Given the equation $P(x) = x^4 + bx^2 + c$ and the points $(3,25)$ and $(1,9).$ How to find $b$ and $c$ such...












0














By using Desmos I've found that the values of $;b = -8;$ and $;c = 16,;$ but that doesn't really help me understand how to get those numbers.



Answer and why I got it wrong:



I've evaluated:



$ 9 = 1^4 + bcdot1^2 + c implies b = 8 - c implies c = 8 - b$



and



$ 25 = 3^4 + bcdot3^2 + c implies c = -56 - 9b $



The part which I forgot to do was to substitute the equations into each other and not the original equation. KM101 provides a more complete breakdown.



$ c = -56 - 9(8 - c) implies c = 16$



$ -9b - 56 = 8 - b implies b = -8$






polynomials







share|cite|improve this question




















  • 1




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Oct 7 at 5:36










  • Two unknowns, two equations!
    – Jyrki Lahtonen
    Oct 7 at 5:56
















0














By using Desmos I've found that the values of $;b = -8;$ and $;c = 16,;$ but that doesn't really help me understand how to get those numbers.



Answer and why I got it wrong:



I've evaluated:



$ 9 = 1^4 + bcdot1^2 + c implies b = 8 - c implies c = 8 - b$



and



$ 25 = 3^4 + bcdot3^2 + c implies c = -56 - 9b $



The part which I forgot to do was to substitute the equations into each other and not the original equation. KM101 provides a more complete breakdown.



$ c = -56 - 9(8 - c) implies c = 16$



$ -9b - 56 = 8 - b implies b = -8$






polynomials







share|cite|improve this question




















  • 1




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Oct 7 at 5:36










  • Two unknowns, two equations!
    – Jyrki Lahtonen
    Oct 7 at 5:56














0












0








0







By using Desmos I've found that the values of $;b = -8;$ and $;c = 16,;$ but that doesn't really help me understand how to get those numbers.



Answer and why I got it wrong:



I've evaluated:



$ 9 = 1^4 + bcdot1^2 + c implies b = 8 - c implies c = 8 - b$



and



$ 25 = 3^4 + bcdot3^2 + c implies c = -56 - 9b $



The part which I forgot to do was to substitute the equations into each other and not the original equation. KM101 provides a more complete breakdown.



$ c = -56 - 9(8 - c) implies c = 16$



$ -9b - 56 = 8 - b implies b = -8$






polynomials







share|cite|improve this question















By using Desmos I've found that the values of $;b = -8;$ and $;c = 16,;$ but that doesn't really help me understand how to get those numbers.



Answer and why I got it wrong:



I've evaluated:



$ 9 = 1^4 + bcdot1^2 + c implies b = 8 - c implies c = 8 - b$



and



$ 25 = 3^4 + bcdot3^2 + c implies c = -56 - 9b $



The part which I forgot to do was to substitute the equations into each other and not the original equation. KM101 provides a more complete breakdown.



$ c = -56 - 9(8 - c) implies c = 16$



$ -9b - 56 = 8 - b implies b = -8$






polynomials




polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 at 22:47









user376343

2,7782822




2,7782822










asked Oct 7 at 5:32









SuchAndSuch

41




41








  • 1




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Oct 7 at 5:36










  • Two unknowns, two equations!
    – Jyrki Lahtonen
    Oct 7 at 5:56














  • 1




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Oct 7 at 5:36










  • Two unknowns, two equations!
    – Jyrki Lahtonen
    Oct 7 at 5:56








1




1




Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Oct 7 at 5:36




Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Oct 7 at 5:36












Two unknowns, two equations!
– Jyrki Lahtonen
Oct 7 at 5:56




Two unknowns, two equations!
– Jyrki Lahtonen
Oct 7 at 5:56










4 Answers
4






active

oldest

votes


















1














Notice that $P(u)-P(v)=u^4-v^4+b(u^2-v^2)$. From $25-9=81-1+b(9-1)$ you’ll find $b=-8$. Now exploit $P(1)=9$, that is $1-8+c=9$ to find $c=16$.






