Trying to simplify $frac{sqrt{8}}{1-sqrt{3x}}$ to be $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$












3














I am asked to simplify $frac{sqrt{8}}{1-sqrt{3x}}$. The solution is provided as $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$



Here is my working:
$frac{sqrt{8}}{1-sqrt{3x}}$ = $frac{sqrt{8}}{1-sqrt{3x}}$ * $frac{1+sqrt{3x}}{1+sqrt{3x}}$ = $frac{1+sqrt{8}sqrt{3x}}{1-3x}$ = $frac{1+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x}$ = $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$



Is $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$?






algebra-precalculus







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  • Please review your computations. There is a least 2 errors.
    – mathcounterexamples.net
    2 days ago
















3














I am asked to simplify $frac{sqrt{8}}{1-sqrt{3x}}$. The solution is provided as $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$



Here is my working:
$frac{sqrt{8}}{1-sqrt{3x}}$ = $frac{sqrt{8}}{1-sqrt{3x}}$ * $frac{1+sqrt{3x}}{1+sqrt{3x}}$ = $frac{1+sqrt{8}sqrt{3x}}{1-3x}$ = $frac{1+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x}$ = $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$



Is $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$?






algebra-precalculus







share|cite|improve this question
























  • Please review your computations. There is a least 2 errors.
    – mathcounterexamples.net
    2 days ago














3












3








3







I am asked to simplify $frac{sqrt{8}}{1-sqrt{3x}}$. The solution is provided as $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$



Here is my working:
$frac{sqrt{8}}{1-sqrt{3x}}$ = $frac{sqrt{8}}{1-sqrt{3x}}$ * $frac{1+sqrt{3x}}{1+sqrt{3x}}$ = $frac{1+sqrt{8}sqrt{3x}}{1-3x}$ = $frac{1+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x}$ = $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$



Is $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$?






algebra-precalculus







share|cite|improve this question















I am asked to simplify $frac{sqrt{8}}{1-sqrt{3x}}$. The solution is provided as $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$



Here is my working:
$frac{sqrt{8}}{1-sqrt{3x}}$ = $frac{sqrt{8}}{1-sqrt{3x}}$ * $frac{1+sqrt{3x}}{1+sqrt{3x}}$ = $frac{1+sqrt{8}sqrt{3x}}{1-3x}$ = $frac{1+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x}$ = $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$



Is $frac{1+2sqrt{2}sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $frac{2sqrt{2}+2sqrt{6x}}{1-3x}$?






algebra-precalculus




algebra-precalculus






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edited 2 days ago







Doug Fir

















asked 2 days ago









Doug FirDoug Fir

2967




2967












  • Please review your computations. There is a least 2 errors.
    – mathcounterexamples.net
    2 days ago


















  • Please review your computations. There is a least 2 errors.
    – mathcounterexamples.net
    2 days ago
















Please review your computations. There is a least 2 errors.
– mathcounterexamples.net
2 days ago




Please review your computations. There is a least 2 errors.
– mathcounterexamples.net
2 days ago










7 Answers
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4














$$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = color{blue}{frac{sqrt{8}cdotleft({1+sqrt{3x}}right)}{1-3x}} = frac{sqrt{8}+sqrt{24x}}{1-3x}$$



$$= frac{sqrt{2^3}+sqrt{2^3cdot3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$



Notice the step I highlighted in blue, which is where you made an error. You have to multiply $sqrt{8}$ to $left(1+sqrt{3x}right)$ completely. You multiplied it by only $sqrt{3x}$ and added $1$, which wasn’t correct.






share|cite|improve this answer





























    2














    Observe that
    $$
    sqrt{8}times (1+sqrt{3x})=sqrt{8}+sqrt{8}times sqrt{3x}
    $$     and
    $$
    sqrt{8}times sqrt{3}=sqrt{24}=sqrt{4times 6}=2sqrt{6}.
    $$






    share|cite|improve this answer





























      2














      Well, in rationalizing the denominator, we arrive at the intermediate step $frac{sqrt{8}+sqrt{24x}}{1-3x}$ which simplifies to the provided solution.



