Computing $sqrt[4]{28+16 sqrt 3}$












1












$begingroup$


I want to compute following radical



$$sqrt[4]{28+16 sqrt 3}$$



For that, I first tried to rewrite this in terms of exponential.



$$(28+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$



We know that $ 28 = 2 cdot 7^{frac{1}{2}}$



$$(2 cdot 7^{frac{1}{2}}+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$



However, I'm stuck at this step. Could you assist me?



Regards






radicals







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  • $begingroup$
    Is there an exponential approach to this question?
    $endgroup$
    – Enzo
    Dec 22 '18 at 10:08






  • 3




    $begingroup$
    The statment $28 = 2 cdot 7^{frac{1}{2}}$ is false. It should be $28=2^2cdot 7$ or $28^{frac12} = 2cdot 7^{frac{1}{2}}$.
    $endgroup$
    – Somos
    Dec 22 '18 at 10:12












  • $begingroup$
    Read about completing a square
    $endgroup$
    – farruhota
    Dec 30 '18 at 2:21
















1












$begingroup$


I want to compute following radical



$$sqrt[4]{28+16 sqrt 3}$$



For that, I first tried to rewrite this in terms of exponential.



$$(28+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$



We know that $ 28 = 2 cdot 7^{frac{1}{2}}$



$$(2 cdot 7^{frac{1}{2}}+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$



However, I'm stuck at this step. Could you assist me?



Regards






radicals







share|cite|improve this question









$endgroup$












  • $begingroup$
    Is there an exponential approach to this question?
    $endgroup$
    – Enzo
    Dec 22 '18 at 10:08






  • 3




    $begingroup$
    The statment $28 = 2 cdot 7^{frac{1}{2}}$ is false. It should be $28=2^2cdot 7$ or $28^{frac12} = 2cdot 7^{frac{1}{2}}$.
    $endgroup$
    – Somos
    Dec 22 '18 at 10:12












  • $begingroup$
    Read about completing a square
    $endgroup$
    – farruhota
    Dec 30 '18 at 2:21














1












1








1





$begingroup$


I want to compute following radical



$$sqrt[4]{28+16 sqrt 3}$$



For that, I first tried to rewrite this in terms of exponential.



$$(28+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$



We know that $ 28 = 2 cdot 7^{frac{1}{2}}$



$$(2 cdot 7^{frac{1}{2}}+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$



However, I'm stuck at this step. Could you assist me?



Regards






radicals







share|cite|improve this question









$endgroup$




I want to compute following radical



$$sqrt[4]{28+16 sqrt 3}$$



For that, I first tried to rewrite this in terms of exponential.



$$(28+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$



We know that $ 28 = 2 cdot 7^{frac{1}{2}}$



$$(2 cdot 7^{frac{1}{2}}+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$



However, I'm stuck at this step. Could you assist me?



Regards






radicals




radicals






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share|cite|improve this question











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asked Dec 22 '18 at 9:52









EnzoEnzo

30617




30617












  • $begingroup$
    Is there an exponential approach to this question?
    $endgroup$
    – Enzo
    Dec 22 '18 at 10:08






  • 3




    $begingroup$
    The statment $28 = 2 cdot 7^{frac{1}{2}}$ is false. It should be $28=2^2cdot 7$ or $28^{frac12} = 2cdot 7^{frac{1}{2}}$.
    $endgroup$
    – Somos
    Dec 22 '18 at 10:12












  • $begingroup$
    Read about completing a square
    $endgroup$
    – farruhota
    Dec 30 '18 at 2:21


















  • $begingroup$
    Is there an exponential approach to this question?
    $endgroup$
    – Enzo
    Dec 22 '18 at 10:08






  • 3




    $begingroup$
    The statment $28 = 2 cdot 7^{frac{1}{2}}$ is false. It should be $28=2^2cdot 7$ or $28^{frac12} = 2cdot 7^{frac{1}{2}}$.
    $endgroup$
    – Somos
    Dec 22 '18 at 10:12












  • $begingroup$
    Read about completing a square
    $endgroup$
    – farruhota
    Dec 30 '18 at 2:21
















$begingroup$
Is there an exponential approach to this question?
$endgroup$
– Enzo
Dec 22 '18 at 10:08




$begingroup$
Is there an exponential approach to this question?
$endgroup$
– Enzo
Dec 22 '18 at 10:08




