How to solve equations of this type [closed]












4














My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:



Q1. If: $x - frac{1}{x} = 3$ then what is $x^2 + frac{1}{x^2}$ equal to?



The answer for this question is 11, but I don't know how, I thought it should be 9, but my answer was wrong.



Q2. If: $frac{x}{x + y} = 5$ then what is $frac{y}{x + y}$ equal to?



The answer for this is -5, I also don't know how.



Q3. If: $x^4 + y^4 = 6 x^2 y^2 land xneq y$ then what is $frac{x^2 + y^2}{x^2 - y^2}$ equal to?



I guess the answer for this was $sqrt{2}$ but I'm not sure.



Any body can explain how to solve these questions, and questions of the same pattern?










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Ameer Taweel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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closed as too broad by José Carlos Santos, user10354138, Holo, Cesareo, Saad Dec 23 at 1:35


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.











  • 2




    The first one should be $x^2 + 1/x^2$.
    – Rebellos
    Dec 22 at 15:53






  • 2




    $frac x{x+y}+frac y{x+y}=1$.
    – Lord Shark the Unknown
    Dec 22 at 15:55












  • Sorry, didn't notice, now it's fixed.
    – Ameer Taweel
    Dec 22 at 15:55










  • @LordSharktheUnknown this solves the second one, thanks. But what about the others.
    – Ameer Taweel
    Dec 22 at 15:56










  • The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
    – Sik Feng Cheong
    Dec 22 at 16:21


















4














My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:



Q1. If: $x - frac{1}{x} = 3$ then what is $x^2 + frac{1}{x^2}$ equal to?



The answer for this question is 11, but I don't know how, I thought it should be 9, but my answer was wrong.



Q2. If: $frac{x}{x + y} = 5$ then what is $frac{y}{x + y}$ equal to?



The answer for this is -5, I also don't know how.



Q3. If: $x^4 + y^4 = 6 x^2 y^2 land xneq y$ then what is $frac{x^2 + y^2}{x^2 - y^2}$ equal to?



I guess the answer for this was $sqrt{2}$ but I'm not sure.



Any body can explain how to solve these questions, and questions of the same pattern?










share|cite|improve this question









New contributor




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











closed as too broad by José Carlos Santos, user10354138, Holo, Cesareo, Saad Dec 23 at 1:35


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.











  • 2




    The first one should be $x^2 + 1/x^2$.
    – Rebellos
    Dec 22 at 15:53






  • 2




    $frac x{x+y}+frac y{x+y}=1$.
    – Lord Shark the Unknown
    Dec 22 at 15:55












  • Sorry, didn't notice, now it's fixed.
    – Ameer Taweel
    Dec 22 at 15:55










  • @LordSharktheUnknown this solves the second one, thanks. But what about the others.
    – Ameer Taweel
    Dec 22 at 15:56










  • The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
    – Sik Feng Cheong
    Dec 22 at 16:21
















4












4








4







My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:



Q1. If: $x - frac{1}{x} = 3$ then what is $x^2 + frac{1}{x^2}$ equal to?



The answer for this question is 11, but I don't know how, I thought it should be 9, but my answer was wrong.



Q2. If: $frac{x}{x + y} = 5$ then what is $frac{y}{x + y}$ equal to?



The answer for this is -5, I also don't know how.



Q3. If: $x^4 + y^4 = 6 x^2 y^2 land xneq y$ then what is $frac{x^2 + y^2}{x^2 - y^2}$ equal to?



I guess the answer for this was $sqrt{2}$ but I'm not sure.



Any body can explain how to solve these questions, and questions of the same pattern?










share|cite|improve this question









New contributor




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











My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:



Q1. If: $x - frac{1}{x} = 3$ then what is $x^2 + frac{1}{x^2}$ equal to?



The answer for this question is 11, but I don't know how, I thought it should be 9, but my answer was wrong.



Q2. If: $frac{x}{x + y} = 5$ then what is $frac{y}{x + y}$ equal to?



The answer for this is -5, I also don't know how.



Q3. If: $x^4 + y^4 = 6 x^2 y^2 land xneq y$ then what is $frac{x^2 + y^2}{x^2 - y^2}$ equal to?



I guess the answer for this was $sqrt{2}$ but I'm not sure.



