{"id":128053,"date":"2021-05-27T07:00:00","date_gmt":"2021-05-27T10:00:00","guid":{"rendered":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/?post_type=ensino_fundamental&#038;p=128053"},"modified":"2021-12-22T12:02:00","modified_gmt":"2021-12-22T14:02:00","slug":"matematica-equacoes-do-segundo-grau","status":"publish","type":"ensino_fundamental","link":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/ensino_fundamental\/matematica-equacoes-do-segundo-grau\/","title":{"rendered":"Matem\u00e1tica &#8211; Equa\u00e7\u00f5es do Segundo Grau"},"content":{"rendered":"\n<p class=\"has-text-align-center has-white-color has-luminous-vivid-orange-background-color has-text-color has-background\" style=\"font-size:23px\">Ol\u00e1, educando (a)! Esta videoaula de<strong> Matem\u00e1tica para o Agrupamento I (9\u00ba ano) do Ciclo da adolesc\u00eancia<\/strong> foi veiculada na TV no dia<strong> 27\/05\/2021 (Quinta-feira).<\/strong> Aqui no Portal Conex\u00e3o Escola, ela est\u00e1 dispon\u00edvel juntamente com a proposta de atividade.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img fetchpriority=\"high\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/1-5-e1622052253890.jpg\" alt=\"\" class=\"wp-image-128054\" width=\"677\" height=\"455\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/1-5-e1622052253890.jpg 377w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/1-5-e1622052253890-300x202.jpg 300w\" sizes=\"(max-width: 677px) 100vw, 677px\" \/><figcaption>Fonte:<br><a href=\"https:\/\/ccnull.de\/foto\/teaching-writing-and-explaining-the-discriminant-formula-in-quadratic-equations\/1063100\">https:\/\/ccnull.de\/foto\/teaching-writing-and-explaining-the-discriminant-formula-in-quadratic-equations\/1063100<\/a><\/figcaption><\/figure><\/div>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-text-align-center has-black-color has-text-color has-background\" style=\"background-color:#e7ce88;font-size:22px\">Nesta atividade, voc\u00ea, estudante do 9\u00ba ano, ir\u00e1 estudar sobre as equa\u00e7\u00f5es polinomiais do segundo grau, compreendendo algumas situa\u00e7\u00f5es problemas que envolvem esse tipo de equa\u00e7\u00e3o, inclusive, realizando algumas demonstra\u00e7\u00f5es alg\u00e9bricas para se chegar a resolu\u00e7\u00e3o de Bhaskara e resolu\u00e7\u00e3o geom\u00e9trica. <\/p>\n\n\n\n<p class=\"has-text-align-center has-vivid-red-color has-text-color has-background\" style=\"background-color:#e7ce88;font-size:32px\">Voc\u00ea n\u00e3o pode ficar fora dessa!<\/p>\n\n\n\n<p class=\"has-text-align-center has-white-color has-black-background-color has-text-color has-background\" style=\"font-size:24px\">Assista a videoaula a seguir com a tem\u00e1tica: \u00c1LGEBRA E FUN\u00c7\u00d5ES<\/p>\n\n\n\n<figure class=\"wp-block-embed is-type-video is-provider-youtube wp-block-embed-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<p class=\"responsive-video-wrap clr\"><iframe title=\"Matem\u00e1tica - 9 ano\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/3GgSbqa6wcQ?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<\/div><figcaption><strong>Agrupamento I|9\u00ba ano|Ciclo da adolesc\u00eancia |Matem\u00e1tica | Prof. Bruno Silva Silvestre<\/strong><\/figcaption><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-cyan-bluish-gray-background-color has-background\" style=\"font-size:22px\">Ol\u00e1 estudante do nono ano da Rede Municipal de Educa\u00e7\u00e3o da Cidade de Goi\u00e2nia. Nesta atividade de matem\u00e1tica voc\u00ea vai estudar sobre as equa\u00e7\u00f5es polinomiais do segundo grau, ressaltando:<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li><strong><span class=\"has-inline-color has-vivid-red-color\">O que \u00e9 uma equa\u00e7\u00e3o do segundo grau;<\/span><\/strong><\/li><li><strong><span class=\"has-inline-color has-vivid-red-color\">Equa\u00e7\u00f5es do segundo grau: completas e incompletas;<\/span><\/strong><\/li><li><strong><span class=\"has-inline-color has-vivid-red-color\">Resolu\u00e7\u00e3o de situa\u00e7\u00f5es problemas envolvendo equa\u00e7\u00f5es do segundo grau;<\/span><\/strong><\/li><li><strong><span class=\"has-inline-color has-vivid-red-color\">Resolu\u00e7\u00e3o alg\u00e9brica da equa\u00e7\u00e3o do segundo grau pelo m\u00e9todo babil\u00f4nico de completar quadrados e m\u00e9todo de Bhaskara;<\/span><\/strong><\/li><\/ul>\n\n\n\n<p style=\"font-size:22px\">Para iniciar seus estudos sobre as equa\u00e7\u00f5es do segundo grau, analise a situa\u00e7\u00e3o abaixo:<\/p>\n\n\n\n<p class=\"has-text-align-center has-white-color has-vivid-cyan-blue-background-color has-text-color has-background\" style=\"font-size:22px\"><em>Certo sat\u00e9lite artificial fotografou uma regi\u00e3o da Terra de formato quadrado, medindo 625 m 2 de \u00e1rea. Qual \u00e9 a medida do comprimento do lado da regi\u00e3o fotografada pelo sat\u00e9lite?