{"id":143130,"date":"2022-05-09T07:00:00","date_gmt":"2022-05-09T10:00:00","guid":{"rendered":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/?post_type=eaja&#038;p=143130"},"modified":"2023-11-24T13:05:00","modified_gmt":"2023-11-24T16:05:00","slug":"matematica-equacoes-do-2o-grau-2","status":"publish","type":"eaja","link":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/eaja\/matematica-equacoes-do-2o-grau-2\/","title":{"rendered":"Matem\u00e1tica &#8211; A F\u00f3rmula de Bhaskara em A\u00e7\u00e3o"},"content":{"rendered":"\n<p class=\"has-text-align-center has-black-color has-white-background-color has-text-color has-background has-medium-font-size\"><strong>Esta proposta de atividade de&nbsp;MATEM\u00c1TICA&nbsp;\u00e9 destinada aos estudantes do 6\u00ba Per\u00edodo&nbsp;(8\u00aa S\u00e9rie)&nbsp;da Educa\u00e7\u00e3o de Jovens e Adultos \u2013 EJA<\/strong><\/p>\n\n\n\n<div class=\"wp-block-buttons has-custom-font-size has-small-font-size is-content-justification-center is-layout-flex wp-container-core-buttons-is-layout-16018d1d wp-block-buttons-is-layout-flex\">\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link wp-element-button\" href=\"https:\/\/drive.google.com\/uc?export=douwnload&amp;id=18LuFvuLBVA-H6B-alXGke8-opH90ERAF\" target=\"_blank\" rel=\"noreferrer noopener\">BAIXAR A ATIVIDADE<\/a><\/div>\n\n\n\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link wp-element-button\" href=\"https:\/\/drive.google.com\/uc?export=douwnload&amp;id=1Mn5SchwmVgcFFz61LGNfw2_IAJqWT1Q8\" target=\"_blank\" rel=\"noreferrer noopener\">BAIXAR OS SLIDES<\/a><\/div>\n\n\n\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link wp-element-button\" href=\"https:\/\/drive.google.com\/uc?export=douwnload&amp;id=1CIAfazH81je0FwdPF5pxMll0Cli59Hpx\" target=\"_blank\" rel=\"noreferrer noopener\">BAIXAR O TEXTO<\/a><\/div>\n<\/div>\n\n\n\n<div class=\"wp-block-media-text is-stacked-on-mobile\" style=\"grid-template-columns:29% auto\"><figure class=\"wp-block-media-text__media\"><img fetchpriority=\"high\" decoding=\"async\" width=\"308\" height=\"205\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2022\/05\/pp2.png\" alt=\"\" class=\"wp-image-173242 size-full\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2022\/05\/pp2.png 308w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2022\/05\/pp2-300x200.png 300w\" sizes=\"(max-width: 308px) 100vw, 308px\" \/><\/figure><div class=\"wp-block-media-text__content\">\n<p class=\"has-text-color has-medium-font-size\" style=\"color:#0b670b\"><strong>O que s\u00e3o equa\u00e7\u00f5es?<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">As equa\u00e7\u00f5es s\u00e3o express\u00f5es matem\u00e1ticas que estabelecem uma igualdade entre duas informa\u00e7\u00f5es. S\u00e3o compostas por vari\u00e1veis (representadas por letras do nosso alfabeto), constantes (n\u00fameros reais), coeficientes (n\u00fameros reais) e um sinal de igualdade. Elas se diferem pelo valor do maior expoente das vari\u00e1veis.