{"id":37999,"date":"2015-05-22T22:08:41","date_gmt":"2015-05-23T01:08:41","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=37999"},"modified":"2023-03-21T22:11:07","modified_gmt":"2023-03-22T01:11:07","slug":"desafio-raizes-em-progressao","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/desafio-raizes-em-progressao\/","title":{"rendered":".Desafio: Ra\u00edzes em progress\u00e3o"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 3\u00ba ano do E. M.)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 10px;border: 0 none\"\/>\n<p style=\"margin:20px\">\n<p>Sabendo que [tex]p[\/tex] \u00e9 um n\u00famero real n\u00e3o nulo e que as tr\u00eas ra\u00edzes distintas da equa\u00e7\u00e3o [tex]x^3 + 2px^2 &#8211; px + 10 = 0[\/tex] formam uma progress\u00e3o aritm\u00e9tica, determine essas ra\u00edzes.\n\t<\/p><\/div>\n<div class=\"cb36 g1n\" style=\"margin:40px;line-height:30px; width:70%; margin-left:auto; margin-right:auto; \"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" style=\"border:none\"  alt=\"explicador_p\" width=\"70\" height=\"84\" class=\"alignright size-full wp-image-1001\" \/> <\/p>\n<p class=\"c3 ti3\" style=\"line-height:25px;margin:10px\"> Lembretes<\/p>\n<p>Dada uma progress\u00e3o aritm\u00e9tica [tex](a_{1}, a_{2}, a_{3}, \\dots)[\/tex] de raz\u00e3o [tex]r[\/tex]:<\/p>\n<ul>[tex]\\textcolor{#800000}{(a)}[\/tex] O termo geral, [tex]a_{n}[\/tex], \u00e9 dado por   [tex]\\boxed{a_{n}=a_{1}+ (n-1)\\cdot r}[\/tex].<br \/>\n\t[tex]\\textcolor{#800000}{(b)}[\/tex]  [tex]r=a_2-a_1=a_3-a_2=a_4-a_3=\\cdots[\/tex].<br \/>\n\t[tex]\\textcolor{#800000}{(c)}[\/tex]  [tex]a_k=\\dfrac{a_{k+1}+a_{k-1}}{2}[\/tex], para todo [tex]k\\geq 2[\/tex].  (Qualquer termo, a partir do segundo, ser\u00e1 a m\u00e9dia aritm\u00e9tica do seu antecessor com o seu sucessor.)<\/ul>\n<\/div>\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Solu\u00e7\u00e3o<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 10px;border: 0 none\"\/>\n<p style=\"margin:20px\">\n<p>Como as tr\u00eas ra\u00edzes distintas formam uma progress\u00e3o aritm\u00e9tica, vamos supor que [tex]a, \\ \\dfrac{a+c}{2}[\/tex] e [tex]c[\/tex], sejam as ra\u00edzes da equa\u00e7\u00e3o. Dessa forma temos:<\/p>\n<p>[tex]\\qquad x^3 + 2px^2-px + 10 = (x-a) \\left(x-\\frac{(a+c)}{2} \\right) (x-c)\\\\<br \/>\n\\qquad x^3 + 2px^2-px + 10 = x^3-\\dfrac{3\\left(a+c\\right)}{2}\\,x^2+\\left(\\dfrac{(a+c)^2}{2}+ac\\right)x\\,-\\,\\dfrac{ac(a+c)}{2},[\/tex]<\/p>\n<p>donde, identificando os coeficientes, obtemos tr\u00eas equa\u00e7\u00f5es:<\/p>\n<p>[tex]\\qquad \\dfrac{3(a+c)}{2} = -2p\\,,[\/tex]  [tex]\\hspace{1.9cm}[\/tex] <strong>(1)<\/strong><\/p>\n<p>[tex]\\qquad \\dfrac{(a+c)^2}{2}+ac = -p\\,,[\/tex] [tex]\\hspace{1.2cm}[\/tex] <strong>(2)<\/strong><\/p>\n<p>[tex]\\qquad \\dfrac{(a+c)ac}{2} = -10\\,.[\/tex] [tex]\\hspace{1.8cm}[\/tex] <strong>(3)<\/strong><\/p>\n<p>Das equa\u00e7\u00f5es <strong>(1)<\/strong> e <strong>(3)<\/strong>, segue  que [tex]ac=\\dfrac{15}{p}\\, [\/tex] e [tex]\\, a+c = \\dfrac{-4p}{3}[\/tex]. Substituindo esses valores na equa\u00e7\u00e3o <strong>(2)<\/strong>, obtemos a igualdade [tex]\\boxed{16p^3 + 18p^2 + 270 = 0} [\/tex], que \u00e9 equivalente a [tex]\\boxed{8p^3 + 9p^2 + 135 = 0}\\, .[\/tex]<br \/>\nUma raiz dessa \u00faltima equa\u00e7\u00e3o \u00e9 [tex]p=-3[\/tex], ent\u00e3o<\/p>\n<p style=\"margin:10px\">[tex]\\qquad 8p^3 + 9p^2 + 135 = (p + 3) ( 8p^2-15p + 45)[\/tex]. <\/p>\n<p>Dessa \u00faltima igualdade, podemos verificar que [tex]\\,p=-3\\,[\/tex] \u00e9 a \u00fanica raiz real da equa\u00e7\u00e3o [tex]\\boxed{8p^3 + 9p^2 + 135 = 0} [\/tex] (Voc\u00ea saberia justificar essa afirma\u00e7\u00e3o?).<br \/>\nSendo assim, temos [tex]ac = -5\\,[\/tex] e [tex]\\,a + c = 4[\/tex], o que nos permite concluir que [tex]a[\/tex] e [tex]c[\/tex] s\u00e3o [tex]5[\/tex] e [tex]-1[\/tex].<br \/>\nPortanto, as ra\u00edzes em progress\u00e3o s\u00e3o [tex]-1[\/tex], [tex]2[\/tex] e [tex]5[\/tex].<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\" style=\"line-height:20px;\">Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 3\u00ba ano do E. M.) Sabendo que [tex]p[\/tex] \u00e9 um n\u00famero real n\u00e3o nulo e que as tr\u00eas ra\u00edzes distintas da equa\u00e7\u00e3o [tex]x^3 + 2px^2 &#8211; px + 10 = 0[\/tex] formam uma progress\u00e3o aritm\u00e9tica, determine essas ra\u00edzes. Lembretes Dada uma progress\u00e3o aritm\u00e9tica [tex](a_{1}, a_{2}, a_{3}, \\dots)[\/tex] de raz\u00e3o [tex]r[\/tex]: &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/desafio-raizes-em-progressao\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":166,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"open","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-37999","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/37999","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\/166"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=37999"}],"version-history":[{"count":19,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/37999\/revisions"}],"predecessor-version":[{"id":185104,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/37999\/revisions\/185104"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=37999"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}