{"id":161895,"date":"2020-08-18T04:09:25","date_gmt":"2020-08-18T07:09:25","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=161895"},"modified":"2023-07-24T05:40:58","modified_gmt":"2023-07-24T08:40:58","slug":"problemao-ali-babao-e-vigesima-setima-de-suas-40-equacoes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-ali-babao-e-vigesima-setima-de-suas-40-equacoes\/","title":{"rendered":".Problem\u00e3o:  Ali Bab\u00e3o e a vig\u00e9sima s\u00e9tima de suas 40 equa\u00e7\u00f5es"},"content":{"rendered":"<p class=\"g2\" ><strong><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=197789\" rel=\"noopener\" target=\"_blank\">&#9999; <span style=\"color:#8D918D;font-size:14px;\">Link do problema para dispositivos da <em>Apple<\/em>.<\/span><\/a><\/strong><\/p>\n<p style=\"margin:10px\">\n<div class=\"g1n\" style=\"line-height:30px;padding-bottom:15px\">\n<p class=\"tit\" style=\"line-height:20px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 9\u00ba ano do E. F.)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nEncontre as solu\u00e7\u00f5es reais da equa\u00e7\u00e3o:<br \/>\n [tex]\\quad \\quad \\sqrt{x+\\sqrt{4x-4}}+\\sqrt{x-\\sqrt{4x-4}}=\\sqrt{x+3}\\,.[\/tex]<\/p>\n<p style=\"font-size: 12px;text-align:right;line-height:1px;\">(<strong>IME-2014\/2015<\/strong>&#8211; 2\u00ba Fase.)<\/p>\n<\/div>\n<p style=\"margin:40px\">\n<div class=\"g1n\" style=\"line-height:32px;\">\n<p class=\"tit\" style=\"line-height:28px;margin:10px\">Solu\u00e7\u00e3o<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 1px; margin-bottom: 30px;border: 0 none\"\/>\n<span style=\"color:#38ACEC;\">&#9658;<\/span> <strong>Vamos come\u00e7ar elevando ao quadrado os dois membros da igualdade dada no problema:<\/strong><br \/>\n<span style=\"font-size:13px\">[tex]\\quad (\\sqrt{x+\\sqrt{4x-4}}+\\sqrt{x-\\sqrt{4x-4}})^2=(\\sqrt{x+3})^2\\\\<br \/>\n\\quad \\bigg(\\sqrt{x+\\sqrt{4x-4}}\\bigg)^2+2\\cdot \\sqrt{x+\\sqrt{4x-4}}\\cdot \\sqrt{x-\\sqrt{4x-4}}+\\bigg(\\sqrt{x-\\sqrt{4x-4}}\\bigg)^2=x+3\\\\<br \/>\n\\quad x+\\cancel{\\sqrt{4x-4}}+2\\sqrt{(x+\\sqrt{4x-4})\\cdot (x-\\sqrt{4x-4})}+x-\\cancel{\\sqrt{4x-4}}=x+3\\\\<br \/>\n\\quad 2x+2\\sqrt{x^2-\\cancel{x\\sqrt{4x-4}}+\\cancel{x\\sqrt{4x-4}}-(\\sqrt{4x-4})^2}=x+3\\\\<br \/>\n\\quad 2x+2\\sqrt{x^2-(4x-4)}=x+3\\\\<br \/>\n\\quad 2\\sqrt{x^2-4x+4}=x+3-2x\\\\<br \/>\n\\quad 2\\sqrt{x^2-4x+4}=3-x.\\\\<br \/>\n[\/tex]<\/span><br \/>\n<span style=\"color:#38ACEC;\">&#9658;<\/span> <strong>Elevando tamb\u00e9m ambos os membros da igualdade obtida ao quadrado, segue que:<\/strong><br \/>\n[tex]\\quad \\bigg(2\\sqrt{x^2-4x+4}\\bigg)^2=(3-x)^2\\\\<br \/>\n\\quad 4(x^2-4x+4)=9-6x+x^2\\\\<br \/>\n\\quad 4x^2-16x+16=9-6x+x^2\\\\<br \/>\n\\quad 4x^2-16x+16-9+6x-x^2=0\\\\<br \/>\n\\quad 3x^2-10x+7=0.\\\\<br \/>\n[\/tex]<br \/>\n<span style=\"color:#38ACEC;\">&#9658;<\/span> <strong>Vamos resolver a equa\u00e7\u00e3o do [tex]2^{\\circ}[\/tex] grau que obtivemos.