Hard Algebra Square Root Equation












1












$begingroup$


Suppose that real number $x$ satisfies $$sqrt{49-x^2}-sqrt{25-x^2}=3$$What is the value of $sqrt{49-x^2}+sqrt{25-x^2}$?




This is what I did:
I try to multiply by the conjugate. Its value I believe is technically the solution. $(sqrt {49-x^2} + sqrt {25-x^2})(sqrt {49-x^2} - sqrt {25-x^2})=24$. Given that $(sqrt {49-x^2} - sqrt {25-x^2}) = 3$, $(sqrt {49-x^2} + sqrt {25-x^2})(sqrt {49-x^2} - sqrt {25-x^2})= 3(sqrt {49-x^2} + sqrt {25-x^2}) =24implies sqrt {49-x^2} + sqrt {25-x^2}=8$




My question is that will this method work for similar problems, and is there a faster method?
Thanks!
Max0815






algebra-precalculus radicals







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  • 2




    $begingroup$
    This looks very quick to me compared with moving one square root to the other side, squaring, moving everything except the cross term to one side, squaring again, and sorting things out. Very slick.
    $endgroup$
    – Ross Millikan
    7 hours ago
















1












$begingroup$


Suppose that real number $x$ satisfies $$sqrt{49-x^2}-sqrt{25-x^2}=3$$What is the value of $sqrt{49-x^2}+sqrt{25-x^2}$?




This is what I did:
I try to multiply by the conjugate. Its value I believe is technically the solution. $(sqrt {49-x^2} + sqrt {25-x^2})(sqrt {49-x^2} - sqrt {25-x^2})=24$. Given that $(sqrt {49-x^2} - sqrt {25-x^2}) = 3$, $(sqrt {49-x^2} + sqrt {25-x^2})(sqrt {49-x^2} - sqrt {25-x^2})= 3(sqrt {49-x^2} + sqrt {25-x^2}) =24implies sqrt {49-x^2} + sqrt {25-x^2}=8$




My question is that will this method work for similar problems, and is there a faster method?
Thanks!
Max0815






algebra-precalculus radicals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    This looks very quick to me compared with moving one square root to the other side, squaring, moving everything except the cross term to one side, squaring again, and sorting things out. Very slick.
    $endgroup$
    – Ross Millikan
    7 hours ago














1












1








1





$begingroup$


Suppose that real number $x$ satisfies $$sqrt{49-x^2}-sqrt{25-x^2}=3$$What is the value of $sqrt{49-x^2}+sqrt{25-x^2}$?




This is what I did:
I try to multiply by the conjugate. Its value I believe is technically the solution. $(sqrt {49-x^2} + sqrt {25-x^2})(sqrt {49-x^2} - sqrt {25-x^2})=24$. Given that $(sqrt {49-x^2} - sqrt {25-x^2}) = 3$, $(sqrt {49-x^2} + sqrt {25-x^2})(sqrt {49-x^2} - sqrt {25-x^2})= 3(sqrt {49-x^2} + sqrt {25-x^2}) =24implies sqrt {49-x^2} + sqrt {25-x^2}=8$




My question is that will this method work for similar problems, and is there a faster method?
Thanks!
Max0815






algebra-precalculus radicals







share|cite|improve this question











$endgroup$




Suppose that real number $x$ satisfies $$sqrt{49-x^2}-sqrt{25-x^2}=3$$What is the value of $sqrt{49-x^2}+sqrt{25-x^2}$?




This is what I did:
I try to multiply by the conjugate. Its value I believe is technically the solution. $(sqrt {49-x^2} + sqrt {25-x^2})(sqrt {49-x^2} - sqrt {25-x^2})=24$. Given that $(sqrt {49-x^2} - sqrt {25-x^2}) = 3$, $(sqrt {49-x^2} + sqrt {25-x^2})(sqrt {49-x^2} - sqrt {25-x^2})= 3(sqrt {49-x^2} + sqrt {25-x^2}) =24implies sqrt {49-x^2} + sqrt {25-x^2}=8$




My question is that will this method work for similar problems, and is there a faster method?
Thanks!
Max0815






algebra-precalculus radicals




algebra-precalculus radicals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









J. W. Tanner

643111




643111










asked 7 hours ago









Max0815Max0815

45016




45016








  • 2




    $begingroup$
    This looks very quick to me compared with moving one square root to the other side, squaring, moving everything except the cross term to one side, squaring again, and sorting things out. Very slick.
    $endgroup$
    – Ross Millikan
    7 hours ago














  • 2




    $begingroup$
    This looks very quick to me compared with moving one square root to the other side, squaring, moving everything except the cross term to one side, squaring again, and sorting things out. Very slick.
    $endgroup$
    – Ross Millikan
    7 hours ago








2




2




$begingroup$
This looks very quick to me compared with moving one square root to the other side, squaring, moving everything except the cross term to one side, squaring again, and sorting things out. Very slick.
$endgroup$
– Ross Millikan
7 hours ago




$begingroup$
This looks very quick to me compared with moving one square root to the other side, squaring, moving everything except the cross term to one side, squaring again, and sorting things out. Very slick.
$endgroup$
– Ross Millikan
7 hours ago










3 Answers
3






active

oldest

votes


















5












$begingroup$

You basically used the formula $$a+b=frac{a^2-b^2}{a-b}$$
in an indirect way.



