How to solve equations of this type [closed]












4












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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







share|cite|improve this question











$endgroup$



closed as too broad by José Carlos Santos, user10354138, Holo, Cesareo, Saad Dec 23 '18 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




    $begingroup$
    The first one should be $x^2 + 1/x^2$.
    $endgroup$
    – Rebellos
    Dec 22 '18 at 15:53






  • 2




    $begingroup$
    $frac x{x+y}+frac y{x+y}=1$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 15:55












  • $begingroup$
    Sorry, didn't notice, now it's fixed.
    $endgroup$
    – Ameer Taweel
    Dec 22 '18 at 15:55










  • $begingroup$
    @LordSharktheUnknown this solves the second one, thanks. But what about the others.
    $endgroup$
    – Ameer Taweel
    Dec 22 '18 at 15:56










  • $begingroup$
    The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
    $endgroup$
    – Sik Feng Cheong
    Dec 22 '18 at 16:21


















4












$begingroup$


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







share|cite|improve this question











$endgroup$



closed as too broad by José Carlos Santos, user10354138, Holo, Cesareo, Saad Dec 23 '18 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




    $begingroup$
    The first one should be $x^2 + 1/x^2$.
    $endgroup$
    – Rebellos
    Dec 22 '18 at 15:53






  • 2




    $begingroup$
    $frac x{x+y}+frac y{x+y}=1$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 15:55












  • $begingroup$
    Sorry, didn't notice, now it's fixed.
    $endgroup$
    – Ameer Taweel
    Dec 22 '18 at 15:55










  • $begingroup$
    @LordSharktheUnknown this solves the second one, thanks. But what about the others.
    $endgroup$
    – Ameer Taweel
    Dec 22 '18 at 15:56










  • $begingroup$
    The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
    $endgroup$
    – Sik Feng Cheong
    Dec 22 '18 at 16:21
















4












4








4





$begingroup$


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







share|cite|improve this question











$endgroup$




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




exponentiation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 20:12









dantopa

6,44942142




6,44942142










asked Dec 22 '18 at 15:47









Ameer TaweelAmeer Taweel

1234




1234




closed as too broad by José Carlos Santos, user10354138, Holo, Cesareo, Saad Dec 23 '18 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 '18 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




    $begingroup$
    The first one should be $x^2 + 1/x^2$.
    $endgroup$
    – Rebellos
    Dec 22 '18 at 15:53






  • 2




    $begingroup$
    $frac x{x+y}+frac y{x+y}=1$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 15:55












  • $begingroup$
    Sorry, didn't notice, now it's fixed.
    $endgroup$
    – Ameer Taweel
    Dec 22 '18 at 15:55










  • $begingroup$
    @LordSharktheUnknown this solves the second one, thanks. But what about the others.
    $endgroup$
    – Ameer Taweel
    Dec 22 '18 at 15:56










  • $begingroup$
    The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
    $endgroup$
    – Sik Feng Cheong
    Dec 22 '18 at 16:21
















  • 2




    $begingroup$
    The first one should be $x^2 + 1/x^2$.
    $endgroup$
    – Rebellos
    Dec 22 '18 at 15:53






  • 2




    $begingroup$
    $frac x{x+y}+frac y{x+y}=1$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 15:55












  • $begingroup$
    Sorry, didn't notice, now it's fixed.
    $endgroup$
    – Ameer Taweel
    Dec 22 '18 at 15:55










  • $begingroup$
    @LordSharktheUnknown this solves the second one, thanks. But what about the others.
    $endgroup$
    – Ameer Taweel
    Dec 22 '18 at 15:56










  • $begingroup$
    The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
    $endgroup$
    – Sik Feng Cheong
    Dec 22 '18 at 16:21










2




2




$begingroup$
The first one should be $x^2 + 1/x^2$.
$endgroup$
– Rebellos
Dec 22 '18 at 15:53




$begingroup$
The first one should be $x^2 + 1/x^2$.
$endgroup$
– Rebellos
Dec 22 '18 at 15:53




2




2




$begingroup$
$frac x{x+y}+frac y{x+y}=1$.
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 15:55






$begingroup$
$frac x{x+y}+frac y{x+y}=1$.
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 15:55














$begingroup$
Sorry, didn't notice, now it's fixed.
$endgroup$
– Ameer Taweel
Dec 22 '18 at 15:55




$begingroup$
Sorry, didn't notice, now it's fixed.
$endgroup$
– Ameer Taweel
Dec 22 '18 at 15:55












$begingroup$
@LordSharktheUnknown this solves the second one, thanks. But what about the others.
$endgroup$
– Ameer Taweel
Dec 22 '18 at 15:56




$begingroup$
@LordSharktheUnknown this solves the second one, thanks. But what about the others.
$endgroup$
– Ameer Taweel
Dec 22 '18 at 15:56












$begingroup$
The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
$endgroup$
– Sik Feng Cheong
Dec 22 '18 at 16:21






$begingroup$
The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
$endgroup$
– Sik Feng Cheong
Dec 22 '18 at 16:21












4 Answers
4






active

oldest

votes


















5












$begingroup$

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











$endgroup$













  • $begingroup$
    @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
    $endgroup$
    – Rebellos
    Dec 22 '18 at 15:59










  • $begingroup$
    Thanks very much, I guess I'll get A+ tomorrow.
    $endgroup$
    – Ameer Taweel
    Dec 22 '18 at 16:15






