What are the possible solutions of the given equation?












2












$begingroup$


I encountered a question in an exam in which we had:




Find all possible solutions of the equation $$x+y+ {1over x}+{1over y}+4=2 (sqrt {2x+1}+sqrt {2y+1}) $$ where $x $ and $y$ are real numbers.




I tried squaring both sides to eliminate the square roots but the number of terms became too many, making the problem very difficult to handle. I am not really able to understand how to find an easier approach or handle the terms efficiently. Would someone please help me to solve this question?






algebra-precalculus







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I encountered a question in an exam in which we had:




    Find all possible solutions of the equation $$x+y+ {1over x}+{1over y}+4=2 (sqrt {2x+1}+sqrt {2y+1}) $$ where $x $ and $y$ are real numbers.




    I tried squaring both sides to eliminate the square roots but the number of terms became too many, making the problem very difficult to handle. I am not really able to understand how to find an easier approach or handle the terms efficiently. Would someone please help me to solve this question?






    algebra-precalculus







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I encountered a question in an exam in which we had:




      Find all possible solutions of the equation $$x+y+ {1over x}+{1over y}+4=2 (sqrt {2x+1}+sqrt {2y+1}) $$ where $x $ and $y$ are real numbers.




      I tried squaring both sides to eliminate the square roots but the number of terms became too many, making the problem very difficult to handle. I am not really able to understand how to find an easier approach or handle the terms efficiently. Would someone please help me to solve this question?






      algebra-precalculus







      share|cite|improve this question











      $endgroup$




      I encountered a question in an exam in which we had:




      Find all possible solutions of the equation $$x+y+ {1over x}+{1over y}+4=2 (sqrt {2x+1}+sqrt {2y+1}) $$ where $x $ and $y$ are real numbers.




      I tried squaring both sides to eliminate the square roots but the number of terms became too many, making the problem very difficult to handle. I am not really able to understand how to find an easier approach or handle the terms efficiently. Would someone please help me to solve this question?






      algebra-precalculus




      algebra-precalculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 3 hours ago









      Thomas Andrews

      130k12147298




      130k12147298










      asked 3 hours ago









      Shashwat1337Shashwat1337

      889




      889






















          2 Answers
          2






          active

          oldest

          votes


















          7












          $begingroup$

          $$x+y+ {1over x}+{1over y}+4 = x+y+{2x+1over x}+{2y+1over y} $$



          By Am-Gm we have $$ x+{2x+1over x}geq 2sqrt{x{2x+1over x}} = 2sqrt{2x+1}$$ and the same for $y$, so we have



          $$x+y+ {1over x}+{1over y}+4 geq 2sqrt{{2x+1}}+2sqrt{{2y+1}}$$



          Since we have equality is achieved when $x={2x+1over x}$ (and the same for $y$) we have $x=y=1+sqrt{2}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Truly amazing!!! [+1]
            $endgroup$
            – Dr. Mathva
            3 hours ago










          • $begingroup$
            @Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
            $endgroup$
            – Michael Rozenberg
            3 hours ago












          • $begingroup$
            AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
            $endgroup$
            – Anurag A
            3 hours ago





















          2












          $begingroup$

          It's $$sum_{cyc}left(x+frac{1}{x}+2-2sqrt{2x+1}right)=0$$ or
          $$sum_{cyc}frac{x^2-2xsqrt{2x+1}+2x+1}{x}=0$$ or
          $$sum_{cyc}frac{(x-sqrt{2x+1})^2}{x}=0,$$
          which for $xy<0$ gives infinitely many solutions.



          But, for $xy>0$ we obtain:
          $$x=sqrt{2x+1}$$ and $$y=sqrt{2y+1},$$ which gives
          $$x=y=1+sqrt2.$$






          share|cite|improve this answer









          $endgroup$













            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%2f3148348%2fwhat-are-the-possible-solutions-of-the-given-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









            7












            $begingroup$

            $$x+y+ {1over x}+{1over y}+4 = x+y+{2x+1over x}+{2y+1over y} $$



            By Am-Gm we have $$ x+{2x+1over x}geq 2sqrt{x{2x+1over x}} = 2sqrt{2x+1}$$ and the same for $y$, so we have



            $$x+y+ {1over x}+{1over y}+4 geq 2sqrt{{2x+1}}+2sqrt{{2y+1}}$$



            Since we have equality is achieved when $x={2x+1over x}$ (and the same for $y$) we have $x=y=1+sqrt{2}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Truly amazing!!! [+1]
              $endgroup$
              – Dr. Mathva
              3 hours ago