The problem gave you $a-b$ and, since $a,b$ are square roots of simple expressions, it is easy to calculate $a^2-b^2$.



And for the same reason, the same trick can be used to calculate
$$sqrt{mbox{nice}}pm sqrt{mbox{nice}}$$
whenever when $sqrt{mbox{nice}}mp sqrt{mbox{nice}}$ is given.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Multiplying both sides by $sqrt {49-x^2} + sqrt {25-x^2}$, yields:



    $$(sqrt {49-x^2} + sqrt {25-x^2})(sqrt {49-x^2} - sqrt {25-x^2})=3(sqrt {49-x^2} + sqrt {25-x^2})$$



    $$(49-x^2 - (25-x^2)=3(sqrt {49-x^2} + sqrt {25-x^2})$$



    Thus,



    $$sqrt {49-x^2} + sqrt {25-x^2}=8.$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Another way:
      $$1) sqrt{49-x^2}=sqrt{25-x^2}+3 Rightarrow \
      49-x^2=25-x^2+9+6sqrt{25-x^2} Rightarrow \
      2.5=sqrt{25-x^2};\
      2) sqrt{25-x^2}=sqrt{49-x^2}-3 Rightarrow \
      25-x^2=49-x^2+9-6sqrt{49-x^2} Rightarrow \
      sqrt{49-x^2}=5.5.$$
      Hence:
      $$sqrt{49-x^2}+sqrt{25-x^2}=5.5+2.5=8.$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        thx for this method
        $endgroup$
        – Max0815
        6 hours ago










      • $begingroup$
        You are welcome. Good luck.
        $endgroup$
        – farruhota
        2 hours ago











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      You basically used the formula $$a+b=frac{a^2-b^2}{a-b}$$
      in an indirect way.



      The problem gave you $a-b$ and, since $a,b$ are square roots of simple expressions, it is easy to calculate $a^2-b^2$.



      And for the same reason, the same trick can be used to calculate
      $$sqrt{mbox{nice}}pm sqrt{mbox{nice}}$$
      whenever when $sqrt{mbox{nice}}mp sqrt{mbox{nice}}$ is given.






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        You basically used the formula $$a+b=frac{a^2-b^2}{a-b}$$
        in an indirect way.



        The problem gave you $a-b$ and, since $a,b$ are square roots of simple expressions, it is easy to calculate $a^2-b^2$.



        And for the same reason, the same trick can be used to calculate
        $$sqrt{mbox{nice}}pm sqrt{mbox{nice}}$$
        whenever when $sqrt{mbox{nice}}mp sqrt{mbox{nice}}$ is given.






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          You basically used the formula $$a+b=frac{a^2-b^2}{a-b}$$
          in an indirect way.



          The problem gave you $a-b$ and, since $a,b$ are square roots of simple expressions, it is easy to calculate $a^2-b^2$.



          And for the same reason, the same trick can be used to calculate
          $$sqrt{mbox{nice}}pm sqrt{mbox{nice}}$$
          whenever when $sqrt{mbox{nice}}mp sqrt{mbox{nice}}$ is given.






          share|cite|improve this answer









          $endgroup$



          You basically used the formula $$a+b=frac{a^2-b^2}{a-b}$$
          in an indirect way.



          The problem gave you $a-b$ and, since $a,b$ are square roots of simple expressions, it is easy to calculate $a^2-b^2$.



          And for the same reason, the same trick can be used to calculate
          $$sqrt{mbox{nice}}pm sqrt{mbox{nice}}$$
          whenever when $sqrt{mbox{nice}}mp sqrt{mbox{nice}}$ is given.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 7 hours ago









          N. S.N. S.

          103k6111208




          103k6111208























              1












              $begingroup$

              Multiplying both sides by $sqrt {49-x^2} + sqrt {25-x^2}$, yields:



              $$(sqrt {49-x^2} + sqrt {25-x^2})(sqrt {49-x^2} - sqrt {25-x^2})=3(sqrt {49-x^2} + sqrt {25-x^2})$$



              $$(49-x^2 - (25-x^2)=3(sqrt {49-x^2} + sqrt {25-x^2})$$



              Thus,



              $$sqrt {49-x^2} + sqrt {25-x^2}=8.$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Multiplying both sides by $sqrt {49-x^2} + sqrt {25-x^2}$, yields:



                $$(sqrt {49-x^2} + sqrt {25-x^2})(sqrt {49-x^2} - sqrt {25-x^2})=3(sqrt {49-x^2} + sqrt {25-x^2})$$



                $$(49-x^2 - (25-x^2)=3(sqrt {49-x^2} + sqrt {25-x^2})$$



                Thus,



                $$sqrt {49-x^2} + sqrt {25-x^2}=8.$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Multiplying both sides by $sqrt {49-x^2} + sqrt {25-x^2}$, yields:



