{"id":14469,"date":"2018-04-14T00:27:37","date_gmt":"2018-04-13T23:27:37","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=14469"},"modified":"2022-01-07T21:45:32","modified_gmt":"2022-01-07T21:45:32","slug":"resolve-as-seguintes-equacoes-em-x","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=14469","title":{"rendered":"Resolve as seguintes equa\u00e7\u00f5es em <em>x<\/em>"},"content":{"rendered":"<p><ul id='GTTabs_ul_14469' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_14469' class='GTTabs_curr'><a  id=\"14469_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_14469' ><a  id=\"14469_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_14469'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Resolve as seguintes equa\u00e7\u00f5es em <em>x<\/em>.<\/p>\n<ol>\n<li>\\({3{x^2} + 5x = 0}\\)<\/li>\n<li>\\({\\sqrt 2 {x^2} + 11x = 0}\\)<\/li>\n<li>\\({{x^2} + 9 = 0}\\)<\/li>\n<li>\\({\\left( {x &#8211; 4} \\right)\\left( {x + 1} \\right) = 10 &#8211; 3x}\\)<\/li>\n<li>\\({{x^2} &#8211; 10x + 24 = 0}\\)<\/li>\n<li>\\({{x^2} &#8211; 4x = &#8211; 3}\\)<\/li>\n<li>\\({\\left( {x + 4} \\right)\\left( {x &#8211; 1} \\right) = 5x &#8211; 20}\\)<\/li>\n<li>\\({5x + {{\\left( {x + 2} \\right)}^2} = 3x\\left( {x + 2} \\right) + x}\\)<\/li>\n<li>\\({\\left( {x + 2} \\right)\\left( {x &#8211; 2} \\right) &#8211; {{\\left( {x &#8211; 1} \\right)}^2} = {x^2} &#8211; 8}\\)<\/li>\n<li>\\({\\frac{{{x^2} &#8211; 1}}{4} = \\frac{{x &#8211; 1}}{3}}\\)<\/li>\n<li>\\(\\frac{{{x^2}}}{2} &#8211; 1 = \\frac{x}{3} + 15\\)<\/li>\n<li>\\({4,8{x^2} &#8211; 8,4x + 2,4 = 0}\\)<\/li>\n<li>\\({\\frac{{x &#8211; 1}}{2} &#8211; \\frac{{x\\left( {3 &#8211; x} \\right)}}{3} = x + \\frac{1}{3}}\\)<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_14469' onClick='GTTabs_show(1,14469)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_14469'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{3{x^2} + 5x = 0}&amp; \\Leftrightarrow &amp;{x\\left( {3x + 5} \\right) = 0}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{3x + 5 = 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{x = &#8211; \\frac{5}{3}}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ { &#8211; \\frac{5}{3},\\;0} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{\\sqrt 2 {x^2} + 11x = 0}&amp; \\Leftrightarrow &amp;{x\\left( {\\sqrt 2 x + 11} \\right) = 0}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{\\sqrt 2 x + 11 = 0}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{x = &#8211; \\frac{{11}}{{\\sqrt 2 }}}\\end{array}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 0}&amp; \\vee &amp;{x = &#8211; \\frac{{11\\sqrt 2 }}{2}}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ { &#8211; \\frac{{11\\sqrt 2 }}{2},\\;0} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{{x^2} + 9 = 0}&amp; \\Leftrightarrow &amp;{{x^2} = &#8211; 9}\\\\{}&amp; \\Leftrightarrow &amp;{x \\in \\emptyset }\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ {} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{\\left( {x &#8211; 4} \\right)\\left( {x + 1} \\right) = 10 &#8211; 3x}&amp; \\Leftrightarrow &amp;{{x^2} + x &#8211; 4x &#8211; 4 &#8211; 10 + 3x = 0}\\\\{}&amp; \\Leftrightarrow &amp;{{x^2} &#8211; 14 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{{x^2} = 14}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\mp \\sqrt {14} }\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = &#8211; \\sqrt {14} }&amp; \\vee &amp;{x = \\sqrt {14} }\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ { &#8211; \\sqrt {14} ,\\;\\sqrt {14} } \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{{x^2} &#8211; 10x + 24 = 0}&amp; \\Leftrightarrow &amp;{x = \\frac{{10 \\mp \\sqrt {100 &#8211; 96} }}{2}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{10 \\mp 2}}{2}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 4}&amp; \\vee &amp;{x = 6}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ {4,\\;6} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{{x^2} &#8211; 4x = &#8211; 3}&amp; \\Leftrightarrow &amp;{{x^2} &#8211; 4x + 3 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{4 \\mp \\sqrt {16 &#8211; 12} }}{2}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{4 \\mp 2}}{2}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = 1}&amp; \\vee &amp;{x = 3}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ {1,\\;3} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{\\left( {x + 4} \\right)\\left( {x &#8211; 1} \\right) = 5x &#8211; 20}&amp; \\Leftrightarrow &amp;{{x^2} &#8211; x + 4x &#8211; 4 &#8211; 5x + 20 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{\\underbrace {{x^2} &#8211; 2x + 16 = 0}_{\\Delta = 4 &#8211; 64 &lt; 0}}\\\\{}&amp; \\Leftrightarrow &amp;{x \\in \\emptyset }\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ {} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{5x + {{\\left( {x + 2} \\right)}^2} = 3x\\left( {x + 2} \\right) + x}&amp; \\Leftrightarrow &amp;{5x + {x^2} + 4x + 4 &#8211; 3{x^2} &#8211; 6x &#8211; x = 0}\\\\{}&amp; \\Leftrightarrow &amp;{ &#8211; 2{x^2} + 2x + 4 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{{x^2} &#8211; x &#8211; 2 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{1 \\mp \\sqrt {1 + 8} }}{2}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{1 \\mp 3}}{2}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = &#8211; 1}&amp; \\vee &amp;{x = 2}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ { &#8211; 1,\\;2} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{\\left( {x + 2} \\right)\\left( {x &#8211; 2} \\right) &#8211; {{\\left( {x &#8211; 1} \\right)}^2} = {x^2} &#8211; 8}&amp; \\Leftrightarrow &amp;{{x^2} &#8211; 4 &#8211; {x^2} + 2x &#8211; 1 &#8211; {x^2} + 8 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{ &#8211; {x^2} + 2x + 3 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 2 \\mp \\sqrt {4 + 12} }}{{ &#8211; 2}}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 2 \\mp 4}}{{ &#8211; 2}}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = &#8211; 1}&amp; \\vee &amp;{x = 3}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ { &#8211; 1,\\;3} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{\\frac{{{x^2} &#8211; 1}}{4} = \\frac{{x &#8211; 1}}{3}}&amp; \\Leftrightarrow &amp;{3{x^2} &#8211; 3 = 4x &#8211; 4}\\\\{}&amp; \\Leftrightarrow &amp;{3{x^2} &#8211; 4x + 1 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{4 \\mp \\sqrt {16 &#8211; 12} }}{6}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{4 \\mp 2}}{6}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{1}{3}}&amp; \\vee &amp;{x = 1}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ {\\frac{1}{3},\\;1} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{\\frac{{{x^2}}}{{\\mathop 2\\limits_{\\left( 3 \\right)} }} &#8211; \\mathop 1\\limits_{\\left( 6 \\right)} = \\frac{x}{{\\mathop 3\\limits_{\\left( 2 \\right)} }} + \\mathop {15}\\limits_{\\left( 6 \\right)} }&amp; \\Leftrightarrow &amp;{3{x^2} &#8211; 6 = 2x + 90}\\\\{}&amp; \\Leftrightarrow &amp;{3{x^2} &#8211; 2x &#8211; 96 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{2 \\mp \\sqrt {4 + 1152} }}{6}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{2 \\mp 34}}{6}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = &#8211; \\frac{{16}}{3}}&amp; \\vee &amp;{x = 6}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ { &#8211; \\frac{{16}}{3},\\;6} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{4,8{x^2} &#8211; 8,4x + 2,4 = 0}&amp; \\Leftrightarrow &amp;{4{x^2} &#8211; 7x + 2 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{7 \\mp \\sqrt {49 &#8211; 32} }}{8}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{7 \\mp \\sqrt {17} }}{8}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = \\frac{{7 &#8211; \\sqrt {17} }}{8}}&amp; \\vee &amp;{x = \\frac{{7 + \\sqrt {17} }}{8}}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ {\\frac{{7 &#8211; \\sqrt {17} }}{8},\\;\\frac{{7 + \\sqrt {17} }}{8}} \\right\\}}\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{20}{l}}{\\frac{{x &#8211; 1}}{2} &#8211; \\frac{{x\\left( {3 &#8211; x} \\right)}}{3} = x + \\frac{1}{3}}&amp; \\Leftrightarrow &amp;{3x &#8211; 3 &#8211; 6x + 2{x^2} &#8211; 6x &#8211; 2 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{2{x^2} &#8211; 9x &#8211; 5 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{9 \\mp \\sqrt {81 + 40} }}{4}}\\\\{}&amp; \\Leftrightarrow &amp;{x = \\frac{{9 \\mp 11}}{4}}\\\\{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}{x = &#8211; \\frac{1}{2}}&amp; \\vee &amp;{x = 5}\\end{array}}\\\\{}&amp;{}&amp;{}\\\\{}&amp;{}&amp;{S = \\left\\{ { &#8211; \\frac{1}{2},\\;5} \\right\\}}\\end{array}\\]<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_14469' onClick='GTTabs_show(0,14469)'>&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 em x. \\({3{x^2} + 5x = 0}\\) \\({\\sqrt 2 {x^2} + 11x = 0}\\) \\({{x^2} + 9 = 0}\\) \\({\\left( {x &#8211; 4} \\right)\\left( {x + 1}&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":14056,"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-14469","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":1645,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/03\/Mat01.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/14469","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=14469"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/14469\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14056"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=14469"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=14469"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=14469"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=14469"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}