      Your error comes in multiplying by the conjugate of the denominator. $sqrt{8}*(1+sqrt{3x})$ becomes $sqrt{8}+sqrt{24x}$.






      share|cite|improve this answer





























        1














        Write $$frac{sqrt{8}(1+sqrt{3x})}{(1-sqrt{3x})(1+sqrt{3x})}$$ and this is
        $$frac{2sqrt{2}(1+sqrt{3x})}{1-3x}$$






        share|cite|improve this answer





























          1














          Hints:$sqrt8=sqrt{4×2}=2sqrt{2}.$



          $sqrt24=2sqrt6$. (Why?)



          $(1+sqrt{3x})×(1-sqrt{3x})=1-3x $.






          share|cite|improve this answer





























            1














            You just made some mistakes in your arithmetic.
            $$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = frac{sqrt{8}+sqrt{8}sqrt{3x}}{1-3x} = frac{sqrt{2}sqrt{2}sqrt{2}+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$






            share|cite|improve this answer





























              1














              The second equality is where you mess up - note that
              $$sqrt{8}cdot(1+sqrt{3x})=sqrt{8}+sqrt{8}cdotsqrt{3x}$$
              by the distributive property of real numbers






              share|cite|improve this answer























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                7 Answers
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                7 Answers
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                4














                $$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = color{blue}{frac{sqrt{8}cdotleft({1+sqrt{3x}}right)}{1-3x}} = frac{sqrt{8}+sqrt{24x}}{1-3x}$$



                $$= frac{sqrt{2^3}+sqrt{2^3cdot3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$



                Notice the step I highlighted in blue, which is where you made an error. You have to multiply $sqrt{8}$ to $left(1+sqrt{3x}right)$ completely. You multiplied it by only $sqrt{3x}$ and added $1$, which wasn’t correct.






                share|cite|improve this answer


























                  4














                  $$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = color{blue}{frac{sqrt{8}cdotleft({1+sqrt{3x}}right)}{1-3x}} = frac{sqrt{8}+sqrt{24x}}{1-3x}$$



                  $$= frac{sqrt{2^3}+sqrt{2^3cdot3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$



                  Notice the step I highlighted in blue, which is where you made an error. You have to multiply $sqrt{8}$ to $left(1+sqrt{3x}right)$ completely. You multiplied it by only $sqrt{3x}$ and added $1$, which wasn’t correct.






                  share|cite|improve this answer
























                    4












                    4








                    4






                    $$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = color{blue}{frac{sqrt{8}cdotleft({1+sqrt{3x}}right)}{1-3x}} = frac{sqrt{8}+sqrt{24x}}{1-3x}$$



                    $$= frac{sqrt{2^3}+sqrt{2^3cdot3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$



                    Notice the step I highlighted in blue, which is where you made an error. You have to multiply $sqrt{8}$ to $left(1+sqrt{3x}right)$ completely. You multiplied it by only $sqrt{3x}$ and added $1$, which wasn’t correct.






                    share|cite|improve this answer












                    $$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = color{blue}{frac{sqrt{8}cdotleft({1+sqrt{3x}}right)}{1-3x}} = frac{sqrt{8}+sqrt{24x}}{1-3x}$$



                    $$= frac{sqrt{2^3}+sqrt{2^3cdot3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$



                    Notice the step I highlighted in blue, which is where you made an error. You have to multiply $sqrt{8}$ to $left(1+sqrt{3x}right)$ completely. You multiplied it by only $sqrt{3x}$ and added $1$, which wasn’t correct.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    KM101KM101

                    5,5611423




                    5,5611423























                        2














                        Observe that
                        $$
                        sqrt{8}times (1+sqrt{3x})=sqrt{8}+sqrt{8}times sqrt{3x}
                        $$                         and
                        $$
                        sqrt{8}times sqrt{3}=sqrt{24}=sqrt{4times 6}=2sqrt{6}.
                        $$






                        share|cite|improve this answer


























                          2














                          Observe that
                          $$
                          sqrt{8}times (1+sqrt{3x})=sqrt{8}+sqrt{8}times sqrt{3x}
                          $$                           and
                          $$
                          sqrt{8}times sqrt{3}=sqrt{24}=sqrt{4times 6}=2sqrt{6}.
                          $$






