Evaluating the modulus of roots of a 6 degree polynomial

Multi tool use
Multi tool use

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
5
down vote

favorite












Let $x_1,x_2,x_3,x_4,x_5,x_6$ be the roots of the polynomial equation



$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0.$$



Then,



(A)$|x_i|=2$ for exactly one value of $i$



(B)$|x_i|=2$ for exactly two values of $i$



(C)$|x_i|=2$ for all values of $i$.



(D)$|x_i|=2$ for no value of $i$.



My attempt: I noticed that the polynomial forms a geometric



progression with common ratio $2/x$.



Then I summed the terms and found this relation $x^7=128$ which gives us



$x=2$ but $x=2$ doesn't satisfy the original polynomial.



I looked at the graph too and found that the polynomial has no real roots .



So now , how to find the modulus of the roots ?










share|cite|improve this question



















  • 1




    You found $x^ 7 = 128$. Note that $x neq 2$, of course. It turns out that, in fact, the above polynomial is a factor of $x^7 - 128$. Consequently, every root must have modulus equal to $2$, and we can write down all the roots explicitly using the seventh roots of unity.
    – Ð°ÑÑ‚он вілла олоф мэллбэрг
    Aug 21 at 12:21















up vote
5
down vote

favorite












Let $x_1,x_2,x_3,x_4,x_5,x_6$ be the roots of the polynomial equation



$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0.$$



Then,



(A)$|x_i|=2$ for exactly one value of $i$



(B)$|x_i|=2$ for exactly two values of $i$



(C)$|x_i|=2$ for all values of $i$.



(D)$|x_i|=2$ for no value of $i$.



My attempt: I noticed that the polynomial forms a geometric



progression with common ratio $2/x$.



Then I summed the terms and found this relation $x^7=128$ which gives us



$x=2$ but $x=2$ doesn't satisfy the original polynomial.



I looked at the graph too and found that the polynomial has no real roots .



So now , how to find the modulus of the roots ?










share|cite|improve this question



















  • 1




    You found $x^ 7 = 128$. Note that $x neq 2$, of course. It turns out that, in fact, the above polynomial is a factor of $x^7 - 128$. Consequently, every root must have modulus equal to $2$, and we can write down all the roots explicitly using the seventh roots of unity.
    – Ð°ÑÑ‚он вілла олоф мэллбэрг
    Aug 21 at 12:21













up vote
5
down vote

favorite









up vote
5
down vote

favorite











Let $x_1,x_2,x_3,x_4,x_5,x_6$ be the roots of the polynomial equation



$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0.$$



Then,



(A)$|x_i|=2$ for exactly one value of $i$



(B)$|x_i|=2$ for exactly two values of $i$



(C)$|x_i|=2$ for all values of $i$.



(D)$|x_i|=2$ for no value of $i$.



My attempt: I noticed that the polynomial forms a geometric



progression with common ratio $2/x$.



Then I summed the terms and found this relation $x^7=128$ which gives us



$x=2$ but $x=2$ doesn't satisfy the original polynomial.



I looked at the graph too and found that the polynomial has no real roots .



So now , how to find the modulus of the roots ?










share|cite|improve this question















Let $x_1,x_2,x_3,x_4,x_5,x_6$ be the roots of the polynomial equation



$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0.$$



Then,



(A)$|x_i|=2$ for exactly one value of $i$



(B)$|x_i|=2$ for exactly two values of $i$



(C)$|x_i|=2$ for all values of $i$.



(D)$|x_i|=2$ for no value of $i$.



My attempt: I noticed that the polynomial forms a geometric



progression with common ratio $2/x$.



Then I summed the terms and found this relation $x^7=128$ which gives us



$x=2$ but $x=2$ doesn't satisfy the original polynomial.



I looked at the graph too and found that the polynomial has no real roots .



So now , how to find the modulus of the roots ?







polynomials roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 21 at 12:18









Clayton

18.3k22883




18.3k22883










asked Aug 21 at 12:15









Alphanerd

8719




8719







  • 1




    You found $x^ 7 = 128$. Note that $x neq 2$, of course. It turns out that, in fact, the above polynomial is a factor of $x^7 - 128$. Consequently, every root must have modulus equal to $2$, and we can write down all the roots explicitly using the seventh roots of unity.
    – Ð°ÑÑ‚он вілла олоф мэллбэрг
    Aug 21 at 12:21













  • 1




    You found $x^ 7 = 128$. Note that $x neq 2$, of course. It turns out that, in fact, the above polynomial is a factor of $x^7 - 128$. Consequently, every root must have modulus equal to $2$, and we can write down all the roots explicitly using the seventh roots of unity.
    – Ð°ÑÑ‚он вілла олоф мэллбэрг
    Aug 21 at 12:21








1




1




You found $x^ 7 = 128$. Note that $x neq 2$, of course. It turns out that, in fact, the above polynomial is a factor of $x^7 - 128$. Consequently, every root must have modulus equal to $2$, and we can write down all the roots explicitly using the seventh roots of unity.
– Ð°ÑÑ‚он вілла олоф мэллбэрг
Aug 21 at 12:21





