Find the roots of $3x^3-4x-8$

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It is given that $alpha$, $beta$ and $gamma$ are the roots of the polynomial $3x^3-4x-8$.



I have been asked to calculate the value of $alpha^2 + beta^2 + gamma^2$.



However I am unsure how to find these roots, seeing as though I haven't been given a root to start with.



I began by identifying that



$alpha + beta + gamma = 0$



$alphabeta + alphagamma + betagamma = -4/3$



$alphabetagamma = 8/3$



However I am unsure how to continue to find $alpha^2 + beta^2 + gamma^2$.




polynomials




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  • Instead of "the given polynomial expression", why not just give the polynomial expression itself? Just like in cinema: show, don't tell.
    – Asaf Karagila♦
    Aug 14 at 13:45














up vote
5
down vote

favorite












It is given that $alpha$, $beta$ and $gamma$ are the roots of the polynomial $3x^3-4x-8$.



I have been asked to calculate the value of $alpha^2 + beta^2 + gamma^2$.



However I am unsure how to find these roots, seeing as though I haven't been given a root to start with.



I began by identifying that



$alpha + beta + gamma = 0$



$alphabeta + alphagamma + betagamma = -4/3$



$alphabetagamma = 8/3$



However I am unsure how to continue to find $alpha^2 + beta^2 + gamma^2$.




polynomials




share|cite|improve this question






















  • Instead of "the given polynomial expression", why not just give the polynomial expression itself? Just like in cinema: show, don't tell.
    – Asaf Karagila♦
    Aug 14 at 13:45












up vote
5
down vote

favorite









up vote
5
down vote

favorite











It is given that $alpha$, $beta$ and $gamma$ are the roots of the polynomial $3x^3-4x-8$.



I have been asked to calculate the value of $alpha^2 + beta^2 + gamma^2$.



However I am unsure how to find these roots, seeing as though I haven't been given a root to start with.



I began by identifying that



$alpha + beta + gamma = 0$



$alphabeta + alphagamma + betagamma = -4/3$



$alphabetagamma = 8/3$



However I am unsure how to continue to find $alpha^2 + beta^2 + gamma^2$.




polynomials




share|cite|improve this question














It is given that $alpha$, $beta$ and $gamma$ are the roots of the polynomial $3x^3-4x-8$.



I have been asked to calculate the value of $alpha^2 + beta^2 + gamma^2$.



However I am unsure how to find these roots, seeing as though I haven't been given a root to start with.



I began by identifying that



$alpha + beta + gamma = 0$



$alphabeta + alphagamma + betagamma = -4/3$



$alphabetagamma = 8/3$



However I am unsure how to continue to find $alpha^2 + beta^2 + gamma^2$.





polynomials





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 14 at 13:45









Asaf Karagila♦

294k31409736




294k31409736










asked Aug 14 at 8:35









Emily

244




244











  • Instead of "the given polynomial expression", why not just give the polynomial expression itself? Just like in cinema: show, don't tell.
    – Asaf Karagila♦
    Aug 14 at 13:45
















  • Instead of "the given polynomial expression", why not just give the polynomial expression itself? Just like in cinema: show, don't tell.
    – Asaf Karagila♦
    Aug 14 at 13:45















Instead of "the given polynomial expression", why not just give the polynomial expression itself? Just like in cinema: show, don't tell.
– Asaf Karagila♦
Aug 14 at 13:45




Instead of "the given polynomial expression", why not just give the polynomial expression itself? Just like in cinema: show, don't tell.
– Asaf Karagila♦
Aug 14 at 13:45










4 Answers
4






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up vote
15
down vote













Use the identity
$(a+b+c)^2=a^2 + b^2 + c^2 + 2ab+ 2bc+2ca$






share|cite|improve this answer



























    up vote
    4
    down vote













    $ x(3x^2-4)=8$



    Squaring we get $x^2(3x^2-4)^2=8^2$



    Let $x^2=yimplies y(3y-4)^2=64iff9y^3-24y^2+16y-64=0$



    whose roots are $a^2,b^2,c^2$



    $implies a^2+b^2+c^2=dfrac249=?$






    share|cite|improve this answer






















    • @ancientmathematician, Shift missed! Thanks
      – lab bhattacharjee
      Aug 14 at 8:57

















    up vote
    3
    down vote













    Alternative approach: for any root we have $3x^3-4x=8$, hence by squaring $9x^6-24x^4+16 x^2 = 64$ and $alpha^2,beta^2,gamma^2$ are the roots of the polynomial $9z^3-24z^2+16z-64$.



