{"id":1992,"date":"2021-08-19T16:07:01","date_gmt":"2021-08-19T16:07:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-separable-equations\/"},"modified":"2022-04-19T20:52:41","modified_gmt":"2022-04-19T20:52:41","slug":"skills-review-for-separable-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-separable-equations\/","title":{"raw":"Skills Review for Separable Equations","rendered":"Skills Review for Separable Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve linear equations<\/li>\r\n \t<li>Use the zero product principle to solve quadratic equations that can be factored<\/li>\r\n \t<li>Use the square root property to solve quadratic equations<\/li>\r\n \t<li>Write function equations using given conditions<\/li>\r\n<\/ul>\r\n<\/div>\r\nThe Separable Equations section will require that you set factors of various degrees equal to zero and solve the resulting equation. We will also be asked to find a function equation using given conditions. Here we review all of these skills.\r\n<h2>Solve Linear Equations<\/h2>\r\n<p id=\"fs-id1169737931596\">A linear equation is an equation whose highest power on the variable is one. If the equation is in the form [latex]ax+b=c[\/latex], where <i>x<\/i> is the variable, to solve the equation as before, first \u201cundo\u201d the addition and subtraction and then \u201cundo\u201d the multiplication and division.<\/p>\r\n\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example: Solving a Linear Equation<\/h3>\r\nSolve [latex]3y+2=11[\/latex].\r\n\r\n[reveal-answer q=\"843520\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"843520\"]\r\n\r\nSubtract 2 from both sides of the equation to get the term with the variable by itself.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}3y+2\\,\\,\\,=\\,\\,11\\\\\\underline{\\,\\,\\,\\,\\,\\,\\,-2\\,\\,\\,\\,\\,\\,\\,\\,-2}\\\\3y\\,\\,\\,\\,=\\,\\,\\,\\,\\,9\\end{array}[\/latex]<\/p>\r\nDivide both sides of the equation by [latex]3[\/latex] to get a coefficient of [latex]1[\/latex] for the variable.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,\\,\\,\\,\\,\\underline{3y}\\,\\,\\,\\,=\\,\\,\\,\\,\\,\\underline{9}\\\\3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y\\,\\,\\,\\,=\\,\\,\\,\\,3\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example: Solving a Linear Equation<\/h3>\r\nSolve [latex]3x+5x+4-x+7=88[\/latex].\r\n\r\n[reveal-answer q=\"455516\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"455516\"]\r\n\r\nThere are three like terms involving a variable: [latex]3x[\/latex], [latex]5x[\/latex], and [latex]\u2013x[\/latex]. Combine these like terms. [latex]4[\/latex] and [latex]7[\/latex] are also like terms and can be added.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,3x+5x+4-x+7=\\,\\,\\,88\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7x+11=\\,\\,\\,88\\end{array}[\/latex]<\/p>\r\nThe equation is now in the form [latex]ax+b=c[\/latex], so we can solve as before.\r\n\r\nSubtract 11 from both sides.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}7x+11\\,\\,\\,=\\,\\,\\,88\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-11\\,\\,\\,\\,\\,\\,\\,-11}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7x\\,\\,\\,=\\,\\,\\,77\\end{array}[\/latex]<\/p>\r\nDivide both sides by 7.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{7x}\\,\\,\\,=\\,\\,\\,\\underline{77}\\\\7\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7\\,\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,=\\,\\,\\,11\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nSome equations may have the variable on both sides of the equal sign, as in this equation: [latex]4x-6=2x+10[\/latex].\r\n\r\nTo solve this equation, we need to \u201cmove\u201d one of the variable terms. This can make it difficult to decide which side to work with. It does not matter which term gets moved, [latex]4x[\/latex] or [latex]2x[\/latex]; however, to avoid negative coefficients, you can move the smaller term.