<\/strong><br \/>\nObserve que considerando os coeficientes da equa\u00e7\u00e3o como [tex]a=3[\/tex], [tex]b=-10[\/tex] e [tex]c=7[\/tex]; ent\u00e3o:<br \/>\n[tex]\\quad \\Delta=b^2-4ac[\/tex]<br \/>\n[tex]\\quad \\Delta=(-10)^2-4\\cdot 3\\cdot 7[\/tex]<br \/>\n[tex]\\quad \\Delta=100-84[\/tex]<br \/>\n[tex]\\quad \\Delta=16.[\/tex]<br \/>\nAssim,<br \/>\n[tex]\\quad x=\\dfrac{-b\\pm \\sqrt{\\Delta}}{2a}\\\\<br \/>\n\\quad x=\\dfrac{-(-10)\\pm \\sqrt{16}}{2\\cdot 3}\\\\<br \/>\n\\quad x=\\dfrac{10\\pm 4}{6}[\/tex]<br \/>\ndonde:<br \/>\n[tex]\\quad  \\begin{cases} x=\\dfrac{10-4}{6}=\\dfrac{6}{6}=1\\\\<br \/>\n\\text{ou} \\\\<br \/>\nx=\\dfrac{10+4}{6}=\\dfrac{14}{6}=\\dfrac{7}{3}<br \/>\n\\end{cases}\\quad .[\/tex] <\/p>\n<div class=\"cb36 g1\" style=\"margin:30px;line-height:30px; width:80%; margin-left:auto; margin-right:auto; \">\nCabe aqui uma palavrinha com rela\u00e7\u00e3o \u00e0s consequ\u00eancias do &quot;algebrismo&quot; que fizemos.<\/p>\n<ul>\n<li>Observe que se [tex]z=5[\/tex], ent\u00e3o podemos concluir que [tex]z^2=25[\/tex], uma vez que <\/li>\n<p>[tex]\\quad z^2=z\\cdot z=5\\cdot 5=25 \\, .[\/tex]<br \/>\nNo entanto, se [tex]z^2=25[\/tex] e se n\u00e3o tivermos outras informa\u00e7\u00f5es sobre [tex]z[\/tex], ent\u00e3o n\u00e3o podemos simplesmente concluir que [tex]z=5[\/tex], pois poder\u00edamos ter [tex]z=-5.[\/tex]<br \/>\nEm termos de s\u00edmbolos, estamos afirmando que<br \/>\n[tex]\\qquad z=5 \\Rightarrow z^2=25 [\/tex],<br \/>\nmas que<br \/>\n[tex]\\qquad z^2=25 \\nRightarrow z=5[\/tex],<br \/>\ne, com isso,<br \/>\n[tex]\\qquad z^2=25 \\nLeftrightarrow z=5[\/tex],<br \/>\no que quer dizer que <strong>as duas equa\u00e7\u00f5es n\u00e3o s\u00e3o equivalentes<\/strong>.<br \/>\n<span class=\"c3\"><strong>E se duas equa\u00e7\u00f5es n\u00e3o s\u00e3o equivalentes, elas n\u00e3o t\u00eam necessariamente as mesmas solu\u00e7\u00f5es!<\/strong><\/span><\/ul>\n<\/div>\n<p>Precisamos verificar que as solu\u00e7\u00f5es da equa\u00e7\u00e3o [tex]\\, 3x^2-10x+7=0[\/tex] s\u00e3o, de fato, solu\u00e7\u00f5es da equa\u00e7\u00e3o dada no problema.<\/p>\n<ul>\n<li>Se [tex]x=1[\/tex] temos que:<br \/>\n[tex]\\quad\\fcolorbox{black}{#DAE4EA}{$\\sqrt{x+\\sqrt{4x-4}}+\\sqrt{x-\\sqrt{4x-4}}$}=\\\\<br \/>\n\\qquad =\\sqrt{1+\\sqrt{4\\cdot 1-4}}+\\sqrt{1-\\sqrt{4\\cdot 1-4}}=\\fcolorbox{black}{#DAE4EA}{$2$}[\/tex]<br \/>\ne<br \/>\n[tex]\\quad \\fcolorbox{black}{#DAE4EA}{$\\sqrt{x+3}$}=\\sqrt{1+3}=\\sqrt{4}=\\fcolorbox{black}{#DAE4EA}{$2$}.[\/tex]<br \/>\nAssim, [tex]x=1[\/tex] \u00e9 uma solu\u00e7\u00e3o da equa\u00e7\u00e3o do problema.