  • 2




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










  • $begingroup$
    thanks for the advice.
    $endgroup$
    – Ameer Taweel
    Dec 22 '18 at 20:10



















1












$begingroup$

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









$endgroup$





















    1












    $begingroup$

    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









    $endgroup$





















      1












      $begingroup$

      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









      $endgroup$













      • $begingroup$
        Thanks for the detailed explanation of the last one.
        $endgroup$
        – Ameer Taweel
        Dec 22 '18 at 16:16


















      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      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











      $endgroup$













      • $begingroup$
        @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
        $endgroup$
        – Rebellos
        Dec 22 '18 at 15:59










      • $begingroup$
        Thanks very much, I guess I'll get A+ tomorrow.
        $endgroup$
        – Ameer Taweel
        Dec 22 '18 at 16:15






      • 2




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










      • $begingroup$
        thanks for the advice.
        $endgroup$
        – Ameer Taweel
        Dec 22 '18 at 20:10
















      5












      $begingroup$

      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











      $endgroup$













      • $begingroup$
        @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
        $endgroup$
        – Rebellos
        Dec 22 '18 at 15:59










      • $begingroup$
        Thanks very much, I guess I'll get A+ tomorrow.
        $endgroup$
        – Ameer Taweel
        Dec 22 '18 at 16:15






      • 2




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










      • $begingroup$
        thanks for the advice.
        $endgroup$
        – Ameer Taweel
        Dec 22 '18 at 20:10














      5












      5








      5





      $begingroup$

      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











      $endgroup$



      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 '18 at 15:58

























      answered Dec 22 '18 at 15:56









      RebellosRebellos

      14.5k31246




      14.5k31246












      • $begingroup$
        @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
        $endgroup$
        – Rebellos
        Dec 22 '18 at 15:59










      • $begingroup$
        Thanks very much, I guess I'll get A+ tomorrow.
        $endgroup$
        – Ameer Taweel
        Dec 22 '18 at 16:15






      • 2




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










      • $begingroup$
        thanks for the advice.
        $endgroup$
        – Ameer Taweel
        Dec 22 '18 at 20:10


















      • $begingroup$
        @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
        $endgroup$
        – Rebellos
        Dec 22 '18 at 15:59










      • $begingroup$
        Thanks very much, I guess I'll get A+ tomorrow.
        $endgroup$
        – Ameer Taweel
        Dec 22 '18 at 16:15






      • 2




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










      • $begingroup$
        thanks for the advice.
        $endgroup$
        – Ameer Taweel
        Dec 22 '18 at 20:10
















      $begingroup$
      @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
      $endgroup$
      – Rebellos
      Dec 22 '18 at 15:59




      $begingroup$
      @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
      $endgroup$
      – Rebellos
      Dec 22 '18 at 15:59












      $begingroup$
      Thanks very much, I guess I'll get A+ tomorrow.
      $endgroup$
      – Ameer Taweel
      Dec 22 '18 at 16:15




      $begingroup$
      Thanks very much, I guess I'll get A+ tomorrow.
      $endgroup$
      – Ameer Taweel
      Dec 22 '18 at 16:15




      2




      2




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




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












      $begingroup$
      thanks for the advice.
      $endgroup$
      – Ameer Taweel
      Dec 22 '18 at 20:10




      $begingroup$
      thanks for the advice.
      $endgroup$
      – Ameer Taweel
      Dec 22 '18 at 20:10











      1












      $begingroup$

      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









      $endgroup$


















        1












        $begingroup$

        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









        $endgroup$
















          1












          1








          1





          $begingroup$

          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









          $endgroup$



          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 '18 at 15:57









          l''''''''ll''''''''l

          2,198726




          2,198726























              1












              $begingroup$

              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









              $endgroup$


















                1












                $begingroup$

                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









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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









                  $endgroup$



                  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 '18 at 15:58









                  Von NeumannVon Neumann

                  16.3k72543




                  16.3k72543























                      1












                      $begingroup$

                      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









                      $endgroup$













                      • $begingroup$
                        Thanks for the detailed explanation of the last one.
                        $endgroup$
                        – Ameer Taweel
                        Dec 22 '18 at 16:16
















                      1












                      $begingroup$

                      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









                      $endgroup$













                      • $begingroup$
                        Thanks for the detailed explanation of the last one.
                        $endgroup$
                        – Ameer Taweel
                        Dec 22 '18 at 16:16














                      1












                      1








                      1





                      $begingroup$

                      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









                      $endgroup$



                      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 '18 at 16:08









                      Shubham JohriShubham Johri

                      4,760717




                      4,760717












                      • $begingroup$
                        Thanks for the detailed explanation of the last one.
                        $endgroup$
                        – Ameer Taweel
                        Dec 22 '18 at 16:16


















                      • $begingroup$
                        Thanks for the detailed explanation of the last one.
                        $endgroup$
                        – Ameer Taweel
                        Dec 22 '18 at 16:16
















                      $begingroup$
                      Thanks for the detailed explanation of the last one.
                      $endgroup$
                      – Ameer Taweel
                      Dec 22 '18 at 16:16




                      $begingroup$
                      Thanks for the detailed explanation of the last one.
                      $endgroup$
                      – Ameer Taweel
                      Dec 22 '18 at 16:16



                      -

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