How to solve $alpha,beta$ in $frac{x}{1-x-x^2}=frac{x}{(1-alpha x)(1-beta x)}$












0












$begingroup$


I'm trying to understand a procedure within solving a generating function problem:



$$G(x) = dfrac{x}{1-x-x^2}=dfrac{x}{(1-alpha x)(1-beta x)}$$



I understand that the reason for this step is to utilize partial fraction s.t.



$$ =dfrac{a}{(1-alpha x)} + dfrac{b}{(1-beta x)}$$



I don't know why turning $(1-x-x^2)$ into $(1-alpha x)(1-beta x)$ is possible, and is there some formula for this case solving $alpha$ and $beta$? The method I'm familiar with is



$$(1-x-x^2)=-(x-r_1)(x-r_2)$$



then I can apply the formula $$r_i=frac{-bpmsqrt{b^2-4ac}}{2a}$$ but I don't know how to deal with the case $(1-alpha x)(1-beta x)$. I need a general method/formula to solve $alpha$ and $beta$...






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    To utilize en.wikipedia.org/wiki/Binomial_series
    $endgroup$
    – lab bhattacharjee
    Dec 21 '18 at 7:18






  • 2




    $begingroup$
    It's just factorisation. Always works over the complex numbers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:20










  • $begingroup$
    @LordSharktheUnknown So this is not related to testing whether $x=0$ is a root (my current idea)?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:21






  • 2




    $begingroup$
    If you have $1-x-x^2=-(x-r_1)(x-r_2)$ then $r_1r_2=1$ and then $1-x-x^2=(1-x/r_2)(1-x/r_1)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:31










  • $begingroup$
    @LordSharktheUnknown: Oh so since $r_1r_2=1implies r_1,r_2not=0$ so I can divide it by $r_1cdot r_2$?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:33
















0












$begingroup$


I'm trying to understand a procedure within solving a generating function problem:



$$G(x) = dfrac{x}{1-x-x^2}=dfrac{x}{(1-alpha x)(1-beta x)}$$



I understand that the reason for this step is to utilize partial fraction s.t.



$$ =dfrac{a}{(1-alpha x)} + dfrac{b}{(1-beta x)}$$



I don't know why turning $(1-x-x^2)$ into $(1-alpha x)(1-beta x)$ is possible, and is there some formula for this case solving $alpha$ and $beta$? The method I'm familiar with is



$$(1-x-x^2)=-(x-r_1)(x-r_2)$$



then I can apply the formula $$r_i=frac{-bpmsqrt{b^2-4ac}}{2a}$$ but I don't know how to deal with the case $(1-alpha x)(1-beta x)$. I need a general method/formula to solve $alpha$ and $beta$...






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    To utilize en.wikipedia.org/wiki/Binomial_series
    $endgroup$
    – lab bhattacharjee
    Dec 21 '18 at 7:18






  • 2




    $begingroup$
    It's just factorisation. Always works over the complex numbers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:20










  • $begingroup$
    @LordSharktheUnknown So this is not related to testing whether $x=0$ is a root (my current idea)?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:21






  • 2




    $begingroup$
    If you have $1-x-x^2=-(x-r_1)(x-r_2)$ then $r_1r_2=1$ and then $1-x-x^2=(1-x/r_2)(1-x/r_1)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:31










  • $begingroup$
    @LordSharktheUnknown: Oh so since $r_1r_2=1implies r_1,r_2not=0$ so I can divide it by $r_1cdot r_2$?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:33














0












0








0





$begingroup$


I'm trying to understand a procedure within solving a generating function problem:



$$G(x) = dfrac{x}{1-x-x^2}=dfrac{x}{(1-alpha x)(1-beta x)}$$



I understand that the reason for this step is to utilize partial fraction s.t.



$$ =dfrac{a}{(1-alpha x)} + dfrac{b}{(1-beta x)}$$



I don't know why turning $(1-x-x^2)$ into $(1-alpha x)(1-beta x)$ is possible, and is there some formula for this case solving $alpha$ and $beta$? The method I'm familiar with is



$$(1-x-x^2)=-(x-r_1)(x-r_2)$$



then I can apply the formula $$r_i=frac{-bpmsqrt{b^2-4ac}}{2a}$$ but I don't know how to deal with the case $(1-alpha x)(1-beta x)$. I need a general method/formula to solve $alpha$ and $beta$...






algebra-precalculus







share|cite|improve this question











$endgroup$




I'm trying to understand a procedure within solving a generating function problem:



$$G(x) = dfrac{x}{1-x-x^2}=dfrac{x}{(1-alpha x)(1-beta x)}$$



I understand that the reason for this step is to utilize partial fraction s.t.



$$ =dfrac{a}{(1-alpha x)} + dfrac{b}{(1-beta x)}$$



I don't know why turning $(1-x-x^2)$ into $(1-alpha x)(1-beta x)$ is possible, and is there some formula for this case solving $alpha$ and $beta$? The method I'm familiar with is



$$(1-x-x^2)=-(x-r_1)(x-r_2)$$



then I can apply the formula $$r_i=frac{-bpmsqrt{b^2-4ac}}{2a}$$ but I don't know how to deal with the case $(1-alpha x)(1-beta x)$. I need a general method/formula to solve $alpha$ and $beta$...






