{"id":203393,"date":"2023-10-28T05:56:20","date_gmt":"2023-10-28T08:56:20","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=203393"},"modified":"2023-10-28T06:07:37","modified_gmt":"2023-10-28T09:07:37","slug":"problema-de-gincana-quantas-raizes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-de-gincana-quantas-raizes\/","title":{"rendered":".Problema de Gincana: Quantas ra\u00edzes?"},"content":{"rendered":"<p class=\"g2\" ><strong><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=204312\" rel=\"noopener\" target=\"_blank\">&#9999; <span style=\"color:#915F6D;font-size:14px;\">Link do problema para dispositivos da <em>Apple<\/em>.<\/span><\/a><\/strong><\/p>\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:28px;margin:10px\">Problema<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nSupondo que  as tr\u00eas ra\u00edzes quadradas abaixo est\u00e3o definidas no conjunto dos n\u00fameros reais, quantas solu\u00e7\u00f5es reais tem a equa\u00e7\u00e3o<br \/>\n [tex]~\\qquad  \\qquad \\sqrt{x+2\\,}+\\sqrt{4\u2013x\\,}=\\sqrt{8+4x\\,}\\,?[\/tex]\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n Observe a seguinte sequ\u00eancia de equa\u00e7\u00f5es equivalentes:<br \/>\n[tex]\\qquad \\sqrt{x+2}+\\sqrt{4\u2013x}=\\sqrt{8+4x}[\/tex]<br \/>\n[tex]\\qquad \\sqrt{x+2}+\\sqrt{4\u2013x}=\\sqrt{4(2+x)}[\/tex]<br \/>\n[tex]\\qquad \\sqrt{x+2}+\\sqrt{4\u2013x}=2\\sqrt{x+2}[\/tex]<br \/>\n[tex]\\qquad \\sqrt{4\u2013x}=\\sqrt{x+2}.[\/tex]<\/p>\n<p>Elevando ambos os membros da \u00faltima igualdade ao quadrado, temos<br \/>\n[tex]\\qquad 4\u2013x=x+2[\/tex]<br \/>\n[tex]\\qquad 2x=2[\/tex]<br \/>\n[tex]\\qquad x=1.[\/tex]<\/p>\n<p>Vamos verificar que [tex]x=1[\/tex], de fato, \u00e9 solu\u00e7\u00e3o da equa\u00e7\u00e3o. Substituindo o valor na equa\u00e7\u00e3o, segue que:<br \/>\n[tex]\\qquad \\sqrt{1+2}+\\sqrt{4\u20131}=\\sqrt{8+4}.[\/tex]<br \/>\n[tex]\\qquad \\sqrt{3}+\\sqrt{3}=\\sqrt{12}.[\/tex]<br \/>\n[tex]\\qquad 2\\sqrt{3}=2\\sqrt{3}.[\/tex]<\/p>\n<p>Logo, [tex]x = 1[\/tex] e <strong>a equa\u00e7\u00e3o tem apenas uma solu\u00e7\u00e3o<\/strong>.<\/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<p style=\"margin:30px\">\n<div class=\"gg\">\nPrimeira Gincana de 2023 &#8211; Clubes de Matem\u00e1tica da OBMEP<br \/>\nN\u00edvel B &#8211; Quest\u00e3o F\u00e1cil<\/div>\n","protected":false},"excerpt":{"rendered":"<p>&#9999; Link do problema para dispositivos da Apple. Problema Supondo que as tr\u00eas ra\u00edzes quadradas abaixo est\u00e3o definidas no conjunto dos n\u00fameros reais, quantas solu\u00e7\u00f5es reais tem a equa\u00e7\u00e3o [tex]~\\qquad \\qquad \\sqrt{x+2\\,}+\\sqrt{4\u2013x\\,}=\\sqrt{8+4x\\,}\\,?[\/tex] Solu\u00e7\u00e3o Observe a seguinte sequ\u00eancia de equa\u00e7\u00f5es equivalentes: [tex]\\qquad \\sqrt{x+2}+\\sqrt{4\u2013x}=\\sqrt{8+4x}[\/tex] [tex]\\qquad \\sqrt{x+2}+\\sqrt{4\u2013x}=\\sqrt{4(2+x)}[\/tex] [tex]\\qquad \\sqrt{x+2}+\\sqrt{4\u2013x}=2\\sqrt{x+2}[\/tex] [tex]\\qquad \\sqrt{4\u2013x}=\\sqrt{x+2}.[\/tex] Elevando ambos os membros da \u00faltima igualdade &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-de-gincana-quantas-raizes\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":254,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-203393","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/203393","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\/254"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=203393"}],"version-history":[{"count":12,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/203393\/revisions"}],"predecessor-version":[{"id":204314,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/203393\/revisions\/204314"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=203393"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}