x-8$













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$begingroup$



The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$frac{alpha+beta}{omega}+frac{alpha+omega}{beta}+frac{beta+omega}{alpha}$$




So far I have found:
$$alpha+beta+omega=frac{-b}{a} = 4 \
alphabeta+betaomega+alphaomega=frac{c}{a} = 1 \
alpha×beta×omega=frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$frac{alpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2}{alphabetaomega}$$



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



Note:

The answer I have been given is $-dfrac{11}{3}$






polynomials roots







share|cite|improve this question











$endgroup$








  • 1




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
















3












$begingroup$



The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$frac{alpha+beta}{omega}+frac{alpha+omega}{beta}+frac{beta+omega}{alpha}$$




So far I have found:
$$alpha+beta+omega=frac{-b}{a} = 4 \
alphabeta+betaomega+alphaomega=frac{c}{a} = 1 \
alpha×beta×omega=frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$frac{alpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2}{alphabetaomega}$$



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



Note:

The answer I have been given is $-dfrac{11}{3}$






polynomials roots







share|cite|improve this question











$endgroup$








  • 1




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














3












3








3


1



$begingroup$



The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$frac{alpha+beta}{omega}+frac{alpha+omega}{beta}+frac{beta+omega}{alpha}$$




So far I have found:
$$alpha+beta+omega=frac{-b}{a} = 4 \
alphabeta+betaomega+alphaomega=frac{c}{a} = 1 \
alpha×beta×omega=frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$frac{alpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2}{alphabetaomega}$$



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



Note:

The answer I have been given is $-dfrac{11}{3}$






polynomials roots







share|cite|improve this question











$endgroup$





The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$frac{alpha+beta}{omega}+frac{alpha+omega}{beta}+frac{beta+omega}{alpha}$$




So far I have found:
$$alpha+beta+omega=frac{-b}{a} = 4 \
alphabeta+betaomega+alphaomega=frac{c}{a} = 1 \
alpha×beta×omega=frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$frac{alpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2}{alphabetaomega}$$



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



Note:

The answer I have been given is $-dfrac{11}{3}$






polynomials roots




polynomials roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 55 mins ago









Lee David Chung Lin

4,51851342




4,51851342










asked 1 hour ago









Alex Alex

236




236








  • 1




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














  • 1




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








1




1




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




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










4 Answers
4






active

oldest

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3












$begingroup$

$$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$



$$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$



$$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$



$$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$



$$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$



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






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
    This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      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:
      $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
      frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
      frac13-5+1=\
      -frac{11}{3}.$$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.






        share|cite|improve this answer









        $endgroup$














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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$



          $$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$



          $$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$



          $$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$



          $$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$



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






          share|cite|improve this answer









          $endgroup$


















            3












            $begingroup$

            $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$



            $$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$



            $$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$



            $$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$



            $$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$



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






            share|cite|improve this answer









            $endgroup$
















              3












              3








              3





              $begingroup$

              $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$



              $$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$



              $$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$



              $$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$



              $$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$



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






              share|cite|improve this answer









              $endgroup$



              $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$



              $$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$



              $$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$



              $$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$



              $$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$



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







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 1 hour ago









              user1952500user1952500

              1,5451016




              1,5451016























                  2












                  $begingroup$

                  Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
                  This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
                    This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
                      This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$






                      share|cite|improve this answer









                      $endgroup$



                      Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
                      This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 1 hour ago









                      Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                      79.7k42867




                      79.7k42867























                          1












                          $begingroup$

                          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:
                          $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
                          frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
                          frac13-5+1=\
                          -frac{11}{3}.$$






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            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:
                            $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
                            frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
                            frac13-5+1=\
                            -frac{11}{3}.$$






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              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:
                              $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
                              frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
                              frac13-5+1=\
                              -frac{11}{3}.$$






                              share|cite|improve this answer









                              $endgroup$



                              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:
                              $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
                              frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
                              frac13-5+1=\
                              -frac{11}{3}.$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 31 mins ago









                              farruhotafarruhota

                              22.5k2942




                              22.5k2942























                                  0












                                  $begingroup$

                                  That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 1 hour ago









                                      Chris CusterChris Custer

                                      14.7k3827




                                      14.7k3827






























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