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 8:45









Lorenzo B.

1,8402520




1,8402520










asked Dec 21 '18 at 7:16









user7813604user7813604

15912




15912












  • $begingroup$
    To utilize en.wikipedia.org/wiki/Binomial_series
    $endgroup$
    – lab bhattacharjee
    Dec 21 '18 at 7:18






  • 2




    $begingroup$
    It's just factorisation. Always works over the complex numbers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:20










  • $begingroup$
    @LordSharktheUnknown So this is not related to testing whether $x=0$ is a root (my current idea)?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:21






  • 2




    $begingroup$
    If you have $1-x-x^2=-(x-r_1)(x-r_2)$ then $r_1r_2=1$ and then $1-x-x^2=(1-x/r_2)(1-x/r_1)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:31










  • $begingroup$
    @LordSharktheUnknown: Oh so since $r_1r_2=1implies r_1,r_2not=0$ so I can divide it by $r_1cdot r_2$?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:33


















  • $begingroup$
    To utilize en.wikipedia.org/wiki/Binomial_series
    $endgroup$
    – lab bhattacharjee
    Dec 21 '18 at 7:18






  • 2




    $begingroup$
    It's just factorisation. Always works over the complex numbers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:20










  • $begingroup$
    @LordSharktheUnknown So this is not related to testing whether $x=0$ is a root (my current idea)?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:21






  • 2




    $begingroup$
    If you have $1-x-x^2=-(x-r_1)(x-r_2)$ then $r_1r_2=1$ and then $1-x-x^2=(1-x/r_2)(1-x/r_1)$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 7:31










  • $begingroup$
    @LordSharktheUnknown: Oh so since $r_1r_2=1implies r_1,r_2not=0$ so I can divide it by $r_1cdot r_2$?
    $endgroup$
    – user7813604
    Dec 21 '18 at 7:33
















$begingroup$
To utilize en.wikipedia.org/wiki/Binomial_series
$endgroup$
– lab bhattacharjee
Dec 21 '18 at 7:18




$begingroup$
To utilize en.wikipedia.org/wiki/Binomial_series
$endgroup$
– lab bhattacharjee
Dec 21 '18 at 7:18




2




2




$begingroup$
It's just factorisation. Always works over the complex numbers.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 7:20




$begingroup$
It's just factorisation. Always works over the complex numbers.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 7:20












$begingroup$
@LordSharktheUnknown So this is not related to testing whether $x=0$ is a root (my current idea)?
$endgroup$
– user7813604
Dec 21 '18 at 7:21




$begingroup$
@LordSharktheUnknown So this is not related to testing whether $x=0$ is a root (my current idea)?
$endgroup$
– user7813604
Dec 21 '18 at 7:21




2




2




$begingroup$
If you have $1-x-x^2=-(x-r_1)(x-r_2)$ then $r_1r_2=1$ and then $1-x-x^2=(1-x/r_2)(1-x/r_1)$.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 7:31




$begingroup$
If you have $1-x-x^2=-(x-r_1)(x-r_2)$ then $r_1r_2=1$ and then $1-x-x^2=(1-x/r_2)(1-x/r_1)$.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 7:31












$begingroup$
@LordSharktheUnknown: Oh so since $r_1r_2=1implies r_1,r_2not=0$ so I can divide it by $r_1cdot r_2$?
$endgroup$
– user7813604
Dec 21 '18 at 7:33




$begingroup$
@LordSharktheUnknown: Oh so since $r_1r_2=1implies r_1,r_2not=0$ so I can divide it by $r_1cdot r_2$?
$endgroup$
– user7813604
Dec 21 '18 at 7:33










4 Answers
4






active

oldest

votes


















1












$begingroup$

Finally I understand @Lord Shark the Unknown's comment, it's



$$(1-x-x^2)=-(x-r_1)(x-r_2)$$



then solving for $r_1,r_2=dfrac{-1pmsqrt5}{2},$ and since $r_1r_2=-1$, simplify to



$$=dfrac{1}{r_1r_2}(x-r_1)(x-r_2)=(dfrac{x}{r_1}-1)(dfrac{x}{r_2}-1)$$



And since $dfrac{1}{r_1}=-r_2$ and $dfrac{1}{r_2}=-r_1$,



begin{align}&=(-r_2x-1)(-r_1x-1)\
&=(dfrac{1+sqrt5}{2}x-1)(dfrac{1-sqrt5}{2}x-1)end{align}



(But why $1+sqrt5over2$ is so strange...?)