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Linear Equation<\/h3>\r\nSolve: [latex]4x-6=2x+10[\/latex]\r\n\r\n[reveal-answer q=\"457216\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"457216\"]\r\n\r\nChoose the variable term to move\u2014to avoid negative terms choose [latex]2x[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\,\\,\\,4x-6=2x+10\\\\\\underline{-2x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2x}\\\\\\,\\,\\,4x-6=10[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now add 6 to both sides to isolate the term with the variable.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4x-6=10\\\\\\underline{\\,\\,\\,\\,+6\\,\\,\\,+6}\\\\2x=16\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now divide each side by 2 to isolate the variable x.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\dfrac{2x}{2}\\normalsize=\\dfrac{16}{2}\\\\\\\\\\normalsize{x=8}\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the linear equation [latex]7x-15=3x+4[\/latex].\r\n\r\n[reveal-answer q=\"7016641\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"7016641\"]\r\n\r\n[latex]x=\\dfrac{19}{4}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solve Quadratic Equations<\/h2>\r\nOften, the easiest method of solving a quadratic equation is by\u00a0<strong>factoring<\/strong>.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a quadratic equation, Solve it by factoring<\/h3>\r\n<ol>\r\n \t<li>Set the quadratic equation equal to 0.<\/li>\r\n \t<li>Factor.<\/li>\r\n \t<li>Set each factor equal to 0 and solve for the variable.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving A Quadratic Equation by Factoring<\/h3>\r\nFactor and solve the equation: [latex]{x}^{2}+x - 6=0[\/latex].\r\n\r\n[reveal-answer q=\"16111\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"16111\"]\r\n\r\nThe equation is already set equal to zero, so we factor. The factored form of the equation is:\r\n<p style=\"text-align: center;\">[latex]\\left(x - 2\\right)\\left(x+3\\right)=0[\/latex]<\/p>\r\nNow, set each factor equal to zero and solve.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x - 2\\right)\\left(x+3\\right)=0\\hfill \\\\ \\left(x - 2\\right)=0\\hfill \\\\ x=2\\hfill \\\\ \\left(x+3\\right)=0\\hfill \\\\ x=-3\\hfill \\end{array}[\/latex]<\/p>\r\nThe two solutions are [latex]x=2[\/latex] and [latex]x=-3[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving A Quadratic Equation by Factoring<\/h3>\r\nFactor and solve the equation: [latex]{8x}^{2}+2x - 3=0[\/latex].\r\n\r\n[reveal-answer q=\"244877\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"244877\"]\r\n\r\nThe equation is already set equal to zero, but [latex]a \\ne{1}[\/latex] so we need to split the middle term to help with grouping:\r\n<p style=\"text-align: center;\">[latex]\\left(8x^{2} +6x\\right)+\\left(-4x-3\\right)=0[\/latex]<\/p>\r\nNext we will pull out the GCF in each term:\r\n<p style=\"text-align: center;\">[latex]2x\\left(4x +3\\right)+(-1)\\left(4x+3\\right)=0[\/latex]<\/p>\r\nThen we will factor the equation:\r\n<p style=\"text-align: center;\">[latex]\\left(4x +3\\right)+\\left(2x-1\\right)=0[\/latex]<\/p>\r\nNow, set each factor equal to zero and solve.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l} \\left(4x +3\\right)=0\\hfill \\\\ 4x=-3\\hfill \\\\ x=-\\frac{3}{4}\\hfill \\\\ \\hfill \\\\ \\left(2x-1\\right)=0\\hfill \\\\ 2x=1\\hfill \\\\ x=\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nThe two solutions are [latex]x=-\\dfrac{3}{4}[\/latex] and [latex]x=\\dfrac{1}{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]2029[\/ohm_question]\r\n\r\n<\/div>\r\nWhen there is no linear term in the equation, another method of solving a quadratic equation is by using the <strong>square root property<\/strong>, in which we isolate the [latex]{x}^{2}[\/latex] term and take the square root of the number on the other side of the equal sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the [latex]{x}^{2}[\/latex] term so that the square root property can be used.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Quadratic Equation Using the Square Root Property<\/h3>\r\nSolve the quadratic using the square root property: [latex]{x}^{2}=8[\/latex].\r\n\r\n[reveal-answer q=\"210107\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"210107\"]\r\n\r\nTake the square root of both sides, and then simplify the radical. Remember to use a [latex]\\pm [\/latex] sign before the radical symbol.