{"id":14392,"date":"2018-04-11T18:00:51","date_gmt":"2018-04-11T17:00:51","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=14392"},"modified":"2022-01-06T23:01:05","modified_gmt":"2022-01-06T23:01:05","slug":"equacoes","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=14392","title":{"rendered":"Equa\u00e7\u00f5es"},"content":{"rendered":"<p><ul id='GTTabs_ul_14392' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_14392' class='GTTabs_curr'><a  id=\"14392_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_14392' ><a  id=\"14392_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_14392'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Resolve as seguintes equa\u00e7\u00f5es:<\/p>\n<ol>\n<li>$3{x^2} &#8211; 7 = 0$<\/li>\n<li>$2\\left( {{x^2} + x} \\right) = x$<\/li>\n<li>$\\frac{{13}}{4}{x^2} = \\frac{{13}}{5}$<\/li>\n<li>$2{x^2} + 3 = 0$<\/li>\n<li>$\\frac{4}{7}\\left( {x &#8211; 2} \\right)(x + 2) + x = \\frac{{9 + 7x}}{7}$<\/li>\n<li>\\(x = 3{x^2}\\)<\/li>\n<li>\\({4{x^2} &#8211; \\frac{1}{4}x = 0}\\)<\/li>\n<li>\\({ &#8211; 2{x^2} &#8211; 5x = 0}\\)<\/li>\n<li>\\({\\frac{1}{2}x &#8211; \\frac{1}{5}{x^2} = 0}\\)<\/li>\n<li>\\({\\frac{4}{5}{{\\left( {x &#8211; 2} \\right)}^2} + x = \\frac{{16 &#8211; 3x}}{5}}\\)<\/li>\n<li>\\({{{\\left( {x &#8211; 1} \\right)}^2} = 3{x^2} + 1}\\)<\/li>\n<li>\\({{{\\left( {x &#8211; 3} \\right)}^2} = \\left( {x &#8211; 3} \\right)\\left( {x + 3} \\right)}\\)<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_14392' onClick='GTTabs_show(1,14392)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_14392'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p><strong>Casos not\u00e1veis<\/strong>:<br \/>\n$${\\left( {A + B} \\right)^2} = {A^2} + 2AB + {B^2}$$<\/p>\n<p>$$\\left( {A + B} \\right)\\left( {A &#8211; B} \\right) = {A^2} &#8211; {B^2}$$<\/p>\n<p><strong>Lei do anulamento do produto<\/strong>:<br \/>\n$$\\begin{array}{*{20}{c}}<br \/>\n{A \\times B = 0}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}<br \/>\n{A = 0}&amp; \\vee &amp;{B = 0}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/p>\n<\/blockquote>\n<p>\u00ad<\/p>\n<ol>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{3{x^2} &#8211; 7 = 0}&amp; \\Leftrightarrow &amp;{{x^2} = \\frac{7}{3}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x = &#8211; \\sqrt {\\frac{7}{3}} }&amp; \\vee &amp;{x = \\sqrt {\\frac{7}{3}} }<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{2\\left( {{x^2} + x} \\right) = x}&amp; \\Leftrightarrow &amp;{2{x^2} + 2x &#8211; x = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x\\left( {2x + 1} \\right) = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x = 0}&amp; \\vee &amp;{2x + 1 = 0}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x = 0}&amp; \\vee &amp;{x = &#8211; \\frac{1}{2}}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{\\frac{{13}}{4}{x^2} = \\frac{{13}}{5}}&amp; \\Leftrightarrow &amp;{\\frac{{{x^2}}}{4} = \\frac{1}{5}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{{x^2} = \\frac{4}{5}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x = &#8211; \\sqrt {\\frac{4}{5}} }&amp; \\vee &amp;{x = \\sqrt {\\frac{4}{5}} }<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{2{x^2} + 3 = 0}&amp; \\Leftrightarrow &amp;{\\overbrace {{x^2} = &#8211; \\frac{3}{2}}^{{\\text{Eq}}{\\text{. impossivel}}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x \\in \\emptyset }<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{\\frac{4}{{\\mathop 7\\limits_{(1)} }}\\left( {x &#8211; 2} \\right)(x + 2) + \\mathop x\\limits_{(7)} = \\frac{{9 + 7x}}{{\\mathop 7\\limits_{(1)} }}}&amp; \\Leftrightarrow &amp;{4\\left( {x &#8211; 2} \\right)(x + 2) + 7x = 9 + 7x} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{4\\left( {{x^2} &#8211; 4} \\right) + 7x = 9 + 7x} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{4{x^2} &#8211; 16 = 9} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{{x^2} = \\frac{{25}}{4}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x = &#8211; \\sqrt {\\frac{{25}}{4}} }&amp; \\vee &amp;{x = \\sqrt {\\frac{{25}}{4}} }<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x = &#8211; \\frac{5}{2}}&amp; \\vee &amp;{x = \\frac{5}{2}}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n\\[\\begin{array}{*{20}{l}}{x = 3{x^2}}&amp; \\Leftrightarrow &amp;{3{x^2} &#8211; x = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x\\left( {3x &#8211; 1} \\right) = 0}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{3x &#8211; 1 = 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{x = \\frac{1}{3}}\\end{array}}\\end{array}\\]<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n\\[\\begin{array}{*{20}{l}}{4{x^2} &#8211; \\frac{1}{4}x = 0}&amp; \\Leftrightarrow &amp;{x\\left( {4x &#8211; \\frac{1}{4}} \\right) = 0}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{4x &#8211; \\frac{1}{4} = 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{x = \\frac{1}{{16}}}\\end{array}}\\end{array}\\]<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n\\[\\begin{array}{*{20}{l}}{ &#8211; 2{x^2} &#8211; 5x = 0}&amp; \\Leftrightarrow &amp;{ &#8211; x\\left( {2x + 5} \\right) = 0}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{ &#8211; x = 0}&amp; \\vee &amp;{2x + 5 = 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{x = &#8211; \\frac{5}{2}}\\end{array}}\\end{array}\\]<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n\\[\\begin{array}{*{20}{l}}{\\frac{1}{2}x &#8211; \\frac{1}{5}{x^2} = 0}&amp; \\Leftrightarrow &amp;{x\\left( {\\frac{1}{2} &#8211; \\frac{1}{5}x} \\right) = 0}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{\\frac{1}{2} &#8211; \\frac{1}{5}x = 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{5 &#8211; 2x = 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{x = \\frac{5}{2}}\\end{array}}\\end{array}\\]<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n\\[\\begin{array}{*{20}{l}}{\\frac{4}{5}{{\\left( {x &#8211; 2} \\right)}^2} + x = \\frac{{16 &#8211; 3x}}{5}}&amp; \\Leftrightarrow &amp;{4{{\\left( {x &#8211; 2} \\right)}^2} + 5x = 16 &#8211; 3x}\\\\{}&amp; \\Leftrightarrow &amp;{4\\left( {{x^2} &#8211; 4x + 4} \\right) + 5x = 16 &#8211; 3x}\\\\{}&amp; \\Leftrightarrow &amp;{4{x^2} &#8211; 8x = 0}\\\\{}&amp; \\Leftrightarrow &amp;{4x\\left( {x &#8211; 2} \\right) = 0}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{4x = 0}&amp; \\vee &amp;{x &#8211; 2 = 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{x = 2}\\end{array}}\\end{array}\\]<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n\\[\\begin{array}{*{20}{l}}{{{\\left( {x &#8211; 1} \\right)}^2} = 3{x^2} + 1}&amp; \\Leftrightarrow &amp;{{x^2} &#8211; 2x + 1 = 3{x^2} + 1}\\\\{}&amp; \\Leftrightarrow &amp;{2{x^2} + 2x = 0}\\\\{}&amp; \\Leftrightarrow &amp;{2x\\left( {x + 1} \\right) = 0}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{2x = 0}&amp; \\vee &amp;{x + 1 = 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{x = &#8211; 1}\\end{array}}\\end{array}\\]<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, vem:<br \/>\n\\[\\begin{array}{*{20}{l}}{{{\\left( {x &#8211; 3} \\right)}^2} = \\left( {x &#8211; 3} \\right)\\left( {x + 3} \\right)}&amp; \\Leftrightarrow &amp;{{{\\left( {x &#8211; 3} \\right)}^2} &#8211; \\left( {x &#8211; 3} \\right)\\left( {x + 3} \\right) = 0}\\\\{}&amp; \\Leftrightarrow &amp;{\\left( {x &#8211; 3} \\right)\\left[ {\\left( {x &#8211; 3} \\right) &#8211; \\left( {x + 3} \\right)} \\right] = 0}\\\\{}&amp; \\Leftrightarrow &amp;{\\left( {x &#8211; 3} \\right)\\left( { &#8211; 6} \\right) = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x = 3}\\end{array}\\]<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_14392' onClick='GTTabs_show(0,14392)'>&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: $3{x^2} &#8211; 7 = 0$ $2\\left( {{x^2} + x} \\right) = x$ $\\frac{{13}}{4}{x^2} = \\frac{{13}}{5}$ $2{x^2} + 3 = 0$ $\\frac{4}{7}\\left( {x &#8211; 2} \\right)(x + 2)&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":14061,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,303],"tags":[426,196,306,198],"series":[],"class_list":["post-14392","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-casos-notaveis","tag-equacao-do-2-o-grau","tag-lei-do-anulamento-do-produto"],"views":2946,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/03\/Mat06.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/14392","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=14392"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/14392\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=14392"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=14392"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=14392"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=14392"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}