How do we solve for all values of $x$?












1












$begingroup$


Suppose



$$(x^2 - 5)^{8} (x+1)^{-16} = 1$$



How do we solve for all values of $x$?



I think we can equate the bases to $1$ in order to get $$1 times 1 = 1$$



$$x^2-5 = 1, x^2-5=-1$$



or



$$x + 1 = 1, x + 1 = -1$$



Could you assist me with this?










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Suppose



    $$(x^2 - 5)^{8} (x+1)^{-16} = 1$$



    How do we solve for all values of $x$?



    I think we can equate the bases to $1$ in order to get $$1 times 1 = 1$$



    $$x^2-5 = 1, x^2-5=-1$$



    or



    $$x + 1 = 1, x + 1 = -1$$



    Could you assist me with this?










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Suppose



      $$(x^2 - 5)^{8} (x+1)^{-16} = 1$$



      How do we solve for all values of $x$?



      I think we can equate the bases to $1$ in order to get $$1 times 1 = 1$$



      $$x^2-5 = 1, x^2-5=-1$$



      or



      $$x + 1 = 1, x + 1 = -1$$



      Could you assist me with this?










      share|cite|improve this question









      $endgroup$




      Suppose



      $$(x^2 - 5)^{8} (x+1)^{-16} = 1$$



      How do we solve for all values of $x$?



      I think we can equate the bases to $1$ in order to get $$1 times 1 = 1$$



      $$x^2-5 = 1, x^2-5=-1$$



      or



      $$x + 1 = 1, x + 1 = -1$$



      Could you assist me with this?







      algebra-precalculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 18 '18 at 21:35









      EnzoEnzo

      19917




      19917






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          $(x^2 - 5)^{8} (x+1)^{-16} = 1$ means



          $(x^2 - 5)^{8} = (x+1)^{16}$.



          Assuming you only want real and not complex answers then



          $ sqrt[8]{(x^2 - 5)^{8}} = sqrt[8]{(x+1)^{16}}$



          $|x^2 - 5| = (x+1)^2$



          or either A) $x^2 - 5 = (x+1)^2$ or B) $5- x^2 = (x+1)^2$



          A) $x^2 - 5 = x^2 + 2x + 1$



          $2x = -6$



          $x = -3$.



          Verification: $(x^2 - 5)^8(x+1)^{-16}=frac{(9-5)^8}{(-2)^{16}}= frac {4^8}{2^{16}}=frac {2^{16}}{2^{16}} = 1$.



          B)$5- x^2 = (x+1)^2$



          $5-x^2 = x^2 + 2x + 1$



          $2x^2 + 2x -4 = 0$



          $x^2 + x -2 = 0$



          $(x +2)(x-1) = 0$ so either C) $x+2 = 0$ and $x=-2$ or $x-1=0$ and $x = 1$.



          So solutions are $x = -3; x = -2;$ or $x = 1$.



          Verification:



          $(x^2 - 5)^8(x+1)^{-16} = frac {(4-5)^8}{(-2 + 1)^{16}}= frac {(-1)^8}{(-1)^{16}} = frac 11 = 1$. so $x = -2$ is a solution.



          $(x^2 - 5)^8(x+1)^{-16} = frac {(1-5)^8}{(1 + 1)^{16}}= frac {(-4)^8}{(2)^{16}} = frac {4^8}{4^8} = 1$. so $x = 1$ is a solution.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            We have $(x^2-5)^8(x+1)^{-16}=1$, so $left(frac{x^2-5}{(x+1)^2}right)^8=1$, so $frac{x^2-5}{(x+1)^2}$ is an $8$'th root of unity. Assuming you need to solve over the reals, we get $frac{x^2-5}{(x+1)^2}in{-1,1}$. Fixing for $-1$ or $1$, you can multiply both sides by $(x+1)^2$ to get a quadratic equation.






            share|cite|improve this answer











            $endgroup$





















              1












              $begingroup$

              $(x^2-5)^8=(x+1)^{16}$



              So



              $[(x^2-5)^4-(x+1)^8] [(x^2-5)^4+(x+1)^8]=0$



              If and only if



              $[(x^2-5)^4-(x+1)^8]=0$



              If and only if



              $[(x^2-5)^2-(x+1)^4]=0$



              If and only if



              $[(x^2-5)-(x+1)^2] [(x^2-5)+(x+1)^2]=0$



              So



              $[-2x-6][2x^2+2x-4]=0$






              share|cite|improve this answer









              $endgroup$













                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%2f3045737%2fhow-do-we-solve-for-all-values-of-x%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









                3












                $begingroup$

                $(x^2 - 5)^{8} (x+1)^{-16} = 1$ means



                $(x^2 - 5)^{8} = (x+1)^{16}$.



