{"id":4781,"date":"2017-06-07T18:57:03","date_gmt":"2017-06-07T18:57:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-quadratic-equations-2\/"},"modified":"2017-08-16T02:34:52","modified_gmt":"2017-08-16T02:34:52","slug":"read-quadratic-equations-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-quadratic-equations-2\/","title":{"raw":"Solving Quadratic Equations","rendered":"Solving Quadratic Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Recognize a quadratic equation<\/li>\r\n \t<li>Use the zero product principle to solve a quadratic equation that can be factored<\/li>\r\n \t<li>Determine when solutions to quadratic equations can be discarded<\/li>\r\n<\/ul>\r\n<\/div>\r\nWhen a polynomial is set equal to a value (whether an integer or another polynomial), the result is an equation. An equation that can be written in the form [latex]ax^{2}+bx+c=0[\/latex]\u00a0is called a <strong>quadratic equation<\/strong>. You can solve a quadratic equation using the rules of algebra, applying factoring techniques where necessary, and by using the <strong>Principle of Zero Products<\/strong>.\r\n\r\nThere are many applications for quadratic equations. When you use the Principle of Zero Products to solve a quadratic equation, you need to make sure that the equation is equal to zero. For example, [latex]12x^{2}+11x+2=7[\/latex] must first be changed to [latex]12x^{2}+11x+-5=0[\/latex]\u00a0by subtracting 7 from both sides.\r\n\r\nThe example below shows a quadratic equation where neither side is originally equal to zero. (Note that the factoring sequence has been shortened.)\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve [latex]5b^{2}+4=\u221212b[\/latex]\u00a0for <i>b.<\/i>\r\n\r\n[reveal-answer q=\"950625\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"950625\"]The original equation has [latex]\u221212b[\/latex] on the right. To make this side equal to 0, add [latex]12b[\/latex] to both sides.\r\n<p style=\"text-align: center\">[latex]5b^{2}+4+12b=\u221212b+12b[\/latex]<\/p>\r\nCombine like terms.\r\n<p style=\"text-align: center\">[latex]5b^{2}+12b+4=0[\/latex]<\/p>\r\nRewrite [latex]12b[\/latex]\u00a0as\u00a0[latex]10b+2b[\/latex].\r\n<p style=\"text-align: center\">[latex]5b^{2}+10b+2b+4=0[\/latex]<\/p>\r\nFactor out [latex]5b[\/latex] from the first pair and 2 from the second pair.\r\n<p style=\"text-align: center\">[latex]5b\\left(b+2\\right)+2\\left(b+2\\right)=0[\/latex]<\/p>\r\nFactor out [latex]b+2[\/latex].\r\n<p style=\"text-align: center\">[latex]\\left(5b+2\\right)\\left(b+2\\right)=0[\/latex]<\/p>\r\nApply the Zero Product Property.\r\n<p style=\"text-align: center\">[latex]5b+2=0\\,\\,\\,\\text{or}\\,\\,\\,b+2=0[\/latex]<\/p>\r\nSolve each equation.\r\n<p style=\"text-align: center\">[latex]b=-\\frac{2}{5}\\,\\,\\,\\text{OR}\\,\\,\\,b=\u22122[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]b=-\\frac{2}{5}\\,\\,\\,\\text{or}\\,\\,\\,b=\u22122[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video contains another example of solving a quadratic equation using factoring with grouping.\r\n\r\nhttps:\/\/youtu.be\/04zEXaOiO4U\r\n\r\nIf you factor out a constant, the constant will never equal 0. So it can essentially be ignored when solving. See the following example.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve for k: [latex]-2k^2+90=-8k[\/latex]\r\n[reveal-answer q=\"831890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"831890\"]\r\n\r\nWe need to move all the terms to one side so we can use the zero product principle.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}-2k^2+90=-8k\\\\\\underline{+8k}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{+8k}\\\\-2k^2+8k+90=0\\end{array}[\/latex]<\/p>\r\nYou will either need to try to factor out a -2, or use the method where we multiply [latex]-2\\cdot{90}[\/latex] and find factors that sum to 8. Each term is divisible by 2, so we can factor out -2.