    By Vièta's formulas it follows that $alpha^2+beta^2+gamma^2 = frac249 = colorredfrac83$.






    share|cite|improve this answer



























      up vote
      -1
      down vote













      Yet another approach: $,dfrac1alpha, dfrac1beta, dfrac1gamma,$ are the roots of $,8x^3+4x^2-3,$, so $,dfrac1alpha+ dfrac1beta+ dfrac1gamma=-dfrac12,$.



      But $,3alpha^3=4alpha+8 iff alpha^2=dfrac43 + dfrac83alpha,$, and therefore: $$alpha^2+beta^2+gamma^2=3 cdot frac43+ frac83left(frac1alpha+ frac1beta+ frac1gammaright)=4-frac43 = frac83$$






      share|cite|improve this answer




















      • Wish the downvoter had left a comment why.
        – dxiv
        Aug 18 at 20:19










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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      15
      down vote













      Use the identity
      $(a+b+c)^2=a^2 + b^2 + c^2 + 2ab+ 2bc+2ca$






      share|cite|improve this answer
























        up vote
        15
        down vote













        Use the identity
        $(a+b+c)^2=a^2 + b^2 + c^2 + 2ab+ 2bc+2ca$






        share|cite|improve this answer






















          up vote
          15
          down vote










          up vote
          15
          down vote









          Use the identity
          $(a+b+c)^2=a^2 + b^2 + c^2 + 2ab+ 2bc+2ca$






          share|cite|improve this answer












          Use the identity
          $(a+b+c)^2=a^2 + b^2 + c^2 + 2ab+ 2bc+2ca$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 14 at 8:46









          Sadil Khan

          3437




          3437




















              up vote
              4
              down vote













              $ x(3x^2-4)=8$



              Squaring we get $x^2(3x^2-4)^2=8^2$



              Let $x^2=yimplies y(3y-4)^2=64iff9y^3-24y^2+16y-64=0$



              whose roots are $a^2,b^2,c^2$



              $implies a^2+b^2+c^2=dfrac249=?$






              share|cite|improve this answer






















              • @ancientmathematician, Shift missed! Thanks
                – lab bhattacharjee
                Aug 14 at 8:57














              up vote
              4
              down vote













              $ x(3x^2-4)=8$



              Squaring we get $x^2(3x^2-4)^2=8^2$



              Let $x^2=yimplies y(3y-4)^2=64iff9y^3-24y^2+16y-64=0$



              whose roots are $a^2,b^2,c^2$



              $implies a^2+b^2+c^2=dfrac249=?$






              share|cite|improve this answer






















              • @ancientmathematician, Shift missed! Thanks
                – lab bhattacharjee
                Aug 14 at 8:57












              up vote
              4
              down vote










              up vote
              4
              down vote









              $ x(3x^2-4)=8$



              Squaring we get $x^2(3x^2-4)^2=8^2$



              Let $x^2=yimplies y(3y-4)^2=64iff9y^3-24y^2+16y-64=0$



              whose roots are $a^2,b^2,c^2$



              $implies a^2+b^2+c^2=dfrac249=?$






              share|cite|improve this answer














              $ x(3x^2-4)=8$



              Squaring we get $x^2(3x^2-4)^2=8^2$



              Let $x^2=yimplies y(3y-4)^2=64iff9y^3-24y^2+16y-64=0$



              whose roots are $a^2,b^2,c^2$



              $implies a^2+b^2+c^2=dfrac249=?$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Aug 14 at 8:56

























              answered Aug 14 at 8:50









              lab bhattacharjee

              216k14153265




              216k14153265











              • @ancientmathematician, Shift missed! Thanks
                – lab bhattacharjee
                Aug 14 at 8:57
















              • @ancientmathematician, Shift missed! Thanks
                – lab bhattacharjee
                Aug 14 at 8:57















              @ancientmathematician, Shift missed! Thanks
              – lab bhattacharjee
              Aug 14 at 8:57




              @ancientmathematician, Shift missed! Thanks
              – lab bhattacharjee
              Aug 14 at 8:57










              up vote
              3
              down vote













              Alternative approach: for any root we have $3x^3-4x=8$, hence by squaring $9x^6-24x^4+16 x^2 = 64$ and $alpha^2,beta^2,gamma^2$ are the roots of the polynomial $9z^3-24z^2+16z-64$.