                Assuming you only want real and not complex answers then



                $ sqrt[8]{(x^2 - 5)^{8}} = sqrt[8]{(x+1)^{16}}$



                $|x^2 - 5| = (x+1)^2$



                or either A) $x^2 - 5 = (x+1)^2$ or B) $5- x^2 = (x+1)^2$



                A) $x^2 - 5 = x^2 + 2x + 1$



                $2x = -6$



                $x = -3$.



                Verification: $(x^2 - 5)^8(x+1)^{-16}=frac{(9-5)^8}{(-2)^{16}}= frac {4^8}{2^{16}}=frac {2^{16}}{2^{16}} = 1$.



                B)$5- x^2 = (x+1)^2$



                $5-x^2 = x^2 + 2x + 1$



                $2x^2 + 2x -4 = 0$



                $x^2 + x -2 = 0$



                $(x +2)(x-1) = 0$ so either C) $x+2 = 0$ and $x=-2$ or $x-1=0$ and $x = 1$.



                So solutions are $x = -3; x = -2;$ or $x = 1$.



                Verification:



                $(x^2 - 5)^8(x+1)^{-16} = frac {(4-5)^8}{(-2 + 1)^{16}}= frac {(-1)^8}{(-1)^{16}} = frac 11 = 1$. so $x = -2$ is a solution.



                $(x^2 - 5)^8(x+1)^{-16} = frac {(1-5)^8}{(1 + 1)^{16}}= frac {(-4)^8}{(2)^{16}} = frac {4^8}{4^8} = 1$. so $x = 1$ is a solution.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  $(x^2 - 5)^{8} (x+1)^{-16} = 1$ means



                  $(x^2 - 5)^{8} = (x+1)^{16}$.



                  Assuming you only want real and not complex answers then



                  $ sqrt[8]{(x^2 - 5)^{8}} = sqrt[8]{(x+1)^{16}}$



                  $|x^2 - 5| = (x+1)^2$



                  or either A) $x^2 - 5 = (x+1)^2$ or B) $5- x^2 = (x+1)^2$



                  A) $x^2 - 5 = x^2 + 2x + 1$



                  $2x = -6$



                  $x = -3$.



                  Verification: $(x^2 - 5)^8(x+1)^{-16}=frac{(9-5)^8}{(-2)^{16}}= frac {4^8}{2^{16}}=frac {2^{16}}{2^{16}} = 1$.



                  B)$5- x^2 = (x+1)^2$



                  $5-x^2 = x^2 + 2x + 1$



                  $2x^2 + 2x -4 = 0$



                  $x^2 + x -2 = 0$



                  $(x +2)(x-1) = 0$ so either C) $x+2 = 0$ and $x=-2$ or $x-1=0$ and $x = 1$.



                  So solutions are $x = -3; x = -2;$ or $x = 1$.



                  Verification:



                  $(x^2 - 5)^8(x+1)^{-16} = frac {(4-5)^8}{(-2 + 1)^{16}}= frac {(-1)^8}{(-1)^{16}} = frac 11 = 1$. so $x = -2$ is a solution.



                  $(x^2 - 5)^8(x+1)^{-16} = frac {(1-5)^8}{(1 + 1)^{16}}= frac {(-4)^8}{(2)^{16}} = frac {4^8}{4^8} = 1$. so $x = 1$ is a solution.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    $(x^2 - 5)^{8} (x+1)^{-16} = 1$ means



                    $(x^2 - 5)^{8} = (x+1)^{16}$.