\r\n<p style=\"text-align: center\">[latex]-2\\left(k^2-4k-45\\right)=0[\/latex]<\/p>\r\nNote how we changed the signs when we factored out a negative number. If we can factor the polynomial, we will be able to solve.\r\n\r\nUsing the shortcut for factoring we will start with the variable and place a plus and a minus sign in the binomials. We do this because 45 is negative and the only way to get a product that is negative is if one of the factors is negative.\r\n<p style=\"text-align: center\">[latex]-2\\left(k-\\,\\,\\,\\right)\\left(k+\\,\\,\\,\\right)=0[\/latex]<\/p>\r\n<p style=\"text-align: left\">We want our factors to have a product of -45 and a sum of -4:<\/p>\r\n\r\n<table style=\"width: 20%\">\r\n<thead>\r\n<tr>\r\n<th>Factors whose product is -45<\/th>\r\n<th>Sum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1\\cdot-45=-45[\/latex]<\/td>\r\n<td>[latex]1-45=-44[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3\\cdot-15=-45[\/latex]<\/td>\r\n<td>[latex]3-15=-12[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]5\\cdot-9=-45[\/latex]<\/td>\r\n<td>[latex]5-9=-4[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThere are more factors that will give -45, but we have found the ones that sum to -4, so we will stop. Fill in the rest of the binomials with the factors we found.\r\n<p style=\"text-align: center\">[latex]-2\\left(k-9\\right)\\left(k+5\\right)=0[\/latex]<\/p>\r\n<p style=\"text-align: left\">Now we can set each factor equal to zero using the zero product rule.<\/p>\r\n<p style=\"text-align: left\">[latex]-2=0[\/latex] This solution is nonsense so we discard it.<\/p>\r\n<p style=\"text-align: left\">[latex]k-9=0, k=9[\/latex]<\/p>\r\n<p style=\"text-align: left\">[latex]k+5=0, k=-5[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left\">Answers<\/h4>\r\n[latex]k=9\\text{ and }k=-5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn this video example, we solve a quadratic equation with a leading coefficient of -1 using the shortcut method of factoring and the zero product principle.\r\n\r\nhttps:\/\/youtu.be\/nZYfgHygXis\r\n<h2><span style=\"color: #4a8bb0\">Area<\/span><\/h2>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nThe area of a rectangular garden is 30 square feet. If the length is 7 feet longer than the width, find the dimensions.\r\n[reveal-answer q=\"948371\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"948371\"]The formula for the area of a rectangle is [latex]\\text{Area}=\\text{length}\\cdot\\text{width}[\/latex], or [latex]A=l\\cdot{w}[\/latex].\r\n\r\n[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,A=l\\cdot{w}\\\\\\,\\text{width}=w\\\\\\text{length}=w+7\\\\\\,\\,\\,\\,\\,\\text{area}=30\\\\\\\\30=\\left(w+7\\right)\\left(w\\right)\\end{array}[\/latex]\r\n\r\nMultiply.\r\n\r\n[latex]30=w^{2}+7w[\/latex]\r\n\r\nSubtract 30 from both sides to set the equation equal to 0.\r\n\r\n[latex]w^{2}+7w\u201330=0[\/latex]\r\n\r\nFind two numbers whose product is [latex]\u221230[\/latex] and whose sum is 7, and write the middle term as\u00a0[latex]10w\u20133w[\/latex].\r\n\r\n[latex]w^{2}+10w\u20133w\u201330=0[\/latex]\r\n\r\nFactor <i>w<\/i> out of the first pair and [latex]\u22123[\/latex] out of the second pair.\r\n\r\n[latex]w\\left(w+10\\right)-3\\left(w+10\\right)=0[\/latex]\r\n\r\nFactor out [latex]w+10[\/latex].\r\n\r\n[latex]\\left(w\u20133\\right)\\left(w+10\\right)=0[\/latex]\r\n\r\nUse the Zero Product Property to solve for <i>w<\/i>.\r\n\r\n[latex]\\begin{array}{l}w-3=0\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,w+10=0\\\\\\,\\,\\,\\,\\,\\,\\,w=3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=-10\\end{array}[\/latex]\r\n\r\nThe solution [latex]w=\u221210[\/latex]\u00a0does not work for this application, as the width cannot be a negative number, we discard the [latex]\u221210[\/latex]. So, the width is 3 feet.