{"id":200673,"date":"2023-10-01T05:28:48","date_gmt":"2023-10-01T08:28:48","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=200673"},"modified":"2023-10-06T02:36:56","modified_gmt":"2023-10-06T05:36:56","slug":"desafio-ali-babao-e-a-trigesima-primeira-de-suas-40-equacoes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/desafio-ali-babao-e-a-trigesima-primeira-de-suas-40-equacoes\/","title":{"rendered":".Desafio: Ali Bab\u00e3o e a Trig\u00e9sima Primeira de suas 40 Equa\u00e7\u00f5es"},"content":{"rendered":"

✏ Link do problema para dispositivos da Apple<\/em>.<\/span><\/a><\/strong><\/p>\n

\n

\n

Problema
\n (Indicado a partir do 2\u00ba s\u00e9rie do E. M.)<\/span><\/p>\n

\nResolva a equa\u00e7\u00e3o [tex]4x^{4}-ax^{3}+bx^{2}-cx+5=0[\/tex], sabendo que todas as ra\u00edzes [tex]x_{1}, x_{2}, x_{3} [\/tex] e [tex] x_{4}[\/tex] s\u00e3o reais e positivas e que [tex]\\dfrac{x_{1}}{2}+\\dfrac{x_{2}}{4}+\\dfrac{x_{3}}{5}+\\dfrac{x_{4}}{8}=1.[\/tex]<\/p>\n

Extra\u00eddo de Ol\u00edmpiada Internacional<\/strong>.<\/p>\n<\/div>\n

\n

\"explicador_p\"<\/p>\n

Lembretes<\/p>\n

\n

(I)<\/strong> <\/span> Dada uma equa\u00e7\u00e3o alg\u00e9brica [tex]ax^{4}+bx^{3}+cx^{2}+dx+e=0[\/tex] cujas ra\u00edzes s\u00e3o [tex]x_{1}, x_{2}, x_{3} [\/tex] e [tex]x_{4}[\/tex] \u00e9 v\u00e1lido o seguinte resultado:
\n [tex]\\qquad x_{1} \\cdot x_{2} \\cdot x_{3} \\cdot x_{4}=\\dfrac{e}{a}[\/tex] (Rela\u00e7\u00e3o de Girard).
\n (II)<\/strong> <\/span> A desigualdade das m\u00e9dias afirma que a m\u00e9dia aritm\u00e9tica \u00e9 maior ou igual \u00e0 m\u00e9dia geom\u00e9trica; assim, dados os n\u00fameros reais positivos [tex]x_{1}, x_{2}, x_{3} [\/tex] e [tex]x_{4}[\/tex] temos:
\n

<\/span> [tex]\\dfrac{x_{1} + x_{2} + x_{3} + x_{4}}{4} \\geq \\sqrt[4]{x_{1} \\cdot x_{2} \\cdot x_{3} \\cdot x_{4}}[\/tex].<\/center>
\nMais do que isso, se [tex]\\dfrac{x_{1} + x_{2} + x_{3} + x_{4}}{4}=\\sqrt[4]{x_{1} \\cdot x_{2} \\cdot x_{3} \\cdot x_{4}}[\/tex] , ent\u00e3o [tex]x_{1}= x_{2}= x_{3}= x_{4} [\/tex].<\/p>\n<\/div>\n

\n

\n

Solu\u00e7\u00e3o<\/p>\n

\nDo enunciado, temos que [tex]\\dfrac{x_{1}}{2}+\\dfrac{x_{2}}{4}+\\dfrac{x_{3}}{5}+\\dfrac{x_{4}}{8}=1[\/tex]; portanto, [tex]\\dfrac{\\dfrac{x_{1}}{2}+\\dfrac{x_{2}}{4}+\\dfrac{x_{3}}{5}+\\dfrac{x_{4}}{8}}{4}=\\dfrac{1}{4}[\/tex]. <\/p>\n