<\/p>\n\n\n\n<p class=\"has-small-font-size\">Imagem: canva.com\/sombra_equa\u00e7\u00e3o_<a href=\"https:\/\/encurtador.com.br\/eoxP6\">https:\/\/encurtador.com.br\/eoxP6<\/a><\/p>\n<\/div><\/div>\n\n\n\n<p class=\"has-medium-font-size\"><strong>As equa\u00e7\u00f5es do 2\u00b0 grau com uma vari\u00e1vel<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">As equa\u00e7\u00f5es do segundo grau com uma vari\u00e1vel s\u00e3o aquelas em que o maior expoente da vari\u00e1vel \u00e9 igual a 2. Elas possuem a seguinte forma geral:<\/p>\n\n\n\n<p class=\"has-text-align-center has-vivid-red-color has-text-color has-medium-font-size\"><strong>ax<\/strong><strong><sup>2<\/sup><\/strong><strong> + bx + c = 0<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li class=\"has-medium-font-size\"><strong><em>x<\/em><\/strong> \u00e9 a vari\u00e1vel.<\/li>\n\n\n\n<li class=\"has-medium-font-size\"><strong><em>a<\/em><\/strong>, <strong><em>b<\/em><\/strong> e <strong><em>c<\/em><\/strong> s\u00e3o os coeficientes, com <strong>a\u22600<\/strong>&nbsp;<\/li>\n<\/ul>\n\n\n\n<p class=\"has-medium-font-size\"><strong>Como resolver uma equa\u00e7\u00e3o do 2\u00b0 grau?<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">A solu\u00e7\u00e3o de uma equa\u00e7\u00e3o&nbsp; do 2\u00ba grau (denominada de <strong>raiz da equa\u00e7\u00e3o<\/strong>) \u00e9 dada pelos valores da vari\u00e1vel (x) que satisfazem a igualdade.<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Existem alguns m\u00e9todos para determinar essa solu\u00e7\u00e3o, neste texto iremos apresentar o m\u00e9todo que se utiliza da <strong>F\u00d3RMULA RESOLUTIVA<\/strong>, mais conhecida como <strong>F\u00d3RMULA DE BHASKARA<\/strong>.<\/p>\n\n\n\n<p class=\"has-medium-font-size\"><strong>M\u00e9todo da F\u00f3rmula Resolutiva (F\u00f3rmula de Bhaskara)<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">A solu\u00e7\u00e3o de uma equa\u00e7\u00e3o do 2\u00b0 grau <strong>ax<\/strong><strong><sup>2<\/sup><\/strong><strong> + bx + c = 0<\/strong> pode ser encontrada pela f\u00f3rmula:<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter\"><img decoding=\"async\" src=\"https:\/\/lh7-us.googleusercontent.com\/wh6n6-JPRDt89owyXSQtILlI_STy7h9Cncj3r2L24DQ1Ye5OUWI0G8tRIM67OI8mQEa48Ma-O9KxUAVO3PmEE5r3j_LIwxLW386x3jezTAaDH3p0X7eM0lGFBTkO9pRjl60a7aO5iEv7\" alt=\"\"\/><\/figure><\/div>\n\n\n<p class=\"has-medium-font-size\">Esta f\u00f3rmula se utiliza dos coeficientes a, b e c.<\/p>\n\n\n\n<p class=\"has-medium-font-size\">O termo <strong>b<sup>2<\/sup> &#8211; 4.a.c<\/strong> ser\u00e1 denominado de discriminante (representado pela letra grega \u0394), este termo ir\u00e1 indicar o n\u00famero de ra\u00edzes que ter\u00e1 a equa\u00e7\u00e3o, ou seja:<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Se <strong>\u0394&gt;0<\/strong>, a equa\u00e7\u00e3o ter\u00e1 <strong>2 ra\u00edzes reais diferentes<\/strong>.<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Se <strong>\u0394&lt;0<\/strong>, a equa\u00e7\u00e3o <strong>n\u00e3o ter\u00e1 ra\u00edzes reais<\/strong>.<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Se <strong>\u0394=0<\/strong>, a equa\u00e7\u00e3o ter\u00e1 <strong>2 ra\u00edzes reais iguais<\/strong>, ou simplesmente, <strong>1 raiz real<\/strong>.