\r\n\r\nThe width = 3 feet\r\n\r\nSubstitute [latex]w=3[\/latex] into the expression [latex]w+7[\/latex] to find the length: [latex]3+7=10[\/latex].\r\n\r\nThe length is [latex]3+7=10[\/latex] feet\r\n<h4>Answer<\/h4>\r\nThe width of the garden is 3 feet, and the length is 10 feet.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the example in the following video, we present another area application of factoring trinomials.\r\n\r\nhttps:\/\/youtu.be\/PvXsWZp588o\r\n<h2><span style=\"color: #4a8bb0\">Summary<\/span><\/h2>\r\nYou can find the solutions, or roots, of quadratic equations by setting one side equal to zero, factoring the polynomial, and then applying the Zero Product Property. The Principle of Zero Products states that if [latex]ab=0[\/latex], then either [latex]a=0[\/latex] or [latex]b=0[\/latex], or both <i>a<\/i> and <i>b<\/i> are 0. Once the polynomial is factored, set each factor equal to zero and solve them separately. The answers will be the set of solutions for the original equation.\r\n\r\nNot all solutions are appropriate for some applications. In many real-world situations, negative solutions are not appropriate and must be discarded.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Recognize a quadratic equation<\/li>\n<li>Use the zero product principle to solve a quadratic equation that can be factored<\/li>\n<li>Determine when solutions to quadratic equations can be discarded<\/li>\n<\/ul>\n<\/div>\n<p>When a polynomial is set equal to a value (whether an integer or another polynomial), the result is an equation. An equation that can be written in the form [latex]ax^{2}+bx+c=0[\/latex]\u00a0is called a <strong>quadratic equation<\/strong>. You can solve a quadratic equation using the rules of algebra, applying factoring techniques where necessary, and by using the <strong>Principle of Zero Products<\/strong>.<\/p>\n<p>There are many applications for quadratic equations. When you use the Principle of Zero Products to solve a quadratic equation, you need to make sure that the equation is equal to zero. For example, [latex]12x^{2}+11x+2=7[\/latex] must first be changed to [latex]12x^{2}+11x+-5=0[\/latex]\u00a0by subtracting 7 from both sides.<\/p>\n<p>The example below shows a quadratic equation where neither side is originally equal to zero. (Note that the factoring sequence has been shortened.)<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve [latex]5b^{2}+4=\u221212b[\/latex]\u00a0for <i>b.<\/i><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q950625\">Show Solution<\/span><\/p>\n<div id=\"q950625\" class=\"hidden-answer\" style=\"display: none\">The original equation has [latex]\u221212b[\/latex] on the right. To make this side equal to 0, add [latex]12b[\/latex] to both sides.<\/p>\n<p style=\"text-align: center\">[latex]5b^{2}+4+12b=\u221212b+12b[\/latex]<\/p>\n<p>Combine like terms.<\/p>\n<p style=\"text-align: center\">[latex]5b^{2}+12b+4=0[\/latex]<\/p>\n<p>Rewrite [latex]12b[\/latex]\u00a0as\u00a0[latex]10b+2b[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]5b^{2}+10b+2b+4=0[\/latex]<\/p>\n<p>Factor out [latex]5b[\/latex] from the first pair and 2 from the second pair.<\/p>\n<p style=\"text-align: center\">[latex]5b\\left(b+2\\right)+2\\left(b+2\\right)=0[\/latex]<\/p>\n<p>Factor out [latex]b+2[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\left(5b+2\\right)\\left(b+2\\right)=0[\/latex]<\/p>\n<p>Apply the Zero Product Property.<\/p>\n<p style=\"text-align: center\">[latex]5b+2=0\\,\\,\\,\\text{or}\\,\\,\\,b+2=0[\/latex]<\/p>\n<p>Solve each equation.<\/p>\n<p style=\"text-align: center\">[latex]b=-\\frac{2}{5}\\,\\,\\,\\text{OR}\\,\\,\\,b=\u22122[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]b=-\\frac{2}{5}\\,\\,\\,\\text{or}\\,\\,\\,b=\u22122[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video contains another example of solving a quadratic equation using factoring with grouping.