share|cite|improve this answer





























    0














    Substitute the $(x,y)$ pairs in to get a system of linear equations, which may be easily solved:
    $$1^4+1^2b+c=9qquad3^4+3^2b+c=25$$
    $$b+c=8qquad 9b+c=-56$$






    share|cite|improve this answer





























      0














      Hint:



      $$25=3^4+b3^2+c implies 9b+c=-56$$



      $$9=1^4+b1^2+c implies b+c=8$$






      share|cite|improve this answer





























        0














        Here’s our function.
        $$f(x) = x^4+bx^2+c$$
        We’re given two points.
        $$A: (1, 9)$$
        $$B: (3, 25)$$
        For point $A$, substitute $x$ and $y$ with $1$ and $9$, respectively. $$9 = 1^4+b(1)^2+c$$
        $$9 = 1+b+c implies boxed{b+c = 8}$$
        For point $B$, substitute $x$ and $y$ with $3$ and $25$, respectively.
        $$25 = 3^4+b(3)^2+c$$
        $$25 = 81+9b+c implies boxed{9b+c = -56}$$
        Now, use elimination to solve for $b$ and $c$.
        $$9b+c-(b+c)= -56-8$$
        $$9b+c-b-c = -56-9 implies 8b = -64$$
        $$boxed{b = -8}$$
        Now, plug in $b = -8$ in either of the two equations to get $c$.
        $$9(-8)+c = -56$$
        $$-72+c = -56 implies boxed{c = 16}$$
        So, the final quartic equation containing these two points is $f(x) = x^4-8x^2+16$.






        share|cite|improve this answer























          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%2f2945317%2fgiven-the-equation-px-x4-bx2-c-and-the-points-3-25-and-1-9%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














          Notice that $P(u)-P(v)=u^4-v^4+b(u^2-v^2)$. From $25-9=81-1+b(9-1)$ you’ll find $b=-8$. Now exploit $P(1)=9$, that is $1-8+c=9$ to find $c=16$.






          share|cite|improve this answer


























            1














            Notice that $P(u)-P(v)=u^4-v^4+b(u^2-v^2)$. From $25-9=81-1+b(9-1)$ you’ll find $b=-8$. Now exploit $P(1)=9$, that is $1-8+c=9$ to find $c=16$.






            share|cite|improve this answer
























              1












              1








              1






              Notice that $P(u)-P(v)=u^4-v^4+b(u^2-v^2)$. From $25-9=81-1+b(9-1)$ you’ll find $b=-8$. Now exploit $P(1)=9$, that is $1-8+c=9$ to find $c=16$.






              share|cite|improve this answer












              Notice that $P(u)-P(v)=u^4-v^4+b(u^2-v^2)$. From $25-9=81-1+b(9-1)$ you’ll find $b=-8$. Now exploit $P(1)=9$, that is $1-8+c=9$ to find $c=16$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Oct 7 at 10:13









              Michael Hoppe

              10.8k31834




              10.8k31834























                  0














                  Substitute the $(x,y)$ pairs in to get a system of linear equations, which may be easily solved:
                  $$1^4+1^2b+c=9qquad3^4+3^2b+c=25$$
                  $$b+c=8qquad 9b+c=-56$$






                  share|cite|improve this answer


























                    0














                    Substitute the $(x,y)$ pairs in to get a system of linear equations, which may be easily solved:
                    $$1^4+1^2b+c=9qquad3^4+3^2b+c=25$$
                    $$b+c=8qquad 9b+c=-56$$






                    share|cite|improve this answer
























                      0












                      0








                      0






                      Substitute the $(x,y)$ pairs in to get a system of linear equations, which may be easily solved:
                      $$1^4+1^2b+c=9qquad3^4+3^2b+c=25$$
                      $$b+c=8qquad 9b+c=-56$$






                      share|cite|improve this answer












                      Substitute the $(x,y)$ pairs in to get a system of linear equations, which may be easily solved:
                      $$1^4+1^2b+c=9qquad3^4+3^2b+c=25$$
                      $$b+c=8qquad 9b+c=-56$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Oct 7 at 5:36