3




3




$begingroup$
The statment $28 = 2 cdot 7^{frac{1}{2}}$ is false. It should be $28=2^2cdot 7$ or $28^{frac12} = 2cdot 7^{frac{1}{2}}$.
$endgroup$
– Somos
Dec 22 '18 at 10:12






$begingroup$
The statment $28 = 2 cdot 7^{frac{1}{2}}$ is false. It should be $28=2^2cdot 7$ or $28^{frac12} = 2cdot 7^{frac{1}{2}}$.
$endgroup$
– Somos
Dec 22 '18 at 10:12














$begingroup$
Read about completing a square
$endgroup$
– farruhota
Dec 30 '18 at 2:21




$begingroup$
Read about completing a square
$endgroup$
– farruhota
Dec 30 '18 at 2:21










5 Answers
5






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5












$begingroup$

$$sqrt[4] {28+16sqrt 3}=sqrt[4] {(sqrt {12})^2 +(sqrt {16})^2 +2sqrt {16cdot 12}}=sqrt {4+2sqrt 3}=sqrt {(sqrt 3)^2 +(1)^2 +2sqrt {3cdot 1}}=sqrt 3 +1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is it a rule or something?
    $endgroup$
    – Enzo
    Dec 22 '18 at 10:05










  • $begingroup$
    @Enzo Since you are taking the fourth root, you would like to express $28+16sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$
    $endgroup$
    – Shubham Johri
    Dec 22 '18 at 10:11





















2












$begingroup$

Hint.




  • $28+16sqrt3=12+16+2cdot2sqrt3cdot4=(2sqrt3)^2+4^2+2cdot2sqrt3cdot4=(4+2sqrt3)^2$


  • $4+2sqrt3=3+1+2cdot1cdotsqrt3=(sqrt3)^2+1^2+2cdot1cdotsqrt3=(sqrt3+1)^2$







share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint: Compute $$(1+sqrt{3})^4$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $a+bsqrt 3$ for integers $a$ and $b$, then you can solve it like this:



      Step 1



      ($a+bsqrt 3)^4=28+16sqrt 3$. So suppose ($a+bsqrt 3)^2=c+dsqrt 3$. Then $(c+dsqrt 3)^2=28+16sqrt 3$, i.e.



      $$(c^2+3d^2)+2cdsqrt 3=28+16sqrt 3$$



      $c^2+3d^2$ and $2cd$ are integers, so this can only be true if $c^2+3d^2=28$ and $2cd=16$.

      Then $d=8/c$, so $c^2+dfrac{192}{c^2}=28$. Multiplying by $c^2$ and rearranging, we get a quadratic in $c^2$:



      $$c^4-28c^2+192=0$$



      This factors as $(c^2-12)(c^2-16)=0$. Since $c$ is an integer, we must have $c=pm4$, and so $d=pm2$. So we have two possible solutions: $pm(4+2sqrt 3)$. (You should try squaring this to check that you get the expected result $28+16sqrt 3$). Pick the positive solution $4+2sqrt 3$ (if you choose the negative solution, you will find that Step 2 fails).



      Step 2



      Now you need to solve $(a+bsqrt 3)^2=4+2sqrt 3$. You should be able to do this yourself now, using exactly the same method as Step 1.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Hint:



        Try to write
        $$28+16 sqrt 3=(a+bsqrt 3)^4$$
        for suitable $a$ and $b$.



        You obtain, reordering the terms
        begin{align}
        (a+bsqrt 3)^4&=(a^4+18a^2b^2+9b^4)+4ab(a^2+3b^2)sqrt 3.\
        end{align}
        Can you choose $a$ and $b$ so that
        $$a^4+18a^2b^2+9b^4=28,quad ab(a^2+3b^2)=4?$$






        share|cite|improve this answer











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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          $$sqrt[4] {28+16sqrt 3}=sqrt[4] {(sqrt {12})^2 +(sqrt {16})^2 +2sqrt {16cdot 12}}=sqrt {4+2sqrt 3}=sqrt {(sqrt 3)^2 +(1)^2 +2sqrt {3cdot 1}}=sqrt 3 +1$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is it a rule or something?
            $endgroup$
            – Enzo
            Dec 22 '18 at 10:05










          • $begingroup$
            @Enzo Since you are taking the fourth root, you would like to express $28+16sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$
            $endgroup$
            – Shubham Johri
            Dec 22 '18 at 10:11


