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex: Solve a Quadratic Equation Using Factor By Grouping\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/04zEXaOiO4U?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>If you factor out a constant, the constant will never equal 0. So it can essentially be ignored when solving. See the following example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve for k: [latex]-2k^2+90=-8k[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q831890\">Show Solution<\/span><\/p>\n<div id=\"q831890\" class=\"hidden-answer\" style=\"display: none\">\n<p>We need to move all the terms to one side so we can use the zero product principle.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}-2k^2+90=-8k\\\\\\underline{+8k}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{+8k}\\\\-2k^2+8k+90=0\\end{array}[\/latex]<\/p>\n<p>You will either need to try to factor out a -2, or use the method where we multiply [latex]-2\\cdot{90}[\/latex] and find factors that sum to 8. Each term is divisible by 2, so we can factor out -2.<\/p>\n<p style=\"text-align: center\">[latex]-2\\left(k^2-4k-45\\right)=0[\/latex]<\/p>\n<p>Note how we changed the signs when we factored out a negative number. If we can factor the polynomial, we will be able to solve.<\/p>\n<p>Using the shortcut for factoring we will start with the variable and place a plus and a minus sign in the binomials. We do this because 45 is negative and the only way to get a product that is negative is if one of the factors is negative.<\/p>\n<p style=\"text-align: center\">[latex]-2\\left(k-\\,\\,\\,\\right)\\left(k+\\,\\,\\,\\right)=0[\/latex]<\/p>\n<p style=\"text-align: left\">We want our factors to have a product of -45 and a sum of -4:<\/p>\n<table style=\"width: 20%\">\n<thead>\n<tr>\n<th>Factors whose product is -45<\/th>\n<th>Sum of the factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1\\cdot-45=-45[\/latex]<\/td>\n<td>[latex]1-45=-44[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3\\cdot-15=-45[\/latex]<\/td>\n<td>[latex]3-15=-12[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]5\\cdot-9=-45[\/latex]<\/td>\n<td>[latex]5-9=-4[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>There are more factors that will give -45, but we have found the ones that sum to -4, so we will stop. Fill in the rest of the binomials with the factors we found.<\/p>\n<p style=\"text-align: center\">[latex]-2\\left(k-9\\right)\\left(k+5\\right)=0[\/latex]<\/p>\n<p style=\"text-align: left\">Now we can set each factor equal to zero using the zero product rule.<\/p>\n<p style=\"text-align: left\">[latex]-2=0[\/latex] This solution is nonsense so we discard it.<\/p>\n<p style=\"text-align: left\">[latex]k-9=0, k=9[\/latex]<\/p>\n<p style=\"text-align: left\">[latex]k+5=0, k=-5[\/latex]<\/p>\n<h4 style=\"text-align: left\">Answers<\/h4>\n<p>[latex]k=9\\text{ and }k=-5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In this video example, we solve a quadratic equation with a leading coefficient of -1 using the shortcut method of factoring and the zero product principle.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex: Factor and Solve Quadratic Equation - Trinomial a = -1\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/nZYfgHygXis?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2><span style=\"color: #4a8bb0\">Area<\/span><\/h2>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>The area of a rectangular garden is 30 square feet. If the length is 7 feet longer than the width, find the dimensions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q948371\">Show Solution<\/span><\/p>\n<div id=\"q948371\" class=\"hidden-answer\" style=\"display: none\">The formula for the area of a rectangle is [latex]\\text{Area}=\\text{length}\\cdot\\text{width}[\/latex], or [latex]A=l\\cdot{w}[\/latex].