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}\\hfill&amp;=8\\hfill \\\\ x\\hfill&amp;=\\pm \\sqrt{8}\\hfill \\\\ \\hfill&amp;=\\pm 2\\sqrt{2}\\hfill \\end{array}[\/latex]<\/div>\r\nThe solutions are [latex]x=2\\sqrt{2}[\/latex], [latex]x=-2\\sqrt{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Quadratic Equation Using the Square Root Property<\/h3>\r\nSolve the quadratic equation: [latex]4{x}^{2}+1=7[\/latex]\r\n\r\n[reveal-answer q=\"885054\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"885054\"]\r\n\r\nFirst, isolate the [latex]{x}^{2}[\/latex] term. Then take the square root of both sides.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}4{x}^{2}+1=7\\hfill \\\\ 4{x}^{2}=6\\hfill \\\\ {x}^{2}=\\frac{6}{4}\\hfill \\\\ x=\\pm \\frac{\\sqrt{6}}{2}\\hfill \\end{array}[\/latex]<\/div>\r\nThe solutions are [latex]x=\\frac{\\sqrt{6}}{2}[\/latex], [latex]x=-\\frac{\\sqrt{6}}{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the quadratic equation using the square root property: [latex]3{\\left(x - 4\\right)}^{2}=15[\/latex].\r\n\r\n[reveal-answer q=\"701664\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"701664\"]\r\n\r\n[latex]x=4\\pm \\sqrt{5}[\/latex]\r\n\r\n[latex]x=4\\sqrt{5}[\/latex],\u00a0[latex]-4\\sqrt{5}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]29172[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Write Function Equations Using Given Conditions<\/h2>\r\n<strong><em>(See <a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-basics-of-differential-equations\/\" target=\"_blank\" rel=\"noopener\">Module 4, Skills Review for Basics of Differential Equations<\/a>)<\/em><\/strong>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve linear equations<\/li>\n<li>Use the zero product principle to solve quadratic equations that can be factored<\/li>\n<li>Use the square root property to solve quadratic equations<\/li>\n<li>Write function equations using given conditions<\/li>\n<\/ul>\n<\/div>\n<p>The Separable Equations section will require that you set factors of various degrees equal to zero and solve the resulting equation. We will also be asked to find a function equation using given conditions. Here we review all of these skills.<\/p>\n<h2>Solve Linear Equations<\/h2>\n<p id=\"fs-id1169737931596\">A linear equation is an equation whose highest power on the variable is one. If the equation is in the form [latex]ax+b=c[\/latex], where <i>x<\/i> is the variable, to solve the equation as before, first \u201cundo\u201d the addition and subtraction and then \u201cundo\u201d the multiplication and division.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example: Solving a Linear Equation<\/h3>\n<p>Solve [latex]3y+2=11[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q843520\">Show Solution<\/span><\/p>\n<div id=\"q843520\" class=\"hidden-answer\" style=\"display: none\">\n<p>Subtract 2 from both sides of the equation to get the term with the variable by itself.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}3y+2\\,\\,\\,=\\,\\,11\\\\\\underline{\\,\\,\\,\\,\\,\\,\\,-2\\,\\,\\,\\,\\,\\,\\,\\,-2}\\\\3y\\,\\,\\,\\,=\\,\\,\\,\\,\\,9\\end{array}[\/latex]<\/p>\n<p>Divide both sides of the equation by [latex]3[\/latex] to get a coefficient of [latex]1[\/latex] for the variable.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,\\,\\,\\,\\,\\underline{3y}\\,\\,\\,\\,=\\,\\,\\,\\,\\,\\underline{9}\\\\3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y\\,\\,\\,\\,=\\,\\,\\,\\,3\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example: Solving a Linear Equation<\/h3>\n<p>Solve [latex]3x+5x+4-x+7=88[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q455516\">Show Solution<\/span><\/p>\n<div id=\"q455516\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are three like terms involving a variable: [latex]3x[\/latex], [latex]5x[\/latex], and [latex]\u2013x[\/latex]. Combine these like terms. [latex]4[\/latex] and [latex]7[\/latex] are also like terms and can be added.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,3x+5x+4-x+7=\\,\\,\\,88\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7x+11=\\,\\,\\,88\\end{array}[\/latex]<\/p>\n<p>The equation is now in the form [latex]ax+b=c[\/latex], so we can solve as before.