                    Assuming you only want real and not complex answers then



                    $ sqrt[8]{(x^2 - 5)^{8}} = sqrt[8]{(x+1)^{16}}$



                    $|x^2 - 5| = (x+1)^2$



                    or either A) $x^2 - 5 = (x+1)^2$ or B) $5- x^2 = (x+1)^2$



                    A) $x^2 - 5 = x^2 + 2x + 1$



                    $2x = -6$



                    $x = -3$.



                    Verification: $(x^2 - 5)^8(x+1)^{-16}=frac{(9-5)^8}{(-2)^{16}}= frac {4^8}{2^{16}}=frac {2^{16}}{2^{16}} = 1$.



                    B)$5- x^2 = (x+1)^2$



                    $5-x^2 = x^2 + 2x + 1$



                    $2x^2 + 2x -4 = 0$



                    $x^2 + x -2 = 0$



                    $(x +2)(x-1) = 0$ so either C) $x+2 = 0$ and $x=-2$ or $x-1=0$ and $x = 1$.



                    So solutions are $x = -3; x = -2;$ or $x = 1$.



                    Verification:



                    $(x^2 - 5)^8(x+1)^{-16} = frac {(4-5)^8}{(-2 + 1)^{16}}= frac {(-1)^8}{(-1)^{16}} = frac 11 = 1$. so $x = -2$ is a solution.



                    $(x^2 - 5)^8(x+1)^{-16} = frac {(1-5)^8}{(1 + 1)^{16}}= frac {(-4)^8}{(2)^{16}} = frac {4^8}{4^8} = 1$. so $x = 1$ is a solution.






                    share|cite|improve this answer









                    $endgroup$



                    $(x^2 - 5)^{8} (x+1)^{-16} = 1$ means



                    $(x^2 - 5)^{8} = (x+1)^{16}$.



                    Assuming you only want real and not complex answers then



                    $ sqrt[8]{(x^2 - 5)^{8}} = sqrt[8]{(x+1)^{16}}$



                    $|x^2 - 5| = (x+1)^2$



                    or either A) $x^2 - 5 = (x+1)^2$ or B) $5- x^2 = (x+1)^2$



                    A) $x^2 - 5 = x^2 + 2x + 1$



                    $2x = -6$



                    $x = -3$.



                    Verification: $(x^2 - 5)^8(x+1)^{-16}=frac{(9-5)^8}{(-2)^{16}}= frac {4^8}{2^{16}}=frac {2^{16}}{2^{16}} = 1$.



                    B)$5- x^2 = (x+1)^2$



                    $5-x^2 = x^2 + 2x + 1$



                    $2x^2 + 2x -4 = 0$



                    $x^2 + x -2 = 0$



                    $(x +2)(x-1) = 0$ so either C) $x+2 = 0$ and $x=-2$ or $x-1=0$ and $x = 1$.



                    So solutions are $x = -3; x = -2;$ or $x = 1$.



                    Verification:



                    $(x^2 - 5)^8(x+1)^{-16} = frac {(4-5)^8}{(-2 + 1)^{16}}= frac {(-1)^8}{(-1)^{16}} = frac 11 = 1$. so $x = -2$ is a solution.



                    $(x^2 - 5)^8(x+1)^{-16} = frac {(1-5)^8}{(1 + 1)^{16}}= frac {(-4)^8}{(2)^{16}} = frac {4^8}{4^8} = 1$. so $x = 1$ is a solution.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 18 '18 at 21:56









                    fleabloodfleablood

                    71.6k22686




                    71.6k22686























                        2












                        $begingroup$

                        We have $(x^2-5)^8(x+1)^{-16}=1$, so $left(frac{x^2-5}{(x+1)^2}right)^8=1$, so $frac{x^2-5}{(x+1)^2}$ is an $8$'th root of unity. Assuming you need to solve over the reals, we get $frac{x^2-5}{(x+1)^2}in{-1,1}$. Fixing for $-1$ or $1$, you can multiply both sides by $(x+1)^2$ to get a quadratic equation.






