{"id":811,"date":"2025-10-09T15:40:39","date_gmt":"2025-10-09T15:40:39","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=811"},"modified":"2025-11-20T18:45:19","modified_gmt":"2025-11-20T18:45:19","slug":"second-order-differential-equations-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/second-order-differential-equations-background-youll-need-1\/","title":{"raw":"Second-Order Differential Equations: Background You'll Need 1","rendered":"Second-Order Differential Equations: Background You&#8217;ll Need 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Solve quadratic equations using factoring, the quadratic formula, and completing the square.<\/p>\r\n<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Solving Quadratic Equations by Factoring<\/h2>\r\nOften the easiest method of solving a quadratic equation is <strong>factoring<\/strong>. Factoring a quadratic equation involves expressing it as a product of simpler polynomials, typically two binomials.\r\n\r\nIf a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property which states that if [latex]a\\cdot b=0[\/latex], then [latex]a=0[\/latex] or [latex]b=0[\/latex], where [latex]a[\/latex] and [latex]b[\/latex] are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.\r\n\r\nThe process of factoring a quadratic equation depends on the leading coefficient, whether it is [latex]1 [\/latex] or another integer.\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a quadratic equation with the leading coefficient of 1, factor it<\/strong>\r\n<ol>\r\n \t<li>Find two numbers whose product equals <em>c<\/em> and whose sum equals [latex]b[\/latex].<\/li>\r\n \t<li>Use those numbers to write two factors of the form [latex]\\left(x+k\\right)\\text{ or }\\left(x-k\\right)[\/latex], where [latex]k[\/latex] is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are [latex]1[\/latex] and [latex]-2[\/latex], the factors are [latex]\\left(x+1\\right)\\left(x - 2\\right)[\/latex].<\/li>\r\n \t<li>Solve using the zero-product property by setting each factor equal to zero and solving for the variable.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\">Factor and solve the equation:<center>[latex]{x}^{2}+x - 6=0[\/latex]<\/center>\r\n<ol style=\"list-style-type: decimal;\">\r\n \t<li>Factor the quadratic expression.\r\n<center>[latex]x^2+x - 6 = (x+2)(x-3)[\/latex]<\/center>\r\n[reveal-answer q=\"451031\"]Factoring steps[\/reveal-answer]\r\n[hidden-answer a=\"451031\"]\r\n<ul>\r\n \t<li>We have a trinomial with leading coefficient [latex]1[\/latex], [latex]b = 1[\/latex], and [latex]c = -6[\/latex].<\/li>\r\n \t<li>We need to find two numbers with a product of [latex]-6[\/latex] and a sum of [latex]1[\/latex].<\/li>\r\n<\/ul>\r\n<table style=\"border-collapse: collapse; width: 100%;\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 50%;\"><strong>Factors of [latex]-6[\/latex]<\/strong><\/td>\r\n<td style=\"width: 50%;\"><strong>Sum of Factors<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">[latex]1, -6[\/latex]<\/td>\r\n<td style=\"width: 50%;\">[latex]-5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">[latex]-1, 6[\/latex]<\/td>\r\n<td style=\"width: 50%;\">[latex]5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">[latex]2, -3[\/latex]<\/td>\r\n<td style=\"width: 50%;\">[latex]-1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">[latex]-2, 3[\/latex]<\/td>\r\n<td style=\"width: 50%;\">[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<ul>\r\n \t<li>Now that we have identified [latex]p[\/latex] and [latex]q[\/latex] as [latex]-2 [\/latex] and [latex]3[\/latex], we can write the factored form as [latex](x-(-2))(x-3) = (x+2)(x-3)[\/latex].[\/hidden-answer]<\/li>\r\n<\/ul>\r\n&nbsp;<\/li>\r\n \t<li>Apply the Zero-Product Property to find [latex]x[\/latex].