{"id":14421,"date":"2018-04-12T19:29:54","date_gmt":"2018-04-12T18:29:54","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=14421"},"modified":"2022-01-06T23:28:09","modified_gmt":"2022-01-06T23:28:09","slug":"resolve-cada-uma-das-seguintes-equacoes","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=14421","title":{"rendered":"Resolve cada uma das seguintes equa\u00e7\u00f5es"},"content":{"rendered":"<p><ul id='GTTabs_ul_14421' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_14421' class='GTTabs_curr'><a  id=\"14421_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_14421' ><a  id=\"14421_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_14421'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Resolve cada uma das seguintes equa\u00e7\u00f5es:<\/p>\n<ol>\n<li>\\(6{x^2} + 5x + 1 = 0\\)<\/li>\n<li>\\({x^2} &#8211; 4x + 4 = 0\\)<\/li>\n<li>\\({x^2} &#8211; 3x + 2 = 0\\)<\/li>\n<li>\\({x^2} &#8211; \\frac{5}{3}x &#8211; \\frac{2}{3} = 0\\)<\/li>\n<li>\\(x\\left( {x &#8211; 8} \\right) = &#8211; 42 + 5x\\)<\/li>\n<li>\\(\\frac{x}{4} &#8211; \\frac{{{{\\left( {x &#8211; 1} \\right)}^2}}}{2} = 0\\)<\/li>\n<li>\\(5\\left( {3 + x} \\right) = \\frac{1}{3}{\\left( { &#8211; 3 &#8211; x} \\right)^2}\\)<\/li>\n<li>\\(4x\\left( {2x &#8211; 5} \\right) = 3x &#8211; 14\\)<\/li>\n<li>\\(2{x^2} + 4x &#8211; 4 = 0\\)<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_14421' onClick='GTTabs_show(1,14421)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_14421'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p><strong>F\u00f3rmula resolvente da equa\u00e7\u00e3o do 2.\u00ba grau<\/strong>:<\/p>\n<p>$$\\begin{array}{*{20}{c}}<br \/>\n{a{x^2} + bx + c = 0}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; b \\pm \\sqrt {{b^2} &#8211; 4ac} }}{{2a}}}&amp;{(a \\ne 0)}<br \/>\n\\end{array}$$<\/p>\n<\/blockquote>\n<p>\u00ad<\/p>\n<ol>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{6{x^2} + 5x + 1 = 0}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 5 \\mp \\sqrt {{5^2} &#8211; 4 \\times 6 \\times 1} }}{{2 \\times 6}}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 5 \\mp 1}}{{12}}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{{ &#8211; 5 &#8211; 1}}{{12}}}&amp; \\vee &amp;{x = \\frac{{ &#8211; 5 + 1}}{{12}}}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = &#8211; \\frac{1}{2}}&amp; \\vee &amp;{x = &#8211; \\frac{1}{3}}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ { &#8211; \\frac{1}{2}, &#8211; \\frac{1}{3}} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{{x^2} &#8211; 4x + 4 = 0}&amp; \\Leftrightarrow &amp;{x = \\frac{{4 \\mp \\sqrt {{{\\left( { &#8211; 4} \\right)}^2} &#8211; 4 \\times 1 \\times 4} }}{{2 \\times 1}}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{4 \\mp 0}}{2}}\\\\{}&amp; \\Leftrightarrow &amp;{x = 2}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ 2 \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{{x^2} &#8211; 3x + 2 = 0}&amp; \\Leftrightarrow &amp;{x = \\frac{{3 \\mp \\sqrt {{{\\left( { &#8211; 3} \\right)}^2} &#8211; 4 \\times 1 \\times 2} }}{{2 \\times 1}}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{3 \\mp 1}}{2}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{{3 &#8211; 1}}{2}}&amp; \\vee &amp;{x = \\frac{{3 + 1}}{2}}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 1}&amp; \\vee &amp;{x = 2}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ {1,\\;2} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{{x^2} &#8211; \\frac{5}{3}x &#8211; \\frac{2}{3} = 0}&amp; \\Leftrightarrow &amp;{3{x^2} &#8211; 5x &#8211; 2 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{5 \\mp \\sqrt {{{\\left( { &#8211; 5} \\right)}^2} &#8211; 4 \\times 3 \\times \\left( { &#8211; 2} \\right)} }}{{2 \\times 3}}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{5 \\mp \\sqrt {49} }}{6}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{{5 &#8211; 7}}{6}}&amp; \\vee &amp;{x = \\frac{{5 + 7}}{6}}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = &#8211; \\frac{1}{3}}&amp; \\vee &amp;{x = 2}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ { &#8211; \\frac{1}{3},\\;2} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{x\\left( {x &#8211; 8} \\right) = &#8211; 42 + 5x}&amp; \\Leftrightarrow &amp;{{x^2} &#8211; 13x + 42 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{13 \\mp \\sqrt {{{\\left( { &#8211; 13} \\right)}^2} &#8211; 4 \\times 1 \\times 42} }}{{2 \\times 1}}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{13 \\mp 1}}{2}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 6}&amp; \\vee &amp;{x = 7}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ {6,\\;7} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{\\frac{x}{4} &#8211; \\frac{{{{\\left( {x &#8211; 1} \\right)}^2}}}{2} = 0}&amp; \\Leftrightarrow &amp;{x &#8211; 2\\left( {{x^2} &#8211; 2x + 1} \\right) = 0}\\\\{}&amp; \\Leftrightarrow &amp;{2{x^2} &#8211; 5x + 2 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{5 \\mp \\sqrt {{{\\left( { &#8211; 5} \\right)}^2} &#8211; 4 \\times 2 \\times 2} }}{{2 \\times 2}}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{5 \\mp \\sqrt 9 }}{4}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{{5 &#8211; 3}}{4}}&amp; \\vee &amp;{x = \\frac{{5 + 3}}{4}}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{1}{2}}&amp; \\vee &amp;{x = 2}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ {\\frac{1}{2},\\;2} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{5\\left( {3 + x} \\right) = \\frac{1}{3}{{\\left( { &#8211; 3 &#8211; x} \\right)}^2}}&amp; \\Leftrightarrow &amp;{45 + 15x = 9 + 6x + {x^2}}\\\\{}&amp; \\Leftrightarrow &amp;{{x^2} &#8211; 9x &#8211; 36 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{9 \\mp \\sqrt {{{\\left( { &#8211; 9} \\right)}^2} &#8211; 4 \\times 1 \\times \\left( { &#8211; 36} \\right)} }}{{2 \\times 1}}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{9 \\mp \\sqrt {225} }}{2}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{{9 &#8211; 15}}{2}}&amp; \\vee &amp;{x = \\frac{{9 + 15}}{2}}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = &#8211; 3}&amp; \\vee &amp;{x = 12}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ { &#8211; 3,\\;12} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{4x\\left( {2x &#8211; 5} \\right) = 3x &#8211; 14}&amp; \\Leftrightarrow &amp;{8{x^2} &#8211; 23x + 14 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{23 \\mp \\sqrt {{{\\left( { &#8211; 23} \\right)}^2} &#8211; 4 \\times 8 \\times 14} }}{{2 \\times 8}}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{23 \\mp \\sqrt {81} }}{{16}}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{{23 &#8211; 9}}{{16}}}&amp; \\vee &amp;{x = \\frac{{23 + 9}}{{16}}}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{7}{8}}&amp; \\vee &amp;{x = 2}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ {\\frac{7}{8},\\;2} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{2{x^2} + 4x &#8211; 4 = 0}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 4 \\mp \\sqrt {{4^2} &#8211; 4 \\times 2 \\times \\left( { &#8211; 4} \\right)} }}{{2 \\times 2}}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 4 \\mp \\sqrt {48} }}{4}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 4 \\mp 4\\sqrt 3 }}{4}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = &#8211; 1 &#8211; \\sqrt 3 }&amp; \\vee &amp;{x = 1 + \\sqrt 3 }\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ {1 &#8211; \\sqrt 3 ,\\;1 + \\sqrt 3 } \\right\\}}\\end{array}\\]<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_14421' onClick='GTTabs_show(0,14421)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Resolve cada uma das seguintes equa\u00e7\u00f5es: \\(6{x^2} + 5x + 1 = 0\\) \\({x^2} &#8211; 4x + 4 = 0\\) \\({x^2} &#8211; 3x + 2 = 0\\) \\({x^2} &#8211; \\frac{5}{3}x &#8211;&#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":[426,306,304],"series":[],"class_list":["post-14421","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","tag-formula-resolvente"],"views":1996,"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\/14421","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=14421"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/14421\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14061"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=14421"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=14421"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=14421"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=14421"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}