Any body can explain how to solve these questions, and questions of the same pattern?







exponentiation






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New contributor




Ameer Taweel 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




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









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edited Dec 22 at 20:12









dantopa

6,41132042




6,41132042






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asked Dec 22 at 15:47









Ameer Taweel

234




234




New contributor




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





New contributor





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






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




closed as too broad by José Carlos Santos, user10354138, Holo, Cesareo, Saad Dec 23 at 1:35


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






closed as too broad by José Carlos Santos, user10354138, Holo, Cesareo, Saad Dec 23 at 1:35


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 2




    The first one should be $x^2 + 1/x^2$.
    – Rebellos
    Dec 22 at 15:53






  • 2




    $frac x{x+y}+frac y{x+y}=1$.
    – Lord Shark the Unknown
    Dec 22 at 15:55












  • Sorry, didn't notice, now it's fixed.
    – Ameer Taweel
    Dec 22 at 15:55










  • @LordSharktheUnknown this solves the second one, thanks. But what about the others.
    – Ameer Taweel
    Dec 22 at 15:56










  • The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
    – Sik Feng Cheong
    Dec 22 at 16:21
















  • 2




    The first one should be $x^2 + 1/x^2$.
    – Rebellos
    Dec 22 at 15:53






  • 2




    $frac x{x+y}+frac y{x+y}=1$.
    – Lord Shark the Unknown
    Dec 22 at 15:55












  • Sorry, didn't notice, now it's fixed.
    – Ameer Taweel
    Dec 22 at 15:55










  • @LordSharktheUnknown this solves the second one, thanks. But what about the others.
    – Ameer Taweel
    Dec 22 at 15:56










  • The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
    – Sik Feng Cheong
    Dec 22 at 16:21










2




2




The first one should be $x^2 + 1/x^2$.
– Rebellos
Dec 22 at 15:53




The first one should be $x^2 + 1/x^2$.
– Rebellos
Dec 22 at 15:53




2




2




$frac x{x+y}+frac y{x+y}=1$.
– Lord Shark the Unknown
Dec 22 at 15:55






$frac x{x+y}+frac y{x+y}=1$.
– Lord Shark the Unknown
Dec 22 at 15:55














Sorry, didn't notice, now it's fixed.
– Ameer Taweel
Dec 22 at 15:55




Sorry, didn't notice, now it's fixed.
– Ameer Taweel
Dec 22 at 15:55












@LordSharktheUnknown this solves the second one, thanks. But what about the others.
– Ameer Taweel
Dec 22 at 15:56




@LordSharktheUnknown this solves the second one, thanks. But what about the others.
– Ameer Taweel
Dec 22 at 15:56












The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
– Sik Feng Cheong
Dec 22 at 16:21






The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
– Sik Feng Cheong
Dec 22 at 16:21












4 Answers
4






active

oldest

votes


















5














For the first one :



$$left(x-frac{1}{x}right)^2 = x^2 - 2 + frac{1}{x^2} Leftrightarrow x^2 + frac{1}{x^2} = left(x-frac{1}{x}right)^2 + 2 implies x^2+ frac{1}{x^2} = 11$$



For the second one, observe that :



$$frac{x}{x+y} + frac{y}{x+y} = 1 Rightarrow 5 + frac{y}{x+y} = 1 Leftrightarrow frac{y}{x+y} = -4 $$



For the third one, a small hint :



$$(x^2-y^2)^2 = x^4 - 2x^2y^2 + y^4$$



Alternativelly, observe that it also is :



$$x^4 + y^4 = 6 x^2 y^2 Leftrightarrow frac{x^2}{y^2} + frac{y^2}{x^2} = 6$$






share|cite|improve this answer























  • @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
    – Rebellos
    Dec 22 at 15:59










  • Thanks very much, I guess I'll get A+ tomorrow.
    – Ameer Taweel
    Dec 22 at 16:15






  • 2




    @AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
    – Rebellos
    Dec 22 at 18:25










  • thanks for the advice.
    – Ameer Taweel
    Dec 22 at 20:10



















1














Hint:

1: $x - dfrac{1}{x} = 3 implies x^2 + dfrac{1}{x^2} - 2 = 9$

Now use, $big(x + dfrac{1}{x}big)^2 = x^2 + dfrac{1}{x^2} + 2$



2: $dfrac{x}{x+y} = dfrac{1}{1 + dfrac{y}{x}}$



3: $x^4 + y^4 = 6 * x^2 * y^2 implies dfrac{x^2}{y^2} + dfrac{y^2}{x^2} = 6$, solve for y/x or x/y use that to get the value of final expression






share|cite|improve this answer





























    1














    1



    Notice that



    $$x^2 + frac{1}{x^2} = left(x -frac{1}{x}right)^2 + 2 = 3^2 + 2 = 11$$



    2



    Notice that the sum of the two fractions gives $1$ hence it's rather trivial to obtain the second value.