[latex]\\begin{align*} \\text{Equation:} &amp;&amp; x^2 + x - 6 &amp;= 0 \\\\ \\text{Factored form:} &amp;&amp; (x - 2)(x + 3) &amp;= 0 \\\\ \\text{Apply zero product property:} &amp;&amp; x - 2 &amp;= 0 \\quad \\text{or} \\quad x + 3 = 0 \\\\ \\text{Solution:} &amp;&amp; x &amp;= 2 \\quad \\text{or} \\quad x = -3\u00a0 \\end{align*}[\/latex]<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]18961[\/ohm2_question]<\/section>When the leading coefficient is not [latex]1[\/latex], we factor a quadratic equation using the method called grouping, which requires four terms.\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>With the equation in standard form, let\u2019s review the grouping procedures:<\/strong>\r\n<ol>\r\n \t<li>With the quadratic in standard form, [latex]ax^2 + bx + c = 0[\/latex], multiply [latex]a \\cdot c[\/latex].<\/li>\r\n \t<li>Find two numbers whose product equals [latex]a \\cdot c[\/latex] and whose sum equals [latex]b[\/latex].<\/li>\r\n \t<li>Rewrite the equation replacing the [latex]bx[\/latex] term with two terms using the numbers found in step 2 as coefficients of [latex]x[\/latex].<\/li>\r\n \t<li>Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.<\/li>\r\n \t<li>Factor out the expression in parentheses.<\/li>\r\n \t<li>Set the expressions equal to zero and solve for the variable.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\">Use grouping to factor and solve the quadratic equation:<center>[latex]4x^2 + 15x + 9 = 0[\/latex]<\/center>[reveal-answer q=\"304864\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"304864\"]First, multiply [latex]a \\cdot c[\/latex]: [latex]4 \\cdot 9 = 36[\/latex]. Then list the factors of [latex]36[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c c}\r\n1 \\cdot 36 \\\\\r\n2 \\cdot 18 \\\\\r\n3 \\cdot 12 \\\\\r\n4 \\cdot 9 \\\\\r\n6 \\cdot 6 \\\\\r\n\\end{array}[\/latex]<\/p>\r\nThe only pair of factors that sums to [latex]15[\/latex] is [latex]3 + 12[\/latex]. Rewrite the equation replacing the [latex]b[\/latex] term, [latex]15x[\/latex], with two terms using [latex]3[\/latex] and [latex]12[\/latex] as coefficients of [latex]x[\/latex]. Factor the first two terms, and then factor the last two terms.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r c l}\r\n4x^2 + 3x + 12x + 9 &amp; = &amp; 0 \\\\\r\nx(4x + 3) + 3(4x + 3) &amp; = &amp; 0 \\\\\r\n(4x + 3)(x + 3) &amp; = &amp; 0 \\\\\r\n\\end{array}\r\n[\/latex]<\/p>\r\nSolve using the zero-product property.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r c l}\r\n(4x + 3)(x + 3) &amp; = &amp; 0 \\\\\r\n4x + 3 &amp; = &amp; 0 \\quad \\Rightarrow \\quad x = -\\frac{3}{4} \\\\\r\nx + 3 &amp; = &amp; 0 \\quad \\Rightarrow \\quad x = -3 \\\\\r\n\\end{array}\r\n[\/latex]<\/p>\r\nThe solutions are [latex]x = -\\frac{3}{4}[\/latex] and [latex]x = -3[\/latex].\r\n\r\n[caption id=\"attachment_3548\" align=\"aligncenter\" width=\"487\"]<img class=\"wp-image-3548 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/06122929\/c8c54dba0d1717875901655bddf937a8b9e3db6a.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 6 to 2 with every other tick mark labeled and the y-axis ranging from negative 6 to 2 with each tick mark numbered. The equation: four x squared plus fifteen x plus nine is graphed with its x-intercepts: (-3\/4,0) and (-3,0) plotted as well.\" width=\"487\" height=\"433\" \/> Coordinate plane with quadratic equation[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18962[\/ohm2_question]<\/section><section>\r\n<h2>Using the Square Root Property<\/h2>\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\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>square root property<\/h3>\r\nWith the [latex]{x}^{2}[\/latex] term isolated, the square root property states that:\r\n<div style=\"text-align: center;\">[latex]\\text{if }{x}^{2}=k,\\text{then }x=\\pm \\sqrt{k}[\/latex]<\/div>\r\nwhere [latex]k[\/latex] is a non-negative real number.\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a quadratic equation with an [latex]{x}^{2}[\/latex] term but no [latex]x[\/latex] term, use the square root property to solve it<\/strong>\r\n<ol>\r\n \t<li>Isolate the [latex]{x}^{2}[\/latex] term on one side of the equal sign.