<\/p>\n<p>Subtract 11 from both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}7x+11\\,\\,\\,=\\,\\,\\,88\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-11\\,\\,\\,\\,\\,\\,\\,-11}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7x\\,\\,\\,=\\,\\,\\,77\\end{array}[\/latex]<\/p>\n<p>Divide both sides by 7.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{7x}\\,\\,\\,=\\,\\,\\,\\underline{77}\\\\7\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7\\,\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,=\\,\\,\\,11\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Some equations may have the variable on both sides of the equal sign, as in this equation: [latex]4x-6=2x+10[\/latex].<\/p>\n<p>To solve this equation, we need to \u201cmove\u201d one of the variable terms. This can make it difficult to decide which side to work with. It does not matter which term gets moved, [latex]4x[\/latex] or [latex]2x[\/latex]; however, to avoid negative coefficients, you can move the smaller term.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Linear Equation<\/h3>\n<p>Solve: [latex]4x-6=2x+10[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q457216\">Show Solution<\/span><\/p>\n<div id=\"q457216\" class=\"hidden-answer\" style=\"display: none\">\n<p>Choose the variable term to move\u2014to avoid negative terms choose [latex]2x[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\,\\,\\,4x-6=2x+10\\\\\\underline{-2x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2x}\\\\\\,\\,\\,4x-6=10[\/latex]<\/p>\n<p style=\"text-align: left;\">Now add 6 to both sides to isolate the term with the variable.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4x-6=10\\\\\\underline{\\,\\,\\,\\,+6\\,\\,\\,+6}\\\\2x=16\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Now divide each side by 2 to isolate the variable x.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\dfrac{2x}{2}\\normalsize=\\dfrac{16}{2}\\\\\\\\\\normalsize{x=8}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the linear equation [latex]7x-15=3x+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q7016641\">Show Solution<\/span><\/p>\n<div id=\"q7016641\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\\dfrac{19}{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Solve Quadratic Equations<\/h2>\n<p>Often, the easiest method of solving a quadratic equation is by\u00a0<strong>factoring<\/strong>.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a quadratic equation, Solve it by factoring<\/h3>\n<ol>\n<li>Set the quadratic equation equal to 0.<\/li>\n<li>Factor.<\/li>\n<li>Set each factor equal to 0 and solve for the variable.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving A Quadratic Equation by Factoring<\/h3>\n<p>Factor and solve the equation: [latex]{x}^{2}+x - 6=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q16111\">Show Solution<\/span><\/p>\n<div id=\"q16111\" class=\"hidden-answer\" style=\"display: none\">\n<p>The equation is already set equal to zero, so we factor. The factored form of the equation is:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x - 2\\right)\\left(x+3\\right)=0[\/latex]<\/p>\n<p>Now, set each factor equal to zero and solve.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x - 2\\right)\\left(x+3\\right)=0\\hfill \\\\ \\left(x - 2\\right)=0\\hfill \\\\ x=2\\hfill \\\\ \\left(x+3\\right)=0\\hfill \\\\ x=-3\\hfill \\end{array}[\/latex]<\/p>\n<p>The two solutions are [latex]x=2[\/latex] and [latex]x=-3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving A Quadratic Equation by Factoring<\/h3>\n<p>Factor and solve the equation: [latex]{8x}^{2}+2x - 3=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q244877\">Show Solution<\/span><\/p>\n<div id=\"q244877\" class=\"hidden-answer\" style=\"display: none\">\n<p>The equation is already set equal to zero, but [latex]a \\ne{1}[\/latex] so we need to split the middle term to help with grouping:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(8x^{2} +6x\\right)+\\left(-4x-3\\right)=0[\/latex]<\/p>\n<p>Next we will pull out the GCF in each term:<\/p>\n<p style=\"text-align: center;\">[latex]2x\\left(4x +3\\right)+(-1)\\left(4x+3\\right)=0[\/latex]<\/p>\n<p>Then we will factor the equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(4x +3\\right)+\\left(2x-1\\right)=0[\/latex]<\/p>\n<p>Now, set each factor equal to zero and solve.