          5












          $begingroup$

          $$sqrt[4] {28+16sqrt 3}=sqrt[4] {(sqrt {12})^2 +(sqrt {16})^2 +2sqrt {16cdot 12}}=sqrt {4+2sqrt 3}=sqrt {(sqrt 3)^2 +(1)^2 +2sqrt {3cdot 1}}=sqrt 3 +1$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is it a rule or something?
            $endgroup$
            – Enzo
            Dec 22 '18 at 10:05










          • $begingroup$
            @Enzo Since you are taking the fourth root, you would like to express $28+16sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$
            $endgroup$
            – Shubham Johri
            Dec 22 '18 at 10:11
















          5












          5








          5





          $begingroup$

          $$sqrt[4] {28+16sqrt 3}=sqrt[4] {(sqrt {12})^2 +(sqrt {16})^2 +2sqrt {16cdot 12}}=sqrt {4+2sqrt 3}=sqrt {(sqrt 3)^2 +(1)^2 +2sqrt {3cdot 1}}=sqrt 3 +1$$






          share|cite|improve this answer









          $endgroup$



          $$sqrt[4] {28+16sqrt 3}=sqrt[4] {(sqrt {12})^2 +(sqrt {16})^2 +2sqrt {16cdot 12}}=sqrt {4+2sqrt 3}=sqrt {(sqrt 3)^2 +(1)^2 +2sqrt {3cdot 1}}=sqrt 3 +1$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 22 '18 at 10:00









          DarkraiDarkrai

          6,4311442




          6,4311442












          • $begingroup$
            Is it a rule or something?
            $endgroup$
            – Enzo
            Dec 22 '18 at 10:05










          • $begingroup$
            @Enzo Since you are taking the fourth root, you would like to express $28+16sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$
            $endgroup$
            – Shubham Johri
            Dec 22 '18 at 10:11




















          • $begingroup$
            Is it a rule or something?
            $endgroup$
            – Enzo
            Dec 22 '18 at 10:05










          • $begingroup$
            @Enzo Since you are taking the fourth root, you would like to express $28+16sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$
            $endgroup$
            – Shubham Johri
            Dec 22 '18 at 10:11


















          $begingroup$
          Is it a rule or something?
          $endgroup$
          – Enzo
          Dec 22 '18 at 10:05




          $begingroup$
          Is it a rule or something?
          $endgroup$
          – Enzo
          Dec 22 '18 at 10:05












          $begingroup$
          @Enzo Since you are taking the fourth root, you would like to express $28+16sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$
          $endgroup$
          – Shubham Johri
          Dec 22 '18 at 10:11






          $begingroup$
          @Enzo Since you are taking the fourth root, you would like to express $28+16sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$
          $endgroup$
          – Shubham Johri
          Dec 22 '18 at 10:11













          2












          $begingroup$

          Hint.




          • $28+16sqrt3=12+16+2cdot2sqrt3cdot4=(2sqrt3)^2+4^2+2cdot2sqrt3cdot4=(4+2sqrt3)^2$


          • $4+2sqrt3=3+1+2cdot1cdotsqrt3=(sqrt3)^2+1^2+2cdot1cdotsqrt3=(sqrt3+1)^2$







          share|cite|improve this answer









          $endgroup$


















            2












            $begingroup$

            Hint.




            • $28+16sqrt3=12+16+2cdot2sqrt3cdot4=(2sqrt3)^2+4^2+2cdot2sqrt3cdot4=(4+2sqrt3)^2$


            • $4+2sqrt3=3+1+2cdot1cdotsqrt3=(sqrt3)^2+1^2+2cdot1cdotsqrt3=(sqrt3+1)^2$







            share|cite|improve this answer









            $endgroup$
















              2












              2








              2





              $begingroup$

              Hint.




              • $28+16sqrt3=12+16+2cdot2sqrt3cdot4=(2sqrt3)^2+4^2+2cdot2sqrt3cdot4=(4+2sqrt3)^2$


              • $4+2sqrt3=3+1+2cdot1cdotsqrt3=(sqrt3)^2+1^2+2cdot1cdotsqrt3=(sqrt3+1)^2$







              share|cite|improve this answer









              $endgroup$



              Hint.