<\/li>\n<li>Se [tex]x=\\dfrac{7}{3}[\/tex] temos que:<br \/>\n[tex]\\quad \\fcolorbox{black}{#FFE5B4}{$\\sqrt{x+\\sqrt{4x-4}}+\\sqrt{x-\\sqrt{4x-4}}$}=\\\\<br \/>\n\\quad = \\sqrt{\\dfrac{7}{3}+\\sqrt{4\\cdot \\dfrac{7}{3}-4}}+\\sqrt{\\dfrac{7}{3}-\\sqrt{4\\cdot \\dfrac{7}{3}-4}}=\\\\<br \/>\n\\quad=\\sqrt{\\dfrac{7}{3}+\\sqrt{\\dfrac{16}{3}}}+\\sqrt{\\dfrac{7}{3}-\\sqrt{\\dfrac{16}{3}}}=\\\\<br \/>\n\\quad=\\sqrt{\\dfrac{7}{3}+\\dfrac{4}{\\sqrt{3}}}+\\sqrt{\\dfrac{7}{3}-\\dfrac{4}{\\sqrt{3}}}=\\\\<br \/>\n\\quad=\\sqrt{\\dfrac{7}{3}+\\dfrac{4\\sqrt{3}}{3}}+\\sqrt{\\dfrac{7}{3}-\\dfrac{4\\sqrt{3}}{3}}=\\\\<br \/>\n\\quad=\\dfrac{\\sqrt{7+4\\sqrt{3}}}{\\sqrt{3}}+\\dfrac{\\sqrt{7-4\\sqrt{3}}}{\\sqrt{3}}=\\\\<br \/>\n\\quad=\\dfrac{\\sqrt{2^2+2\\cdot 2\\cdot \\sqrt{3}+(\\sqrt{3})^2}+\\sqrt{2^2-2\\cdot 2\\cdot \\sqrt{3}+(\\sqrt{3})^2}}{\\sqrt{3}}=\\\\<br \/>\n\\quad=\\dfrac{\\sqrt{(2+\\sqrt{3})^2}+\\sqrt{(2-\\sqrt{3})^2}}{\\sqrt{3}}=\\\\<br \/>\n\\quad=\\dfrac{2+\\sqrt{3}+2-\\sqrt{3}}{\\sqrt{3}}=\\fcolorbox{black}{#FFE5B4}{$\\dfrac{4}{\\sqrt{3}}$}[\/tex]<br \/>\n e<br \/>\n[tex]\\quad \\fcolorbox{black}{#FFE5B4}{$\\sqrt{x+3}$}=\\sqrt{\\dfrac{7}{3}+3}=\\sqrt{\\dfrac{7+9}{3}}=\\sqrt{\\dfrac{16}{3}}= \\fcolorbox{black}{#FFE5B4}{$\\dfrac{4}{\\sqrt{3}}$}.[\/tex]<br \/>\nAssim, [tex]x=\\dfrac{7}{3}[\/tex] \u00e9 tamb\u00e9m uma solu\u00e7\u00e3o da equa\u00e7\u00e3o do problema.<\/li>\n<\/ul>\n<p>Portanto, as solu\u00e7\u00f5es da equa\u00e7\u00e3o [tex]\\;\\sqrt{x+\\sqrt{4x-4}}+\\sqrt{x-\\sqrt{4x-4}}=\\sqrt{x+3}\\;\\;[\/tex]s\u00e3o [tex]\\;\\fcolorbox{black}{#eee0e5}{$1 \\text{ e } \\dfrac{7}{3}$}\\,.[\/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<p style=\"margin:10px\">\n","protected":false},"excerpt":{"rendered":"<p>&#9999; Link do problema para dispositivos da Apple. Problema (Indicado a partir do 9\u00ba ano do E. F.) Encontre as solu\u00e7\u00f5es reais da equa\u00e7\u00e3o: [tex]\\quad \\quad \\sqrt{x+\\sqrt{4x-4}}+\\sqrt{x-\\sqrt{4x-4}}=\\sqrt{x+3}\\,.[\/tex] (IME-2014\/2015&#8211; 2\u00ba Fase.) Solu\u00e7\u00e3o &#9658; Vamos come\u00e7ar elevando ao quadrado os dois membros da igualdade dada no problema: [tex]\\quad (\\sqrt{x+\\sqrt{4x-4}}+\\sqrt{x-\\sqrt{4x-4}})^2=(\\sqrt{x+3})^2\\\\ \\quad \\bigg(\\sqrt{x+\\sqrt{4x-4}}\\bigg)^2+2\\cdot \\sqrt{x+\\sqrt{4x-4}}\\cdot \\sqrt{x-\\sqrt{4x-4}}+\\bigg(\\sqrt{x-\\sqrt{4x-4}}\\bigg)^2=x+3\\\\ \\quad x+\\cancel{\\sqrt{4x-4}}+2\\sqrt{(x+\\sqrt{4x-4})\\cdot (x-\\sqrt{4x-4})}+x-\\cancel{\\sqrt{4x-4}}=x+3\\\\ \\quad &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problemao-ali-babao-e-vigesima-setima-de-suas-40-equacoes\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":721,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-161895","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/161895","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\/721"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=161895"}],"version-history":[{"count":15,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/161895\/revisions"}],"predecessor-version":[{"id":197792,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/161895\/revisions\/197792"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=161895"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}