You found $x^ 7 = 128$. Note that $x neq 2$, of course. It turns out that, in fact, the above polynomial is a factor of $x^7 - 128$. Consequently, every root must have modulus equal to $2$, and we can write down all the roots explicitly using the seventh roots of unity.
– Ð°ÑÑ‚он вілла олоф мэллбэрг
Aug 21 at 12:21











2 Answers
2






active

oldest

votes

















up vote
4
down vote



accepted










$$f(x)=sum_r=0^6(x^r2^6-r)=2^6dfracleft(dfrac x2right)^7-1dfrac x2-1=dfracx^7-128x-2$$



Clearly, $f(2)ne0$



$implies x^7=128$ with $xne2$



$iffleft(dfrac x2right)^7=1$



Now take modulus in both sides using $|z^m|=|z|^m$






share|cite|improve this answer



























    up vote
    3
    down vote













    As $x=2$ is not a root, the equation is



    $$fracx^7-128x-2=0.$$



    From this we draw



    $$|x^7|=|x|^7=128$$ and the only real solution $$|x|=2.$$






    share|cite|improve this answer




















      Your Answer




      StackExchange.ifUsing("editor", function ()
      return StackExchange.using("mathjaxEditing", function ()
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      );
      );
      , "mathjax-editing");

      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',
      convertImagesToLinks: true,
      noModals: false,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      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%2f2889795%2fevaluating-the-modulus-of-roots-of-a-6-degree-polynomial%23new-answer', 'question_page');

      );

      Post as a guest






























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      4
      down vote



      accepted










      $$f(x)=sum_r=0^6(x^r2^6-r)=2^6dfracleft(dfrac x2right)^7-1dfrac x2-1=dfracx^7-128x-2$$



      Clearly, $f(2)ne0$



      $implies x^7=128$ with $xne2$



      $iffleft(dfrac x2right)^7=1$



      Now take modulus in both sides using $|z^m|=|z|^m$






      share|cite|improve this answer
























        up vote
        4
        down vote



        accepted










        $$f(x)=sum_r=0^6(x^r2^6-r)=2^6dfracleft(dfrac x2right)^7-1dfrac x2-1=dfracx^7-128x-2$$



        Clearly, $f(2)ne0$



        $implies x^7=128$ with $xne2$



        $iffleft(dfrac x2right)^7=1$



        Now take modulus in both sides using $|z^m|=|z|^m$






        share|cite|improve this answer






















          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          $$f(x)=sum_r=0^6(x^r2^6-r)=2^6dfracleft(dfrac x2right)^7-1dfrac x2-1=dfracx^7-128x-2$$



          Clearly, $f(2)ne0$



          $implies x^7=128$ with $xne2$



          $iffleft(dfrac x2right)^7=1$



          Now take modulus in both sides using $|z^m|=|z|^m$






          share|cite|improve this answer












          $$f(x)=sum_r=0^6(x^r2^6-r)=2^6dfracleft(dfrac x2right)^7-1dfrac x2-1=dfracx^7-128x-2$$



          Clearly, $f(2)ne0$



          $implies x^7=128$ with $xne2$



          $iffleft(dfrac x2right)^7=1$



          Now take modulus in both sides using $|z^m|=|z|^m$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 21 at 12:19









          lab bhattacharjee

          216k14153266




          216k14153266




















              up vote
              3
              down vote













              As $x=2$ is not a root, the equation is



              $$fracx^7-128x-2=0.$$



              From this we draw



              $$|x^7|=|x|^7=128$$ and the only real solution $$|x|=2.$$






              share|cite|improve this answer
























                up vote
                3
                down vote













                As $x=2$ is not a root, the equation is



                $$fracx^7-128x-2=0.$$



                From this we draw



                $$|x^7|=|x|^7=128$$ and the only real solution $$|x|=2.$$






                share|cite|improve this answer






















                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  As $x=2$ is not a root, the equation is



                  $$fracx^7-128x-2=0.$$



                  From this we draw



                  $$|x^7|=|x|^7=128$$ and the only real solution $$|x|=2.$$






                  share|cite|improve this answer












                  As $x=2$ is not a root, the equation is



                  $$fracx^7-128x-2=0.$$



                  From this we draw



                  $$|x^7|=|x|^7=128$$ and the only real solution $$|x|=2.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 21 at 12:20









                  Yves Daoust

                  115k666209




                  115k666209



























                       

                      draft saved


                      draft discarded















































                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2889795%2fevaluating-the-modulus-of-roots-of-a-6-degree-polynomial%23new-answer', 'question_page');

                      );

                      Post as a guest













































































                      30Ec2y sZX52bx9,PVnJJMa,LbHsUIFmeN661tbL0W0 j,FHRripnkNOcYg,VFWGQ kG5kx,rbjTkYgEi A8 VgpeAfUtdm
                      gWcKMN4U4OUVrD6aEtfT vaQf,Gq3gPLzlg,37ttiI plpt6YsVbm,tFL zV HyKuzjCM gOfp7D fQzxpqL

                      Popular posts from this blog

                      How to check contact read email or not when send email to Individual?

                      How many registers does an x86_64 CPU actually have?

                      Displaying single band from multi-band raster using QGIS