    3



    You can transform the equation and solve for $frac{y}{x}$ for example, and find everything you need.






    share|cite|improve this answer





























      1














      Hints.



      $1) x-frac1x=3implies x^2+frac1{x^2}-2=9\2)frac y{x+y}+frac x{x+y}=1\3)x^4+y^4=6x^2y^2\implies x^4+y^4+2x^2y^2=(x^2+y^2)^2=8x^2y^2\implies x^4+y^4-2x^2y^2=(x^2-y^2)^2=4x^2y^2$



      Divide the two and take the square root.






      share|cite|improve this answer





















      • Thanks for the detailed explanation of the last one.
        – Ameer Taweel
        Dec 22 at 16:16


















      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5














      For the first one :



      $$left(x-frac{1}{x}right)^2 = x^2 - 2 + frac{1}{x^2} Leftrightarrow x^2 + frac{1}{x^2} = left(x-frac{1}{x}right)^2 + 2 implies x^2+ frac{1}{x^2} = 11$$



      For the second one, observe that :



      $$frac{x}{x+y} + frac{y}{x+y} = 1 Rightarrow 5 + frac{y}{x+y} = 1 Leftrightarrow frac{y}{x+y} = -4 $$



      For the third one, a small hint :



      $$(x^2-y^2)^2 = x^4 - 2x^2y^2 + y^4$$



      Alternativelly, observe that it also is :



      $$x^4 + y^4 = 6 x^2 y^2 Leftrightarrow frac{x^2}{y^2} + frac{y^2}{x^2} = 6$$






      share|cite|improve this answer























      • @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
        – Rebellos
        Dec 22 at 15:59










      • Thanks very much, I guess I'll get A+ tomorrow.
        – Ameer Taweel
        Dec 22 at 16:15






      • 2




        @AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
        – Rebellos
        Dec 22 at 18:25










      • thanks for the advice.
        – Ameer Taweel
        Dec 22 at 20:10
















      5














      For the first one :



      $$left(x-frac{1}{x}right)^2 = x^2 - 2 + frac{1}{x^2} Leftrightarrow x^2 + frac{1}{x^2} = left(x-frac{1}{x}right)^2 + 2 implies x^2+ frac{1}{x^2} = 11$$



      For the second one, observe that :



      $$frac{x}{x+y} + frac{y}{x+y} = 1 Rightarrow 5 + frac{y}{x+y} = 1 Leftrightarrow frac{y}{x+y} = -4 $$



      For the third one, a small hint :



      $$(x^2-y^2)^2 = x^4 - 2x^2y^2 + y^4$$



      Alternativelly, observe that it also is :



      $$x^4 + y^4 = 6 x^2 y^2 Leftrightarrow frac{x^2}{y^2} + frac{y^2}{x^2} = 6$$






      share|cite|improve this answer























      • @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
        – Rebellos
        Dec 22 at 15:59










      • Thanks very much, I guess I'll get A+ tomorrow.
        – Ameer Taweel
        Dec 22 at 16:15






      • 2




        @AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
        – Rebellos
        Dec 22 at 18:25










      • thanks for the advice.
        – Ameer Taweel
        Dec 22 at 20:10














      5












      5








      5






      For the first one :



      $$left(x-frac{1}{x}right)^2 = x^2 - 2 + frac{1}{x^2} Leftrightarrow x^2 + frac{1}{x^2} = left(x-frac{1}{x}right)^2 + 2 implies x^2+ frac{1}{x^2} = 11$$



      For the second one, observe that :



      $$frac{x}{x+y} + frac{y}{x+y} = 1 Rightarrow 5 + frac{y}{x+y} = 1 Leftrightarrow frac{y}{x+y} = -4 $$



      For the third one, a small hint :



      $$(x^2-y^2)^2 = x^4 - 2x^2y^2 + y^4$$



      Alternativelly, observe that it also is :



      $$x^4 + y^4 = 6 x^2 y^2 Leftrightarrow frac{x^2}{y^2} + frac{y^2}{x^2} = 6$$






      share|cite|improve this answer














      For the first one :



      $$left(x-frac{1}{x}right)^2 = x^2 - 2 + frac{1}{x^2} Leftrightarrow x^2 + frac{1}{x^2} = left(x-frac{1}{x}right)^2 + 2 implies x^2+ frac{1}{x^2} = 11$$