<\/em><\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/2-4-e1622052273169.jpg\" alt=\"\" class=\"wp-image-128055\" width=\"654\" height=\"514\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/2-4-e1622052273169.jpg 363w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/2-4-e1622052273169-300x236.jpg 300w\" sizes=\"(max-width: 654px) 100vw, 654px\" \/><figcaption>Fonte: (PATARO &amp; BALESTRI, 2018, p. 35) PNLD<\/figcaption><\/figure><\/div>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-white-color has-vivid-cyan-blue-background-color has-text-color has-background\" style=\"font-size:22px\"><em>Para resolver o problema, representamos por x a medida do comprimento do lado da regi\u00e3o fotografada e escrevemos a equa\u00e7\u00e3o:<\/em><\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/3-6-e1622052285276.jpg\" alt=\"\" class=\"wp-image-128056\" width=\"248\" height=\"296\"\/><\/figure><\/div>\n\n\n\n<p class=\"has-medium-font-size\"><em>Nesse caso, x corresponde \u00e0 medida do lado da regi\u00e3o, que deve ser positiva. Assim, desconsideramos o valor negativo \u2212 25 .&nbsp;<\/em><\/p>\n\n\n\n<p class=\"has-medium-font-size\"><em>Portanto, a regi\u00e3o fotografada tem 25 m de medida de comprimento de lado. Na equa\u00e7\u00e3o, o maior expoente da inc\u00f3gnita \u00e9 2.&nbsp;<\/em><\/p>\n\n\n\n<p class=\"has-text-align-center has-white-color has-luminous-vivid-orange-background-color has-text-color has-background\" style=\"font-size:23px\"><em>Dizemos que essa \u00e9 uma <\/em><strong><em>equa\u00e7\u00e3o do 2o grau com uma inc\u00f3gnita<\/em><\/strong><em>.<\/em><\/p>\n\n\n\n<p class=\"has-medium-font-size\">Uma equa\u00e7\u00e3o do segundo grau completa possui tr\u00eas termos, conforme mostra a imagem abaixo:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/4-6-e1622052298128.jpg\" alt=\"\" class=\"wp-image-128057\" width=\"308\" height=\"80\"\/><\/figure><\/div>\n\n\n\n<p class=\"has-medium-font-size\">No entanto, as equa\u00e7\u00f5es do segundo grau, tamb\u00e9m podem aparecer em sua forma incompleta, conforme mostra o quadro abaixo:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/5-5-e1622052311925.jpg\" alt=\"\" class=\"wp-image-128058\" width=\"883\" height=\"598\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/5-5-e1622052311925.jpg 636w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/5-5-e1622052311925-300x203.jpg 300w\" sizes=\"(max-width: 883px) 100vw, 883px\" \/><figcaption>Fonte: (PATARO &amp; BALESTRI, 2018, p. 35) PNLD<\/figcaption><\/figure><\/div>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-white-color has-vivid-cyan-blue-background-color has-text-color has-background\" style=\"font-size:23px\"><strong>Quest\u00e3o 01<\/strong>. Sobre as equa\u00e7\u00f5es do segundo grau, e, conforme estudado at\u00e9 o momento, quais das equa\u00e7\u00f5es abaixo s\u00e3o do segundo grau?<\/p>\n\n\n\n<figure class=\"wp-block-image size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/6-5-e1622052325835.jpg\" alt=\"\" class=\"wp-image-128059\" width=\"691\" height=\"215\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/6-5-e1622052325835.jpg 514w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/6-5-e1622052325835-300x93.jpg 300w\" sizes=\"(max-width: 691px) 100vw, 691px\" \/><figcaption>Fonte: (PATARO &amp; BALESTRI, 2018, p. 35) PNLD<\/figcaption><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-white-color has-vivid-cyan-blue-background-color has-text-color has-background\" style=\"font-size:22px\"><strong>Quest\u00e3o 02<\/strong>. Escreva uma equa\u00e7\u00e3o do 2\u00ba grau na forma reduzida que representa a medida da \u00e1rea de um quadrado cujo comprimento do lado mede 2x + 1 cm \u00e9 igual a 16 cent\u00edmetros quadrados.