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

156k22125227




156k22125227










asked Jan 1 at 17:53









Doug FirDoug Fir

3227




3227








  • 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











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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

    156k22125227




    156k22125227












    • $begingroup$
      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

        1,884219




        1,884219























            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

            69.2k22685




            69.2k22685












            • $begingroup$
              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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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2
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