<\/li>\r\n \t<li>Take the square root of both sides of the equation, putting a [latex]\\pm [\/latex] sign before the expression on the side opposite the squared term.<\/li>\r\n \t<li>Simplify the numbers on the side with the [latex]\\pm [\/latex] sign.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\">Solve the quadratic using the square root property:<center>[latex]{x}^{2}=8[\/latex].<\/center>[latex]\\begin{align*} \\text{Given equation:} &amp; \\quad x^2 = 8 \\\\ \\text{Apply the square root property:} &amp; \\quad x = \\pm \\sqrt{8} \\\\ \\text{Simplify the square root:} &amp; \\quad \\sqrt{8} = \\sqrt{4 \\times 2} = 2\\sqrt{2} \\\\ \\text{Thus, the solution is:} &amp; \\quad x = \\pm 2\\sqrt{2} \\end{align*}[\/latex]<\/section><section class=\"textbox example\">Solve the quadratic equation using the square root property:<center>[latex]3{\\left(x - 4\\right)}^{2}=15[\/latex]<\/center>[reveal-answer q=\"352957\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"352957\"][latex]\\begin{align*} \\text{Isolate the squared term:} &amp; \\quad (x - 4)^2 = \\frac{15}{3} \\\\ &amp; \\quad (x - 4)^2 = 5 \\\\ \\text{Apply the square root property:} &amp; \\quad x - 4 = \\pm \\sqrt{5} \\\\ \\text{Solve for } x: &amp; \\quad x = 4 \\pm \\sqrt{5} \\\\ &amp; \\quad x = 4 + \\sqrt{5} \\quad \\text{or} \\quad x = 4 - \\sqrt{5} \\end{align*}[\/latex][\/hidden-answer]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18963[\/ohm2_question]<\/section><section>\r\n<h2>Completing the Square<\/h2>\r\nNot all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use other methods for solving a quadratic equation. One method of solving quadratic equation is known as <strong>completing the square<\/strong>.\r\n\r\nUsing this process, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, [latex]a[\/latex],\u00a0must equal [latex]1[\/latex]. If it does not, then divide the entire equation by [latex]a[\/latex]. Then, we can use the following procedures to solve a quadratic equation by completing the square.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>completing the square<\/h3>\r\nThe goal of <strong>completing the square<\/strong> is to transform a quadratic equation of the form [latex]ax^2+bx+c = 0[\/latex] into a perfect square trinomial:\r\n<p style=\"text-align: center;\">[latex](x-h)^2 = k[\/latex],<\/p>\r\nwhich can easily be solved by taking square roots.\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to: Solve a quadratic equation by completing the square<\/strong>\r\n<ol>\r\n \t<li>Rearrange your equation so that it is in standard form: [latex]ax^2 + bx + c = 0[\/latex]. Divide by [latex]a[\/latex] if [latex]a \\neq 1[\/latex].<\/li>\r\n \t<li>Isolate the constant term by moving it to the right side of the equation.<\/li>\r\n \t<li>Add [latex](\\frac{b}{2})^2[\/latex] on both sides of the equation.<\/li>\r\n \t<li>Form the perfect square trinomial.<\/li>\r\n \t<li>Solve for [latex]x[\/latex] using the square root property.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Solve the quadratic equation by completing the square:<center>[latex]{x}^{2}-3x - 5=0[\/latex]<\/center>[reveal-answer q=\"349481\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"349481\"]First, move the constant term to the right side of the equal sign by adding [latex]5[\/latex] to both sides of the equation.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}-3x=5[\/latex]<\/div>\r\nThen, take [latex]\\frac{1}{2}[\/latex] of the [latex]b[\/latex] term and square it.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}\\left(-3\\right)=-\\frac{3}{2}\\hfill \\\\ {\\left(-\\frac{3}{2}\\right)}^{2}=\\frac{9}{4}\\hfill \\end{array}[\/latex]<\/div>\r\nAdd the result to both sides of the equal sign.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }{x}^{2}-3x+{\\left(-\\frac{3}{2}\\right)}^{2}=5+{\\left(-\\frac{3}{2}\\right)}^{2}\\hfill \\\\ {x}^{2}-3x+\\frac{9}{4}=5+\\frac{9}{4}\\hfill \\end{array}[\/latex]<\/div>\r\nFactor the left side as a perfect square and simplify the right side.