Rationalizing denominator of $frac{7}{2+sqrt{3}}$. Cannot match textbook solution












2












$begingroup$


I am given this expression and asked to simplify by rationalizing the denominator:



$$frac{7}{2+sqrt{3}}$$



The solution is provided:



$14 - 7sqrt{3}$



I was able to get to this in the numerator but am left with a 4 in the denominator. Here are my steps:



$$frac{7}{2+sqrt{3}} * frac{2-sqrt{3}}{2-sqrt{3}}$$



=



$$frac{14 - 7sqrt{3}}{4}$$



Presumably I should not have a denominator here since the solution I'm given is just whats in the numerator. Presumably my numerator calculation is correct, where did I go wrong on the denominator?






algebra-precalculus rationalising-denominator







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    How did you compute $4$ in the denominator?
    $endgroup$
    – Martin R
    Jan 1 at 18:15










  • $begingroup$
    @MartinR my train of thought was 2 * 2 and then that $+sqrt{3}$ cancels out $-sqrt{3}$
    $endgroup$
    – Doug Fir
    Jan 1 at 18:18






  • 1




    $begingroup$
    Why the downvote???
    $endgroup$
    – Randall
    Jan 1 at 18:22
















2












$begingroup$


I am given this expression and asked to simplify by rationalizing the denominator:



$$frac{7}{2+sqrt{3}}$$



The solution is provided:



$14 - 7sqrt{3}$



I was able to get to this in the numerator but am left with a 4 in the denominator. Here are my steps:



$$frac{7}{2+sqrt{3}} * frac{2-sqrt{3}}{2-sqrt{3}}$$



=



$$frac{14 - 7sqrt{3}}{4}$$



Presumably I should not have a denominator here since the solution I'm given is just whats in the numerator. Presumably my numerator calculation is correct, where did I go wrong on the denominator?






algebra-precalculus rationalising-denominator







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    How did you compute $4$ in the denominator?
    $endgroup$
    – Martin R
    Jan 1 at 18:15










  • $begingroup$
    @MartinR my train of thought was 2 * 2 and then that $+sqrt{3}$ cancels out $-sqrt{3}$
    $endgroup$
    – Doug Fir
    Jan 1 at 18:18






  • 1




    $begingroup$
    Why the downvote???
    $endgroup$
    – Randall
    Jan 1 at 18:22














2












2








2





$begingroup$


I am given this expression and asked to simplify by rationalizing the denominator:



$$frac{7}{2+sqrt{3}}$$



The solution is provided:



$14 - 7sqrt{3}$



I was able to get to this in the numerator but am left with a 4 in the denominator. Here are my steps:



$$frac{7}{2+sqrt{3}} * frac{2-sqrt{3}}{2-sqrt{3}}$$



=



$$frac{14 - 7sqrt{3}}{4}$$



Presumably I should not have a denominator here since the solution I'm given is just whats in the numerator. Presumably my numerator calculation is correct, where did I go wrong on the denominator?






algebra-precalculus rationalising-denominator







share|cite|improve this question











$endgroup$




I am given this expression and asked to simplify by rationalizing the denominator:



$$frac{7}{2+sqrt{3}}$$



The solution is provided:



$14 - 7sqrt{3}$



I was able to get to this in the numerator but am left with a 4 in the denominator. Here are my steps:



$$frac{7}{2+sqrt{3}} * frac{2-sqrt{3}}{2-sqrt{3}}$$



=



$$frac{14 - 7sqrt{3}}{4}$$



Presumably I should not have a denominator here since the solution I'm given is just whats in the numerator. Presumably my numerator calculation is correct, where did I go wrong on the denominator?






algebra-precalculus rationalising-denominator




algebra-precalculus rationalising-denominator






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 18:27









José Carlos Santos

164k22131234




164k22131234










asked Jan 1 at 17:53









Doug FirDoug Fir

3828




3828








  • 1




    $begingroup$
    How did you compute $4$ in the denominator?
    $endgroup$
    – Martin R
    Jan 1 at 18:15










  • $begingroup$
    @MartinR my train of thought was 2 * 2 and then that $+sqrt{3}$ cancels out $-sqrt{3}$
    $endgroup$
    – Doug Fir
    Jan 1 at 18:18






  • 1




    $begingroup$
    Why the downvote???
    $endgroup$
    – Randall
    Jan 1 at 18:22