            • $begingroup$
              @Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
              $endgroup$
              – Michael Rozenberg
              3 hours ago












            • $begingroup$
              AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
              $endgroup$
              – Anurag A
              3 hours ago


















            7












            $begingroup$

            $$x+y+ {1over x}+{1over y}+4 = x+y+{2x+1over x}+{2y+1over y} $$



            By Am-Gm we have $$ x+{2x+1over x}geq 2sqrt{x{2x+1over x}} = 2sqrt{2x+1}$$ and the same for $y$, so we have



            $$x+y+ {1over x}+{1over y}+4 geq 2sqrt{{2x+1}}+2sqrt{{2y+1}}$$



            Since we have equality is achieved when $x={2x+1over x}$ (and the same for $y$) we have $x=y=1+sqrt{2}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Truly amazing!!! [+1]
              $endgroup$
              – Dr. Mathva
              3 hours ago










            • $begingroup$
              @Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
              $endgroup$
              – Michael Rozenberg
              3 hours ago












            • $begingroup$
              AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
              $endgroup$
              – Anurag A
              3 hours ago
















            7












            7








            7





            $begingroup$

            $$x+y+ {1over x}+{1over y}+4 = x+y+{2x+1over x}+{2y+1over y} $$



            By Am-Gm we have $$ x+{2x+1over x}geq 2sqrt{x{2x+1over x}} = 2sqrt{2x+1}$$ and the same for $y$, so we have



            $$x+y+ {1over x}+{1over y}+4 geq 2sqrt{{2x+1}}+2sqrt{{2y+1}}$$



            Since we have equality is achieved when $x={2x+1over x}$ (and the same for $y$) we have $x=y=1+sqrt{2}$






            share|cite|improve this answer











            $endgroup$



            $$x+y+ {1over x}+{1over y}+4 = x+y+{2x+1over x}+{2y+1over y} $$



            By Am-Gm we have $$ x+{2x+1over x}geq 2sqrt{x{2x+1over x}} = 2sqrt{2x+1}$$ and the same for $y$, so we have



            $$x+y+ {1over x}+{1over y}+4 geq 2sqrt{{2x+1}}+2sqrt{{2y+1}}$$



            Since we have equality is achieved when $x={2x+1over x}$ (and the same for $y$) we have $x=y=1+sqrt{2}$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 3 hours ago

























            answered 3 hours ago









            Maria MazurMaria Mazur

            46.9k1260120




            46.9k1260120












            • $begingroup$
              Truly amazing!!! [+1]
              $endgroup$
              – Dr. Mathva
              3 hours ago










            • $begingroup$
              @Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
              $endgroup$
              – Michael Rozenberg
              3 hours ago












            • $begingroup$
              AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
              $endgroup$
              – Anurag A
              3 hours ago




















            • $begingroup$
              Truly amazing!!! [+1]
              $endgroup$
              – Dr. Mathva
              3 hours ago










            • $begingroup$
              @Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
              $endgroup$
              – Michael Rozenberg
              3 hours ago












            • $begingroup$
              AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
              $endgroup$
              – Anurag A
              3 hours ago


















            $begingroup$
            Truly amazing!!! [+1]
            $endgroup$
            – Dr. Mathva
            3 hours ago




            $begingroup$
            Truly amazing!!! [+1]
            $endgroup$
            – Dr. Mathva
            3 hours ago












            $begingroup$
            @Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
            $endgroup$
            – Michael Rozenberg
            3 hours ago






            $begingroup$
            @Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
            $endgroup$
            – Michael Rozenberg
            3 hours ago














            $begingroup$
            AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
            $endgroup$
            – Anurag A
            3 hours ago






            $begingroup$
            AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
            $endgroup$
            – Anurag A
            3 hours ago













            2












            $begingroup$

            It's $$sum_{cyc}left(x+frac{1}{x}+2-2sqrt{2x+1}right)=0$$ or
            $$sum_{cyc}frac{x^2-2xsqrt{2x+1}+2x+1}{x}=0$$ or
            $$sum_{cyc}frac{(x-sqrt{2x+1})^2}{x}=0,$$
            which for $xy<0$ gives infinitely many solutions.