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/7-6-e1622052336961.jpg\" alt=\"\" class=\"wp-image-128060\" width=\"274\" height=\"298\"\/><figcaption>Fonte: (PATARO &amp; BALESTRI, 2018, p. 36) PNLD<\/figcaption><\/figure><\/div>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-medium-font-size\">Para resolver esse problema, basta multiplicar o lado do quadrado por ele mesmo, colocando essa multiplica\u00e7\u00e3o igual ao valor da \u00e1rea do quadrado:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"381\" height=\"211\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/8-5-e1622052351774.jpg\" alt=\"\" class=\"wp-image-128061\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/8-5-e1622052351774.jpg 381w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/8-5-e1622052351774-300x166.jpg 300w\" sizes=\"(max-width: 381px) 100vw, 381px\" \/><\/figure><\/div>\n\n\n\n<p class=\"has-white-color has-luminous-vivid-orange-background-color has-text-color has-background\" style=\"font-size:23px\">Resolvendo uma equa\u00e7\u00e3o do segundo grau incompleta em b<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Uma equa\u00e7\u00e3o do segundo grau, incompleta em b, tem a forma geral:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/9-5-e1622052364758.jpg\" alt=\"\" class=\"wp-image-128062\" width=\"308\" height=\"118\"\/><\/figure><\/div>\n\n\n\n<p class=\"has-medium-font-size\">Agora, observe como procedemos para sua resolu\u00e7\u00e3o:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/10-4-e1622052377628.jpg\" alt=\"\" class=\"wp-image-128063\" width=\"571\" height=\"343\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/10-4-e1622052377628.jpg 376w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/10-4-e1622052377628-300x180.jpg 300w\" sizes=\"(max-width: 571px) 100vw, 571px\" \/><figcaption><strong>Fonte: produ\u00e7\u00e3o do NE<\/strong>C<\/figcaption><\/figure><\/div>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-medium-font-size\">Assim, as ra\u00edzes (solu\u00e7\u00e3o) da equa\u00e7\u00e3o \u00e9 {-8 e +8}<\/p>\n\n\n\n<p class=\"has-white-color has-luminous-vivid-orange-background-color has-text-color has-background\" style=\"font-size:23px\">Resolvendo uma equa\u00e7\u00e3o do segundo grau incompleta em c<\/p>\n\n\n\n<p style=\"font-size:24px\">Uma equa\u00e7\u00e3o do segundo grau incompleta em c, tem a forma geral:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/11-4-e1622052391244.jpg\" alt=\"\" class=\"wp-image-128064\" width=\"305\" height=\"101\"\/><\/figure><\/div>\n\n\n\n<p class=\"has-vivid-red-color has-text-color\" style=\"font-size:24px\">Agora observe o seu processo de resolu\u00e7\u00e3o:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/12-4-e1622052411975.jpg\" alt=\"\" class=\"wp-image-128065\" width=\"543\" height=\"545\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/12-4-e1622052411975.jpg 384w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/12-4-e1622052411975-298x300.jpg 298w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/12-4-e1622052411975-150x150.jpg 150w\" sizes=\"(max-width: 543px) 100vw, 543px\" \/><figcaption>Fonte: produ\u00e7\u00e3o do NEC<\/figcaption><\/figure><\/div>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-medium-font-size\">Assim, as ra\u00edzes (solu\u00e7\u00e3o) da equa\u00e7\u00e3o \u00e9 {0 e 8}<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Agora que voc\u00ea j\u00e1 sabe como resolver equa\u00e7\u00f5es incompletas em b ou c, resolva a quest\u00e3o abaixo:<\/p>\n\n\n\n<p class=\"has-white-color has-vivid-cyan-blue-background-color has-text-color has-background\" style=\"font-size:22px\"><strong>Quest\u00e3o 03<\/strong>. Determine a medida do comprimento dos lados de cada ret\u00e2ngulo.