<\/p>\n<p>[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,A=l\\cdot{w}\\\\\\,\\text{width}=w\\\\\\text{length}=w+7\\\\\\,\\,\\,\\,\\,\\text{area}=30\\\\\\\\30=\\left(w+7\\right)\\left(w\\right)\\end{array}[\/latex]<\/p>\n<p>Multiply.<\/p>\n<p>[latex]30=w^{2}+7w[\/latex]<\/p>\n<p>Subtract 30 from both sides to set the equation equal to 0.<\/p>\n<p>[latex]w^{2}+7w\u201330=0[\/latex]<\/p>\n<p>Find two numbers whose product is [latex]\u221230[\/latex] and whose sum is 7, and write the middle term as\u00a0[latex]10w\u20133w[\/latex].<\/p>\n<p>[latex]w^{2}+10w\u20133w\u201330=0[\/latex]<\/p>\n<p>Factor <i>w<\/i> out of the first pair and [latex]\u22123[\/latex] out of the second pair.<\/p>\n<p>[latex]w\\left(w+10\\right)-3\\left(w+10\\right)=0[\/latex]<\/p>\n<p>Factor out [latex]w+10[\/latex].<\/p>\n<p>[latex]\\left(w\u20133\\right)\\left(w+10\\right)=0[\/latex]<\/p>\n<p>Use the Zero Product Property to solve for <i>w<\/i>.<\/p>\n<p>[latex]\\begin{array}{l}w-3=0\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,w+10=0\\\\\\,\\,\\,\\,\\,\\,\\,w=3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=-10\\end{array}[\/latex]<\/p>\n<p>The solution [latex]w=\u221210[\/latex]\u00a0does not work for this application, as the width cannot be a negative number, we discard the [latex]\u221210[\/latex]. So, the width is 3 feet.<\/p>\n<p>The width = 3 feet<\/p>\n<p>Substitute [latex]w=3[\/latex] into the expression [latex]w+7[\/latex] to find the length: [latex]3+7=10[\/latex].<\/p>\n<p>The length is [latex]3+7=10[\/latex] feet<\/p>\n<h4>Answer<\/h4>\n<p>The width of the garden is 3 feet, and the length is 10 feet.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the example in the following video, we present another area application of factoring trinomials.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 2: Quadratic Equation App - Find the Dimensions of a Rectangle Given Area (Factoring)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/PvXsWZp588o?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2><span style=\"color: #4a8bb0\">Summary<\/span><\/h2>\n<p>You can find the solutions, or roots, of quadratic equations by setting one side equal to zero, factoring the polynomial, and then applying the Zero Product Property. The Principle of Zero Products states that if [latex]ab=0[\/latex], then either [latex]a=0[\/latex] or [latex]b=0[\/latex], or both <i>a<\/i> and <i>b<\/i> are 0. Once the polynomial is factored, set each factor equal to zero and solve them separately. The answers will be the set of solutions for the original equation.<\/p>\n<p>Not all solutions are appropriate for some applications. In many real-world situations, negative solutions are not appropriate and must be discarded.<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-4781\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex 2: Quadratic Equation App - Find the Dimensions of a Rectangle Given Area (Factoring). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/PvXsWZp588o\">https:\/\/youtu.be\/PvXsWZp588o<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Solve a Quadratic Equation Using Factor By Grouping. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/04zEXaOiO4U\">https:\/\/youtu.be\/04zEXaOiO4U<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Factor and Solve Quadratic Equation - Trinomial a = -1. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/nZYfgHygXis\">https:\/\/youtu.be\/nZYfgHygXis<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Unit 12: Factoring, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":25,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex 2: Quadratic Equation App - 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