\r\n<div style=\"text-align: center;\">[latex]{\\left(x-\\frac{3}{2}\\right)}^{2}=\\frac{29}{4}[\/latex]<\/div>\r\nUse the square root property and solve.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{{\\left(x-\\frac{3}{2}\\right)}^{2}}=\\pm \\sqrt{\\frac{29}{4}}\\hfill \\\\ \\left(x-\\frac{3}{2}\\right)=\\pm \\frac{\\sqrt{29}}{2}\\hfill \\\\ x=\\frac{3}{2}\\pm \\frac{\\sqrt{29}}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nThe solutions are [latex]x=\\frac{3}{2}+\\frac{\\sqrt{29}}{2}[\/latex], [latex]x=\\frac{3}{2}-\\frac{\\sqrt{29}}{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">Solve the quadratic equation using completing the square:<center>[latex]3x^2-6x-9=0[\/latex]<\/center>[reveal-answer q=\"479550\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"479550\"][latex]\\begin{align*} \\text{Divide by the coefficient of } x^2: &amp; \\quad \\frac{3x^2 - 6x - 9}{3} = 0 \\\\ &amp; \\quad x^2 - 2x - 3 = 0 \\\\ \\text{Isolate the constant term:} &amp; \\quad x^2 - 2x = 3 \\\\ \\text{Add } \\left(\\frac{-2}{2}\\right)^2 = (-1)^2 = 1 \\text{ on both sides}: &amp; \\quad x^2 - 2x + 1 = 3 +1 \\\\ \\text{Form the perfect square trinomial:} &amp; \\quad (x - 1)^2 = 3 +1 \\\\ \\text{Simplify and solve:} &amp; \\quad (x - 1)^2 = 4 \\\\ &amp; \\quad x - 1 = \\pm 2 \\\\ &amp; \\quad x = 1 \\pm 2 \\\\ &amp; \\quad x = 1 + 2 = 3 \\text{ and } x = 1 - 2 = -1 \\end{align*}[\/latex][\/hidden-answer]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18967[\/ohm2_question]<\/section><section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18965[\/ohm2_question]<\/section><section>\r\n<h2>Quadratic Formula<\/h2>\r\nThe fourth method of solving a <strong>quadratic equation<\/strong> is by using the <strong>quadratic formula<\/strong>, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>quadratic formula<\/h3>\r\nWritten in standard form, [latex]a{x}^{2}+bx+c=0[\/latex], any quadratic equation can be solved using the <strong>quadratic formula<\/strong>:\r\n<div style=\"text-align: center;\">[latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/div>\r\nwhere [latex]a, b,[\/latex] and [latex]c[\/latex] are real numbers and [latex]a\\ne 0[\/latex].\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a quadratic equation, solve it using the quadratic formula<\/strong>\r\n<ol>\r\n \t<li>Make sure the equation is in standard form: [latex]a{x}^{2}+bx+c=0[\/latex].<\/li>\r\n \t<li>Make note of the values of the coefficients and constant term, [latex]a,b[\/latex], and [latex]c[\/latex].<\/li>\r\n \t<li>Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.<\/li>\r\n \t<li>Calculate and solve.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\">Solve the quadratic equation:<center>[latex]{x}^{2}+5x+1=0[\/latex]<\/center>[reveal-answer q=\"641400\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"641400\"]Identify the coefficients: [latex]a=1,b=5,c=1[\/latex]. Then use the quadratic formula.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x\\hfill&amp;=\\frac{-\\left(5\\right)\\pm \\sqrt{{\\left(5\\right)}^{2}-4\\left(1\\right)\\left(1\\right)}}{2\\left(1\\right)}\\hfill \\\\ \\hfill&amp;=\\frac{-5\\pm \\sqrt{25 - 4}}{2}\\hfill \\\\ \\hfill&amp;=\\frac{-5\\pm \\sqrt{21}}{2}\\hfill \\end{array}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/section>\r\n<div><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]293772[\/ohm_question]<\/section><\/div>\r\n<\/section><\/section><\/section><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Solve quadratic equations using factoring, the quadratic formula, and completing the square.<\/p>\n<\/li>\n<\/ul>\n<\/section>\n<h2>Solving Quadratic Equations by Factoring<\/h2>\n<p>Often the easiest method of solving a quadratic equation is <strong>factoring<\/strong>. Factoring a quadratic equation involves expressing it as a product of simpler polynomials, typically two binomials.<\/p>\n<p>If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property which states that if [latex]a\\cdot b=0[\/latex], then [latex]a=0[\/latex] or [latex]b=0[\/latex], where [latex]a[\/latex] and [latex]b[\/latex] are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.