  • 1




    $begingroup$
    How did you compute $4$ in the denominator?
    $endgroup$
    – Martin R
    Jan 1 at 18:15










  • $begingroup$
    @MartinR my train of thought was 2 * 2 and then that $+sqrt{3}$ cancels out $-sqrt{3}$
    $endgroup$
    – Doug Fir
    Jan 1 at 18:18






  • 1




    $begingroup$
    Why the downvote???
    $endgroup$
    – Randall
    Jan 1 at 18:22








1




1




$begingroup$
How did you compute $4$ in the denominator?
$endgroup$
– Martin R
Jan 1 at 18:15




$begingroup$
How did you compute $4$ in the denominator?
$endgroup$
– Martin R
Jan 1 at 18:15












$begingroup$
@MartinR my train of thought was 2 * 2 and then that $+sqrt{3}$ cancels out $-sqrt{3}$
$endgroup$
– Doug Fir
Jan 1 at 18:18




$begingroup$
@MartinR my train of thought was 2 * 2 and then that $+sqrt{3}$ cancels out $-sqrt{3}$
$endgroup$
– Doug Fir
Jan 1 at 18:18




1




1




$begingroup$
Why the downvote???
$endgroup$
– Randall
Jan 1 at 18:22




$begingroup$
Why the downvote???
$endgroup$
– Randall
Jan 1 at 18:22










3 Answers
3






active

oldest

votes


















3












$begingroup$

Note that $left(2+sqrt3right)timesleft(2-sqrt3right)=4-3=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part?
    $endgroup$
    – Doug Fir
    Jan 1 at 17:57






  • 4




    $begingroup$
    @DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 17:58



















3












$begingroup$

note that:
$$(a+b)(c+d)=ac+ad+bc+bd$$
so in some cases it simplifies to:
$$(a+b)(a-b)=a^2-b^2$$
for you, $a=2$ and $b=sqrt{3}$ so $a^2-b^2=4-3=1$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Trick to remember forever and use again and again.



    $(a + b)(a-b) = a(a-b) + b(a-b)=$



    $a^2 - ab + ab - b^2 = a^2 - b^2$.



    So



    1) Whenever you need to factor $a^2 - b^2$ it always factor to to $(a+b)(a-b)$



    and



    2) If you ever need to get rid of a radical sign in $a+sqrt b$ you can always multiple by $(a + sqrt b)(a - sqrt b) = a^2 - sqrt b^2 = a^2 -b$.



    So



    3) to deradicalize a $frac m{sqrt a + sqrt b} = frac {m(sqrt a - sqrt b)}{(sqrt a - sqrt b)} = frac {m(sqrt a - sqrt b)}{a - b}$.



    So:



    $frac {7}{2 + sqrt 3} = $



    $frac {7(2 - sqrt 3)}{(2+sqrt 3)(2 - sqrt 3)} =$



    $frac {7(2-sqrt 3)}{2^2 - sqrt 3^2} =$



    $frac {7(2-sqrt 3)}{4-3} =$



    $frac {7(2-sqrt 3)}{1} =$



    $7(2-sqrt 3)$.



    Learn to recognize $a^2 -b^2 = (a+b)(a-b)$ in all its forms, for all its uses and in all its directions.



    You will be using it for the REST OF YOUR LIFE!






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for the tip!
      $endgroup$
      – Doug Fir
      Jan 1 at 20:32











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Note that $left(2+sqrt3right)timesleft(2-sqrt3right)=4-3=1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part?
      $endgroup$
      – Doug Fir
      Jan 1 at 17:57






    • 4




      $begingroup$
      @DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$.
      $endgroup$
      – Lord Shark the Unknown
      Jan 1 at 17:58
















    3












    $begingroup$

    Note that $left(2+sqrt3right)timesleft(2-sqrt3right)=4-3=1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part?
      $endgroup$
      – Doug Fir
      Jan 1 at 17:57






    • 4




      $begingroup$
      @DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$.
      $endgroup$
      – Lord Shark the Unknown
      Jan 1 at 17:58














    3












    3








    3





    $begingroup$

    Note that $left(2+sqrt3right)timesleft(2-sqrt3right)=4-3=1$.






    share|cite|improve this answer









    $endgroup$



    Note that $left(2+sqrt3right)timesleft(2-sqrt3right)=4-3=1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 1 at 17:54