            But, for $xy>0$ we obtain:
            $$x=sqrt{2x+1}$$ and $$y=sqrt{2y+1},$$ which gives
            $$x=y=1+sqrt2.$$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              It's $$sum_{cyc}left(x+frac{1}{x}+2-2sqrt{2x+1}right)=0$$ or
              $$sum_{cyc}frac{x^2-2xsqrt{2x+1}+2x+1}{x}=0$$ or
              $$sum_{cyc}frac{(x-sqrt{2x+1})^2}{x}=0,$$
              which for $xy<0$ gives infinitely many solutions.



              But, for $xy>0$ we obtain:
              $$x=sqrt{2x+1}$$ and $$y=sqrt{2y+1},$$ which gives
              $$x=y=1+sqrt2.$$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                It's $$sum_{cyc}left(x+frac{1}{x}+2-2sqrt{2x+1}right)=0$$ or
                $$sum_{cyc}frac{x^2-2xsqrt{2x+1}+2x+1}{x}=0$$ or
                $$sum_{cyc}frac{(x-sqrt{2x+1})^2}{x}=0,$$
                which for $xy<0$ gives infinitely many solutions.



                But, for $xy>0$ we obtain:
                $$x=sqrt{2x+1}$$ and $$y=sqrt{2y+1},$$ which gives
                $$x=y=1+sqrt2.$$






                share|cite|improve this answer









                $endgroup$



                It's $$sum_{cyc}left(x+frac{1}{x}+2-2sqrt{2x+1}right)=0$$ or
                $$sum_{cyc}frac{x^2-2xsqrt{2x+1}+2x+1}{x}=0$$ or
                $$sum_{cyc}frac{(x-sqrt{2x+1})^2}{x}=0,$$
                which for $xy<0$ gives infinitely many solutions.



                But, for $xy>0$ we obtain:
                $$x=sqrt{2x+1}$$ and $$y=sqrt{2y+1},$$ which gives
                $$x=y=1+sqrt2.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 3 hours ago









                Michael RozenbergMichael Rozenberg

                108k1895200




                108k1895200






























                    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%2f3148348%2fwhat-are-the-possible-solutions-of-the-given-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

                    What are all the squawk codes?

                    Image
                    2 $begingroup$ I love aviation and I love learning new things about it. I was wondering if anyone knew all of the squawk codes and in particular what squawk 7500 means. squawk-codes share | improve this question edited 15 mins ago Jamiec ♦ 14.8k 4 60 91 asked 5 hours ago Zac Callaghan Zac Callaghan 11 2 New contributor Zac Callaghan is a new co
                    Read more

                    What are differences between VBoxVGA, VMSVGA and VBoxSVGA in VirtualBox?

                    Image
                    1 VirtualBox 6.0 provides a new setting called Graphics Controller. There are four options to choose from: None VBoxVGA VMSVGA VBoxSVGA In the user manual I have found this information: Graphics Controller: Specifies the graphics adapter type used by the guest VM. Note that you must install the Guest Additions on the guest VM to specify the VBoxSVGA or VMSVGA graphics controller. The following options are available: – VBoxSVGA: The default graphics controller for new VMs that use Linux or Windows 7 or later. This graphics controller improves performance and 3D support when compared to the legacy VBoxVGA option. – VBoxVGA: Use this graphics controller for legacy guest OSes. This is the default graphics controller for Windows versions before Windows 7. – VMSVGA: Use this graphics controller to emul
                    Read more

                    Hudsonelva

                    Image
                    Hudson elv Land   USA Område New York Sideelvar Mohawk Lengd 507 km Middelvassføring 606 m³/s Munning Upper New York Bay i Atlanterhavet Elvene Hudson og Mohawk med nedbørsområdet markert med gult. Wikimedia Commons: Hudson River Hudsonelva ( Hudson River ) mohikansk Muh-he-kun-ne-tuk ) er ei elv i USA. Ho renn hovudsakleg gjennom delstaten New York, men markerer også ein del av grensa mellom New York og New Jersey. Hudsonelva har namn etter Henry Hudson, ein engelskmann som utforska elva i 1609. Hudsonelva sett frå Bear Mountain Bridge. Kjelda til elva er Tear-of-the-Clouds-sjøen i Adirondackfjella. Ved Albany renn Mohawk-elva saman med henne, og deretter renn ho nesten rett sørover til ho munnar ut i Atlanterhavet mellom Manhattan i New York og New Jersey. Eriekanalen, som blei opna den 26 oktober 1825, bind saman Hudsonelva og Eriesjøen, slik at byane ved Dei store sjøane fekk vasstilgang til A
                    Read more