                          share|cite|improve this answer
























                            2












                            2








                            2






                            Observe that
                            $$
                            sqrt{8}times (1+sqrt{3x})=sqrt{8}+sqrt{8}times sqrt{3x}
                            $$                             and
                            $$
                            sqrt{8}times sqrt{3}=sqrt{24}=sqrt{4times 6}=2sqrt{6}.
                            $$






                            share|cite|improve this answer












                            Observe that
                            $$
                            sqrt{8}times (1+sqrt{3x})=sqrt{8}+sqrt{8}times sqrt{3x}
                            $$                             and
                            $$
                            sqrt{8}times sqrt{3}=sqrt{24}=sqrt{4times 6}=2sqrt{6}.
                            $$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 days ago









                            Olivier OloaOlivier Oloa

                            108k17176293




                            108k17176293























                                2














                                Well, in rationalizing the denominator, we arrive at the intermediate step $frac{sqrt{8}+sqrt{24x}}{1-3x}$ which simplifies to the provided solution.



                                Your error comes in multiplying by the conjugate of the denominator. $sqrt{8}*(1+sqrt{3x})$ becomes $sqrt{8}+sqrt{24x}$.






                                share|cite|improve this answer


























                                  2














                                  Well, in rationalizing the denominator, we arrive at the intermediate step $frac{sqrt{8}+sqrt{24x}}{1-3x}$ which simplifies to the provided solution.



                                  Your error comes in multiplying by the conjugate of the denominator. $sqrt{8}*(1+sqrt{3x})$ becomes $sqrt{8}+sqrt{24x}$.






                                  share|cite|improve this answer
























                                    2












                                    2








                                    2






                                    Well, in rationalizing the denominator, we arrive at the intermediate step $frac{sqrt{8}+sqrt{24x}}{1-3x}$ which simplifies to the provided solution.



                                    Your error comes in multiplying by the conjugate of the denominator. $sqrt{8}*(1+sqrt{3x})$ becomes $sqrt{8}+sqrt{24x}$.






                                    share|cite|improve this answer












                                    Well, in rationalizing the denominator, we arrive at the intermediate step $frac{sqrt{8}+sqrt{24x}}{1-3x}$ which simplifies to the provided solution.



                                    Your error comes in multiplying by the conjugate of the denominator. $sqrt{8}*(1+sqrt{3x})$ becomes $sqrt{8}+sqrt{24x}$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 2 days ago









                                    GnumbertesterGnumbertester

                                    1285




                                    1285























                                        1














                                        Write $$frac{sqrt{8}(1+sqrt{3x})}{(1-sqrt{3x})(1+sqrt{3x})}$$ and this is
                                        $$frac{2sqrt{2}(1+sqrt{3x})}{1-3x}$$






                                        share|cite|improve this answer


























                                          1














                                          Write $$frac{sqrt{8}(1+sqrt{3x})}{(1-sqrt{3x})(1+sqrt{3x})}$$ and this is
                                          $$frac{2sqrt{2}(1+sqrt{3x})}{1-3x}$$






                                          share|cite|improve this answer
























                                            1












                                            1








                                            1






                                            Write $$frac{sqrt{8}(1+sqrt{3x})}{(1-sqrt{3x})(1+sqrt{3x})}$$ and this is
                                            $$frac{2sqrt{2}(1+sqrt{3x})}{1-3x}$$






                                            share|cite|improve this answer












                                            Write $$frac{sqrt{8}(1+sqrt{3x})}{(1-sqrt{3x})(1+sqrt{3x})}$$ and this is
                                            $$frac{2sqrt{2}(1+sqrt{3x})}{1-3x}$$







                                            share|cite|improve this answer












                                            share|cite|improve this answer



                                            share|cite|improve this answer










                                            answered 2 days ago









                                            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                                            73.5k42865




                                            73.5k42865























                                                1














                                                Hints:$sqrt8=sqrt{4×2}=2sqrt{2}.$



                                                $sqrt24=2sqrt6$. (Why?)



                                                $(1+sqrt{3x})×(1-sqrt{3x})=1-3x $.






                                                share|cite|improve this answer


























                                                  1














                                                  Hints:$sqrt8=sqrt{4×2}=2sqrt{2}.$



                                                  $sqrt24=2sqrt6$. (Why?)