<\/p>\n<p>The process of factoring a quadratic equation depends on the leading coefficient, whether it is [latex]1[\/latex] or another integer.<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a quadratic equation with the leading coefficient of 1, factor it<\/strong><\/p>\n<ol>\n<li>Find two numbers whose product equals <em>c<\/em> and whose sum equals [latex]b[\/latex].<\/li>\n<li>Use those numbers to write two factors of the form [latex]\\left(x+k\\right)\\text{ or }\\left(x-k\\right)[\/latex], where [latex]k[\/latex] is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are [latex]1[\/latex] and [latex]-2[\/latex], the factors are [latex]\\left(x+1\\right)\\left(x - 2\\right)[\/latex].<\/li>\n<li>Solve using the zero-product property by setting each factor equal to zero and solving for the variable.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">Factor and solve the equation:<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}+x - 6=0[\/latex]<\/div>\n<ol style=\"list-style-type: decimal;\">\n<li>Factor the quadratic expression.\n<div style=\"text-align: center;\">[latex]x^2+x - 6 = (x+2)(x-3)[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q451031\">Factoring steps<\/button><\/p>\n<div id=\"q451031\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li>We have a trinomial with leading coefficient [latex]1[\/latex], [latex]b = 1[\/latex], and [latex]c = -6[\/latex].<\/li>\n<li>We need to find two numbers with a product of [latex]-6[\/latex] and a sum of [latex]1[\/latex].<\/li>\n<\/ul>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 50%;\"><strong>Factors of [latex]-6[\/latex]<\/strong><\/td>\n<td style=\"width: 50%;\"><strong>Sum of Factors<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">[latex]1, -6[\/latex]<\/td>\n<td style=\"width: 50%;\">[latex]-5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">[latex]-1, 6[\/latex]<\/td>\n<td style=\"width: 50%;\">[latex]5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">[latex]2, -3[\/latex]<\/td>\n<td style=\"width: 50%;\">[latex]-1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">[latex]-2, 3[\/latex]<\/td>\n<td style=\"width: 50%;\">[latex]1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul>\n<li>Now that we have identified [latex]p[\/latex] and [latex]q[\/latex] as [latex]-2[\/latex] and [latex]3[\/latex], we can write the factored form as [latex](x-(-2))(x-3) = (x+2)(x-3)[\/latex].<\/div>\n<\/div>\n<\/li>\n<\/ul>\n<p>&nbsp;<\/li>\n<li>Apply the Zero-Product Property to find [latex]x[\/latex].[latex]\\begin{align*} \\text{Equation:} && x^2 + x - 6 &= 0 \\\\ \\text{Factored form:} && (x - 2)(x + 3) &= 0 \\\\ \\text{Apply zero product property:} && x - 2 &= 0 \\quad \\text{or} \\quad x + 3 = 0 \\\\ \\text{Solution:} && x &= 2 \\quad \\text{or} \\quad x = -3\u00a0 \\end{align*}[\/latex]<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm18961\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18961&theme=lumen&iframe_resize_id=ohm18961&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<p>When the leading coefficient is not [latex]1[\/latex], we factor a quadratic equation using the method called grouping, which requires four terms.<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>With the equation in standard form, let\u2019s review the grouping procedures:<\/strong><\/p>\n<ol>\n<li>With the quadratic in standard form, [latex]ax^2 + bx + c = 0[\/latex], multiply [latex]a \\cdot c[\/latex].<\/li>\n<li>Find two numbers whose product equals [latex]a \\cdot c[\/latex] and whose sum equals [latex]b[\/latex].<\/li>\n<li>Rewrite the equation replacing the [latex]bx[\/latex] term with two terms using the numbers found in step 2 as coefficients of [latex]x[\/latex].<\/li>\n<li>Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.<\/li>\n<li>Factor out the expression in parentheses.<\/li>\n<li>Set the expressions equal to zero and solve for the variable.