                      Parcly Taxel

                      41.2k137199




                      41.2k137199























                          0














                          Hint:



                          $$25=3^4+b3^2+c implies 9b+c=-56$$



                          $$9=1^4+b1^2+c implies b+c=8$$






                          share|cite|improve this answer


























                            0














                            Hint:



                            $$25=3^4+b3^2+c implies 9b+c=-56$$



                            $$9=1^4+b1^2+c implies b+c=8$$






                            share|cite|improve this answer
























                              0












                              0








                              0






                              Hint:



                              $$25=3^4+b3^2+c implies 9b+c=-56$$



                              $$9=1^4+b1^2+c implies b+c=8$$






                              share|cite|improve this answer












                              Hint:



                              $$25=3^4+b3^2+c implies 9b+c=-56$$



                              $$9=1^4+b1^2+c implies b+c=8$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Oct 7 at 5:36









                              Carl Schildkraut

                              11.1k11441




                              11.1k11441























                                  0














                                  Here’s our function.
                                  $$f(x) = x^4+bx^2+c$$
                                  We’re given two points.
                                  $$A: (1, 9)$$
                                  $$B: (3, 25)$$
                                  For point $A$, substitute $x$ and $y$ with $1$ and $9$, respectively. $$9 = 1^4+b(1)^2+c$$
                                  $$9 = 1+b+c implies boxed{b+c = 8}$$
                                  For point $B$, substitute $x$ and $y$ with $3$ and $25$, respectively.
                                  $$25 = 3^4+b(3)^2+c$$
                                  $$25 = 81+9b+c implies boxed{9b+c = -56}$$
                                  Now, use elimination to solve for $b$ and $c$.
                                  $$9b+c-(b+c)= -56-8$$
                                  $$9b+c-b-c = -56-9 implies 8b = -64$$
                                  $$boxed{b = -8}$$
                                  Now, plug in $b = -8$ in either of the two equations to get $c$.
                                  $$9(-8)+c = -56$$
                                  $$-72+c = -56 implies boxed{c = 16}$$
                                  So, the final quartic equation containing these two points is $f(x) = x^4-8x^2+16$.






                                  share|cite|improve this answer




























                                    0














                                    Here’s our function.
                                    $$f(x) = x^4+bx^2+c$$
                                    We’re given two points.
                                    $$A: (1, 9)$$
                                    $$B: (3, 25)$$
                                    For point $A$, substitute $x$ and $y$ with $1$ and $9$, respectively. $$9 = 1^4+b(1)^2+c$$
                                    $$9 = 1+b+c implies boxed{b+c = 8}$$
                                    For point $B$, substitute $x$ and $y$ with $3$ and $25$, respectively.
                                    $$25 = 3^4+b(3)^2+c$$
                                    $$25 = 81+9b+c implies boxed{9b+c = -56}$$
                                    Now, use elimination to solve for $b$ and $c$.
                                    $$9b+c-(b+c)= -56-8$$
                                    $$9b+c-b-c = -56-9 implies 8b = -64$$
                                    $$boxed{b = -8}$$
                                    Now, plug in $b = -8$ in either of the two equations to get $c$.
                                    $$9(-8)+c = -56$$
                                    $$-72+c = -56 implies boxed{c = 16}$$
                                    So, the final quartic equation containing these two points is $f(x) = x^4-8x^2+16$.