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l} \\left(4x +3\\right)=0\\hfill \\\\ 4x=-3\\hfill \\\\ x=-\\frac{3}{4}\\hfill \\\\ \\hfill \\\\ \\left(2x-1\\right)=0\\hfill \\\\ 2x=1\\hfill \\\\ x=\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>The two solutions are [latex]x=-\\dfrac{3}{4}[\/latex] and [latex]x=\\dfrac{1}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm2029\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2029&theme=oea&iframe_resize_id=ohm2029&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>When there is no linear term in the equation, another method of solving a quadratic equation is by using the <strong>square root property<\/strong>, in which we isolate the [latex]{x}^{2}[\/latex] term and take the square root of the number on the other side of the equal sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the [latex]{x}^{2}[\/latex] term so that the square root property can be used.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Quadratic Equation Using the Square Root Property<\/h3>\n<p>Solve the quadratic using the square root property: [latex]{x}^{2}=8[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q210107\">Show Solution<\/span><\/p>\n<div id=\"q210107\" class=\"hidden-answer\" style=\"display: none\">\n<p>Take the square root of both sides, and then simplify the radical. Remember to use a [latex]\\pm[\/latex] sign before the radical symbol.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}\\hfill&=8\\hfill \\\\ x\\hfill&=\\pm \\sqrt{8}\\hfill \\\\ \\hfill&=\\pm 2\\sqrt{2}\\hfill \\end{array}[\/latex]<\/div>\n<p>The solutions are [latex]x=2\\sqrt{2}[\/latex], [latex]x=-2\\sqrt{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Quadratic Equation Using the Square Root Property<\/h3>\n<p>Solve the quadratic equation: [latex]4{x}^{2}+1=7[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q885054\">Show Solution<\/span><\/p>\n<div id=\"q885054\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, isolate the [latex]{x}^{2}[\/latex] term. Then take the square root of both sides.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}4{x}^{2}+1=7\\hfill \\\\ 4{x}^{2}=6\\hfill \\\\ {x}^{2}=\\frac{6}{4}\\hfill \\\\ x=\\pm \\frac{\\sqrt{6}}{2}\\hfill \\end{array}[\/latex]<\/div>\n<p>The solutions are [latex]x=\\frac{\\sqrt{6}}{2}[\/latex], [latex]x=-\\frac{\\sqrt{6}}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the quadratic equation using the square root property: [latex]3{\\left(x - 4\\right)}^{2}=15[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q701664\">Show Solution<\/span><\/p>\n<div id=\"q701664\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=4\\pm \\sqrt{5}[\/latex]<\/p>\n<p>[latex]x=4\\sqrt{5}[\/latex],\u00a0[latex]-4\\sqrt{5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm29172\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29172&theme=oea&iframe_resize_id=ohm29172&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Write Function Equations Using Given Conditions<\/h2>\n<p><strong><em>(See <a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-basics-of-differential-equations\/\" target=\"_blank\" rel=\"noopener\">Module 4, Skills Review for Basics of Differential Equations<\/a>)<\/em><\/strong><\/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-1992\">\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>Modification and Revision. <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>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/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>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/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":349141,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra Corequisite\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"\",\"organization\":\"Lumen 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