{"id":1245,"date":"2024-05-07T21:46:17","date_gmt":"2024-05-07T21:46:17","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=1245"},"modified":"2025-08-13T15:39:16","modified_gmt":"2025-08-13T15:39:16","slug":"quadratic-equations-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/quadratic-equations-learn-it-1\/","title":{"raw":"Quadratic Equations: Learn It 1","rendered":"Quadratic Equations: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n \t<li>Solve quadratic equations by factoring.<\/li>\r\n \t<li>Solve quadratic equations by square root property.<\/li>\r\n \t<li>Solve quadratic equations by completing the square.<\/li>\r\n \t<li>Solve quadratic equations by using quadratic formula.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Quadratic Equations<\/h2>\r\nAn equation containing a second-degree polynomial is called a <strong>quadratic equation<\/strong>. For example, equations such as [latex]2{x}^{2}+3x - 1=0[\/latex] and [latex]{x}^{2}-4=0[\/latex] are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>quadratic equations<\/h3>\r\nA <strong>quadratic equation<\/strong> is a type of polynomial equation of the second degree, which means it involves at least one term that is squared (i.e., raised to the power of two).\r\n\r\n&nbsp;\r\n\r\nThe standard form of a quadratic equation is:\r\n<p style=\"text-align: center;\">[latex]ax^2+bx+c = 0[\/latex]<\/p>\r\n<p style=\"text-align: left;\">where [latex]a, b, \\text{ and }c[\/latex] are real numbers and [latex]a \\ne 0[\/latex].<\/p>\r\n\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 <em>a <\/em>and <em>b <\/em>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\nMultiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>zero-product property<\/h3>\r\nThe <strong>zero-product property<\/strong> states\r\n<div style=\"text-align: center;\">[latex]\\text{If }a\\cdot b=0,\\text{ then }a=0\\text{ or }b=0[\/latex],<\/div>\r\nwhere [latex]a[\/latex] and [latex]b[\/latex] are real numbers or algebraic expressions.\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">For example, expand the factored expression [latex]\\left(x - 2\\right)\\left(x+3\\right)[\/latex] by multiplying the two factors together.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x - 2\\right)\\left(x+3\\right)\\hfill&amp;={x}^{2}+3x - 2x - 6\\hfill \\\\ \\hfill&amp;={x}^{2}+x - 6\\hfill \\end{array}[\/latex]<\/div>\r\nThe product is a quadratic expression. Set equal to zero, [latex]{x}^{2}+x - 6=0[\/latex] is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.\r\n\r\n<\/section>The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 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>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Solve quadratic equations by factoring.<\/li>\n<li>Solve quadratic equations by square root property.<\/li>\n<li>Solve quadratic equations by completing the square.<\/li>\n<li>Solve quadratic equations by using quadratic formula.<\/li>\n<\/ul>\n<\/section>\n<h2>Quadratic Equations<\/h2>\n<p>An equation containing a second-degree polynomial is called a <strong>quadratic equation<\/strong>. For example, equations such as [latex]2{x}^{2}+3x - 1=0[\/latex] and [latex]{x}^{2}-4=0[\/latex] are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>quadratic equations<\/h3>\n<p>A <strong>quadratic equation<\/strong> is a type of polynomial equation of the second degree, which means it involves at least one term that is squared (i.e., raised to the power of two).<\/p>\n<p>&nbsp;<\/p>\n<p>The standard form of a quadratic equation is:<\/p>\n<p style=\"text-align: center;\">[latex]ax^2+bx+c = 0[\/latex]<\/p>\n<p style=\"text-align: left;\">where [latex]a, b, \\text{ and }c[\/latex] are real numbers and [latex]a \\ne 0[\/latex].<\/p>\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 <em>a <\/em>and <em>b <\/em>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>Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>zero-product property<\/h3>\n<p>The <strong>zero-product property<\/strong> states<\/p>\n<div style=\"text-align: center;\">[latex]\\text{If }a\\cdot b=0,\\text{ then }a=0\\text{ or }b=0[\/latex],<\/div>\n<p>where [latex]a[\/latex] and [latex]b[\/latex] are real numbers or algebraic expressions.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">For example, expand the factored expression [latex]\\left(x - 2\\right)\\left(x+3\\right)[\/latex] by multiplying the two factors together.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x - 2\\right)\\left(x+3\\right)\\hfill&={x}^{2}+3x - 2x - 6\\hfill \\\\ \\hfill&={x}^{2}+x - 6\\hfill \\end{array}[\/latex]<\/div>\n<p>The product is a quadratic expression. Set equal to zero, [latex]{x}^{2}+x - 6=0[\/latex] is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.<\/p>\n<\/section>\n<p>The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 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\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/06122929\/c8c54dba0d1717875901655bddf937a8b9e3db6a.jpg 487w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/06122929\/c8c54dba0d1717875901655bddf937a8b9e3db6a-300x267.jpg 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/06122929\/c8c54dba0d1717875901655bddf937a8b9e3db6a-65x58.jpg 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/06122929\/c8c54dba0d1717875901655bddf937a8b9e3db6a-225x200.jpg 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/06122929\/c8c54dba0d1717875901655bddf937a8b9e3db6a-350x311.jpg 350w\" sizes=\"(max-width: 487px) 100vw, 487px\" \/><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","protected":false},"author":12,"menu_order":5,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":92,"module-header":"learn_it","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\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1245"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":22,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1245\/revisions"}],"predecessor-version":[{"id":7610,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1245\/revisions\/7610"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/92"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1245\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=1245"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1245"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=1245"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=1245"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}