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How to simplify $sqrt{2+sqrt{3}}$ $?$



The Next CEO of Stack OverflowSimple doubt about complex numbersRationalising the denominator: $frac{11}{3sqrt{3}+7}$Rationalize and simplify $frac{x-1}{sqrt{2sqrt{x}} + 1 - sqrt[4]{x}}$How to simplify $frac{4 + 2sqrt6}{sqrt{5 + 2sqrt{6}}}$?Proof by contradiction that $sqrt{2} + sqrt{6} < sqrt{15}$Simplify $frac {sqrt5}{sqrt3+1} - sqrtfrac{30}{8} + frac {sqrt {45}}{2}$Difficulty in “expressing radical as square”Proof that $sqrt6 - sqrt2 - sqrt3$ is irrational.Simplify $sqrt xleft(sqrt x+frac1{sqrt x}right)$Simplifying $frac { sqrt2 + sqrt 6}{sqrt2 + sqrt3}$?Simplify $frac{sqrt{24}}{8}$












0












$begingroup$



Simplify $left(frac{2(sqrt2 + sqrt6)}{3(sqrt{2+sqrt3}}right)$




The answer to this question is $frac{4}{3}$ in a workbook.



How would I simplify $sqrt{2+sqrt3}$ $?$ If it was something like $sqrt{3 + 2sqrt2}$ , I would have simplified it as follows:
$sqrt{3 + 2sqrt2}$
$=$ $sqrt{(sqrt2)^2 + 2(sqrt2)(1) + (1)^2}$
$=$
$sqrt{(sqrt2 + 1)^2}$
$=$
$sqrt2 + 1$

But I can't simplify $sqrt{2+sqrt3}$  like that as $2+sqrt3$ is can't be written as squares of two numbers. Is there any other method?






algebra-precalculus radicals nested-radicals







share|cite|improve this question











$endgroup$












  • $begingroup$
    See the latter half of this answer.
    $endgroup$
    – Theo Bendit
    Mar 17 at 7:45
















0












$begingroup$



Simplify $left(frac{2(sqrt2 + sqrt6)}{3(sqrt{2+sqrt3}}right)$




The answer to this question is $frac{4}{3}$ in a workbook.



How would I simplify $sqrt{2+sqrt3}$ $?$ If it was something like $sqrt{3 + 2sqrt2}$ , I would have simplified it as follows:
$sqrt{3 + 2sqrt2}$
$=$ $sqrt{(sqrt2)^2 + 2(sqrt2)(1) + (1)^2}$
$=$
$sqrt{(sqrt2 + 1)^2}$
$=$
$sqrt2 + 1$

But I can't simplify $sqrt{2+sqrt3}$  like that as $2+sqrt3$ is can't be written as squares of two numbers. Is there any other method?






algebra-precalculus radicals nested-radicals







share|cite|improve this question











$endgroup$












  • $begingroup$
    See the latter half of this answer.
    $endgroup$
    – Theo Bendit
    Mar 17 at 7:45














0












0








0





$begingroup$



Simplify $left(frac{2(sqrt2 + sqrt6)}{3(sqrt{2+sqrt3}}right)$




The answer to this question is $frac{4}{3}$ in a workbook.



How would I simplify $sqrt{2+sqrt3}$ $?$ If it was something like $sqrt{3 + 2sqrt2}$ , I would have simplified it as follows:
$sqrt{3 + 2sqrt2}$
$=$ $sqrt{(sqrt2)^2 + 2(sqrt2)(1) + (1)^2}$
$=$
$sqrt{(sqrt2 + 1)^2}$
$=$
$sqrt2 + 1$

But I can't simplify $sqrt{2+sqrt3}$  like that as $2+sqrt3$ is can't be written as squares of two numbers. Is there any other method?






algebra-precalculus radicals nested-radicals







share|cite|improve this question











$endgroup$





Simplify $left(frac{2(sqrt2 + sqrt6)}{3(sqrt{2+sqrt3}}right)$




The answer to this question is $frac{4}{3}$ in a workbook.



How would I simplify $sqrt{2+sqrt3}$ $?$ If it was something like $sqrt{3 + 2sqrt2}$ , I would have simplified it as follows:
$sqrt{3 + 2sqrt2}$
$=$ $sqrt{(sqrt2)^2 + 2(sqrt2)(1) + (1)^2}$
$=$
$sqrt{(sqrt2 + 1)^2}$
$=$
$sqrt2 + 1$

But I can't simplify $sqrt{2+sqrt3}$  like that as $2+sqrt3$ is can't be written as squares of two numbers. Is there any other method?






algebra-precalculus radicals nested-radicals




algebra-precalculus radicals nested-radicals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 17 at 9:14









José Carlos Santos

171k23132240




171k23132240










asked Mar 17 at 7:30









Sanjay SinghSanjay Singh

32




32












  • $begingroup$
    See the latter half of this answer.
    $endgroup$
    – Theo Bendit
    Mar 17 at 7:45


















  • $begingroup$
    See the latter half of this answer.
    $endgroup$
    – Theo Bendit
    Mar 17 at 7:45
















$begingroup$
See the latter half of this answer.
$endgroup$
– Theo Bendit
Mar 17 at 7:45




$begingroup$
See the latter half of this answer.
$endgroup$
– Theo Bendit
Mar 17 at 7:45