                        share|cite|improve this answer











                        $endgroup$


















                          2












                          $begingroup$

                          We have $(x^2-5)^8(x+1)^{-16}=1$, so $left(frac{x^2-5}{(x+1)^2}right)^8=1$, so $frac{x^2-5}{(x+1)^2}$ is an $8$'th root of unity. Assuming you need to solve over the reals, we get $frac{x^2-5}{(x+1)^2}in{-1,1}$. Fixing for $-1$ or $1$, you can multiply both sides by $(x+1)^2$ to get a quadratic equation.






                          share|cite|improve this answer











                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            We have $(x^2-5)^8(x+1)^{-16}=1$, so $left(frac{x^2-5}{(x+1)^2}right)^8=1$, so $frac{x^2-5}{(x+1)^2}$ is an $8$'th root of unity. Assuming you need to solve over the reals, we get $frac{x^2-5}{(x+1)^2}in{-1,1}$. Fixing for $-1$ or $1$, you can multiply both sides by $(x+1)^2$ to get a quadratic equation.






                            share|cite|improve this answer











                            $endgroup$



                            We have $(x^2-5)^8(x+1)^{-16}=1$, so $left(frac{x^2-5}{(x+1)^2}right)^8=1$, so $frac{x^2-5}{(x+1)^2}$ is an $8$'th root of unity. Assuming you need to solve over the reals, we get $frac{x^2-5}{(x+1)^2}in{-1,1}$. Fixing for $-1$ or $1$, you can multiply both sides by $(x+1)^2$ to get a quadratic equation.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 18 '18 at 22:46

























                            answered Dec 18 '18 at 21:47









                            SmileyCraftSmileyCraft

                            3,611517




                            3,611517























                                1












                                $begingroup$

                                $(x^2-5)^8=(x+1)^{16}$



                                So



                                $[(x^2-5)^4-(x+1)^8] [(x^2-5)^4+(x+1)^8]=0$



                                If and only if



                                $[(x^2-5)^4-(x+1)^8]=0$



                                If and only if



                                $[(x^2-5)^2-(x+1)^4]=0$



                                If and only if



                                $[(x^2-5)-(x+1)^2] [(x^2-5)+(x+1)^2]=0$



                                So



                                $[-2x-6][2x^2+2x-4]=0$






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  $(x^2-5)^8=(x+1)^{16}$



                                  So



                                  $[(x^2-5)^4-(x+1)^8] [(x^2-5)^4+(x+1)^8]=0$



                                  If and only if



                                  $[(x^2-5)^4-(x+1)^8]=0$



                                  If and only if



                                  $[(x^2-5)^2-(x+1)^4]=0$



                                  If and only if



                                  $[(x^2-5)-(x+1)^2] [(x^2-5)+(x+1)^2]=0$



                                  So



                                  $[-2x-6][2x^2+2x-4]=0$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    $(x^2-5)^8=(x+1)^{16}$



                                    So



                                    $[(x^2-5)^4-(x+1)^8] [(x^2-5)^4+(x+1)^8]=0$



                                    If and only if



                                    $[(x^2-5)^4-(x+1)^8]=0$



                                    If and only if



                                    $[(x^2-5)^2-(x+1)^4]=0$



                                    If and only if



                                    $[(x^2-5)-(x+1)^2] [(x^2-5)+(x+1)^2]=0$



                                    So



                                    $[-2x-6][2x^2+2x-4]=0$






                                    share|cite|improve this answer









                                    $endgroup$



                                    $(x^2-5)^8=(x+1)^{16}$



                                    So



                                    $[(x^2-5)^4-(x+1)^8] [(x^2-5)^4+(x+1)^8]=0$



                                    If and only if



                                    $[(x^2-5)^4-(x+1)^8]=0$



                                    If and only if



                                    $[(x^2-5)^2-(x+1)^4]=0$



                                    If and only if



                                    $[(x^2-5)-(x+1)^2] [(x^2-5)+(x+1)^2]=0$



                                    So



                                    $[-2x-6][2x^2+2x-4]=0$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 18 '18 at 21:40









                                    Federico FalluccaFederico Fallucca

                                    2,191210




                                    2,191210






























                                        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%2f3045737%2fhow-do-we-solve-for-all-values-of-x%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

                                        Probability when a professor distributes a quiz and homework assignment to a class of n students.

                                        Aardman Animations

                                        Are they similar matrix