<\/p>\n\n\n\n<p class=\"has-medium-font-size\"><strong>Tr\u00eas exemplos para finalizar<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\"><strong>Exemplo 1:<\/strong>&nbsp;<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Determinar a solu\u00e7\u00e3o da equa\u00e7\u00e3o do 2\u00ba grau <strong>x<\/strong><strong><sup>2<\/sup><\/strong><strong> &#8211; 9x + 14 = 0<\/strong>.<\/p>\n\n\n\n<p class=\"has-medium-font-size\"><strong>1\u00ba Passo: Destacar os coeficientes.<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">a = 1, b = &#8211; 9 e c = 14<\/p>\n\n\n\n<p class=\"has-medium-font-size\"><strong>2\u00b0 Passo: Determinar o valor do discriminante e verificar o n\u00famero de ra\u00edzes.<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">\u0394 = b<sup>2<\/sup> &#8211; 4 . a . c = (-9)<sup>2<\/sup> &#8211; 4.1.14 = 81 &#8211; 56 = 25<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Como \u0394 \u00e9 maior do que zero, a equa\u00e7\u00e3o ter\u00e1 2 ra\u00edzes reais e diferentes.<\/p>\n\n\n\n<p class=\"has-medium-font-size\"><strong>3\u00ba Passo: Determinar as ra\u00edzes (solu\u00e7\u00e3o)<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">Utilizando a f\u00f3rmula de Bhaskara e indicando as ra\u00edzes por x<sub>1<\/sub> e x<sub>2<\/sub> teremos:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh7-us.googleusercontent.com\/H67UkWmav09Adq_X9fuxOlg2dv4CEYPcKVPZYDyHRduTIdVJLHvgnAwtXed-bU13yjekc5HASdQZGHJa2R6KsQXEPPjXtYXfgIIffWZL-2zn5WykFiniRjKHKS5_1Asa140WNznAn2XQ\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"has-medium-font-size\"><strong>Exemplo 2<\/strong>:&nbsp;<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Determinar a solu\u00e7\u00e3o da equa\u00e7\u00e3o do 2\u00ba grau x<sup>2<\/sup> + 5x + 8 = 0.<\/p>\n\n\n\n<p class=\"has-medium-font-size\"><strong>1\u00ba Passo: Destacar os coeficientes.<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">a = 1, b = 5 e c = 8<\/p>\n\n\n\n<p class=\"has-medium-font-size\"><strong>2\u00b0 Passo: Determinar o valor do discriminante e verificar o n\u00famero de ra\u00edzes.<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">\u0394 = b<sup>2<\/sup> &#8211; 4 . a . c = 5<sup>2<\/sup> &#8211; 4.1.8 = 25 &#8211; 32 = &#8211; 7<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Como \u0394 \u00e9 menor do que zero, a equa\u00e7\u00e3o n\u00e3o ter\u00e1 ra\u00edzes reais.<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Neste caso, a solu\u00e7\u00e3o ser\u00e1 um conjunto vazio, representada por S = { &nbsp; } ou S =\u00d8.<\/p>\n\n\n\n<p class=\"has-medium-font-size\"><strong>Exemplo 3<\/strong>:&nbsp;<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Determinar a solu\u00e7\u00e3o da equa\u00e7\u00e3o do 2\u00ba grau x<sup>2<\/sup> + 8x + 16 = 0<\/p>\n\n\n\n<p class=\"has-medium-font-size\"><strong>1\u00ba Passo: Destacar os coeficientes.<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">a = 1, b = 8 e c = 16<\/p>\n\n\n\n<p class=\"has-medium-font-size\">2\u00b0 Passo: Determinar o valor do discriminante e verificar o n\u00famero de ra\u00edzes.