{"id":14420,"date":"2018-04-12T19:18:10","date_gmt":"2018-04-12T18:18:10","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=14420"},"modified":"2022-01-06T23:24:33","modified_gmt":"2022-01-06T23:24:33","slug":"resolve-as-seguintes-equacoes-do-2-o-grau-utilizando-o-completamento-do-quadrado","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=14420","title":{"rendered":"Resolve as seguintes equa\u00e7\u00f5es do 2.\u00ba grau, utilizando o completamento do quadrado"},"content":{"rendered":"<p><ul id='GTTabs_ul_14420' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_14420' class='GTTabs_curr'><a  id=\"14420_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_14420' ><a  id=\"14420_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_14420'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Resolve as seguintes equa\u00e7\u00f5es do 2.\u00ba grau, utilizando o completamento do quadrado.<\/p>\n<ol>\n<li>\\({x^2} + 2x &#8211; 3 = 0\\)<\/li>\n<li>\\({x^2} &#8211; 13x + 42 = 0\\)<\/li>\n<li>\\( &#8211; {x^2} &#8211; 5x + 3 = 0\\)<\/li>\n<li>\\(3{x^2} + 5x = 2\\)<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_14420' onClick='GTTabs_show(1,14420)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_14420'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{{x^2} + 2x &#8211; 3 = 0}&amp; \\Leftrightarrow &amp;{{{\\left( {x + 1} \\right)}^2} &#8211; 1 &#8211; 3 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{{{\\left( {x + 1} \\right)}^2} = 4}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x + 1 = &#8211; 2}&amp; \\vee &amp;{x + 1 = 2}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = &#8211; 3}&amp; \\vee &amp;{x = 1}\\end{array}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{{x^2} &#8211; 13x + 42 = 0}&amp; \\Leftrightarrow &amp;{{{\\left( {x &#8211; \\frac{{13}}{2}} \\right)}^2} &#8211; \\frac{{169}}{4} + 42 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{{{\\left( {x &#8211; \\frac{{13}}{2}} \\right)}^2} = \\frac{1}{4}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x &#8211; \\frac{{13}}{2} = &#8211; \\frac{1}{2}}&amp; \\vee &amp;{x &#8211; \\frac{{13}}{2} = \\frac{1}{2}}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 6}&amp; \\vee &amp;{x = 7}\\end{array}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{ &#8211; {x^2} &#8211; 5x + 3 = 0}&amp; \\Leftrightarrow &amp;{{x^2} + 5x &#8211; 3 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{{{\\left( {x + \\frac{5}{2}} \\right)}^2} &#8211; \\frac{{25}}{4} &#8211; 3 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{{{\\left( {x + \\frac{5}{2}} \\right)}^2} = \\frac{{37}}{4}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x + \\frac{5}{2} = &#8211; \\frac{{\\sqrt {37} }}{2}}&amp; \\vee &amp;{x + \\frac{5}{2} = \\frac{{\\sqrt {37} }}{2}}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{{ &#8211; 5 &#8211; \\sqrt {37} }}{2}}&amp; \\vee &amp;{x = \\frac{{ &#8211; 5 + \\sqrt {37} }}{2}}\\end{array}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{3{x^2} + 5x = 2}&amp; \\Leftrightarrow &amp;{{x^2} + \\frac{5}{3}x &#8211; \\frac{2}{3} = 0}\\\\{}&amp; \\Leftrightarrow &amp;{{{\\left( {x + \\frac{5}{6}} \\right)}^2} &#8211; \\frac{{25}}{{36}} &#8211; \\frac{2}{3} = 0}\\\\{}&amp; \\Leftrightarrow &amp;{{{\\left( {x + \\frac{5}{6}} \\right)}^2} = \\frac{{49}}{{36}}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x + \\frac{5}{6} = &#8211; \\frac{7}{6}}&amp; \\vee &amp;{x + \\frac{5}{6} = \\frac{7}{6}}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = &#8211; 2}&amp; \\vee &amp;{x = \\frac{1}{3}}\\end{array}}\\end{array}\\]<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_14420' onClick='GTTabs_show(0,14420)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Resolve as seguintes equa\u00e7\u00f5es do 2.\u00ba grau, utilizando o completamento do quadrado. \\({x^2} + 2x &#8211; 3 = 0\\) \\({x^2} &#8211; 13x + 42 = 0\\) \\( &#8211; {x^2} &#8211; 5x&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":14115,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,303],"tags":[426,306],"series":[],"class_list":["post-14420","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-9--ano","category-aplicando","category-equacoes-do-2-o-grau","tag-9-o-ano","tag-equacao-do-2-o-grau"],"views":1901,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/03\/Mat57.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/14420","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=14420"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/14420\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14115"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=14420"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=14420"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=14420"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=14420"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}