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?






algebra-precalculus







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?






    algebra-precalculus







    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?






      algebra-precalculus







      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




      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
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          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









              $endgroup$














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                3 Answers
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                3 Answers
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                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

                    73.6k22891




                    73.6k22891























                        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

                            3,761519




                            3,761519























                                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

                                    2,280210




                                    2,280210






























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