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







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












$begingroup$


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





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090437%2fhard-algebra-square-root-equation%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      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


















                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090437%2fhard-algebra-square-root-equation%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      404 Error Contact Form 7 ajax form submitting

                      Image
                      0 The contact form does not work, and affects loading after one try. The chrome console shows the error message: POST http://example.com/wp-json/contact-form-7/v1/contact-forms/50/feedback 404 (Not Found) send @ jquery.js?ver=1.12.4:4 ajax @ jquery.js?ver=1.12.4:4 wpcf7.submit @ scripts.js?ver=5.0.3:346 (anonymous) @ scripts.js?ver=5.0.3:53 dispatch @ jquery.js?ver=1.12.4:3 r.handle @ jquery.js?ver=1.12.4:3 How to solve this issue? thanks in advance javascript wordpress contact-form-7 share | improve this question edited Jul 17 '18 at 5:39 Kiran Shahi 4,745 3 16 43 ...
                      Read more

                      How to know if a Active Directory user can login interactively

                      Image
                      1 I would like to know if and how is it possible to know if a AD user can login interactively (on a server) in a domain. I need to know if I can find it out using a LDAP search. Thanks. windows active-directory login user-permissions user-profile share | improve this question edited 1 hour ago yagmoth555 ♦ 11.6k 3 17 42 asked 1 hour ago Luigi Luigi 6 1 New contributor ...
                      Read more

                      Refactoring coordinates for Minecraft Pi buildings written in Python

                      Image
                      0 $begingroup$ I'm fairly new to coding and I wrote this script in Python for a house with a roof in Minecraft Pi. I've found different ways to write the coordinates to make the code base shorter, but I'm wondering if it is possible for me to refactor it any further so I don't have to list each coordinate line by line. import mcpi.minecraft as minecraft import mcpi.block as block mc = minecraft.Minecraft.create() p = mc.player.getTilePos() mc.setBlocks(p.x+1, p.y, p.z+1, #Main structure starting coordinates (DO NOT ALTER) p.x+19, p.y+4, p.z+6, #Structure size/coordinates (Alters cube size) block.STONE_BRICK) mc.setBlocks(p.x+2, p.y+1, p.z+2, #Hollowing out main structure interior p.x+18, p.y+3, p.z+5, block.AIR) mc.setBlocks(p.x, p.y...
                      Read more