{"id":8540,"date":"2012-04-19T01:10:43","date_gmt":"2012-04-19T00:10:43","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=8540"},"modified":"2022-01-06T19:14:39","modified_gmt":"2022-01-06T19:14:39","slug":"resolve-as-seguintes-equacoes-3","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=8540","title":{"rendered":"Resolve as seguintes equa\u00e7\u00f5es"},"content":{"rendered":"<p><ul id='GTTabs_ul_8540' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_8540' class='GTTabs_curr'><a  id=\"8540_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_8540' ><a  id=\"8540_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_8540'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Resolve as seguintes equa\u00e7\u00f5es pelo processo mais adequado:<\/p>\n<ol>\n<li>${x^2} &#8211; 2x + 1 = 0$<\/li>\n<li>$9{x^2} + 12x + 4 = 0$<\/li>\n<li>$4{x^2} &#8211; 20x + 25 = 0$<\/li>\n<li>${x^2} &#8211; 8x = 4$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_8540' onClick='GTTabs_show(1,8540)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_8540'>\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<ol>\n<li>Resolvendo a equa\u00e7\u00e3o, temos:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{{x^2} &#8211; 2x + 1 = 0}&amp; \\Leftrightarrow &amp;{{{\\left( {x &#8211; 1} \\right)}^2} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x &#8211; 1 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = 1}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, temos:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{9{x^2} + 12x + 4 = 0}&amp; \\Leftrightarrow &amp;{{{\\left( {3x + 2} \\right)}^2} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{3x + 2 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x =\u00a0 &#8211; \\frac{2}{3}}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, temos:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{4{x^2} &#8211; 20x + 25 = 0}&amp; \\Leftrightarrow &amp;{{{\\left( {2x &#8211; 5} \\right)}^2} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{2x &#8211; 5 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{5}{2}}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, temos:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{{x^2} &#8211; 8x = 4}&amp; \\Leftrightarrow &amp;{{x^2} &#8211; 8x &#8211; 4 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{{{\\left( {x &#8211; 4} \\right)}^2} &#8211; 16 &#8211; 4 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{{{\\left( {x &#8211; 4} \\right)}^2} = 20} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x &#8211; 4 =\u00a0 &#8211; \\sqrt {20} }&amp; \\vee &amp;{x &#8211; 4 = \\sqrt {20} }<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x = 4 &#8211; \\sqrt {20} }&amp; \\vee &amp;{x = 4 + \\sqrt {20} }<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<br \/>\n<strong>Alternativa<\/strong>:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{{x^2} &#8211; 8x = 4}&amp; \\Leftrightarrow &amp;{{x^2} &#8211; 8x &#8211; 4 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{{{\\left( {x &#8211; 4} \\right)}^2} &#8211; 16 &#8211; 4 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{{{\\left( {x &#8211; 4} \\right)}^2} &#8211; 20 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{{{\\left( {x &#8211; 4} \\right)}^2} &#8211; {{\\left( {\\sqrt {20} } \\right)}^2} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\left[ {\\left( {x &#8211; 4} \\right) + \\sqrt {20} } \\right]\\left[ {\\left( {x &#8211; 4} \\right) &#8211; \\sqrt {20} } \\right] = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x &#8211; 4 + \\sqrt {20}\u00a0 = 0}&amp; \\vee &amp;{x &#8211; 4 &#8211; \\sqrt {20} }<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x = 4 &#8211; \\sqrt {20} }&amp; \\vee &amp;{x = 4 + \\sqrt {20} }<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_8540' onClick='GTTabs_show(0,8540)'>&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 pelo processo mais adequado: ${x^2} &#8211; 2x + 1 = 0$ $9{x^2} + 12x + 4 = 0$ $4{x^2} &#8211; 20x + 25 = 0$ ${x^2} &#8211;&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19178,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,303],"tags":[426,196,160,198],"series":[],"class_list":["post-8540","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","tag-lei-do-anulamento-do-produto"],"views":2894,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat69.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/8540","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=8540"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/8540\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=8540"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=8540"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=8540"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=8540"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}