Por outro lado, dada a equa\u00e7\u00e3o [tex]4x^{4}-ax^{3}+bx^{2}-cx+5=0[\/tex], pelo Lembrete (I)<\/strong> <\/span>, podemos afirmar que [tex]x_{1} \\cdot x_{2} \\cdot x_{3} \\cdot x_{4}=\\dfrac{5}{4}.[\/tex]
\nAssim, segue que:
\n[tex]\\qquad \\sqrt[4]{\\dfrac{x_{1}}{2} \\cdot \\dfrac{x_{2}}{4} \\cdot \\dfrac{x_{3}}{5} \\cdot \\dfrac{x_{4}}{8}}=[\/tex] [tex]\\sqrt[4]{\\dfrac{\\dfrac{5}{4}}{2 \\cdot 4 \\cdot 5 \\cdot 8}}=\\sqrt[4]{\\dfrac{1}{256}}=\\dfrac{1}{4}.[\/tex]
\nDa\u00ed, conclu\u00edmos que a m\u00e9dia aritm\u00e9tica dos elementos [tex]\\dfrac{x_{1}}{2}[\/tex], [tex]\\dfrac{x_{2}}{4}[\/tex], [tex]\\dfrac{x_{3}}{5}[\/tex] e [tex]\\dfrac{x_{4}}{8}[\/tex] \u00e9 igual \u00e0 m\u00e9dia geom\u00e9trica e, portanto, pelo Lembrete (II)<\/strong> <\/span>, temos que [tex]\\dfrac{x_{1}}{2}=\\dfrac{x_{2}}{4}=\\dfrac{x_{3}}{5}=\\dfrac{x_{4}}{8}[\/tex].
\nDessa forma, podemos deduzir que [tex]\\boxed{x_{2}=2x_{1}}[\/tex], [tex]~\\boxed{x_{3}=\\dfrac{5x_{1}}{2}}~[\/tex] e [tex]~\\boxed{x_{4}=4x_{1}}.[\/tex]
\nSubstituindo essas tr\u00eas informa\u00e7\u00f5es na equa\u00e7\u00e3o [tex]x_{1} \\cdot x_{2} \\cdot x_{3} \\cdot x_{4}=\\dfrac{5}{4}[\/tex], temos que:
\n [tex]\\qquad x_{1} \\cdot 2x_{1} \\cdot \\dfrac{5x_{1}}{2} \\cdot 4x_{1}=\\dfrac{5}{4}\\\\
\n\\qquad {x_{1}}^4=\\dfrac{1}{16}.[\/tex]
\nComo as ra\u00edzes s\u00e3o n\u00fameros reais positivos, segue que [tex]\\fcolorbox{black}{#eee0e5}{$x_{1}=\\dfrac{1}{2}$}\\,[\/tex] e, com isso, [tex]\\fcolorbox{black}{#eee0e5}{$x_{2}=1$}\\,[\/tex], [tex]\\,\\fcolorbox{black}{#eee0e5}{$x_{3}=\\dfrac{5}{4}$}\\,[\/tex] e [tex]\\,\\fcolorbox{black}{#eee0e5}{$x_{4}=2$}\\,.[\/tex]<\/p>\n

Uma observa\u00e7\u00e3o importante \u00e9 que, com os valores das ra\u00edzes, podemos encontrar os valores dos coeficientes [tex]a[\/tex], [tex]b[\/tex] e [tex]c[\/tex] utilizando as outras rela\u00e7\u00f5es de Girard.<\/p>\n

\n

Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

✏ Link do problema para dispositivos da Apple. Problema (Indicado a partir do 2\u00ba s\u00e9rie do E. M.) Resolva a equa\u00e7\u00e3o [tex]4x^{4}-ax^{3}+bx^{2}-cx+5=0[\/tex], sabendo que todas as ra\u00edzes [tex]x_{1}, x_{2}, x_{3} [\/tex] e [tex] x_{4}[\/tex] s\u00e3o reais e positivas e que [tex]\\dfrac{x_{1}}{2}+\\dfrac{x_{2}}{4}+\\dfrac{x_{3}}{5}+\\dfrac{x_{4}}{8}=1.[\/tex] Extra\u00eddo de Ol\u00edmpiada Internacional. Lembretes (I) Dada uma equa\u00e7\u00e3o alg\u00e9brica [tex]ax^{4}+bx^{3}+cx^{2}+dx+e=0[\/tex] cujas ra\u00edzes s\u00e3o … <\/p>\n

Continue lendo<\/a><\/p>\n","protected":false},"author":668,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-200673","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/200673","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/668"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=200673"}],"version-history":[{"count":24,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/200673\/revisions"}],"predecessor-version":[{"id":202200,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/200673\/revisions\/202200"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=200673"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}