Solving polynominals equations (relationship of roots)












2












$begingroup$



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 $-dfrac{11}{3}$










share|cite|improve this question











$endgroup$








  • 1




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
















2












$begingroup$



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 $-dfrac{11}{3}$










share|cite|improve this question











$endgroup$








  • 1




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














2












2








2


0



$begingroup$



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 $-dfrac{11}{3}$










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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 43 mins ago









Lee David Chung Lin

4,51851342




4,51851342










asked 1 hour ago









Alex Alex

186




186








  • 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

votes


















2












$begingroup$

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



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



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



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



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



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









        $endgroup$














          Your Answer








          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3204072%2fsolving-polynominals-equations-relationship-of-roots%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

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



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



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



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



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



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






          share|cite|improve this answer









          $endgroup$


















            2












            $begingroup$

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



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



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



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



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



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






            share|cite|improve this answer









            $endgroup$
















              2












              2








              2





              $begingroup$

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



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



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



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



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



              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

              1,5351016




              1,5351016























                  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 19 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






























                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Mathematics Stack Exchange!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          Use MathJax to format equations. MathJax reference.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3204072%2fsolving-polynominals-equations-relationship-of-roots%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          Popular posts from this blog

                                          How to label and detect the document text images

                                          Vallis Paradisi

                                          Tabula Rosettana