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">Use grouping to factor and solve the quadratic equation:<\/p>\n<div style=\"text-align: center;\">[latex]4x^2 + 15x + 9 = 0[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q304864\">Show Answer<\/button><\/p>\n<div id=\"q304864\" class=\"hidden-answer\" style=\"display: none\">First, multiply [latex]a \\cdot c[\/latex]: [latex]4 \\cdot 9 = 36[\/latex]. Then list the factors of [latex]36[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c c}  1 \\cdot 36 \\\\  2 \\cdot 18 \\\\  3 \\cdot 12 \\\\  4 \\cdot 9 \\\\  6 \\cdot 6 \\\\  \\end{array}[\/latex]<\/p>\n<p>The only pair of factors that sums to [latex]15[\/latex] is [latex]3 + 12[\/latex]. Rewrite the equation replacing the [latex]b[\/latex] term, [latex]15x[\/latex], with two terms using [latex]3[\/latex] and [latex]12[\/latex] as coefficients of [latex]x[\/latex]. Factor the first two terms, and then factor the last two terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r c l}  4x^2 + 3x + 12x + 9 & = & 0 \\\\  x(4x + 3) + 3(4x + 3) & = & 0 \\\\  (4x + 3)(x + 3) & = & 0 \\\\  \\end{array}[\/latex]<\/p>\n<p>Solve using the zero-product property.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r c l}  (4x + 3)(x + 3) & = & 0 \\\\  4x + 3 & = & 0 \\quad \\Rightarrow \\quad x = -\\frac{3}{4} \\\\  x + 3 & = & 0 \\quad \\Rightarrow \\quad x = -3 \\\\  \\end{array}[\/latex]<\/p>\n<p>The solutions are [latex]x = -\\frac{3}{4}[\/latex] and [latex]x = -3[\/latex].<\/p>\n<figure id=\"attachment_3548\" aria-describedby=\"caption-attachment-3548\" style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-3548 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/06122929\/c8c54dba0d1717875901655bddf937a8b9e3db6a.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 6 to 2 with every other tick mark labeled and the y-axis ranging from negative 6 to 2 with each tick mark numbered. The equation: four x squared plus fifteen x plus nine is graphed with its x-intercepts: (-3\/4,0) and (-3,0) plotted as well.\" width=\"487\" height=\"433\" \/><figcaption id=\"caption-attachment-3548\" class=\"wp-caption-text\">Coordinate plane with quadratic equation<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18962\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18962&theme=lumen&iframe_resize_id=ohm18962&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section>\n<h2>Using the Square Root Property<\/h2>\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<section class=\"textbox keyTakeaway\">\n<h3>square root property<\/h3>\n<p>With the [latex]{x}^{2}[\/latex] term isolated, the square root property states that:<\/p>\n<div style=\"text-align: center;\">[latex]\\text{if }{x}^{2}=k,\\text{then }x=\\pm \\sqrt{k}[\/latex]<\/div>\n<p>where [latex]k[\/latex] is a non-negative real number.<\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a quadratic equation with an [latex]{x}^{2}[\/latex] term but no [latex]x[\/latex] term, use the square root property to solve it<\/strong><\/p>\n<ol>\n<li>Isolate the [latex]{x}^{2}[\/latex] term on one side of the equal sign.<\/li>\n<li>Take the square root of both sides of the equation, putting a [latex]\\pm[\/latex] sign before the expression on the side opposite the squared term.<\/li>\n<li>Simplify the numbers on the side with the [latex]\\pm[\/latex] sign.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">Solve the quadratic using the square root property:<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}=8[\/latex].