<\/p>\n\n\n\n<p class=\"has-medium-font-size\">\u0394 = b<sup>2<\/sup> &#8211; 4 . a . c = 8<sup>2<\/sup> &#8211; 4.1.16 = 64 &#8211; 64 = 0<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Como \u0394 \u00e9 igual a zero, a equa\u00e7\u00e3o ter\u00e1 2 ra\u00edzes reais e igual.<\/p>\n\n\n\n<p class=\"has-medium-font-size\"><strong>3\u00ba Passo: Determinar as ra\u00edzes (solu\u00e7\u00e3o)<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">Utilizando a f\u00f3rmula de Bh\u00e1skara e indicando as ra\u00edzes por x<sub>1<\/sub> e x<sub>2<\/sub> teremos:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh7-us.googleusercontent.com\/e2zOLG55QP-XCDUtuqBcciAg1GYlJueS1dPxJ90LuX66ckOkTvT7ILj0sv3_F6WPioqKKF5_t50MitY8H82o7nFAsKK-bY32wtccu4gDMxh1Uxw26kHuBQG44YsdYoZm739Q6BbzUP3m\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"has-medium-font-size\">Neste caso n\u00e3o \u00e9 necess\u00e1rio calcular o x<sub>2<\/sub>.<\/p>\n\n\n\n<p class=\"has-medium-font-size\">S = {-4,-4} ou S={-4}<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Ficamos por aqui, at\u00e9 o pr\u00f3ximo.<\/p>\n\n\n\n<p class=\"has-vivid-green-cyan-background-color has-background has-medium-font-size\"><strong>Quest\u00e3o 01<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">Um agricultor est\u00e1 planejando o layout de um campo retangular para cultivar duas variedades diferentes de plantas. A \u00e1rea total do campo \u00e9 de 2400 metros quadrados. Ele deseja que o comprimento do campo seja 10 metros a mais que o dobro da largura. Qual deve ser a largura e o comprimento do campo? Veja a imagem ilustrando o problema.<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/lh7-us.googleusercontent.com\/0sXAvL8tLHxyEuGgabRJ9A27rUqkMA0hBrqNkoSbuIrJLHx7jAyDWb9DgSbDnCs7VcDxHyDohySPWvl-vqw87bxXIbzKmpq-88mhwk2oG1_AOIRmBfOz-sNd2MqakrAZb_U1SLxZ9ay8\" alt=\"\" width=\"128\" height=\"179\"\/><\/figure><\/div>\n\n\n<p class=\"has-text-align-center\">Imagem do autor produzida no Geogebra<\/p>\n\n\n\n<p class=\"has-vivid-green-cyan-background-color has-background has-medium-font-size\"><strong>Quest\u00e3o 02<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">Determine a largura e o comprimento de um terreno com \u00e1rea de 30 metros quadrados e o comprimento medindo 3m a mais que a largura.<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/lh7-us.googleusercontent.com\/3_OynZtyZ5_KrHQl5fklMKOdDivTLEy-5t_UqcVbom0-kSDS6Dc3ENr9G8W00GV9sQVj6dFwysmy8OUA4NNQQtimMszNRSegYRke1GBOomuQs_xtJWGlChz0ENsz7vdq5wUd2C42Ccjk\" alt=\"\" width=\"220\" height=\"104\"\/><\/figure><\/div>\n\n\n<p class=\"has-text-align-center has-small-font-size\">Imagem do autor produzida no Geogebra<\/p>\n\n\n\n<p class=\"has-vivid-green-cyan-background-color has-background has-medium-font-size\"><strong>Quest\u00e3o 03<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">Dada a equa\u00e7\u00e3o quadr\u00e1tica 3x<sup>2<\/sup> +7x\u22122=0, qual \u00e9 a solu\u00e7\u00e3o correta para x usando a f\u00f3rmula de Bhaskara?