Solving polynominals equations (relationship of roots)Quadratic equation - $alpha$ and $beta$ RootsTechnique...

Should the Death Curse affect an undead PC in the Tomb of Annihilation adventure?

Get consecutive integer number ranges from list of int

Partitioning the Reals into two Locally Uncountable, Dense Sets

Minor Revision with suggestion of an alternative proof by reviewer

How exactly does Hawking radiation decrease the mass of black holes?

"The cow" OR "a cow" OR "cows" in this context

Why does Mind Blank stop the Feeblemind spell?

Alignment of various blocks in tikz

Can SQL Server create collisions in system generated constraint names?

How can Republicans who favour free markets, consistently express anger when they don't like the outcome of that choice?

Why was the Spitfire's elliptical wing almost uncopied by other aircraft of World War 2?

Do I have an "anti-research" personality?

What makes accurate emulation of old systems a difficult task?

How did Captain America manage to do this?

Does tea made with boiling water cool faster than tea made with boiled (but still hot) water?

Does a large simulator bay have standard public address announcements?

Check if a string is entirely made of the same substring

How to fry ground beef so it is well-browned

A strange hotel

Critique of timeline aesthetic

I preordered a game on my Xbox while on the home screen of my friend's account. Which of us owns the game?

Size of electromagnet needed to replicate Earth's magnetic field

Phrase for the opposite of "foolproof"

How do I check if a string is entirely made of the same substring?



Solving polynominals equations (relationship of roots)


Quadratic equation - $alpha$ and $beta$ RootsTechnique to simplify algebraic calculations on roots of polynomialInterval of Polynomial Root FindingFind $alpha^3 + beta^3$ which are roots of a quadratic equation.sum and product of roots of polynomials: finding equations for rootsSolving two Cubic Equation on their Roots.Finding an equation with related rootsFind the roots of $acx^2-b(c+a)x+(c+a)^2=0$If $3x^2-6x+p=0$ has roots $alpha$ and $beta$, then find a quadratic with roots $(alpha+beta)/alpha$ and $(alpha+beta)/beta$Find the roots of $3x^3-4x-8$













2












$begingroup$



The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$frac{alpha+beta}{omega}+frac{alpha+omega}{beta}+frac{beta+omega}{alpha}$$




So far I have found:
$$alpha+beta+omega=frac{-b}{a} = 4 \
alphabeta+betaomega+alphaomega=frac{c}{a} = 1 \
alpha×beta×omega=frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$frac{alpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2}{alphabetaomega}$$



I don't know how to continue evaluating the question.



Note:

The answer I have been given is $-dfrac{11}{3}$






polynomials roots







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For latex, you use instead of /.
    $endgroup$
    – BadAtGeometry
    1 hour ago
















2












$begingroup$



The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$frac{alpha+beta}{omega}+frac{alpha+omega}{beta}+frac{beta+omega}{alpha}$$




So far I have found:
$$alpha+beta+omega=frac{-b}{a} = 4 \
alphabeta+betaomega+alphaomega=frac{c}{a} = 1 \
alpha×beta×omega=frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$frac{alpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2}{alphabetaomega}$$



I don't know how to continue evaluating the question.



Note:

The answer I have been given is $-dfrac{11}{3}$






polynomials roots







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For latex, you use instead of /.
    $endgroup$
    – BadAtGeometry
    1 hour ago














2












2








2


0



$begingroup$



The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$frac{alpha+beta}{omega}+frac{alpha+omega}{beta}+frac{beta+omega}{alpha}$$




So far I have found:
$$alpha+beta+omega=frac{-b}{a} = 4 \
alphabeta+betaomega+alphaomega=frac{c}{a} = 1 \
alpha×beta×omega=frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$frac{alpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2}{alphabetaomega}$$



I don't know how to continue evaluating the question.



