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

Does holding a wand and speaking its command word count as V/S/M spell components?

How to reduce LED flash rate (frequency)

How can Republicans who favour free markets, consistently express anger when they don't like the outcome of that choice?

Why does academia still use scientific journals and not peer-reviewed government funded alternatives?

What was the first Intel x86 processor with "Base + Index * Scale + Displacement" addressing mode?

Binary Numbers Magic Trick

Unexpected email from Yorkshire Bank

Critique of timeline aesthetic

What does KSP mean?

Error message with tabularx

How to get a plain text file version of a CP/M .BAS (M-BASIC) program?

Don’t seats that recline flat defeat the purpose of having seatbelts?

Can SQL Server create collisions in system generated constraint names?

Noun clause (singular all the time?)

Pulling the rope with one hand is as heavy as with two hands?

The Defining Moment

How do I deal with a coworker that keeps asking to make small superficial changes to a report, and it is seriously triggering my anxiety?

Please, smoke with good manners

what is the sudo password for a --disabled-password user

US visa is under administrative processing, I need the passport back ASAP

Can I spend a night at Vancouver then take a flight to my college in Toronto as an international student?

Why does nature favour the Laplacian?

Why was the Spitfire's elliptical wing almost uncopied by other aircraft of World War 2?

Rivers without rain



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 $3x^3-4x-8$













4












$begingroup$



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
    17 hours ago
















4












$begingroup$



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
    17 hours ago














4












4








4


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 16 hours ago









Lee David Chung Lin

4,54351342




4,54351342










asked 17 hours ago









Alex Alex

286




286








  • 1




    $begingroup$
    For latex, you use instead of /.
    $endgroup$
    – BadAtGeometry
    17 hours ago














  • 1




    $begingroup$
    For latex, you use instead of /.
    $endgroup$
    – BadAtGeometry
    17 hours ago








1




1




$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
17 hours ago




$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
17 hours ago










4 Answers
4






active

oldest

votes


















6












$begingroup$

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



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



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



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



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



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






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    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









    $endgroup$





















      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









      $endgroup$





















        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









          6












          $begingroup$

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



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



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



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



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



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






          share|cite|improve this answer









          $endgroup$


















            6












            $begingroup$

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



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



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



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



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



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






            share|cite|improve this answer









            $endgroup$
















              6












              6








              6





              $begingroup$

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



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



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



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



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



              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 16 hours ago









              user1952500user1952500

              1,6241016




              1,6241016























                  3












                  $begingroup$

                  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









                  $endgroup$


















                    3












                    $begingroup$

                    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









                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      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 15 hours ago









                      farruhotafarruhota

                      22.5k2942




                      22.5k2942























                          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 16 hours ago









                              Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                              79.9k42867




                              79.9k42867























                                  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 16 hours 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

                                          Why do type traits not work with types in namespace scope?What are POD types in C++?Why can templates only be...

                                          Image
                                          What does it mean to express a gate in Dirac notation? A Strange Latex Symbol How can Republicans who favour free markets, consistently express anger when they don't like the outcome of that choice? What happened to Captain America in Endgame? How to solve constants out of the internal energy equation? How can I place the product on a social media post better? Will tsunami waves travel forever if there was no land? Unexpected email from Yorkshire Bank What is the relationship between spectral sequences and obstruction theory? How can I practically buy stocks? Will a top journal at least re
                                          Read more

                                          Will tsunami waves travel forever if there was no land?Why do tsunami waves begin with the water flowing away...

                                          Image
                                          How much cash can I safely carry into the USA and avoid civil forfeiture? Is there really no use for MD5 anymore? Is contacting this expert in the field something acceptable or would it be discourteous? How to pronounce 'C++' in Spanish How to creep the reader out with what seems like a normal person? Critique of timeline aesthetic Sci fi novel series with instant travel between planets through gates. A river runs through the gates How did Captain America manage to do this? Are Boeing 737-800’s grounded? Error message with tabularx How to make a pipeline wait for end-of-file or stop af
                                          Read more

                                          Should I use Docker or LXD?How to cache (more) data on SSD/RAM to avoid spin up?Unable to get Windows File...

                                          Image
                                          kill -0 <PID> は何をするのでしょうか？ Why avoid shared user accounts? Using only 1s, make 29 with the minimum number of digits Cookies - Should the toggles be on? Can I string the DnD Starter Set campaign into another modules, keeping the same characters? Why do neural networks need so many training examples to perform? Intern applicant asking for compensation equivalent to that of permanent employee Why would space fleets be aligned? It took me a lot of time to make this, pls like. (YouTube Comments #1) Can a long polymer chain interact with itself via van der Waals forces? What is the purpose of
                                          Read more