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    If you can factor a quadratic trinomial as



    $$(x-alpha')(x-beta')$$ then you can also solve for



    $$(alpha x-1)(beta x-1).$$



    Because



    $$(alpha x-1)(beta x-1)=alphabetaleft(x-frac1alpharight)left(x-frac1betaright)$$



    and the roots in the second factorization are the inverses of those in the first one.



    In other words,



    $$alpha,beta=frac1{alpha'},frac1{beta'}=frac{2a}{-bpmsqrt{b^2-4ac}}=frac{-bpmsqrt{b^2-4ac}}{2c}.$$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      Since $1-x-x^2=0$ has two real roots, which are$$alpha=frac{-1+sqrt{5}}{2},beta=frac{-1-sqrt{5}}{2},$$We can denote$$1-x-x^2=-(x-alpha)(x-beta).$$Thus, we may assume $$frac{x}{1-x-x^2}=frac{A}{x-alpha}+frac{B}{x-beta}.tag1$$
      Multiply $x-alpha$ to the both sides of $(1)$ and then let $x=alpha$. We have$$A=frac{alpha}{beta-alpha}.$$Likewise, multiply $x-beta$ to the both sides of $(1)$ and then let $x=beta$. We have$$B=frac{beta}{alpha-beta}.$$As a result,$$frac{x}{1-x-x^2}=frac{alpha}{(beta-alpha)(x-alpha)}+frac{beta}{(alpha-beta)(x-beta)}.$$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Hint: we have $$1-x-x^2=1-(alpha+beta)x+alphabeta x^2$$therefore $$begin{cases}alpha+beta=1\alphabeta=-1end{cases}$$can you finish now?






        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%2f3048262%2fhow-to-solve-alpha-beta-in-fracx1-x-x2-fracx1-alpha-x1-beta%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Finally I understand @Lord Shark the Unknown's comment, it's



          $$(1-x-x^2)=-(x-r_1)(x-r_2)$$



          then solving for $r_1,r_2=dfrac{-1pmsqrt5}{2},$ and since $r_1r_2=-1$, simplify to



          $$=dfrac{1}{r_1r_2}(x-r_1)(x-r_2)=(dfrac{x}{r_1}-1)(dfrac{x}{r_2}-1)$$



          And since $dfrac{1}{r_1}=-r_2$ and $dfrac{1}{r_2}=-r_1$,



          begin{align}&=(-r_2x-1)(-r_1x-1)\
          &=(dfrac{1+sqrt5}{2}x-1)(dfrac{1-sqrt5}{2}x-1)end{align}



          (But why $1+sqrt5over2$ is so strange...?)






          share|cite|improve this answer











          $endgroup$


















            1












            $begingroup$

            Finally I understand @Lord Shark the Unknown's comment, it's



            $$(1-x-x^2)=-(x-r_1)(x-r_2)$$



            then solving for $r_1,r_2=dfrac{-1pmsqrt5}{2},$ and since $r_1r_2=-1$, simplify to



            $$=dfrac{1}{r_1r_2}(x-r_1)(x-r_2)=(dfrac{x}{r_1}-1)(dfrac{x}{r_2}-1)$$



            And since $dfrac{1}{r_1}=-r_2$ and $dfrac{1}{r_2}=-r_1$,



            begin{align}&=(-r_2x-1)(-r_1x-1)\
            &=(dfrac{1+sqrt5}{2}x-1)(dfrac{1-sqrt5}{2}x-1)end{align}



            (But why $1+sqrt5over2$ is so strange...?)






            share|cite|improve this answer











            $endgroup$
















              1












              1








              1





              $begingroup$

              Finally I understand @Lord Shark the Unknown's comment, it's



              $$(1-x-x^2)=-(x-r_1)(x-r_2)$$



              then solving for $r_1,r_2=dfrac{-1pmsqrt5}{2},$ and since $r_1r_2=-1$, simplify to



              $$=dfrac{1}{r_1r_2}(x-r_1)(x-r_2)=(dfrac{x}{r_1}-1)(dfrac{x}{r_2}-1)$$



              And since $dfrac{1}{r_1}=-r_2$ and $dfrac{1}{r_2}=-r_1$,



              begin{align}&=(-r_2x-1)(-r_1x-1)\
              &=(dfrac{1+sqrt5}{2}x-1)(dfrac{1-sqrt5}{2}x-1)end{align}



