What are the possible solutions of the given equation?When do we get extraneous roots?Algebraic manipulation with square rootsFind all real numbers such that $sqrtx-frac1x + sqrt1 - frac1x = x$Transposing an equation: x = F/k + sqrt(F/c) to get F as the subjectFinding all real roots of the equation $(x+1) sqrtx+2 + (x+6)sqrtx+7 = x^2+7x+12$Justify the algorithm used to create a polynomial whose roots are squares of the roots of the given polynomial (over $mathbb C$)Integral solutions to the equation $left(frac1nright)^-1/2=sqrta+sqrt15-sqrta-sqrt15.$What rule governs $x^4=10,000$ having complex solutions?Why do I keep getting this incorrect solution when trying to find all the real solutions for $sqrt2x-3 +x=3$.Radical equation - can I square both sides with more than 1 radical on one side?

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What are the possible solutions of the given equation?


When do we get extraneous roots?Algebraic manipulation with square rootsFind all real numbers such that $sqrtx-frac1x + sqrt1 - frac1x = x$Transposing an equation: x = F/k + sqrt(F/c) to get F as the subjectFinding all real roots of the equation $(x+1) sqrtx+2 + (x+6)sqrtx+7 = x^2+7x+12$Justify the algorithm used to create a polynomial whose roots are squares of the roots of the given polynomial (over $mathbb C$)Integral solutions to the equation $left(frac1nright)^-1/2=sqrta+sqrt15-sqrta-sqrt15.$What rule governs $x^4=10,000$ having complex solutions?Why do I keep getting this incorrect solution when trying to find all the real solutions for $sqrt2x-3 +x=3$.Radical equation - can I square both sides with more than 1 radical on one side?













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


















          6












          $begingroup$

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



          By Am-Gm we have $$ x+2x+1over xgeq 2sqrtx2x+1over x = 2sqrt2x+1$$ and the same for $y$, so we have



          $$x+y+ 1over x+1over y+4 geq 2sqrt2x+1+2sqrt2y+1$$



          Since we have equality is achieved when $x=2x+1over x$ (and the same for $y$) we have $x=y=1+sqrt2$






          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
            2 hours ago



















          1












          $begingroup$

          It's $$sum_cycleft(x+frac1x+2-2sqrt2x+1right)=0$$ or
          $$sum_cycfracx^2-2xsqrt2x+1+2x+1x=0$$ or
          $$sum_cycfrac(x-sqrt2x+1)^2x=0,$$
          which for $xy<0$ gives infinitely many solutions.



          But, for $xy>0$ we obtain:
          $$x=sqrt2x+1$$ and $$y=sqrt2y+1,$$ which gives
          $$x=y=1+sqrt2.$$






          share|cite|improve this answer









          $endgroup$












            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6












            $begingroup$

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



            By Am-Gm we have $$ x+2x+1over xgeq 2sqrtx2x+1over x = 2sqrt2x+1$$ and the same for $y$, so we have



            $$x+y+ 1over x+1over y+4 geq 2sqrt2x+1+2sqrt2y+1$$



            Since we have equality is achieved when $x=2x+1over x$ (and the same for $y$) we have $x=y=1+sqrt2$






            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
              2 hours ago
















            6












            $begingroup$

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



            By Am-Gm we have $$ x+2x+1over xgeq 2sqrtx2x+1over x = 2sqrt2x+1$$ and the same for $y$, so we have



            $$x+y+ 1over x+1over y+4 geq 2sqrt2x+1+2sqrt2y+1$$



            Since we have equality is achieved when $x=2x+1over x$ (and the same for $y$) we have $x=y=1+sqrt2$






            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
              2 hours ago














            6












            6








            6





            $begingroup$

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



            By Am-Gm we have $$ x+2x+1over xgeq 2sqrtx2x+1over x = 2sqrt2x+1$$ and the same for $y$, so we have



            $$x+y+ 1over x+1over y+4 geq 2sqrt2x+1+2sqrt2y+1$$



            Since we have equality is achieved when $x=2x+1over x$ (and the same for $y$) we have $x=y=1+sqrt2$






            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 $$ x+2x+1over xgeq 2sqrtx2x+1over x = 2sqrt2x+1$$ and the same for $y$, so we have



            $$x+y+ 1over x+1over y+4 geq 2sqrt2x+1+2sqrt2y+1$$



            Since we have equality is achieved when $x=2x+1over x$ (and the same for $y$) we have $x=y=1+sqrt2$







            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
              2 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
              2 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
            2 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
            2 hours ago












            1












            $begingroup$

            It's $$sum_cycleft(x+frac1x+2-2sqrt2x+1right)=0$$ or
            $$sum_cycfracx^2-2xsqrt2x+1+2x+1x=0$$ or
            $$sum_cycfrac(x-sqrt2x+1)^2x=0,$$
            which for $xy<0$ gives infinitely many solutions.



            But, for $xy>0$ we obtain:
            $$x=sqrt2x+1$$ and $$y=sqrt2y+1,$$ which gives
            $$x=y=1+sqrt2.$$






            share|cite|improve this answer









            $endgroup$

















              1












              $begingroup$

              It's $$sum_cycleft(x+frac1x+2-2sqrt2x+1right)=0$$ or
              $$sum_cycfracx^2-2xsqrt2x+1+2x+1x=0$$ or
              $$sum_cycfrac(x-sqrt2x+1)^2x=0,$$
              which for $xy<0$ gives infinitely many solutions.



              But, for $xy>0$ we obtain:
              $$x=sqrt2x+1$$ and $$y=sqrt2y+1,$$ which gives
              $$x=y=1+sqrt2.$$






              share|cite|improve this answer









              $endgroup$















                1












                1








                1





                $begingroup$

                It's $$sum_cycleft(x+frac1x+2-2sqrt2x+1right)=0$$ or
                $$sum_cycfracx^2-2xsqrt2x+1+2x+1x=0$$ or
                $$sum_cycfrac(x-sqrt2x+1)^2x=0,$$
                which for $xy<0$ gives infinitely many solutions.



                But, for $xy>0$ we obtain:
                $$x=sqrt2x+1$$ and $$y=sqrt2y+1,$$ which gives
                $$x=y=1+sqrt2.$$






                share|cite|improve this answer









                $endgroup$



                It's $$sum_cycleft(x+frac1x+2-2sqrt2x+1right)=0$$ or
                $$sum_cycfracx^2-2xsqrt2x+1+2x+1x=0$$ or
                $$sum_cycfrac(x-sqrt2x+1)^2x=0,$$
                which for $xy<0$ gives infinitely many solutions.



                But, for $xy>0$ we obtain:
                $$x=sqrt2x+1$$ and $$y=sqrt2y+1,$$ which gives
                $$x=y=1+sqrt2.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Michael RozenbergMichael Rozenberg

                108k1895200




                108k1895200



























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