              By Vièta's formulas it follows that $alpha^2+beta^2+gamma^2 = frac249 = colorredfrac83$.






              share|cite|improve this answer
























                up vote
                3
                down vote













                Alternative approach: for any root we have $3x^3-4x=8$, hence by squaring $9x^6-24x^4+16 x^2 = 64$ and $alpha^2,beta^2,gamma^2$ are the roots of the polynomial $9z^3-24z^2+16z-64$.



                By Vièta's formulas it follows that $alpha^2+beta^2+gamma^2 = frac249 = colorredfrac83$.






                share|cite|improve this answer






















                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  Alternative approach: for any root we have $3x^3-4x=8$, hence by squaring $9x^6-24x^4+16 x^2 = 64$ and $alpha^2,beta^2,gamma^2$ are the roots of the polynomial $9z^3-24z^2+16z-64$.



                  By Vièta's formulas it follows that $alpha^2+beta^2+gamma^2 = frac249 = colorredfrac83$.






                  share|cite|improve this answer












                  Alternative approach: for any root we have $3x^3-4x=8$, hence by squaring $9x^6-24x^4+16 x^2 = 64$ and $alpha^2,beta^2,gamma^2$ are the roots of the polynomial $9z^3-24z^2+16z-64$.



                  By Vièta's formulas it follows that $alpha^2+beta^2+gamma^2 = frac249 = colorredfrac83$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 14 at 9:25









                  Jack D'Aurizio♦

                  273k32268636




                  273k32268636




















                      up vote
                      -1
                      down vote













                      Yet another approach: $,dfrac1alpha, dfrac1beta, dfrac1gamma,$ are the roots of $,8x^3+4x^2-3,$, so $,dfrac1alpha+ dfrac1beta+ dfrac1gamma=-dfrac12,$.



                      But $,3alpha^3=4alpha+8 iff alpha^2=dfrac43 + dfrac83alpha,$, and therefore: $$alpha^2+beta^2+gamma^2=3 cdot frac43+ frac83left(frac1alpha+ frac1beta+ frac1gammaright)=4-frac43 = frac83$$






                      share|cite|improve this answer




















                      • Wish the downvoter had left a comment why.
                        – dxiv
                        Aug 18 at 20:19














                      up vote
                      -1
                      down vote













                      Yet another approach: $,dfrac1alpha, dfrac1beta, dfrac1gamma,$ are the roots of $,8x^3+4x^2-3,$, so $,dfrac1alpha+ dfrac1beta+ dfrac1gamma=-dfrac12,$.



                      But $,3alpha^3=4alpha+8 iff alpha^2=dfrac43 + dfrac83alpha,$, and therefore: $$alpha^2+beta^2+gamma^2=3 cdot frac43+ frac83left(frac1alpha+ frac1beta+ frac1gammaright)=4-frac43 = frac83$$






                      share|cite|improve this answer




















                      • Wish the downvoter had left a comment why.
                        – dxiv
                        Aug 18 at 20:19












                      up vote
                      -1
                      down vote










                      up vote
                      -1
                      down vote









                      Yet another approach: $,dfrac1alpha, dfrac1beta, dfrac1gamma,$ are the roots of $,8x^3+4x^2-3,$, so $,dfrac1alpha+ dfrac1beta+ dfrac1gamma=-dfrac12,$.



                      But $,3alpha^3=4alpha+8 iff alpha^2=dfrac43 + dfrac83alpha,$, and therefore: $$alpha^2+beta^2+gamma^2=3 cdot frac43+ frac83left(frac1alpha+ frac1beta+ frac1gammaright)=4-frac43 = frac83$$






                      share|cite|improve this answer












                      Yet another approach: $,dfrac1alpha, dfrac1beta, dfrac1gamma,$ are the roots of $,8x^3+4x^2-3,$, so $,dfrac1alpha+ dfrac1beta+ dfrac1gamma=-dfrac12,$.



                      But $,3alpha^3=4alpha+8 iff alpha^2=dfrac43 + dfrac83alpha,$, and therefore: $$alpha^2+beta^2+gamma^2=3 cdot frac43+ frac83left(frac1alpha+ frac1beta+ frac1gammaright)=4-frac43 = frac83$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Aug 15 at 0:50









                      dxiv

                      55.5k64798




                      55.5k64798











                      • Wish the downvoter had left a comment why.
                        – dxiv
                        Aug 18 at 20:19
















                      • Wish the downvoter had left a comment why.
                        – dxiv
                        Aug 18 at 20:19















                      Wish the downvoter had left a comment why.
                      – dxiv
                      Aug 18 at 20:19




                      Wish the downvoter had left a comment why.
                      – dxiv
                      Aug 18 at 20:19

















                       

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