              • $28+16sqrt3=12+16+2cdot2sqrt3cdot4=(2sqrt3)^2+4^2+2cdot2sqrt3cdot4=(4+2sqrt3)^2$


              • $4+2sqrt3=3+1+2cdot1cdotsqrt3=(sqrt3)^2+1^2+2cdot1cdotsqrt3=(sqrt3+1)^2$








              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 22 '18 at 10:01









              Shubham JohriShubham Johri

              5,668918




              5,668918























                  0












                  $begingroup$

                  Hint: Compute $$(1+sqrt{3})^4$$






                  share|cite|improve this answer









                  $endgroup$


















                    0












                    $begingroup$

                    Hint: Compute $$(1+sqrt{3})^4$$






                    share|cite|improve this answer









                    $endgroup$
















                      0












                      0








                      0





                      $begingroup$

                      Hint: Compute $$(1+sqrt{3})^4$$






                      share|cite|improve this answer









                      $endgroup$



                      Hint: Compute $$(1+sqrt{3})^4$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 22 '18 at 9:58









                      Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                      79.1k42867




                      79.1k42867























                          0












                          $begingroup$

                          In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $a+bsqrt 3$ for integers $a$ and $b$, then you can solve it like this:



                          Step 1



                          ($a+bsqrt 3)^4=28+16sqrt 3$. So suppose ($a+bsqrt 3)^2=c+dsqrt 3$. Then $(c+dsqrt 3)^2=28+16sqrt 3$, i.e.



                          $$(c^2+3d^2)+2cdsqrt 3=28+16sqrt 3$$



                          $c^2+3d^2$ and $2cd$ are integers, so this can only be true if $c^2+3d^2=28$ and $2cd=16$.

                          Then $d=8/c$, so $c^2+dfrac{192}{c^2}=28$. Multiplying by $c^2$ and rearranging, we get a quadratic in $c^2$:



                          $$c^4-28c^2+192=0$$



                          This factors as $(c^2-12)(c^2-16)=0$. Since $c$ is an integer, we must have $c=pm4$, and so $d=pm2$. So we have two possible solutions: $pm(4+2sqrt 3)$. (You should try squaring this to check that you get the expected result $28+16sqrt 3$). Pick the positive solution $4+2sqrt 3$ (if you choose the negative solution, you will find that Step 2 fails).



                          Step 2



                          Now you need to solve $(a+bsqrt 3)^2=4+2sqrt 3$. You should be able to do this yourself now, using exactly the same method as Step 1.






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $a+bsqrt 3$ for integers $a$ and $b$, then you can solve it like this:



                            Step 1



                            ($a+bsqrt 3)^4=28+16sqrt 3$. So suppose ($a+bsqrt 3)^2=c+dsqrt 3$. Then $(c+dsqrt 3)^2=28+16sqrt 3$, i.e.



                            $$(c^2+3d^2)+2cdsqrt 3=28+16sqrt 3$$



                            $c^2+3d^2$ and $2cd$ are integers, so this can only be true if $c^2+3d^2=28$ and $2cd=16$.

                            Then $d=8/c$, so $c^2+dfrac{192}{c^2}=28$. Multiplying by $c^2$ and rearranging, we get a quadratic in $c^2$:



                            $$c^4-28c^2+192=0$$



                            This factors as $(c^2-12)(c^2-16)=0$. Since $c$ is an integer, we must have $c=pm4$, and so $d=pm2$. So we have two possible solutions: $pm(4+2sqrt 3)$. (You should try squaring this to check that you get the expected result $28+16sqrt 3$). Pick the positive solution $4+2sqrt 3$ (if you choose the negative solution, you will find that Step 2 fails).



                            Step 2



                            Now you need to solve $(a+bsqrt 3)^2=4+2sqrt 3$. You should be able to do this yourself now, using exactly the same method as Step 1.






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $a+bsqrt 3$ for integers $a$ and $b$, then you can solve it like this:



                              Step 1



                              ($a+bsqrt 3)^4=28+16sqrt 3$. So suppose ($a+bsqrt 3)^2=c+dsqrt 3$. Then $(c+dsqrt 3)^2=28+16sqrt 3$, i.e.



                              $$(c^2+3d^2)+2cdsqrt 3=28+16sqrt 3$$



                              $c^2+3d^2$ and $2cd$ are integers, so this can only be true if $c^2+3d^2=28$ and $2cd=16$.

                              Then $d=8/c$, so $c^2+dfrac{192}{c^2}=28$. Multiplying by $c^2$ and rearranging, we get a quadratic in $c^2$:



                              $$c^4-28c^2+192=0$$



                              This factors as $(c^2-12)(c^2-16)=0$. Since $c$ is an integer, we must have $c=pm4$, and so $d=pm2$. So we have two possible solutions: $pm(4+2sqrt 3)$. (You should try squaring this to check that you get the expected result $28+16sqrt 3$). Pick the positive solution $4+2sqrt 3$ (if you choose the negative solution, you will find that Step 2 fails).