<\/div>\n<p>[latex]\\begin{align*} \\text{Given equation:} & \\quad x^2 = 8 \\\\ \\text{Apply the square root property:} & \\quad x = \\pm \\sqrt{8} \\\\ \\text{Simplify the square root:} & \\quad \\sqrt{8} = \\sqrt{4 \\times 2} = 2\\sqrt{2} \\\\ \\text{Thus, the solution is:} & \\quad x = \\pm 2\\sqrt{2} \\end{align*}[\/latex]<\/section>\n<section class=\"textbox example\">Solve the quadratic equation using the square root property:<\/p>\n<div style=\"text-align: center;\">[latex]3{\\left(x - 4\\right)}^{2}=15[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q352957\">Show Answer<\/button><\/p>\n<div id=\"q352957\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{align*} \\text{Isolate the squared term:} & \\quad (x - 4)^2 = \\frac{15}{3} \\\\ & \\quad (x - 4)^2 = 5 \\\\ \\text{Apply the square root property:} & \\quad x - 4 = \\pm \\sqrt{5} \\\\ \\text{Solve for } x: & \\quad x = 4 \\pm \\sqrt{5} \\\\ & \\quad x = 4 + \\sqrt{5} \\quad \\text{or} \\quad x = 4 - \\sqrt{5} \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18963\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18963&theme=lumen&iframe_resize_id=ohm18963&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section>\n<h2>Completing the Square<\/h2>\n<p>Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use other methods for solving a quadratic equation. One method of solving quadratic equation is known as <strong>completing the square<\/strong>.<\/p>\n<p>Using this process, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, [latex]a[\/latex],\u00a0must equal [latex]1[\/latex]. If it does not, then divide the entire equation by [latex]a[\/latex]. Then, we can use the following procedures to solve a quadratic equation by completing the square.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>completing the square<\/h3>\n<p>The goal of <strong>completing the square<\/strong> is to transform a quadratic equation of the form [latex]ax^2+bx+c = 0[\/latex] into a perfect square trinomial:<\/p>\n<p style=\"text-align: center;\">[latex](x-h)^2 = k[\/latex],<\/p>\n<p>which can easily be solved by taking square roots.<\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to: Solve a quadratic equation by completing the square<\/strong><\/p>\n<ol>\n<li>Rearrange your equation so that it is in standard form: [latex]ax^2 + bx + c = 0[\/latex]. Divide by [latex]a[\/latex] if [latex]a \\neq 1[\/latex].<\/li>\n<li>Isolate the constant term by moving it to the right side of the equation.<\/li>\n<li>Add [latex](\\frac{b}{2})^2[\/latex] on both sides of the equation.<\/li>\n<li>Form the perfect square trinomial.<\/li>\n<li>Solve for [latex]x[\/latex] using the square root property.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Solve the quadratic equation by completing the square:<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}-3x - 5=0[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q349481\">Show Answer<\/button><\/p>\n<div id=\"q349481\" class=\"hidden-answer\" style=\"display: none\">First, move the constant term to the right side of the equal sign by adding [latex]5[\/latex] to both sides of the equation.<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}-3x=5[\/latex]<\/div>\n<p>Then, take [latex]\\frac{1}{2}[\/latex] of the [latex]b[\/latex] term and square it.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}\\left(-3\\right)=-\\frac{3}{2}\\hfill \\\\ {\\left(-\\frac{3}{2}\\right)}^{2}=\\frac{9}{4}\\hfill \\end{array}[\/latex]<\/div>\n<p>Add the result to both sides of the equal sign.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }{x}^{2}-3x+{\\left(-\\frac{3}{2}\\right)}^{2}=5+{\\left(-\\frac{3}{2}\\right)}^{2}\\hfill \\\\ {x}^{2}-3x+\\frac{9}{4}=5+\\frac{9}{4}\\hfill \\end{array}[\/latex]<\/div>\n<p>Factor the left side as a perfect square and simplify the right side.<\/p>\n<div style=\"text-align: center;\">[latex]{\\left(x-\\frac{3}{2}\\right)}^{2}=\\frac{29}{4}[\/latex]<\/div>\n<p>Use the square root property and solve.