                  $$(sqrt {49-x^2} + sqrt {25-x^2})(sqrt {49-x^2} - sqrt {25-x^2})=3(sqrt {49-x^2} + sqrt {25-x^2})$$



                  $$(49-x^2 - (25-x^2)=3(sqrt {49-x^2} + sqrt {25-x^2})$$



                  Thus,



                  $$sqrt {49-x^2} + sqrt {25-x^2}=8.$$






                  share|cite|improve this answer









                  $endgroup$



                  Multiplying both sides by $sqrt {49-x^2} + sqrt {25-x^2}$, yields:



                  $$(sqrt {49-x^2} + sqrt {25-x^2})(sqrt {49-x^2} - sqrt {25-x^2})=3(sqrt {49-x^2} + sqrt {25-x^2})$$



                  $$(49-x^2 - (25-x^2)=3(sqrt {49-x^2} + sqrt {25-x^2})$$



                  Thus,



                  $$sqrt {49-x^2} + sqrt {25-x^2}=8.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 6 hours ago









                  Ali AbbasinasabAli Abbasinasab

                  947717




                  947717























                      1












                      $begingroup$

                      Another way:
                      $$1) sqrt{49-x^2}=sqrt{25-x^2}+3 Rightarrow \
                      49-x^2=25-x^2+9+6sqrt{25-x^2} Rightarrow \
                      2.5=sqrt{25-x^2};\
                      2) sqrt{25-x^2}=sqrt{49-x^2}-3 Rightarrow \
                      25-x^2=49-x^2+9-6sqrt{49-x^2} Rightarrow \
                      sqrt{49-x^2}=5.5.$$
                      Hence:
                      $$sqrt{49-x^2}+sqrt{25-x^2}=5.5+2.5=8.$$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        thx for this method
                        $endgroup$
                        – Max0815
                        6 hours ago










                      • $begingroup$
                        You are welcome. Good luck.
                        $endgroup$
                        – farruhota
                        2 hours ago
















                      1












                      $begingroup$

                      Another way:
                      $$1) sqrt{49-x^2}=sqrt{25-x^2}+3 Rightarrow \
                      49-x^2=25-x^2+9+6sqrt{25-x^2} Rightarrow \
                      2.5=sqrt{25-x^2};\
                      2) sqrt{25-x^2}=sqrt{49-x^2}-3 Rightarrow \
                      25-x^2=49-x^2+9-6sqrt{49-x^2} Rightarrow \
                      sqrt{49-x^2}=5.5.$$
                      Hence:
                      $$sqrt{49-x^2}+sqrt{25-x^2}=5.5+2.5=8.$$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        thx for this method
                        $endgroup$
                        – Max0815
                        6 hours ago










                      • $begingroup$
                        You are welcome. Good luck.
                        $endgroup$
                        – farruhota
                        2 hours ago














                      1












                      1








                      1





                      $begingroup$

                      Another way:
                      $$1) sqrt{49-x^2}=sqrt{25-x^2}+3 Rightarrow \
                      49-x^2=25-x^2+9+6sqrt{25-x^2} Rightarrow \
                      2.5=sqrt{25-x^2};\
                      2) sqrt{25-x^2}=sqrt{49-x^2}-3 Rightarrow \
                      25-x^2=49-x^2+9-6sqrt{49-x^2} Rightarrow \
                      sqrt{49-x^2}=5.5.$$
                      Hence:
                      $$sqrt{49-x^2}+sqrt{25-x^2}=5.5+2.5=8.$$






                      share|cite|improve this answer









                      $endgroup$



                      Another way:
                      $$1) sqrt{49-x^2}=sqrt{25-x^2}+3 Rightarrow \
                      49-x^2=25-x^2+9+6sqrt{25-x^2} Rightarrow \
                      2.5=sqrt{25-x^2};\
                      2) sqrt{25-x^2}=sqrt{49-x^2}-3 Rightarrow \
                      25-x^2=49-x^2+9-6sqrt{49-x^2} Rightarrow \
                      sqrt{49-x^2}=5.5.$$
                      Hence:
                      $$sqrt{49-x^2}+sqrt{25-x^2}=5.5+2.5=8.$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 6 hours ago









                      farruhotafarruhota

                      19.9k2738




                      19.9k2738












                      • $begingroup$
                        thx for this method
                        $endgroup$
                        – Max0815
                        6 hours ago










                      • $begingroup$
                        You are welcome. Good luck.
                        $endgroup$
                        – farruhota
                        2 hours ago


















                      • $begingroup$
                        thx for this method
                        $endgroup$
                        – Max0815
                        6 hours ago










                      • $begingroup$
                        You are welcome. Good luck.
                        $endgroup$
                        – farruhota
                        2 hours ago
















                      $begingroup$
                      thx for this method
                      $endgroup$
                      – Max0815
                      6 hours ago




                      $begingroup$
                      thx for this method
                      $endgroup$
                      – Max0815
                      6 hours ago












                      $begingroup$
                      You are welcome. Good luck.
                      $endgroup$
                      – farruhota
                      2 hours ago




                      $begingroup$
                      You are welcome. Good luck.
                      $endgroup$
                      – farruhota
                      2 hours ago


















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