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/13-4-e1622076689154.jpg\" alt=\"\" class=\"wp-image-128066\" width=\"880\" height=\"313\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/13-4-e1622076689154.jpg 639w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/13-4-e1622076689154-300x107.jpg 300w\" sizes=\"(max-width: 880px) 100vw, 880px\" \/><figcaption><strong>Fonte: (PATARO &amp; BALESTRI, 2018, p. 39) PNLD<\/strong><\/figcaption><\/figure><\/div>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-white-color has-luminous-vivid-orange-background-color has-text-color has-background\" style=\"font-size:23px\">Resolvendo equa\u00e7\u00f5es completas<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Para resolver equa\u00e7\u00f5es completas na forma:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/14-3-e1622076839324.jpg\" alt=\"\" class=\"wp-image-128067\" width=\"475\" height=\"108\"\/><\/figure><\/div>\n\n\n\n<p class=\"has-medium-font-size\">Existem v\u00e1rios m\u00e9todos, nesta atividade voc\u00ea aprender\u00e1 dois m\u00e9todos de resolu\u00e7\u00e3o: o m\u00e9todo de completar quadrados, tamb\u00e9m conhecido como m\u00e9todo babil\u00f4nico de resolver equa\u00e7\u00f5es do segundo grau e m\u00e9todo de Bhaskara.<\/p>\n\n\n\n<p class=\"has-white-color has-luminous-vivid-orange-background-color has-text-color has-background\" style=\"font-size:23px\">Resolvendo uma equa\u00e7\u00e3o do segundo grau completa: m\u00e9todo de completar quadrados<\/p>\n\n\n\n<p class=\"has-medium-font-size\">H\u00e1 equa\u00e7\u00f5es do 2o grau em que o 1o membro n\u00e3o \u00e9 um trin\u00f4mio quadrado perfeito. Nesses casos, podemos determinar as ra\u00edzes da equa\u00e7\u00e3o utilizando o m\u00e9todo de completar quadrados.<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Tomamos como exemplo a equa\u00e7\u00e3o:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/15-3-e1622076957288.jpg\" alt=\"\" class=\"wp-image-128068\" width=\"732\" height=\"710\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/15-3-e1622076957288.jpg 407w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/15-3-e1622076957288-300x292.jpg 300w\" sizes=\"(max-width: 732px) 100vw, 732px\" \/><\/figure><\/div>\n\n\n\n<p class=\"has-text-align-center has-large-font-size\">Logo, a solu\u00e7\u00e3o da equa\u00e7\u00e3o \u00e9 {-7 e -1}<\/p>\n\n\n\n<p class=\"has-white-color has-luminous-vivid-orange-background-color has-text-color has-background\" style=\"font-size:23px\">Resolvendo uma equa\u00e7\u00e3o do segundo grau completa: m\u00e9todo de Bhaskara<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Outra maneira de resolver uma equa\u00e7\u00e3o do 2\u00ba grau \u00e9 por meio da f\u00f3rmula resolutiva, que consiste na generaliza\u00e7\u00e3o do m\u00e9todo de completar quadrados.&nbsp;<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Utilizando essa f\u00f3rmula, \u00e9 poss\u00edvel obter as ra\u00edzes de uma equa\u00e7\u00e3o do 2\u00ba grau por meio de seus coeficientes.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/17-3-e1622076988195.jpg\" alt=\"\" class=\"wp-image-128070\" width=\"455\" height=\"153\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/17-3-e1622076988195.jpg 322w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/17-3-e1622076988195-300x101.jpg 300w\" sizes=\"(max-width: 455px) 100vw, 455px\" \/><\/figure><\/div>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/21-2-e1622077328829.jpg\" alt=\"\" class=\"wp-image-128080\" width=\"862\" height=\"343\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/21-2-e1622077328829.jpg 642w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/21-2-e1622077328829-300x119.jpg 300w\" sizes=\"(max-width: 862px) 100vw, 862px\" \/><figcaption><strong>Fonte: (PATARO &amp; BALESTRI, 2018, p. 46) PNLD<\/strong><\/figcaption><\/figure><\/div>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-medium-font-size\">Agora \u00e9 com voc\u00ea, resolva as quest\u00f5es abaixo envolvendo equa\u00e7\u00f5es do segundo grau completas:<\/p>\n\n\n\n<p class=\"has-white-color has-vivid-cyan-blue-background-color has-text-color has-background\" style=\"font-size:22px\"><strong>Quest\u00e3o 04<\/strong>. Determine as ra\u00edzes de cada equa\u00e7\u00e3o pelo m\u00e9todo de completar quadrados.