    José Carlos SantosJosé Carlos Santos

    164k22131234




    164k22131234












    • $begingroup$
      Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part?
      $endgroup$
      – Doug Fir
      Jan 1 at 17:57






    • 4




      $begingroup$
      @DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$.
      $endgroup$
      – Lord Shark the Unknown
      Jan 1 at 17:58


















    • $begingroup$
      Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part?
      $endgroup$
      – Doug Fir
      Jan 1 at 17:57






    • 4




      $begingroup$
      @DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$.
      $endgroup$
      – Lord Shark the Unknown
      Jan 1 at 17:58
















    $begingroup$
    Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part?
    $endgroup$
    – Doug Fir
    Jan 1 at 17:57




    $begingroup$
    Hi, thanks for answering and sorry if my follow up sounds very low level but - why is it -3? I see how you get 4, from the product of both 2's in the denominator of both fractions being multiplied, but how do you get the -3 part?
    $endgroup$
    – Doug Fir
    Jan 1 at 17:57




    4




    4




    $begingroup$
    @DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 17:58




    $begingroup$
    @DougFir Difference of Two Squares: $a^2-b^2=(a+b)(a-b)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 1 at 17:58











    3












    $begingroup$

    note that:
    $$(a+b)(c+d)=ac+ad+bc+bd$$
    so in some cases it simplifies to:
    $$(a+b)(a-b)=a^2-b^2$$
    for you, $a=2$ and $b=sqrt{3}$ so $a^2-b^2=4-3=1$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      note that:
      $$(a+b)(c+d)=ac+ad+bc+bd$$
      so in some cases it simplifies to:
      $$(a+b)(a-b)=a^2-b^2$$
      for you, $a=2$ and $b=sqrt{3}$ so $a^2-b^2=4-3=1$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        note that:
        $$(a+b)(c+d)=ac+ad+bc+bd$$
        so in some cases it simplifies to:
        $$(a+b)(a-b)=a^2-b^2$$
        for you, $a=2$ and $b=sqrt{3}$ so $a^2-b^2=4-3=1$






        share|cite|improve this answer









        $endgroup$



        note that:
        $$(a+b)(c+d)=ac+ad+bc+bd$$
        so in some cases it simplifies to:
        $$(a+b)(a-b)=a^2-b^2$$
        for you, $a=2$ and $b=sqrt{3}$ so $a^2-b^2=4-3=1$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 18:31









        Henry LeeHenry Lee

        2,054219




        2,054219























            3












            $begingroup$

            Trick to remember forever and use again and again.



            $(a + b)(a-b) = a(a-b) + b(a-b)=$



            $a^2 - ab + ab - b^2 = a^2 - b^2$.



            So



            1) Whenever you need to factor $a^2 - b^2$ it always factor to to $(a+b)(a-b)$



            and



            2) If you ever need to get rid of a radical sign in $a+sqrt b$ you can always multiple by $(a + sqrt b)(a - sqrt b) = a^2 - sqrt b^2 = a^2 -b$.



            So



            3) to deradicalize a $frac m{sqrt a + sqrt b} = frac {m(sqrt a - sqrt b)}{(sqrt a - sqrt b)} = frac {m(sqrt a - sqrt b)}{a - b}$.



            So:



            $frac {7}{2 + sqrt 3} = $



            $frac {7(2 - sqrt 3)}{(2+sqrt 3)(2 - sqrt 3)} =$



            $frac {7(2-sqrt 3)}{2^2 - sqrt 3^2} =$



            $frac {7(2-sqrt 3)}{4-3} =$



            $frac {7(2-sqrt 3)}{1} =$



            $7(2-sqrt 3)$.



            Learn to recognize $a^2 -b^2 = (a+b)(a-b)$ in all its forms, for all its uses and in all its directions.



            You will be using it for the REST OF YOUR LIFE!






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you for the tip!
              $endgroup$
              – Doug Fir
              Jan 1 at 20:32
















            3












            $begingroup$

            Trick to remember forever and use again and again.



            $(a + b)(a-b) = a(a-b) + b(a-b)=$



            $a^2 - ab + ab - b^2 = a^2 - b^2$.



            So



            1) Whenever you need to factor $a^2 - b^2$ it always factor to to $(a+b)(a-b)$



            and



            2) If you ever need to get rid of a radical sign in $a+sqrt b$ you can always multiple by $(a + sqrt b)(a - sqrt b) = a^2 - sqrt b^2 = a^2 -b$.



