{"id":101832,"date":"2018-04-23T11:42:47","date_gmt":"2018-04-23T14:42:47","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=101832"},"modified":"2023-04-16T02:48:20","modified_gmt":"2023-04-16T05:48:20","slug":"problemao-uma-equacao-geometrica","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-uma-equacao-geometrica\/","title":{"rendered":".Problem\u00e3o: Uma equa\u00e7\u00e3o geom\u00e9trica"},"content":{"rendered":"

\n

\n

Problema
\n (Indicado a partir do 3\u00ba ano do E. M.)<\/span><\/p>\n

\nSejam [tex]x_1, x_2, x_3[\/tex] e [tex]x_4[\/tex] as ra\u00edzes complexas da equa\u00e7\u00e3o [tex]16x^4+8x^3+4x^2+2x+1=0[\/tex].
\nDetermine o valor da soma [tex]|x_1|+|x_2|+|x_3|+|x_4|[\/tex].<\/p>\n

\n<\/div>\n

\n

\n

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

\n Observe que as parcelas do lado esquerdo da equa\u00e7\u00e3o [tex]16x^4+8x^3+4x^2+2x+1=0[\/tex] formam uma progress\u00e3o geom\u00e9trica de cinco termos, com raz\u00e3o [tex]2x[\/tex]. Utilizando a f\u00f3rmula da soma dos termos de uma P. G. encontramos que
\n[tex]\\qquad \\qquad 16x^4+8x^3+4x^2+2x+1=\\dfrac{32x^5-1}{2x-1}.[\/tex]
\nObservando essa \u00faltima igualdade podemos constatar que qualquer solu\u00e7\u00e3o diferente de [tex]\\frac{1}{2}[\/tex] da equa\u00e7\u00e3o [tex]32x^5-1=0[\/tex] \u00e9 tamb\u00e9m solu\u00e7\u00e3o da equa\u00e7\u00e3o original.
\nPor outro lado, perceba que
\n[tex]\\qquad \\qquad 32x^5-1=0 \\quad \\Longleftrightarrow \\quad x=\\sqrt[5]{\\dfrac{1}{32}} \\, [\/tex];
\nassim, as solu\u00e7\u00f5es da equa\u00e7\u00e3o [tex]32x^5-1=0[\/tex] s\u00e3o as cinco ra\u00edzes quintas de [tex] \\dfrac{1}{32}[\/tex] (ou ainda cada uma das cinco ra\u00edzes da unidade multiplicada por [tex]\\dfrac{1}{2}).[\/tex]
\nVamos, ent\u00e3o, utilizar a f\u00f3rmula de radicia\u00e7\u00e3o de Moivre para n\u00fameros complexos para encontr\u00e1-las (Se voc\u00ea n\u00e3o conhece as f\u00f3rmulas de Moivre<\/strong>, clique AQUI<\/strong><\/a>.)<\/span>:<\/p>\n

\n

[tex]\\qquad \\qquad x_0=\\dfrac{1}{2}[\/tex];<\/p>\n

\n

[tex]\\qquad \\qquad x_1=\\dfrac{1}{2}\\left(\\cos 72^{\\circ}+i\\, \\text{sen} 72^{\\circ} \\right)\\ne \\dfrac{1}{2}[\/tex];<\/p>\n

\n

[tex]\\qquad \\qquad x_2=\\dfrac{1}{2}\\left(\\cos 144^{\\circ}+i\\, \\text{sen} 144^{\\circ} \\right)\\ne \\dfrac{1}{2}[\/tex];<\/p>\n

\n

[tex]\\qquad \\qquad x_3=\\dfrac{1}{2}\\left(\\cos 216^{\\circ}+i\\, \\text{sen} 216^{\\circ} \\right)\\ne \\dfrac{1}{2}[\/tex];<\/p>\n

\n

[tex]\\qquad \\qquad x_4=\\dfrac{1}{2}\\left(\\cos 288^{\\circ}+i\\, \\text{sen} 288^{\\circ} \\right)\\ne \\dfrac{1}{2}[\/tex].<\/p>\n