                              Step 2



                              Now you need to solve $(a+bsqrt 3)^2=4+2sqrt 3$. You should be able to do this yourself now, using exactly the same method as Step 1.






                              share|cite|improve this answer









                              $endgroup$



                              In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $a+bsqrt 3$ for integers $a$ and $b$, then you can solve it like this:



                              Step 1



                              ($a+bsqrt 3)^4=28+16sqrt 3$. So suppose ($a+bsqrt 3)^2=c+dsqrt 3$. Then $(c+dsqrt 3)^2=28+16sqrt 3$, i.e.



                              $$(c^2+3d^2)+2cdsqrt 3=28+16sqrt 3$$



                              $c^2+3d^2$ and $2cd$ are integers, so this can only be true if $c^2+3d^2=28$ and $2cd=16$.

                              Then $d=8/c$, so $c^2+dfrac{192}{c^2}=28$. Multiplying by $c^2$ and rearranging, we get a quadratic in $c^2$:



                              $$c^4-28c^2+192=0$$



                              This factors as $(c^2-12)(c^2-16)=0$. Since $c$ is an integer, we must have $c=pm4$, and so $d=pm2$. So we have two possible solutions: $pm(4+2sqrt 3)$. (You should try squaring this to check that you get the expected result $28+16sqrt 3$). Pick the positive solution $4+2sqrt 3$ (if you choose the negative solution, you will find that Step 2 fails).



                              Step 2



                              Now you need to solve $(a+bsqrt 3)^2=4+2sqrt 3$. You should be able to do this yourself now, using exactly the same method as Step 1.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 22 '18 at 10:23









                              TonyKTonyK

                              44.1k358137




                              44.1k358137























                                  0












                                  $begingroup$

                                  Hint:



                                  Try to write
                                  $$28+16 sqrt 3=(a+bsqrt 3)^4$$
                                  for suitable $a$ and $b$.



                                  You obtain, reordering the terms
                                  begin{align}
                                  (a+bsqrt 3)^4&=(a^4+18a^2b^2+9b^4)+4ab(a^2+3b^2)sqrt 3.\
                                  end{align}
                                  Can you choose $a$ and $b$ so that
                                  $$a^4+18a^2b^2+9b^4=28,quad ab(a^2+3b^2)=4?$$






                                  share|cite|improve this answer











                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Hint:



                                    Try to write
                                    $$28+16 sqrt 3=(a+bsqrt 3)^4$$
                                    for suitable $a$ and $b$.



                                    You obtain, reordering the terms
                                    begin{align}
                                    (a+bsqrt 3)^4&=(a^4+18a^2b^2+9b^4)+4ab(a^2+3b^2)sqrt 3.\
                                    end{align}
                                    Can you choose $a$ and $b$ so that
                                    $$a^4+18a^2b^2+9b^4=28,quad ab(a^2+3b^2)=4?$$






                                    share|cite|improve this answer











                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Hint:



                                      Try to write
                                      $$28+16 sqrt 3=(a+bsqrt 3)^4$$
                                      for suitable $a$ and $b$.



                                      You obtain, reordering the terms
                                      begin{align}
                                      (a+bsqrt 3)^4&=(a^4+18a^2b^2+9b^4)+4ab(a^2+3b^2)sqrt 3.\
                                      end{align}
                                      Can you choose $a$ and $b$ so that
                                      $$a^4+18a^2b^2+9b^4=28,quad ab(a^2+3b^2)=4?$$






                                      share|cite|improve this answer











                                      $endgroup$



                                      Hint:



                                      Try to write
                                      $$28+16 sqrt 3=(a+bsqrt 3)^4$$
                                      for suitable $a$ and $b$.



                                      You obtain, reordering the terms
                                      begin{align}
                                      (a+bsqrt 3)^4&=(a^4+18a^2b^2+9b^4)+4ab(a^2+3b^2)sqrt 3.\
                                      end{align}
                                      Can you choose $a$ and $b$ so that
                                      $$a^4+18a^2b^2+9b^4=28,quad ab(a^2+3b^2)=4?$$







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Dec 22 '18 at 10:33

























                                      answered Dec 22 '18 at 10:23









                                      BernardBernard

                                      124k742117




                                      124k742117






























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