Square roots in quadratic equation












1














$${sqrt x}+ frac{8}{sqrt x} = 6$$



I'm trying to get this to the standard quadratic form so I can factorise it, but not sure how to eliminate the square roots/work around them.
Any help would be appreciated :)






quadratics







share|cite|improve this question







New contributor




Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Substitute $t=sqrt{x}$
    – Matko
    26 mins ago










  • What do you get by squaring both sides?
    – Dietrich Burde
    25 mins ago










  • Just pay attention to the fact that when you substitute $;t:=sqrt x;$ , the solutions you take must be non-negative, as we have by agreement that $;sqrt xge0;$ .
    – DonAntonio
    10 mins ago










  • But if it was a cubic, would it be ok to have a negative answer?
    – Ma Ha
    6 mins ago
















1














$${sqrt x}+ frac{8}{sqrt x} = 6$$



I'm trying to get this to the standard quadratic form so I can factorise it, but not sure how to eliminate the square roots/work around them.
Any help would be appreciated :)






quadratics







share|cite|improve this question







New contributor




Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Substitute $t=sqrt{x}$
    – Matko
    26 mins ago










  • What do you get by squaring both sides?
    – Dietrich Burde
    25 mins ago










  • Just pay attention to the fact that when you substitute $;t:=sqrt x;$ , the solutions you take must be non-negative, as we have by agreement that $;sqrt xge0;$ .
    – DonAntonio
    10 mins ago










  • But if it was a cubic, would it be ok to have a negative answer?
    – Ma Ha
    6 mins ago














1












1








1







$${sqrt x}+ frac{8}{sqrt x} = 6$$



I'm trying to get this to the standard quadratic form so I can factorise it, but not sure how to eliminate the square roots/work around them.
Any help would be appreciated :)






quadratics







share|cite|improve this question







New contributor




Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











$${sqrt x}+ frac{8}{sqrt x} = 6$$



I'm trying to get this to the standard quadratic form so I can factorise it, but not sure how to eliminate the square roots/work around them.
Any help would be appreciated :)






quadratics




quadratics






share|cite|improve this question







New contributor




Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 27 mins ago









Ma Ha

163




163




New contributor




Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Substitute $t=sqrt{x}$
    – Matko
    26 mins ago










  • What do you get by squaring both sides?
    – Dietrich Burde
    25 mins ago










  • Just pay attention to the fact that when you substitute $;t:=sqrt x;$ , the solutions you take must be non-negative, as we have by agreement that $;sqrt xge0;$ .
    – DonAntonio
    10 mins ago










  • But if it was a cubic, would it be ok to have a negative answer?
    – Ma Ha
    6 mins ago


















  • Substitute $t=sqrt{x}$
    – Matko
    26 mins ago










  • What do you get by squaring both sides?
    – Dietrich Burde
    25 mins ago










  • Just pay attention to the fact that when you substitute $;t:=sqrt x;$ , the solutions you take must be non-negative, as we have by agreement that $;sqrt xge0;$ .
    – DonAntonio
    10 mins ago










  • But if it was a cubic, would it be ok to have a negative answer?
    – Ma Ha
    6 mins ago
















Substitute $t=sqrt{x}$
– Matko
26 mins ago




Substitute $t=sqrt{x}$
– Matko
26 mins ago












What do you get by squaring both sides?
– Dietrich Burde
25 mins ago




What do you get by squaring both sides?
– Dietrich Burde
25 mins ago












Just pay attention to the fact that when you substitute $;t:=sqrt x;$ , the solutions you take must be non-negative, as we have by agreement that $;sqrt xge0;$ .
– DonAntonio
10 mins ago




Just pay attention to the fact that when you substitute $;t:=sqrt x;$ , the solutions you take must be non-negative, as we have by agreement that $;sqrt xge0;$ .
– DonAntonio
10 mins ago












But if it was a cubic, would it be ok to have a negative answer?
– Ma Ha
6 mins ago




But if it was a cubic, would it be ok to have a negative answer?
– Ma Ha
6 mins ago










2 Answers
2






active

oldest

votes


















4














Hint: change variable $y = sqrt{x}$ .
$$y + frac{8}{y} = 6 Rightarrow y^2- 6y + 8 = 0 Rightarrow (y-2)(y-4) = 0 Rightarrow y in {2, 4} Rightarrow sqrt{x} = 2, 4Rightarrow x in {4,16}$$






share|cite|improve this answer





























    3














    Yor equation is equivalent to $$x+8=6sqrt{x}$$ with $$t=sqrt{x}$$ we get $$t^2-6t+8=0$$






    share|cite|improve this answer





















      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
      });


      }
      });






      Ma Ha is a new contributor. Be nice, and check out our Code of Conduct.