3 Answers
3






active

oldest

votes


















5












$begingroup$

Note that$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)^2=frac{4left(8+4sqrt3right)}{9left(2+sqrt3right)}=frac{16}9=left(frac43right)^2.$$






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    Hints:



    $$sqrt 2+sqrt 6=sqrt 2(1+sqrt 3)=sqrt{2(1+sqrt 3)^2}=sqrt{2(4+2sqrt 3)}=2sqrt{2+sqrt 3}$$



    $$sqrt{2+sqrt 3}=frac 1{sqrt 2}sqrt{4+2sqrt 3}=frac 1{sqrt 2}sqrt{(sqrt 3+1)^2}=frac{sqrt 3+1}{sqrt 2}=frac{sqrt 6+sqrt 2}2$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      $$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)= frac{(2√2+2√6)sqrt{2+sqrt{3}}}{3(2+√3)}=frac{8+4√3}{2+√3}=frac{4}{3}.$$






      share|cite|improve this answer









      $endgroup$














        Your Answer





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        3 Answers
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        3 Answers
        3






        active

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        active

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        active

        oldest

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        5












        $begingroup$

        Note that$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)^2=frac{4left(8+4sqrt3right)}{9left(2+sqrt3right)}=frac{16}9=left(frac43right)^2.$$






        share|cite|improve this answer









        $endgroup$


















          5












          $begingroup$

          Note that$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)^2=frac{4left(8+4sqrt3right)}{9left(2+sqrt3right)}=frac{16}9=left(frac43right)^2.$$






          share|cite|improve this answer









          $endgroup$
















            5












            5








            5





            $begingroup$

            Note that$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)^2=frac{4left(8+4sqrt3right)}{9left(2+sqrt3right)}=frac{16}9=left(frac43right)^2.$$






            share|cite|improve this answer









            $endgroup$



            Note that$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)^2=frac{4left(8+4sqrt3right)}{9left(2+sqrt3right)}=frac{16}9=left(frac43right)^2.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 17 at 7:35









            José Carlos SantosJosé Carlos Santos

            171k23132240




            171k23132240























                4












                $begingroup$

                Hints:



                $$sqrt 2+sqrt 6=sqrt 2(1+sqrt 3)=sqrt{2(1+sqrt 3)^2}=sqrt{2(4+2sqrt 3)}=2sqrt{2+sqrt 3}$$



                $$sqrt{2+sqrt 3}=frac 1{sqrt 2}sqrt{4+2sqrt 3}=frac 1{sqrt 2}sqrt{(sqrt 3+1)^2}=frac{sqrt 3+1}{sqrt 2}=frac{sqrt 6+sqrt 2}2$$






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  Hints:



                  $$sqrt 2+sqrt 6=sqrt 2(1+sqrt 3)=sqrt{2(1+sqrt 3)^2}=sqrt{2(4+2sqrt 3)}=2sqrt{2+sqrt 3}$$



                  $$sqrt{2+sqrt 3}=frac 1{sqrt 2}sqrt{4+2sqrt 3}=frac 1{sqrt 2}sqrt{(sqrt 3+1)^2}=frac{sqrt 3+1}{sqrt 2}=frac{sqrt 6+sqrt 2}2$$






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    Hints:



                    $$sqrt 2+sqrt 6=sqrt 2(1+sqrt 3)=sqrt{2(1+sqrt 3)^2}=sqrt{2(4+2sqrt 3)}=2sqrt{2+sqrt 3}$$



                    $$sqrt{2+sqrt 3}=frac 1{sqrt 2}sqrt{4+2sqrt 3}=frac 1{sqrt 2}sqrt{(sqrt 3+1)^2}=frac{sqrt 3+1}{sqrt 2}=frac{sqrt 6+sqrt 2}2$$






                    share|cite|improve this answer









                    $endgroup$



                    Hints:



                    $$sqrt 2+sqrt 6=sqrt 2(1+sqrt 3)=sqrt{2(1+sqrt 3)^2}=sqrt{2(4+2sqrt 3)}=2sqrt{2+sqrt 3}$$



                    $$sqrt{2+sqrt 3}=frac 1{sqrt 2}sqrt{4+2sqrt 3}=frac 1{sqrt 2}sqrt{(sqrt 3+1)^2}=frac{sqrt 3+1}{sqrt 2}=frac{sqrt 6+sqrt 2}2$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 17 at 7:40









                    learnerlearner

                    1,110417




                    1,110417























                        1












                        $begingroup$

                        $$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)= frac{(2√2+2√6)sqrt{2+sqrt{3}}}{3(2+√3)}=frac{8+4√3}{2+√3}=frac{4}{3}.$$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          $$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)= frac{(2√2+2√6)sqrt{2+sqrt{3}}}{3(2+√3)}=frac{8+4√3}{2+√3}=frac{4}{3}.$$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            $$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)= frac{(2√2+2√6)sqrt{2+sqrt{3}}}{3(2+√3)}=frac{8+4√3}{2+√3}=frac{4}{3}.$$






                            share|cite|improve this answer









                            $endgroup$



                            $$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)= frac{(2√2+2√6)sqrt{2+sqrt{3}}}{3(2+√3)}=frac{8+4√3}{2+√3}=frac{4}{3}.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 17 at 8:11









                            saket kumarsaket kumar

                            168113




                            168113






























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