      For the second one, observe that :



      $$frac{x}{x+y} + frac{y}{x+y} = 1 Rightarrow 5 + frac{y}{x+y} = 1 Leftrightarrow frac{y}{x+y} = -4 $$



      For the third one, a small hint :



      $$(x^2-y^2)^2 = x^4 - 2x^2y^2 + y^4$$



      Alternativelly, observe that it also is :



      $$x^4 + y^4 = 6 x^2 y^2 Leftrightarrow frac{x^2}{y^2} + frac{y^2}{x^2} = 6$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 22 at 15:58

























      answered Dec 22 at 15:56









      Rebellos

      14.4k31245




      14.4k31245












      • @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
        – Rebellos
        Dec 22 at 15:59










      • Thanks very much, I guess I'll get A+ tomorrow.
        – Ameer Taweel
        Dec 22 at 16:15






      • 2




        @AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
        – Rebellos
        Dec 22 at 18:25










      • thanks for the advice.
        – Ameer Taweel
        Dec 22 at 20:10


















      • @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
        – Rebellos
        Dec 22 at 15:59










      • Thanks very much, I guess I'll get A+ tomorrow.
        – Ameer Taweel
        Dec 22 at 16:15






      • 2




        @AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
        – Rebellos
        Dec 22 at 18:25










      • thanks for the advice.
        – Ameer Taweel
        Dec 22 at 20:10
















      @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
      – Rebellos
      Dec 22 at 15:59




      @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
      – Rebellos
      Dec 22 at 15:59












      Thanks very much, I guess I'll get A+ tomorrow.
      – Ameer Taweel
      Dec 22 at 16:15




      Thanks very much, I guess I'll get A+ tomorrow.
      – Ameer Taweel
      Dec 22 at 16:15




      2




      2




      @AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
      – Rebellos
      Dec 22 at 18:25




      @AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
      – Rebellos
      Dec 22 at 18:25












      thanks for the advice.
      – Ameer Taweel
      Dec 22 at 20:10




      thanks for the advice.
      – Ameer Taweel
      Dec 22 at 20:10











      1














      Hint:

      1: $x - dfrac{1}{x} = 3 implies x^2 + dfrac{1}{x^2} - 2 = 9$

      Now use, $big(x + dfrac{1}{x}big)^2 = x^2 + dfrac{1}{x^2} + 2$



      2: $dfrac{x}{x+y} = dfrac{1}{1 + dfrac{y}{x}}$



      3: $x^4 + y^4 = 6 * x^2 * y^2 implies dfrac{x^2}{y^2} + dfrac{y^2}{x^2} = 6$, solve for y/x or x/y use that to get the value of final expression






      share|cite|improve this answer


























        1














        Hint:

        1: $x - dfrac{1}{x} = 3 implies x^2 + dfrac{1}{x^2} - 2 = 9$

        Now use, $big(x + dfrac{1}{x}big)^2 = x^2 + dfrac{1}{x^2} + 2$



        2: $dfrac{x}{x+y} = dfrac{1}{1 + dfrac{y}{x}}$



        3: $x^4 + y^4 = 6 * x^2 * y^2 implies dfrac{x^2}{y^2} + dfrac{y^2}{x^2} = 6$, solve for y/x or x/y use that to get the value of final expression






        share|cite|improve this answer
























          1












          1








          1






          Hint:

          1: $x - dfrac{1}{x} = 3 implies x^2 + dfrac{1}{x^2} - 2 = 9$

          Now use, $big(x + dfrac{1}{x}big)^2 = x^2 + dfrac{1}{x^2} + 2$



          2: $dfrac{x}{x+y} = dfrac{1}{1 + dfrac{y}{x}}$



          3: $x^4 + y^4 = 6 * x^2 * y^2 implies dfrac{x^2}{y^2} + dfrac{y^2}{x^2} = 6$, solve for y/x or x/y use that to get the value of final expression






          share|cite|improve this answer












          Hint:

          1: $x - dfrac{1}{x} = 3 implies x^2 + dfrac{1}{x^2} - 2 = 9$

          Now use, $big(x + dfrac{1}{x}big)^2 = x^2 + dfrac{1}{x^2} + 2$



          2: $dfrac{x}{x+y} = dfrac{1}{1 + dfrac{y}{x}}$



          3: $x^4 + y^4 = 6 * x^2 * y^2 implies dfrac{x^2}{y^2} + dfrac{y^2}{x^2} = 6$, solve for y/x or x/y use that to get the value of final expression







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 22 at 15:57









          l''''''''l

          2,188726




          2,188726























              1














              1



              Notice that



              $$x^2 + frac{1}{x^2} = left(x -frac{1}{x}right)^2 + 2 = 3^2 + 2 = 11$$



              2



              Notice that the sum of the two fractions gives $1$ hence it's rather trivial to obtain the second value.