      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049387%2fsquare-roots-in-quadratic-equation%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      Hint: change variable $y = sqrt{x}$ .
      $$y + frac{8}{y} = 6 Rightarrow y^2- 6y + 8 = 0 Rightarrow (y-2)(y-4) = 0 Rightarrow y in {2, 4} Rightarrow sqrt{x} = 2, 4Rightarrow x in {4,16}$$






      share|cite|improve this answer


























        4














        Hint: change variable $y = sqrt{x}$ .
        $$y + frac{8}{y} = 6 Rightarrow y^2- 6y + 8 = 0 Rightarrow (y-2)(y-4) = 0 Rightarrow y in {2, 4} Rightarrow sqrt{x} = 2, 4Rightarrow x in {4,16}$$






        share|cite|improve this answer
























          4












          4








          4






          Hint: change variable $y = sqrt{x}$ .
          $$y + frac{8}{y} = 6 Rightarrow y^2- 6y + 8 = 0 Rightarrow (y-2)(y-4) = 0 Rightarrow y in {2, 4} Rightarrow sqrt{x} = 2, 4Rightarrow x in {4,16}$$






          share|cite|improve this answer












          Hint: change variable $y = sqrt{x}$ .
          $$y + frac{8}{y} = 6 Rightarrow y^2- 6y + 8 = 0 Rightarrow (y-2)(y-4) = 0 Rightarrow y in {2, 4} Rightarrow sqrt{x} = 2, 4Rightarrow x in {4,16}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 26 mins ago









          OmG

          2,052617




          2,052617























              3














              Yor equation is equivalent to $$x+8=6sqrt{x}$$ with $$t=sqrt{x}$$ we get $$t^2-6t+8=0$$






              share|cite|improve this answer


























                3














                Yor equation is equivalent to $$x+8=6sqrt{x}$$ with $$t=sqrt{x}$$ we get $$t^2-6t+8=0$$






                share|cite|improve this answer
























                  3












                  3








                  3






                  Yor equation is equivalent to $$x+8=6sqrt{x}$$ with $$t=sqrt{x}$$ we get $$t^2-6t+8=0$$






                  share|cite|improve this answer












                  Yor equation is equivalent to $$x+8=6sqrt{x}$$ with $$t=sqrt{x}$$ we get $$t^2-6t+8=0$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 21 mins ago









                  Dr. Sonnhard Graubner

                  72.5k32865




                  72.5k32865






















                      Ma Ha is a new contributor. Be nice, and check out our Code of Conduct.










                      draft saved

                      draft discarded


















                      Ma Ha is a new contributor. Be nice, and check out our Code of Conduct.













                      Ma Ha is a new contributor. Be nice, and check out our Code of Conduct.












                      Ma Ha is a new contributor. Be nice, and check out our Code of Conduct.
















                      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.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • 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.


                      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%2f3049387%2fsquare-roots-in-quadratic-equation%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

                      Futebolista

                      Image
                      Ronaldo primeiro jogador a ser eleito por três vezes o melhor jogador do mundo pela FIFA. Ronaldinho Gaúcho atuando pelo Barcelona onde foi duas vezes eleito o melhor jogador do mundo pela FIFA. Kaká atuando pelo Milan onde foi eleito em 2007 o melhor jogador do mundo pela FIFA. Lothar Matthäus primeiro jogador a ser eleito o melhor jogador do mundo pela FIFA em 1991 pela Inter de Milão. Neymar durante partida contra a Escócia, em 27 de março de 2011. Crianças praticando futebol. Robinho, futebolista brasileiro, em 15 de novembro de 2006. Atletas do Chelsea, da Inglaterra. A venda de camisas com nome e número dos jogadores é uma grande fonte de arrecadação. Jürgen Klinsmann atuou como jogador e foi treinador da Alemanha. Estádio de futebol, principal local onde laboram os jogadores. O jogador de futebol , ou Futebolista , é um atleta profissional de futebol, e cuja prática do desporto coletivo é fundamental. Estes trabalhadores sã...
                      Read more

                      Feedback on college project

                      Image
                      5 0 $begingroup$ I would appreciate some feedback regarding best practices on the following code for a college project. What should go in the controller and what in the views? How to attach and remove the views to the single main frame? Change the Controller after successful login? Studentenverwaltung.java package com.studentenverwaltung; import com.studentenverwaltung.controller.LoginController; import com.studentenverwaltung.model.User; import com.studentenverwaltung.view.LoginView; import javax.swing.*; import java.awt.*; class Studentenverwaltung implements Runnable { public static void main(String args) { EventQueue.invokeLater(new Studentenverwaltung()); } @Override public void run() { User user = new User(); LoginView loginView = new LoginView(u...
                      Read more

                      Albești (Vaslui)

                      Image
                      Coordenadas: 46 42 ºN 27 33 ° L Albeşti      Comuna    Região histórica Moldávia Distrito Vaslui Área  - Total 59,10 km² População (2007)  - Total 3 303     • Densidade 55,9 hab./km² Sítio www.albesti.ro/ Albeşti é uma comuna romena localizada no distrito de Vaslui, na região de Moldávia. A comuna possui uma área de 59.10 km² e sua população era de 3303 habitantes segundo o censo de 2007. [ 1 ] Referências ↑ «População em 1 de janeiro de 2009». INSSE   Distrito de Vaslui da Romênia Municípios Vaslui | Bârlad | Huşi Cidades Murgeni | Negreşti Comunas Albeşti | Alexandru Vlahuţă | Arsura | Banca | Băcani | Băceşti | Bălteni | Berezeni | Blăgeşti | Bogdana | Bogdăneşti | Bogdăniţa | Boţeşti | Buneşti-Avereşti | Ciocani | Codăeşti | Coroieşti | Costeşti | Cozmeşti | Creţeşti | Dăneşti | Deleni | Deleşti | Dimitrie Cantemir | Dodeşti | Dragomireşti...
                      Read more