\n

Dessa forma, os n\u00fameros [tex]x_1, x_2, x_3[\/tex] e [tex]x_4[\/tex] acima definidos s\u00e3o as ra\u00edzes complexas da equa\u00e7\u00e3o [tex]16x^4+8x^3+4x^2+2x+1=0.[\/tex]
\nPara finalizar, com a defini\u00e7\u00e3o de m\u00f3dulo de um n\u00famero complexo e lembrando que se [tex]a[\/tex] \u00e9 a medida em graus de um \u00e2ngulo, ent\u00e3o [tex]\\cos^2 a+ \\text{sen} ^2a=1[\/tex], podemos calcular o m\u00f3dulo de cada uma dessas ra\u00edzes: <\/p>\n

\n

[tex]\\qquad \\qquad |x_1|=\\sqrt{\\dfrac{1}{4} \\cos^2 72^{\\circ}+\\dfrac{1}{4} \\text{sen}^2 72^{\\circ}}=\\dfrac{1}{2}\\sqrt{\\cos^2 72^{\\circ}+\\text{sen}^2 72^{\\circ}}=\\dfrac{1}{2};[\/tex]<\/p>\n

\n

[tex]\\qquad \\qquad |x_2|=\\sqrt{\\dfrac{1}{4} \\cos^2 144^{\\circ}+\\dfrac{1}{4} \\text{sen}^2 144^{\\circ}}=\\dfrac{1}{2}\\sqrt{\\cos^2 144^{\\circ}+\\text{sen}^2 144^{\\circ}}=\\dfrac{1}{2};[\/tex]<\/p>\n

\n

[tex]\\qquad \\qquad |x_3|=\\sqrt{\\dfrac{1}{4} \\cos^2 216^{\\circ}+\\dfrac{1}{4} \\text{sen}^2 216^{\\circ}}=\\dfrac{1}{2}\\sqrt{\\cos^2 216^{\\circ}+\\text{sen}^2 216^{\\circ}}=\\dfrac{1}{2};[\/tex]<\/p>\n

\n

[tex]\\qquad \\qquad |x_4|=\\sqrt{\\dfrac{1}{4} \\cos^2 288^{\\circ}+\\dfrac{1}{4} \\text{sen}^2 288^{\\circ}}=\\dfrac{1}{2}\\sqrt{\\cos^2 288^{\\circ}+\\text{sen}^2 288^{\\circ}}=\\dfrac{1}{2}.[\/tex]<\/p>\n

\n

Portanto, [tex] \\, \\fcolorbox{black}{#eee0e5}{$|x_1|+|x_2|+|x_3|+|x_4|=2$} \\, .[\/tex]<\/p>\n

\n

Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n

\n","protected":false},"excerpt":{"rendered":"

Problema (Indicado a partir do 3\u00ba ano do E. M.) Sejam [tex]x_1, x_2, x_3[\/tex] e [tex]x_4[\/tex] as ra\u00edzes complexas da equa\u00e7\u00e3o [tex]16x^4+8x^3+4x^2+2x+1=0[\/tex]. Determine o valor da soma [tex]|x_1|+|x_2|+|x_3|+|x_4|[\/tex]. Solu\u00e7\u00e3o Observe que as parcelas do lado esquerdo da equa\u00e7\u00e3o [tex]16x^4+8x^3+4x^2+2x+1=0[\/tex] formam uma progress\u00e3o geom\u00e9trica de cinco termos, com raz\u00e3o [tex]2x[\/tex]. Utilizando a f\u00f3rmula da soma dos … <\/p>\n

Continue lendo<\/a><\/p>\n","protected":false},"author":667,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-101832","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/101832","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\/667"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=101832"}],"version-history":[{"count":11,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/101832\/revisions"}],"predecessor-version":[{"id":187345,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/101832\/revisions\/187345"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=101832"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}