            So



            3) to deradicalize a $frac m{sqrt a + sqrt b} = frac {m(sqrt a - sqrt b)}{(sqrt a - sqrt b)} = frac {m(sqrt a - sqrt b)}{a - b}$.



            So:



            $frac {7}{2 + sqrt 3} = $



            $frac {7(2 - sqrt 3)}{(2+sqrt 3)(2 - sqrt 3)} =$



            $frac {7(2-sqrt 3)}{2^2 - sqrt 3^2} =$



            $frac {7(2-sqrt 3)}{4-3} =$



            $frac {7(2-sqrt 3)}{1} =$



            $7(2-sqrt 3)$.



            Learn to recognize $a^2 -b^2 = (a+b)(a-b)$ in all its forms, for all its uses and in all its directions.



            You will be using it for the REST OF YOUR LIFE!






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you for the tip!
              $endgroup$
              – Doug Fir
              Jan 1 at 20:32














            3












            3








            3





            $begingroup$

            Trick to remember forever and use again and again.



            $(a + b)(a-b) = a(a-b) + b(a-b)=$



            $a^2 - ab + ab - b^2 = a^2 - b^2$.



            So



            1) Whenever you need to factor $a^2 - b^2$ it always factor to to $(a+b)(a-b)$



            and



            2) If you ever need to get rid of a radical sign in $a+sqrt b$ you can always multiple by $(a + sqrt b)(a - sqrt b) = a^2 - sqrt b^2 = a^2 -b$.



            So



            3) to deradicalize a $frac m{sqrt a + sqrt b} = frac {m(sqrt a - sqrt b)}{(sqrt a - sqrt b)} = frac {m(sqrt a - sqrt b)}{a - b}$.



            So:



            $frac {7}{2 + sqrt 3} = $



            $frac {7(2 - sqrt 3)}{(2+sqrt 3)(2 - sqrt 3)} =$



            $frac {7(2-sqrt 3)}{2^2 - sqrt 3^2} =$



            $frac {7(2-sqrt 3)}{4-3} =$



            $frac {7(2-sqrt 3)}{1} =$



            $7(2-sqrt 3)$.



            Learn to recognize $a^2 -b^2 = (a+b)(a-b)$ in all its forms, for all its uses and in all its directions.



            You will be using it for the REST OF YOUR LIFE!






            share|cite|improve this answer









            $endgroup$



            Trick to remember forever and use again and again.



            $(a + b)(a-b) = a(a-b) + b(a-b)=$



            $a^2 - ab + ab - b^2 = a^2 - b^2$.



            So



            1) Whenever you need to factor $a^2 - b^2$ it always factor to to $(a+b)(a-b)$



            and



            2) If you ever need to get rid of a radical sign in $a+sqrt b$ you can always multiple by $(a + sqrt b)(a - sqrt b) = a^2 - sqrt b^2 = a^2 -b$.



            So



            3) to deradicalize a $frac m{sqrt a + sqrt b} = frac {m(sqrt a - sqrt b)}{(sqrt a - sqrt b)} = frac {m(sqrt a - sqrt b)}{a - b}$.



            So:



            $frac {7}{2 + sqrt 3} = $



            $frac {7(2 - sqrt 3)}{(2+sqrt 3)(2 - sqrt 3)} =$



            $frac {7(2-sqrt 3)}{2^2 - sqrt 3^2} =$



            $frac {7(2-sqrt 3)}{4-3} =$



            $frac {7(2-sqrt 3)}{1} =$



            $7(2-sqrt 3)$.



            Learn to recognize $a^2 -b^2 = (a+b)(a-b)$ in all its forms, for all its uses and in all its directions.



            You will be using it for the REST OF YOUR LIFE!







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 1 at 18:45









            fleabloodfleablood

            71.6k22686




            71.6k22686












            • $begingroup$
              Thank you for the tip!
              $endgroup$
              – Doug Fir
              Jan 1 at 20:32


















            • $begingroup$
              Thank you for the tip!
              $endgroup$
              – Doug Fir
              Jan 1 at 20:32
















            $begingroup$
            Thank you for the tip!
            $endgroup$
            – Doug Fir
            Jan 1 at 20:32




            $begingroup$
            Thank you for the tip!
            $endgroup$
            – Doug Fir
            Jan 1 at 20:32


















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            Mont Emei

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            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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