Why is $frac{15sqrt[4]{125}}{sqrt[4]{5}}$ $15sqrt{5}$ and not $15sqrt[4]{25}$?












1












$begingroup$


I have an expression I am to simplify:



$$frac{15sqrt[4]{125}}{sqrt[4]{5}}$$



I arrived at $15sqrt[4]{25}$. My textbook tells me that the answer is in fact $15sqrt{5}$. Here is my thought process:



$frac{15sqrt[4]{125}}{sqrt[4]{5}}$ = $frac{15*sqrt[4]{5}*sqrt[4]{25}}{sqrt[4]{5}}$ =



(cancel out $sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15sqrt[4]{25}$$



Where did I go wrong and how can I arrive at $15sqrt{5}$?






algebra-precalculus







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The fourth root is the square root of the square root ...
    $endgroup$
    – Ethan Bolker
    Jan 3 at 16:54
















1












$begingroup$


I have an expression I am to simplify:



$$frac{15sqrt[4]{125}}{sqrt[4]{5}}$$



I arrived at $15sqrt[4]{25}$. My textbook tells me that the answer is in fact $15sqrt{5}$. Here is my thought process:



$frac{15sqrt[4]{125}}{sqrt[4]{5}}$ = $frac{15*sqrt[4]{5}*sqrt[4]{25}}{sqrt[4]{5}}$ =



(cancel out $sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15sqrt[4]{25}$$



Where did I go wrong and how can I arrive at $15sqrt{5}$?






algebra-precalculus







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The fourth root is the square root of the square root ...
    $endgroup$
    – Ethan Bolker
    Jan 3 at 16:54














1












1








1





$begingroup$


I have an expression I am to simplify:



$$frac{15sqrt[4]{125}}{sqrt[4]{5}}$$



I arrived at $15sqrt[4]{25}$. My textbook tells me that the answer is in fact $15sqrt{5}$. Here is my thought process:



$frac{15sqrt[4]{125}}{sqrt[4]{5}}$ = $frac{15*sqrt[4]{5}*sqrt[4]{25}}{sqrt[4]{5}}$ =



(cancel out $sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15sqrt[4]{25}$$



Where did I go wrong and how can I arrive at $15sqrt{5}$?






algebra-precalculus







share|cite|improve this question









$endgroup$




I have an expression I am to simplify:



$$frac{15sqrt[4]{125}}{sqrt[4]{5}}$$



I arrived at $15sqrt[4]{25}$. My textbook tells me that the answer is in fact $15sqrt{5}$. Here is my thought process:



$frac{15sqrt[4]{125}}{sqrt[4]{5}}$ = $frac{15*sqrt[4]{5}*sqrt[4]{25}}{sqrt[4]{5}}$ =



(cancel out $sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15sqrt[4]{25}$$



Where did I go wrong and how can I arrive at $15sqrt{5}$?






algebra-precalculus




algebra-precalculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 3 at 16:52









Doug FirDoug Fir

3227




3227








  • 2




    $begingroup$
    The fourth root is the square root of the square root ...
    $endgroup$
    – Ethan Bolker
    Jan 3 at 16:54














  • 2




    $begingroup$
    The fourth root is the square root of the square root ...
    $endgroup$
    – Ethan Bolker
    Jan 3 at 16:54








2




2




$begingroup$
The fourth root is the square root of the square root ...
$endgroup$
– Ethan Bolker
Jan 3 at 16:54




$begingroup$
The fourth root is the square root of the square root ...
$endgroup$
– Ethan Bolker
Jan 3 at 16:54










3 Answers
3






active

oldest

votes


















1












$begingroup$

$$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
        $endgroup$
        – Doug Fir
        Jan 3 at 17:09











      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060767%2fwhy-is-frac15-sqrt4125-sqrt45-15-sqrt5-and-not-15-sqrt425%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      $$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



      Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        $$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



        Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          $$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



          Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.






          share|cite|improve this answer









          $endgroup$



          $$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



          Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 17:06









          KM101KM101

          5,9481524




          5,9481524























              2












              $begingroup$

              Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$






                  share|cite|improve this answer









                  $endgroup$



                  Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 3 at 16:54









                  Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                  74.7k42865




                  74.7k42865























                      2












                      $begingroup$

                      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                        $endgroup$
                        – Doug Fir
                        Jan 3 at 17:09
















                      2












                      $begingroup$

                      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                        $endgroup$
                        – Doug Fir
                        Jan 3 at 17:09














                      2












                      2








                      2





                      $begingroup$

                      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.






                      share|cite|improve this answer









                      $endgroup$



                      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 3 at 16:55









                      Antonios-Alexandros RobotisAntonios-Alexandros Robotis

                      9,91741640




                      9,91741640












                      • $begingroup$
                        Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                        $endgroup$
                        – Doug Fir
                        Jan 3 at 17:09


















                      • $begingroup$
                        Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                        $endgroup$
                        – Doug Fir
                        Jan 3 at 17:09
















                      $begingroup$
                      Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                      $endgroup$
                      – Doug Fir
                      Jan 3 at 17:09




                      $begingroup$
                      Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                      $endgroup$
                      – Doug Fir
                      Jan 3 at 17:09


















                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060767%2fwhy-is-frac15-sqrt4125-sqrt45-15-sqrt5-and-not-15-sqrt425%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Human spaceflight

                      Image
                      Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
                      Read more

                      Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                      Image
                      0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
                      Read more

                      張江高科駅

                      Image
                      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
                      Read more