Note:

The answer I have been given is $-dfrac{11}{3}$






polynomials roots







share|cite|improve this question











$endgroup$





The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$frac{alpha+beta}{omega}+frac{alpha+omega}{beta}+frac{beta+omega}{alpha}$$




So far I have found:
$$alpha+beta+omega=frac{-b}{a} = 4 \
alphabeta+betaomega+alphaomega=frac{c}{a} = 1 \
alpha×beta×omega=frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$frac{alpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2}{alphabetaomega}$$



I don't know how to continue evaluating the question.



Note:

The answer I have been given is $-dfrac{11}{3}$






polynomials roots




polynomials roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 34 mins ago









Lee David Chung Lin

4,51851342




4,51851342










asked 1 hour ago









Alex Alex

186




186








  • 1




    $begingroup$
    For latex, you use instead of /.
    $endgroup$
    – BadAtGeometry
    1 hour ago














  • 1




    $begingroup$
    For latex, you use instead of /.
    $endgroup$
    – BadAtGeometry
    1 hour ago








1




1




$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
1 hour ago




$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
1 hour ago










4 Answers
4






active

oldest

votes


















1












$begingroup$

$$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$



$$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$



$$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$



$$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$



$$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$



I think you should be able to take it from there.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
    This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Alternatively, you can solve the equation:
        $$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
        alpha =-1, beta =2,omega=3.$$
        Hence:
        $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
        frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
        frac13-5+1=\
        -frac{11}{3}.$$






        share|cite|improve this answer









        $endgroup$














          Your Answer








          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%2f3204072%2fsolving-polynominals-equations-relationship-of-roots%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












          $begingroup$

          $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$



          $$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$



          $$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$



          $$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$



          $$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$



          I think you should be able to take it from there.






          share|cite|improve this answer









          $endgroup$


















            1












            $begingroup$

            $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$



            $$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$



            $$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$



            $$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$



            $$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$



            I think you should be able to take it from there.






            share|cite|improve this answer









            $endgroup$
















              1












              1








              1





              $begingroup$

              $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$



              $$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$



              $$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$



              $$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$



              $$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$



              I think you should be able to take it from there.






              share|cite|improve this answer









              $endgroup$



              $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$



              $$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$



              $$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$



              $$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$



              $$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$



              I think you should be able to take it from there.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 1 hour ago









              user1952500user1952500

              1,5251016




              1,5251016























                  2












                  $begingroup$

                  Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
                  This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
                    This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
                      This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$






                      share|cite|improve this answer









                      $endgroup$



                      Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
                      This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 1 hour ago









                      Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                      79.7k42867




                      79.7k42867























                          0












                          $begingroup$

                          That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.






                              share|cite|improve this answer









                              $endgroup$



                              That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 1 hour ago









                              Chris CusterChris Custer

                              14.7k3827




                              14.7k3827























                                  0












                                  $begingroup$

                                  Alternatively, you can solve the equation:
                                  $$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
                                  alpha =-1, beta =2,omega=3.$$
                                  Hence:
                                  $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
                                  frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
                                  frac13-5+1=\
                                  -frac{11}{3}.$$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Alternatively, you can solve the equation:
                                    $$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
                                    alpha =-1, beta =2,omega=3.$$
                                    Hence:
                                    $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
                                    frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
                                    frac13-5+1=\
                                    -frac{11}{3}.$$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Alternatively, you can solve the equation:
                                      $$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
                                      alpha =-1, beta =2,omega=3.$$
                                      Hence:
                                      $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
                                      frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
                                      frac13-5+1=\
                                      -frac{11}{3}.$$






                                      share|cite|improve this answer









                                      $endgroup$



                                      Alternatively, you can solve the equation:
                                      $$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
                                      alpha =-1, beta =2,omega=3.$$
                                      Hence:
                                      $$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
                                      frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
                                      frac13-5+1=\
                                      -frac{11}{3}.$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 11 mins ago









                                      farruhotafarruhota

                                      22.5k2942




                                      22.5k2942






























                                          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%2f3204072%2fsolving-polynominals-equations-relationship-of-roots%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

                                          El tren de la libertad Índice Antecedentes "Porque yo decido" Desarrollo de la...