<\/p>\n\n\n\n<p class=\"has-medium-font-size\">(A) x = 1<\/p>\n\n\n\n<p class=\"has-medium-font-size\">(B) x = -1<\/p>\n\n\n\n<p class=\"has-medium-font-size\">(C) x = 2<\/p>\n\n\n\n<p class=\"has-medium-font-size\">(D) x = -2<\/p>\n\n\n\n<p class=\"has-vivid-green-cyan-background-color has-background has-medium-font-size\"><strong>Quest\u00e3o 04<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">Considere a equa\u00e7\u00e3o quadr\u00e1tica x<sup>2<\/sup> &#8211; 6x + 8 = 0. Qual das seguintes op\u00e7\u00f5es representa corretamente as solu\u00e7\u00f5es para x usando a f\u00f3rmula de Bhaskara?<\/p>\n\n\n\n<p class=\"has-medium-font-size\">(A) x=2 e x=4<\/p>\n\n\n\n<p class=\"has-medium-font-size\">(B) x=3 e x=5<\/p>\n\n\n\n<p class=\"has-medium-font-size\">(C) x=1 e x=8<\/p>\n\n\n\n<p class=\"has-medium-font-size\">(D) x=2 e x=8<\/p>\n\n\n\n<p class=\"has-white-color has-text-color has-background has-medium-font-size\" style=\"background-color:#336f2f\"><strong>SAIBA MAIS<\/strong><\/p>\n\n\n\n<p class=\"has-medium-font-size\">Quer aprender um pouco mais sobre equa\u00e7\u00f5es do 2\u00ba grau? Ent\u00e3o assista aos v\u00eddeos no canal do YouTube do prof. H\u00e9lio<\/p>\n\n\n\n<p class=\"has-medium-font-size\">Como resolver equa\u00e7\u00f5es do 2\u00b0 grau completas<\/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=\"#1 Equa\u00e7\u00f5es de 2\u00ba Grau Completa\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/MMtVtmOOWcU?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" allowfullscreen><\/iframe><\/p>\n<\/div><figcaption class=\"wp-element-caption\">Canal Professor Helio Roberto da Rocha &#8220;#1 Equa\u00e7\u00f5es de 2\u00ba Grau Completa&#8221;. Dispon\u00edvel em: &lt;https:\/\/youtu.be\/MMtVtmOOWcU&gt;. Acesso em: 09 mai 2022.<\/figcaption><\/figure>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table><tbody><tr><td>Autoria<\/td><td>Prof. H\u00e9lio Roberto da Rocha, Mestre em Matem\u00e1tica<\/td><\/tr><tr><td>Componente Curricular<\/td><td>Matem\u00e1tica<\/td><\/tr><tr><td>Objetivos de aprendizagem e desenvolvimento<\/td><td>(EAJAMA0810) Reconhecer uma equa\u00e7\u00e3o do 2\u00b0grau, identificando seus coeficientes na forma completa (ax<sup>2<\/sup> + bx + c = 0, a \u2260 0) e nas formas incompletas (ax<sup>2<\/sup> + bx = 0, ax<sup>2<\/sup> + c = 0, ax<sup>2<\/sup> = 0 com a \u2260 0) quando apresentada em situa\u00e7\u00f5es-problema, bem como determinar as suas ra\u00edzes por meio da fatora\u00e7\u00e3o ou f\u00f3rmula resolutiva.<\/td><\/tr><tr><td>Refer\u00eancias<\/td><td>SOUZA, Joamir Roberto de: Matem\u00e1tica realidade &amp; tecnologia: 9\u00ba ano: ensino fundamental: anos finais \/ Joamir Roberto de Souza. \u2013 1. ed. \u2013 S\u00e3o Paulo: FTD, 2018.<br>GIOVANNI J\u00daNIOR, Jos\u00e9 Ruy &#8211; A conquista da matem\u00e1tica: 9\u00b0 ano: ensino fundamental: anos finais \/ Jos\u00e9 Ruy Giovanni J\u00fanior, Benedicto Castrucci. \u2014 4. ed. \u2014 S\u00e3o Paulo: FTD, 2018.<br>PATARO, Patricia Moreno Matem\u00e1tica essencial 9\u00b0 ano: ensino fundamental, anos finais \/ Patricia Moreno Pataro, Rodrigo Balestri. &#8211; 1. ed. &#8211; S\u00e3o Paulo: Scipione, 2018.<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-css-opacity\"\/>\n\n\n\n<hr class=\"wp-block-separator 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