How to solve equations of this type












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: $x^4 + y^4 = 6 * x^2 * y^2$ , $xneq y$ 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









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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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  • 2




    The first one should be $x^2 + 1/x^2$.
    – Rebellos
    3 hours ago






  • 2




    $frac x{x+y}+frac y{x+y}=1$.
    – Lord Shark the Unknown
    3 hours ago












  • Sorry, didn't notice, now it's fixed.
    – Ameer Taweel
    3 hours ago










  • @LordSharktheUnknown this solves the second one, thanks. But what about the others.
    – Ameer Taweel
    3 hours ago










  • The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
    – Sik Feng Cheong
    2 hours ago


















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: $x^4 + y^4 = 6 * x^2 * y^2$ , $xneq y$ 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









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.
















  • 2




    The first one should be $x^2 + 1/x^2$.
    – Rebellos
    3 hours ago






  • 2




    $frac x{x+y}+frac y{x+y}=1$.
    – Lord Shark the Unknown
    3 hours ago












  • Sorry, didn't notice, now it's fixed.
    – Ameer Taweel
    3 hours ago










  • @LordSharktheUnknown this solves the second one, thanks. But what about the others.
    – Ameer Taweel
    3 hours ago










  • The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
    – Sik Feng Cheong
    2 hours ago
















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: $x^4 + y^4 = 6 * x^2 * y^2$ , $xneq y$ 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









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: $x^4 + y^4 = 6 * x^2 * y^2$ , $xneq y$ 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









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.









share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Ankit Kumar

1,07917




1,07917






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asked 3 hours ago









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.








  • 2




    The first one should be $x^2 + 1/x^2$.
    – Rebellos
    3 hours ago






  • 2




    $frac x{x+y}+frac y{x+y}=1$.
    – Lord Shark the Unknown
    3 hours ago












  • Sorry, didn't notice, now it's fixed.
    – Ameer Taweel
    3 hours ago










  • @LordSharktheUnknown this solves the second one, thanks. But what about the others.
    – Ameer Taweel
    3 hours ago










  • The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
    – Sik Feng Cheong
    2 hours ago
















  • 2




    The first one should be $x^2 + 1/x^2$.
    – Rebellos
    3 hours ago






  • 2




    $frac x{x+y}+frac y{x+y}=1$.
    – Lord Shark the Unknown
    3 hours ago












  • Sorry, didn't notice, now it's fixed.
    – Ameer Taweel
    3 hours ago










  • @LordSharktheUnknown this solves the second one, thanks. But what about the others.
    – Ameer Taweel
    3 hours ago










  • The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
    – Sik Feng Cheong
    2 hours ago










2




2




The first one should be $x^2 + 1/x^2$.
– Rebellos
3 hours ago




The first one should be $x^2 + 1/x^2$.
– Rebellos
3 hours ago




2




2




$frac x{x+y}+frac y{x+y}=1$.
– Lord Shark the Unknown
3 hours ago






$frac x{x+y}+frac y{x+y}=1$.
– Lord Shark the Unknown
3 hours ago














Sorry, didn't notice, now it's fixed.
– Ameer Taweel
3 hours ago




Sorry, didn't notice, now it's fixed.
– Ameer Taweel
3 hours ago












@LordSharktheUnknown this solves the second one, thanks. But what about the others.
– Ameer Taweel
3 hours ago




@LordSharktheUnknown this solves the second one, thanks. But what about the others.
– Ameer Taweel
3 hours ago












The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
– Sik Feng Cheong
2 hours ago






The first one: $(x - frac{1}{x})^2=x^2-2+frac{1}{x^2} = 9$
– Sik Feng Cheong
2 hours ago












4 Answers
4






active

oldest

votes


















3














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
    3 hours ago










  • Thanks very much, I guess I'll get A+ tomorrow.
    – Ameer Taweel
    2 hours ago










  • @AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
    – Rebellos
    40 mins ago



















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





























    0














    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





























      0














      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
        2 hours ago











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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3














      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
        3 hours ago










      • Thanks very much, I guess I'll get A+ tomorrow.
        – Ameer Taweel
        2 hours ago










      • @AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
        – Rebellos
        40 mins ago
















      3














      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
        3 hours ago










      • Thanks very much, I guess I'll get A+ tomorrow.
        – Ameer Taweel
        2 hours ago










      • @AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
        – Rebellos
        40 mins ago














      3












      3








      3






      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 3 hours ago

























      answered 3 hours ago









      Rebellos

      14.3k31244




      14.3k31244












      • @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
        – Rebellos
        3 hours ago










      • Thanks very much, I guess I'll get A+ tomorrow.
        – Ameer Taweel
        2 hours ago










      • @AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
        – Rebellos
        40 mins ago


















      • @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
        – Rebellos
        3 hours ago










      • Thanks very much, I guess I'll get A+ tomorrow.
        – Ameer Taweel
        2 hours ago










      • @AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
        – Rebellos
        40 mins ago
















      @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
      – Rebellos
      3 hours ago




      @ÍgjøgnumMeg Thanks. You may as well edit to correct those.
      – Rebellos
      3 hours ago












      Thanks very much, I guess I'll get A+ tomorrow.
      – Ameer Taweel
      2 hours ago




      Thanks very much, I guess I'll get A+ tomorrow.
      – Ameer Taweel
      2 hours ago












      @AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
      – Rebellos
      40 mins ago




      @AmeerTaweel Your number one priority should be understanding and learning, not grades ! Just some advice.
      – Rebellos
      40 mins ago











      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 3 hours ago









          Von Neumann

          16.3k72543




          16.3k72543























              0














              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


























                0














                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
























                  0












                  0








                  0






                  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 3 hours ago









                  l''''''''l

                  2,178726




                  2,178726























                      0














                      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
                        2 hours ago
















                      0














                      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.
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                      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.







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                      share|cite|improve this answer










                      answered 2 hours ago









                      Shubham Johri

                      3,378716




                      3,378716












                      • Thanks for the detailed explanation of the last one.
                        – Ameer Taweel
                        2 hours ago


















                      • Thanks for the detailed explanation of the last one.
                        – Ameer Taweel
                        2 hours ago
















                      Thanks for the detailed explanation of the last one.
                      – Ameer Taweel
                      2 hours ago




                      Thanks for the detailed explanation of the last one.
                      – Ameer Taweel
                      2 hours ago










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