                                          Image
                                          Feminismo en EspañaAborto en EspañaManifestaciones y protestas en EspañaManifestaciones y protestas en 2014España en 2014Historia de las mujeres en España movimiento socialfeministaderechos sexuales y reproductivosmujeresLey del aborto en España1 de febrero2014MadridAlberto Ruíz Gallardónla ley del aborto vigente del 2010AsturiasTertulia Feminista Les ComadresCongreso de DiputadosAlicia MiyaresAna MatoAlberto Ruiz GallardónEdimburgoRomaParísLuxemburgoBuenos AiresGracia QuerejetaÁngeles González-SindeMabel LozanoIcíar BollaínIsabel CoixetGeorgina CisquellaTeresa FontVirginia Yagüe2010José Luis Rodríguez ZapateroCongreso de Diputados19 de diciembre20095 de julio2010Partido PopularTribunal Constitucional201120 de diciembre2013Partido PopularMariano RajoyGijónTertulia Feminista Les ComadresBegoña PiñeroRenfeAlicia MiyaresTertulia Feminista Les Comadres1 de febrero2014Congreso de DiputadosMariano RajoyAlberto Ruiz GallardónAna MatoAsturiasAndalucíaCanariasCataluñaComunidad ValencianaCasti
                                          Read more

                                          Puerta de Hutt Referencias Enlaces externos Menú de navegación15°58′00″S 5°42′00″O /...

                                          Image
                                          Localidades de Santa Elena inglésasentamientoisla Santa ElenaLongwoodIglesia de San MateoPico de DianaNapoleón BonaparteHenri Gratien BertrandAlarma ForestalLevelwoodSan Pablo Puerta de Hutt Hutt's Gate Asentamiento Puerta de Hutt Hutt's Gate Coordenadas 15°58′00″S 5°42′00″O  /  -15.96666667, -5.7 Coordenadas: 15°58′00″S 5°42′00″O  /  -15.96666667, -5.7 Entidad Asentamiento  • País   Reino Unido  • Territorio de ultramar   Santa Elena, Ascensión y Tristán de Acuña  • Isla Santa Elena Huso horario UTC+0 [editar datos en Wikidata] Puerta de Hutt (en inglés: Hutt's Gate ) es un pequeño asentamiento ubicado en la isla Santa Elena, que forma parte del distrito de Longwood. Se ubica en las coordenadas 15°58′00″S 5°42′00″O  /  -15.96667, -5.70000 . [ 1 ] ​ Aquí se localiza la Iglesia de San Mateo y en sus cercanías el Pico de Diana [ 2 ] ​ y la tumba de Napoleón Bonaparte. Es una amplia zona residencial y cuent
                                          Read more

                                          Castillo d'Acher Características Menú de navegación

                                          Image
                                          Montañas de los Pirineos de AragónMontañas de la provincia de Huesca msnmSelva de OzaPirineo aragonésEspañabosque Este artículo o sección necesita referencias que aparezcan en una publicación acreditada. Este aviso fue puesto el 11 de agosto de 2011. El pico desde Selva de Oza. El Castillo d' Acher es una cima de 2.384msnm en la zona de la Selva de Oza en la parte occidental del Pirineo aragonés, en España. Debe su nombre a su aspecto de fortaleza que presenta hacia todas sus vertientes. Características En orden ascendente, la cima pasa de un bosque frondoso de tonalidades verde-amarillenta a laderas rojizas. Su cima, que suele presentarse mucha nieve, es un pequeño valle rectangular rodeado en sus cuatro caras por una muralla que le da un aspecto de fortaleza. This page is only for reference, If you need detailed information, please check here
                                          Read more