                                    share|cite|improve this answer


























                                      0












                                      0








                                      0






                                      Here’s our function.
                                      $$f(x) = x^4+bx^2+c$$
                                      We’re given two points.
                                      $$A: (1, 9)$$
                                      $$B: (3, 25)$$
                                      For point $A$, substitute $x$ and $y$ with $1$ and $9$, respectively. $$9 = 1^4+b(1)^2+c$$
                                      $$9 = 1+b+c implies boxed{b+c = 8}$$
                                      For point $B$, substitute $x$ and $y$ with $3$ and $25$, respectively.
                                      $$25 = 3^4+b(3)^2+c$$
                                      $$25 = 81+9b+c implies boxed{9b+c = -56}$$
                                      Now, use elimination to solve for $b$ and $c$.
                                      $$9b+c-(b+c)= -56-8$$
                                      $$9b+c-b-c = -56-9 implies 8b = -64$$
                                      $$boxed{b = -8}$$
                                      Now, plug in $b = -8$ in either of the two equations to get $c$.
                                      $$9(-8)+c = -56$$
                                      $$-72+c = -56 implies boxed{c = 16}$$
                                      So, the final quartic equation containing these two points is $f(x) = x^4-8x^2+16$.






                                      share|cite|improve this answer














                                      Here’s our function.
                                      $$f(x) = x^4+bx^2+c$$
                                      We’re given two points.
                                      $$A: (1, 9)$$
                                      $$B: (3, 25)$$
                                      For point $A$, substitute $x$ and $y$ with $1$ and $9$, respectively. $$9 = 1^4+b(1)^2+c$$
                                      $$9 = 1+b+c implies boxed{b+c = 8}$$
                                      For point $B$, substitute $x$ and $y$ with $3$ and $25$, respectively.
                                      $$25 = 3^4+b(3)^2+c$$
                                      $$25 = 81+9b+c implies boxed{9b+c = -56}$$
                                      Now, use elimination to solve for $b$ and $c$.
                                      $$9b+c-(b+c)= -56-8$$
                                      $$9b+c-b-c = -56-9 implies 8b = -64$$
                                      $$boxed{b = -8}$$
                                      Now, plug in $b = -8$ in either of the two equations to get $c$.
                                      $$9(-8)+c = -56$$
                                      $$-72+c = -56 implies boxed{c = 16}$$
                                      So, the final quartic equation containing these two points is $f(x) = x^4-8x^2+16$.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Oct 7 at 5:58

























                                      answered Oct 7 at 5:46









                                      KM101

                                      4,428418




                                      4,428418






























                                          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.





                                          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                          Please pay close attention to the following guidance:


                                          • 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.


                                          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%2f2945317%2fgiven-the-equation-px-x4-bx2-c-and-the-points-3-25-and-1-9%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

                                          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

                                          Province de Neuquén

                                          Image
                                          Province de Neuquén Provincia del Neuquén .mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Héraldique Drapeau Localisation de la province de Neuquén Administration Pays   Argentine Capitale Neuquén Gouverneur Omar Gutiérrez ISO 3166-2 AR-Q Démographie Gentilé Neuquino/a Population 550 344  hab. (2010) Densité 5,8  hab./km 2 Géographie Superficie 94 078  km 2 Liens Site web http://www.neuquen.gov.ar modifier   La province de Neuquén est une province d'Argentine, qui se trouve à l'extrémité nord-ouest de la Patagonie argentine. Avec une superficie de 94 378  km 2 , elle est la plus petite province des régions australes du pays. Sa géographie s'abîme sur l'imposante Cordillère des Andes. Sommaire 1 Histoire 1.1 Avant la découverte par les Espagnols 1.2 ...
                                          Read more

                                          Journaliste

                                          Image
                                          Pour les articles homonymes, voir Journaliste (homonymie). Journaliste Journalistes Appellation Journaliste Secteur d'activité Média - Communication Professions voisines Journaliste sportif - Reporter-photographe - Éditorialiste Codes CITP 2451, 2642 IDEO (France) 10197 ROME (France) E1106 modifier - modifier le code - modifier Wikidata Un journaliste est une personne qui recueille, écrit ou distribue des informations. L'activité d'un journaliste est le journalisme. Un journaliste peut travailler sur des sujets généraux ou se spécialiser dans certaines questions. La plupart des journalistes ont tendance à se spécialiser et, en collaborant avec d'autres journalistes, rédigent un article ou mettent en forme un reportage qui couvre l'actualité. Par exemple, un journaliste sportif couvre les nouvelles dans le monde du sport, mais ce journaliste peut aussi faire partie d'un journal qui couvre d'autres su...
                                          Read more