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{{\\left(x-\\frac{3}{2}\\right)}^{2}}=\\pm \\sqrt{\\frac{29}{4}}\\hfill \\\\ \\left(x-\\frac{3}{2}\\right)=\\pm \\frac{\\sqrt{29}}{2}\\hfill \\\\ x=\\frac{3}{2}\\pm \\frac{\\sqrt{29}}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>The solutions are [latex]x=\\frac{3}{2}+\\frac{\\sqrt{29}}{2}[\/latex], [latex]x=\\frac{3}{2}-\\frac{\\sqrt{29}}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Solve the quadratic equation using completing the square:<\/p>\n<div style=\"text-align: center;\">[latex]3x^2-6x-9=0[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q479550\">Show Answer<\/button><\/p>\n<div id=\"q479550\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{align*} \\text{Divide by the coefficient of } x^2: & \\quad \\frac{3x^2 - 6x - 9}{3} = 0 \\\\ & \\quad x^2 - 2x - 3 = 0 \\\\ \\text{Isolate the constant term:} & \\quad x^2 - 2x = 3 \\\\ \\text{Add } \\left(\\frac{-2}{2}\\right)^2 = (-1)^2 = 1 \\text{ on both sides}: & \\quad x^2 - 2x + 1 = 3 +1 \\\\ \\text{Form the perfect square trinomial:} & \\quad (x - 1)^2 = 3 +1 \\\\ \\text{Simplify and solve:} & \\quad (x - 1)^2 = 4 \\\\ & \\quad x - 1 = \\pm 2 \\\\ & \\quad x = 1 \\pm 2 \\\\ & \\quad x = 1 + 2 = 3 \\text{ and } x = 1 - 2 = -1 \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18967\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18967&theme=lumen&iframe_resize_id=ohm18967&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18965\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18965&theme=lumen&iframe_resize_id=ohm18965&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section>\n<h2>Quadratic Formula<\/h2>\n<p>The fourth method of solving a <strong>quadratic equation<\/strong> is by using the <strong>quadratic formula<\/strong>, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>quadratic formula<\/h3>\n<p>Written in standard form, [latex]a{x}^{2}+bx+c=0[\/latex], any quadratic equation can be solved using the <strong>quadratic formula<\/strong>:<\/p>\n<div style=\"text-align: center;\">[latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/div>\n<p>where [latex]a, b,[\/latex] and [latex]c[\/latex] are real numbers and [latex]a\\ne 0[\/latex].<\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a quadratic equation, solve it using the quadratic formula<\/strong><\/p>\n<ol>\n<li>Make sure the equation is in standard form: [latex]a{x}^{2}+bx+c=0[\/latex].<\/li>\n<li>Make note of the values of the coefficients and constant term, [latex]a,b[\/latex], and [latex]c[\/latex].<\/li>\n<li>Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.<\/li>\n<li>Calculate and solve.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">Solve the quadratic equation:<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}+5x+1=0[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q641400\">Show Solution<\/button><\/p>\n<div id=\"q641400\" class=\"hidden-answer\" style=\"display: none\">Identify the coefficients: [latex]a=1,b=5,c=1[\/latex]. Then use the quadratic formula.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x\\hfill&=\\frac{-\\left(5\\right)\\pm \\sqrt{{\\left(5\\right)}^{2}-4\\left(1\\right)\\left(1\\right)}}{2\\left(1\\right)}\\hfill \\\\ \\hfill&=\\frac{-5\\pm \\sqrt{25 - 4}}{2}\\hfill \\\\ \\hfill&=\\frac{-5\\pm \\sqrt{21}}{2}\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/section>\n<div>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm293772\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=293772&theme=lumen&iframe_resize_id=ohm293772&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/div>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\n","protected":false},"author":15,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-811","chapter","type-chapter","status-publish","hentry"],"part":440,"module-header":"background_you_need","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/811","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":4,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/811\/revisions"}],"predecessor-version":[{"id":2685,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/811\/revisions\/2685"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/440"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/811\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=811"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=811"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=811"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=811"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}