                                                  $(1+sqrt{3x})×(1-sqrt{3x})=1-3x $.






                                                  share|cite|improve this answer
























                                                    1












                                                    1








                                                    1






                                                    Hints:$sqrt8=sqrt{4×2}=2sqrt{2}.$



                                                    $sqrt24=2sqrt6$. (Why?)



                                                    $(1+sqrt{3x})×(1-sqrt{3x})=1-3x $.






                                                    share|cite|improve this answer












                                                    Hints:$sqrt8=sqrt{4×2}=2sqrt{2}.$



                                                    $sqrt24=2sqrt6$. (Why?)



                                                    $(1+sqrt{3x})×(1-sqrt{3x})=1-3x $.







                                                    share|cite|improve this answer












                                                    share|cite|improve this answer



                                                    share|cite|improve this answer










                                                    answered 2 days ago









                                                    Thomas ShelbyThomas Shelby

                                                    1,907217




                                                    1,907217























                                                        1














                                                        You just made some mistakes in your arithmetic.
                                                        $$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = frac{sqrt{8}+sqrt{8}sqrt{3x}}{1-3x} = frac{sqrt{2}sqrt{2}sqrt{2}+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$






                                                        share|cite|improve this answer


























                                                          1














                                                          You just made some mistakes in your arithmetic.
                                                          $$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = frac{sqrt{8}+sqrt{8}sqrt{3x}}{1-3x} = frac{sqrt{2}sqrt{2}sqrt{2}+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$






                                                          share|cite|improve this answer
























                                                            1












                                                            1








                                                            1






                                                            You just made some mistakes in your arithmetic.
                                                            $$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = frac{sqrt{8}+sqrt{8}sqrt{3x}}{1-3x} = frac{sqrt{2}sqrt{2}sqrt{2}+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$






                                                            share|cite|improve this answer












                                                            You just made some mistakes in your arithmetic.
                                                            $$frac{sqrt{8}}{1-sqrt{3x}} = frac{sqrt{8}}{1-sqrt{3x}} cdot frac{1+sqrt{3x}}{1+sqrt{3x}} = frac{sqrt{8}+sqrt{8}sqrt{3x}}{1-3x} = frac{sqrt{2}sqrt{2}sqrt{2}+sqrt{2}sqrt{2}sqrt{2}sqrt{3x}}{1-3x} = frac{2sqrt{2}+2sqrt{6x}}{1-3x}$$







                                                            share|cite|improve this answer












                                                            share|cite|improve this answer



                                                            share|cite|improve this answer










                                                            answered 2 days ago









                                                            greeliousgreelious

                                                            15510




                                                            15510























                                                                1














                                                                The second equality is where you mess up - note that
                                                                $$sqrt{8}cdot(1+sqrt{3x})=sqrt{8}+sqrt{8}cdotsqrt{3x}$$
                                                                by the distributive property of real numbers






                                                                share|cite|improve this answer




























                                                                  1














                                                                  The second equality is where you mess up - note that
                                                                  $$sqrt{8}cdot(1+sqrt{3x})=sqrt{8}+sqrt{8}cdotsqrt{3x}$$
                                                                  by the distributive property of real numbers






                                                                  share|cite|improve this answer


























                                                                    1












                                                                    1








                                                                    1






                                                                    The second equality is where you mess up - note that
                                                                    $$sqrt{8}cdot(1+sqrt{3x})=sqrt{8}+sqrt{8}cdotsqrt{3x}$$
                                                                    by the distributive property of real numbers






                                                                    share|cite|improve this answer














                                                                    The second equality is where you mess up - note that
                                                                    $$sqrt{8}cdot(1+sqrt{3x})=sqrt{8}+sqrt{8}cdotsqrt{3x}$$
                                                                    by the distributive property of real numbers







                                                                    share|cite|improve this answer














                                                                    share|cite|improve this answer



                                                                    share|cite|improve this answer








                                                                    edited 2 days ago









                                                                    KM101

                                                                    5,5611423




                                                                    5,5611423










                                                                    answered 2 days ago









                                                                    dromastyxdromastyx

                                                                    2,2901517




                                                                    2,2901517






























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