              (But why $1+sqrt5over2$ is so strange...?)






              share|cite|improve this answer











              $endgroup$



              Finally I understand @Lord Shark the Unknown's comment, it's



              $$(1-x-x^2)=-(x-r_1)(x-r_2)$$



              then solving for $r_1,r_2=dfrac{-1pmsqrt5}{2},$ and since $r_1r_2=-1$, simplify to



              $$=dfrac{1}{r_1r_2}(x-r_1)(x-r_2)=(dfrac{x}{r_1}-1)(dfrac{x}{r_2}-1)$$



              And since $dfrac{1}{r_1}=-r_2$ and $dfrac{1}{r_2}=-r_1$,



              begin{align}&=(-r_2x-1)(-r_1x-1)\
              &=(dfrac{1+sqrt5}{2}x-1)(dfrac{1-sqrt5}{2}x-1)end{align}



              (But why $1+sqrt5over2$ is so strange...?)







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              answered Dec 21 '18 at 8:04


























              community wiki





              user7813604
























                  1












                  $begingroup$

                  If you can factor a quadratic trinomial as



                  $$(x-alpha')(x-beta')$$ then you can also solve for



                  $$(alpha x-1)(beta x-1).$$



                  Because



                  $$(alpha x-1)(beta x-1)=alphabetaleft(x-frac1alpharight)left(x-frac1betaright)$$



                  and the roots in the second factorization are the inverses of those in the first one.



                  In other words,



                  $$alpha,beta=frac1{alpha'},frac1{beta'}=frac{2a}{-bpmsqrt{b^2-4ac}}=frac{-bpmsqrt{b^2-4ac}}{2c}.$$






                  share|cite|improve this answer











                  $endgroup$


















                    1












                    $begingroup$

                    If you can factor a quadratic trinomial as



                    $$(x-alpha')(x-beta')$$ then you can also solve for



                    $$(alpha x-1)(beta x-1).$$



                    Because



                    $$(alpha x-1)(beta x-1)=alphabetaleft(x-frac1alpharight)left(x-frac1betaright)$$



                    and the roots in the second factorization are the inverses of those in the first one.



                    In other words,



                    $$alpha,beta=frac1{alpha'},frac1{beta'}=frac{2a}{-bpmsqrt{b^2-4ac}}=frac{-bpmsqrt{b^2-4ac}}{2c}.$$






                    share|cite|improve this answer











                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      If you can factor a quadratic trinomial as



                      $$(x-alpha')(x-beta')$$ then you can also solve for



                      $$(alpha x-1)(beta x-1).$$



                      Because



                      $$(alpha x-1)(beta x-1)=alphabetaleft(x-frac1alpharight)left(x-frac1betaright)$$



                      and the roots in the second factorization are the inverses of those in the first one.



                      In other words,



                      $$alpha,beta=frac1{alpha'},frac1{beta'}=frac{2a}{-bpmsqrt{b^2-4ac}}=frac{-bpmsqrt{b^2-4ac}}{2c}.$$






                      share|cite|improve this answer











                      $endgroup$



                      If you can factor a quadratic trinomial as



                      $$(x-alpha')(x-beta')$$ then you can also solve for



                      $$(alpha x-1)(beta x-1).$$



                      Because



                      $$(alpha x-1)(beta x-1)=alphabetaleft(x-frac1alpharight)left(x-frac1betaright)$$



                      and the roots in the second factorization are the inverses of those in the first one.



                      In other words,



                      $$alpha,beta=frac1{alpha'},frac1{beta'}=frac{2a}{-bpmsqrt{b^2-4ac}}=frac{-bpmsqrt{b^2-4ac}}{2c}.$$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 21 '18 at 8:29

























                      answered Dec 21 '18 at 8:23









                      Yves DaoustYves Daoust

                      128k675227




                      128k675227























                          0












                          $begingroup$

                          Since $1-x-x^2=0$ has two real roots, which are$$alpha=frac{-1+sqrt{5}}{2},beta=frac{-1-sqrt{5}}{2},$$We can denote$$1-x-x^2=-(x-alpha)(x-beta).$$Thus, we may assume $$frac{x}{1-x-x^2}=frac{A}{x-alpha}+frac{B}{x-beta}.tag1$$
                          Multiply $x-alpha$ to the both sides of $(1)$ and then let $x=alpha$. We have$$A=frac{alpha}{beta-alpha}.$$Likewise, multiply $x-beta$ to the both sides of $(1)$ and then let $x=beta$. We have$$B=frac{beta}{alpha-beta}.$$As a result,$$frac{x}{1-x-x^2}=frac{alpha}{(beta-alpha)(x-alpha)}+frac{beta}{(alpha-beta)(x-beta)}.$$