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/19-3-e1622077012856.jpg\" alt=\"\" class=\"wp-image-128072\" width=\"403\" height=\"256\"\/><\/figure>\n\n\n\n<p>Fonte: (PATARO &amp; BALESTRI, 2018, p. 47) PNLD<\/p>\n\n\n\n<p class=\"has-white-color has-vivid-cyan-blue-background-color has-text-color has-background\" style=\"font-size:22px\"><strong>Quest\u00e3o 05<\/strong>. Sabendo que a medida da \u00e1rea do quadrado \u00e9 8 cent\u00edmetros quadrados maior do que a medida da \u00e1rea do ret\u00e2ngulo, calcule o valor de x.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/05\/20-2-e1622077024346.jpg\" alt=\"\" class=\"wp-image-128073\" width=\"304\" height=\"399\"\/><\/figure><\/div>\n\n\n\n<p class=\"has-vivid-red-color has-text-color has-large-font-size\"><strong>RELEMBRANDO!!!!!!!<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">Em s\u00edntese, nesta atividade de matem\u00e1tica voc\u00ea estudou alguns conhecimentos alg\u00e9bricos, ressaltando as equa\u00e7\u00f5es polinomiais do segundo grau, estudando e identificando uma equa\u00e7\u00e3o do segundo grau completa ou incompleta, desenvolvendo c\u00e1lculos necess\u00e1rios para a sua resolu\u00e7\u00e3o.<\/p>\n\n\n\n<p class=\"has-text-align-center has-luminous-vivid-orange-color has-text-color has-background\" style=\"background-color:#eed1d1;font-size:30px\"><strong>Parab\u00e9ns, continue com empenho em seus estudos matem\u00e1ticos. At\u00e9 a pr\u00f3xima atividade!<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<figure class=\"wp-block-table\"><table><tbody><tr><td><strong>Habilidades<\/strong>&nbsp;<\/td><td><strong>Habilidade Estruturante<\/strong><br><strong>(EF09MA06-E) <\/strong>Aplicar a f\u00f3rmula de Bhaskara para resolver equa\u00e7\u00f5es do 2\u00b0 grau associadas \u00e0s fun\u00e7\u00f5es quadr\u00e1ticas<br><strong>Habilidades Complementares<\/strong><br><strong>EF09MA06<\/strong><br><strong>C&nbsp;EF09MA06<\/strong><br><strong>D&nbsp;EF09MA06<\/strong><br><strong>F&nbsp;EF09MA06-G<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<figure class=\"wp-block-table\"><table><tbody><tr><td><strong>Refer\u00eancias:<\/strong><\/td><td>IEZZI, Gelson., MACHADO, Ant\u00f4nio., DOLCE, Osvaldo. <strong>Matem\u00e1tica e realidade 9\u00ba ano <\/strong>&#8211; 9. ed. &#8211; S\u00e3o Paulo : Atual Editora, 2018.&nbsp;PATARO, Patr\u00edcia Moreno., BALESTRI, Rodrigo. <strong>Matem\u00e1tica essencial 9o ano<\/strong> : ensino fundamental, anos finais &#8211; 1. ed. &#8211; S\u00e3o Paulo : Scipione, 2018. PNLD<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"has-cyan-bluish-gray-background-color has-background\" style=\"font-size:25px\">Professor, essa aula segue a Matriz Curricular das Habilidades Estruturantes 2021-2021. Foi elaborada no ano de 2020, com a suspens\u00e3o das aulas presenciais devido \u00e0 pandemia da Covid-19 e segue as orienta\u00e7\u00f5es de flexibiliza\u00e7\u00e3o curricular para o bi\u00eanio 2020\/2021 (Of\u00edcio Circular 147\/2020 Dirped).<\/p>\n","protected":false},"author":25,"featured_media":128054,"template":"","ef_categoria":[16,35],"ef_ano":[92],"ef_componente":[94],"class_list":["post-128053","ensino_fundamental","type-ensino_fundamental","status-publish","has-post-thumbnail","hentry","ef_categoria-ciclo-da-adolescencia-hi","ef_categoria-educacao-financeira-e-empreendedorismo-ciclo-da-adolescencia-hi","ef_ano-9o-ano","ef_componente-matematica","entry","has-media"],"acf":[],"_links":{"self":[{"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/ensino_fundamental\/128053","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/ensino_fundamental"}],"about":[{"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/types\/ensino_fundamental"}],"author":[{"embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/users\/25"}],"version-history":[{"count":0,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/ensino_fundamental\/128053\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/media\/128054"}],"wp:attachment":[{"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/media?parent=128053"}],"wp:term":[{"taxonomy":"ef_categoria","embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/ef_categoria?post=128053"},{"taxonomy":"ef_ano","embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/ef_ano?post=128053"},{"taxonomy":"ef_componente","embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/ef_componente?post=128053"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}