              3



              You can transform the equation and solve for $frac{y}{x}$ for example, and find everything you need.






              share|cite|improve this answer


























                1














                1



                Notice that



                $$x^2 + frac{1}{x^2} = left(x -frac{1}{x}right)^2 + 2 = 3^2 + 2 = 11$$



                2



                Notice that the sum of the two fractions gives $1$ hence it's rather trivial to obtain the second value.



                3



                You can transform the equation and solve for $frac{y}{x}$ for example, and find everything you need.






                share|cite|improve this answer
























                  1












                  1








                  1






                  1



                  Notice that



                  $$x^2 + frac{1}{x^2} = left(x -frac{1}{x}right)^2 + 2 = 3^2 + 2 = 11$$



                  2



                  Notice that the sum of the two fractions gives $1$ hence it's rather trivial to obtain the second value.



                  3



                  You can transform the equation and solve for $frac{y}{x}$ for example, and find everything you need.






                  share|cite|improve this answer












                  1



                  Notice that



                  $$x^2 + frac{1}{x^2} = left(x -frac{1}{x}right)^2 + 2 = 3^2 + 2 = 11$$



                  2



                  Notice that the sum of the two fractions gives $1$ hence it's rather trivial to obtain the second value.



                  3



                  You can transform the equation and solve for $frac{y}{x}$ for example, and find everything you need.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 22 at 15:58









                  Von Neumann

                  16.3k72543




                  16.3k72543























                      1














                      Hints.



                      $1) x-frac1x=3implies x^2+frac1{x^2}-2=9\2)frac y{x+y}+frac x{x+y}=1\3)x^4+y^4=6x^2y^2\implies x^4+y^4+2x^2y^2=(x^2+y^2)^2=8x^2y^2\implies x^4+y^4-2x^2y^2=(x^2-y^2)^2=4x^2y^2$



                      Divide the two and take the square root.






                      share|cite|improve this answer





















                      • Thanks for the detailed explanation of the last one.
                        – Ameer Taweel
                        Dec 22 at 16:16
















                      1














                      Hints.



                      $1) x-frac1x=3implies x^2+frac1{x^2}-2=9\2)frac y{x+y}+frac x{x+y}=1\3)x^4+y^4=6x^2y^2\implies x^4+y^4+2x^2y^2=(x^2+y^2)^2=8x^2y^2\implies x^4+y^4-2x^2y^2=(x^2-y^2)^2=4x^2y^2$



                      Divide the two and take the square root.






                      share|cite|improve this answer





















                      • Thanks for the detailed explanation of the last one.
                        – Ameer Taweel
                        Dec 22 at 16:16














                      1












                      1








                      1






                      Hints.



                      $1) x-frac1x=3implies x^2+frac1{x^2}-2=9\2)frac y{x+y}+frac x{x+y}=1\3)x^4+y^4=6x^2y^2\implies x^4+y^4+2x^2y^2=(x^2+y^2)^2=8x^2y^2\implies x^4+y^4-2x^2y^2=(x^2-y^2)^2=4x^2y^2$



                      Divide the two and take the square root.






                      share|cite|improve this answer












                      Hints.



                      $1) x-frac1x=3implies x^2+frac1{x^2}-2=9\2)frac y{x+y}+frac x{x+y}=1\3)x^4+y^4=6x^2y^2\implies x^4+y^4+2x^2y^2=(x^2+y^2)^2=8x^2y^2\implies x^4+y^4-2x^2y^2=(x^2-y^2)^2=4x^2y^2$



                      Divide the two and take the square root.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 22 at 16:08









                      Shubham Johri

                      3,826716




                      3,826716












                      • Thanks for the detailed explanation of the last one.
                        – Ameer Taweel
                        Dec 22 at 16:16


















                      • Thanks for the detailed explanation of the last one.
                        – Ameer Taweel
                        Dec 22 at 16:16
















                      Thanks for the detailed explanation of the last one.
                      – Ameer Taweel
                      Dec 22 at 16:16




                      Thanks for the detailed explanation of the last one.
                      – Ameer Taweel
                      Dec 22 at 16:16



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