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            Since $1-x-x^2=0$ has two real roots, which are$$alpha=frac{-1+sqrt{5}}{2},beta=frac{-1-sqrt{5}}{2},$$We can denote$$1-x-x^2=-(x-alpha)(x-beta).$$Thus, we may assume $$frac{x}{1-x-x^2}=frac{A}{x-alpha}+frac{B}{x-beta}.tag1$$
                            Multiply $x-alpha$ to the both sides of $(1)$ and then let $x=alpha$. We have$$A=frac{alpha}{beta-alpha}.$$Likewise, multiply $x-beta$ to the both sides of $(1)$ and then let $x=beta$. We have$$B=frac{beta}{alpha-beta}.$$As a result,$$frac{x}{1-x-x^2}=frac{alpha}{(beta-alpha)(x-alpha)}+frac{beta}{(alpha-beta)(x-beta)}.$$






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Since $1-x-x^2=0$ has two real roots, which are$$alpha=frac{-1+sqrt{5}}{2},beta=frac{-1-sqrt{5}}{2},$$We can denote$$1-x-x^2=-(x-alpha)(x-beta).$$Thus, we may assume $$frac{x}{1-x-x^2}=frac{A}{x-alpha}+frac{B}{x-beta}.tag1$$
                              Multiply $x-alpha$ to the both sides of $(1)$ and then let $x=alpha$. We have$$A=frac{alpha}{beta-alpha}.$$Likewise, multiply $x-beta$ to the both sides of $(1)$ and then let $x=beta$. We have$$B=frac{beta}{alpha-beta}.$$As a result,$$frac{x}{1-x-x^2}=frac{alpha}{(beta-alpha)(x-alpha)}+frac{beta}{(alpha-beta)(x-beta)}.$$






                              share|cite|improve this answer









                              $endgroup$



                              Since $1-x-x^2=0$ has two real roots, which are$$alpha=frac{-1+sqrt{5}}{2},beta=frac{-1-sqrt{5}}{2},$$We can denote$$1-x-x^2=-(x-alpha)(x-beta).$$Thus, we may assume $$frac{x}{1-x-x^2}=frac{A}{x-alpha}+frac{B}{x-beta}.tag1$$
                              Multiply $x-alpha$ to the both sides of $(1)$ and then let $x=alpha$. We have$$A=frac{alpha}{beta-alpha}.$$Likewise, multiply $x-beta$ to the both sides of $(1)$ and then let $x=beta$. We have$$B=frac{beta}{alpha-beta}.$$As a result,$$frac{x}{1-x-x^2}=frac{alpha}{(beta-alpha)(x-alpha)}+frac{beta}{(alpha-beta)(x-beta)}.$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 21 '18 at 8:09









                              mengdie1982mengdie1982

                              4,897618




                              4,897618























                                  0












                                  $begingroup$

                                  Hint: we have $$1-x-x^2=1-(alpha+beta)x+alphabeta x^2$$therefore $$begin{cases}alpha+beta=1\alphabeta=-1end{cases}$$can you finish now?






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Hint: we have $$1-x-x^2=1-(alpha+beta)x+alphabeta x^2$$therefore $$begin{cases}alpha+beta=1\alphabeta=-1end{cases}$$can you finish now?






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Hint: we have $$1-x-x^2=1-(alpha+beta)x+alphabeta x^2$$therefore $$begin{cases}alpha+beta=1\alphabeta=-1end{cases}$$can you finish now?






                                      share|cite|improve this answer









                                      $endgroup$



                                      Hint: we have $$1-x-x^2=1-(alpha+beta)x+alphabeta x^2$$therefore $$begin{cases}alpha+beta=1\alphabeta=-1end{cases}$$can you finish now?







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 21 '18 at 10:30









                                      Mostafa AyazMostafa Ayaz

                                      15.6k3939




                                      15.6k3939






























                                          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%2f3048262%2fhow-to-solve-alpha-beta-in-fracx1-x-x2-fracx1-alpha-x1-beta%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

                                          Ellipse (mathématiques)

                                          Image
                                          Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti
                                          Read more

                                          Quarter-circle Tiles

                                          Image
                                          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
                                          Read more

                                          Mont Emei

                                          Image
                                          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
                                          Read more