{"id":8550,"date":"2012-04-19T17:48:05","date_gmt":"2012-04-19T16:48:05","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=8550"},"modified":"2022-01-06T19:17:44","modified_gmt":"2022-01-06T19:17:44","slug":"resolve-as-seguintes-equacoes-usando-a-formula-resolvente","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=8550","title":{"rendered":"Resolve as seguintes equa\u00e7\u00f5es usando a f\u00f3rmula resolvente"},"content":{"rendered":"<p><ul id='GTTabs_ul_8550' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_8550' class='GTTabs_curr'><a  id=\"8550_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_8550' ><a  id=\"8550_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_8550'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Resolve as seguintes equa\u00e7\u00f5es usando a f\u00f3rmula resolvente:<\/p>\n<ol>\n<li>$2{x^2} + 4x &#8211; 4 = 0$<\/li>\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>$x\\left( {x &#8211; 8} \\right) =\u00a0 &#8211; 42 + 5x$<\/li>\n<li>$4x\\left( {2x &#8211; 5} \\right) = 3x &#8211; 14$<\/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<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_8550' onClick='GTTabs_show(1,8550)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_8550'>\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<ol>\n<li>Resolvendo a equa\u00e7\u00e3o, temos:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{2{x^2} + 4x &#8211; 4 = 0}&amp; \\Leftrightarrow &amp;{{x^2} + 2x &#8211; 2 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 2 \\pm \\sqrt {{2^2} &#8211; 4 \\times 1 \\times ( &#8211; 2)} }}{{2 \\times 1}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 2 \\pm \\sqrt {12} }}{2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x = \\frac{{ &#8211; 2 &#8211; \\sqrt {12} }}{2}}&amp; \\vee &amp;{x = \\frac{{ &#8211; 2 + \\sqrt {12} }}{2}}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, temos:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{6{x^2} + 5x + 1 = 0}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 5 \\pm \\sqrt {{{( &#8211; 5)}^2} &#8211; 4 \\times 6 \\times 1} }}{{2 \\times 6}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 5 \\pm \\sqrt 1 }}{{2 \\times 6}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 5 \\pm 1}}{{12}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x =\u00a0 &#8211; \\frac{1}{2}}&amp; \\vee &amp;{x =\u00a0 &#8211; \\frac{1}{3}}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, temos:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{{x^2} &#8211; 4x + 4 = 0}&amp; \\Leftrightarrow &amp;{x = \\frac{{4 \\pm \\sqrt {{{( &#8211; 4)}^2} &#8211; 4 \\times 1 \\times 4} }}{{2 \\times 1}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{4 \\pm \\sqrt 0 }}{2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{4 \\pm 0}}{2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = 2}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, temos:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{{x^2} &#8211; 3x + 2 = 0}&amp; \\Leftrightarrow &amp;{x = \\frac{{3 \\pm \\sqrt {{{( &#8211; 3)}^2} &#8211; 4 \\times 1 \\times 2} }}{{2 \\times 1}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{3 \\pm \\sqrt 1 }}{2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{3 \\pm 1}}{2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x = 1}&amp; \\vee &amp;{x = 2}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, temos:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{{x^2} &#8211; \\frac{5}{3}x &#8211; \\frac{2}{3} = 0}&amp; \\Leftrightarrow &amp;{3{x^2} &#8211; 5x &#8211; 2 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{5 \\pm \\sqrt {{{( &#8211; 5)}^2} &#8211; 4 \\times 3 \\times ( &#8211; 2)} }}{{2 \\times 3}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{5 \\pm \\sqrt {49} }}{6}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{5 \\pm 7}}{6}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x =\u00a0 &#8211; \\frac{1}{3}}&amp; \\vee &amp;{x = 2}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, temos:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{x\\left( {x &#8211; 8} \\right) =\u00a0 &#8211; 42 + 5x}&amp; \\Leftrightarrow &amp;{{x^2} &#8211; 8x &#8211; 5x + 42 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{{x^2} &#8211; 13x + 42 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{13 \\pm \\sqrt {{{( &#8211; 13)}^2} &#8211; 4 \\times 1 \\times 42} }}{{2 \\times 1}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{13 \\pm \\sqrt 1 }}{2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{13 \\pm 1}}{2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x = 6}&amp; \\vee &amp;{x = 7}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, temos:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{4x\\left( {2x &#8211; 5} \\right) = 3x &#8211; 14}&amp; \\Leftrightarrow &amp;{8{x^2} &#8211; 20x &#8211; 3x + 14 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{8{x^2} &#8211; 23x + 14 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{23 \\pm \\sqrt {{{( &#8211; 23)}^2} &#8211; 4 \\times 8 \\times 14} }}{{2 \\times 8}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{23 \\pm \\sqrt {81} }}{{16}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{23 \\pm 9}}{{16}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x = \\frac{7}{8}}&amp; \\vee &amp;{x = 2}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, temos:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{\\frac{x}{{\\mathop 4\\limits_{(1)} }} &#8211; \\frac{{{{\\left( {x &#8211; 1} \\right)}^2}}}{{\\mathop 2\\limits_{(2)} }} = \\mathop 0\\limits_{(4)} }&amp; \\Leftrightarrow &amp;{x &#8211; 2{{\\left( {x &#8211; 1} \\right)}^2} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x &#8211; 2\\left( {{x^2} &#8211; 2x + 1} \\right) = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{ &#8211; 2{x^2} + 5x &#8211; 2 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 5 \\pm \\sqrt {{5^2} &#8211; 4 \\times ( &#8211; 2) \\times ( &#8211; 2)} }}{{2 \\times ( &#8211; 2)}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 5 \\pm \\sqrt 9 }}{{ &#8211; 4}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{ &#8211; 5 \\pm 3}}{{ &#8211; 4}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x = 2}&amp; \\vee &amp;{x = \\frac{1}{2}}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/li>\n<li>Resolvendo a equa\u00e7\u00e3o, temos:<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{\\mathop 5\\limits_{(3)} \\left( {3 + x} \\right) = \\frac{1}{{\\mathop 3\\limits_{(1)} }}{{\\left( { &#8211; 3 &#8211; x} \\right)}^2}}&amp; \\Leftrightarrow &amp;{15\\left( {3 + x} \\right) = {{\\left( { &#8211; 3 &#8211; x} \\right)}^2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{45 + 15x = 9 + 6x + {x^2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{{x^2} &#8211; 9x &#8211; 36 = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{9 \\pm \\sqrt {{{( &#8211; 9)}^2} &#8211; 4 \\times 1 \\times ( &#8211; 36)} }}{{2 \\times 1}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{9 \\pm \\sqrt {225} }}{2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = \\frac{{9 \\pm 15}}{2}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x =\u00a0 &#8211; 3}&amp; \\vee &amp;{x = 12}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_8550' onClick='GTTabs_show(0,8550)'>&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 usando a f\u00f3rmula resolvente: $2{x^2} + 4x &#8211; 4 = 0$ $6{x^2} + 5x + 1 = 0$ ${x^2} &#8211; 4x + 4 = 0$ ${x^2} &#8211;&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19189,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,303],"tags":[426,304],"series":[],"class_list":["post-8550","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-formula-resolvente"],"views":5489,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat75.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/8550","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=8550"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/8550\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=8550"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=8550"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=8550"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=8550"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}