{"id":2582,"date":"2025-08-13T18:07:23","date_gmt":"2025-08-13T18:07:23","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=2582"},"modified":"2026-01-13T18:04:37","modified_gmt":"2026-01-13T18:04:37","slug":"polynomial-equations-background-youll-need-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/polynomial-equations-background-youll-need-4\/","title":{"raw":"Polynomial Equations: Background You'll Need 4","rendered":"Polynomial Equations: Background You&#8217;ll Need 4"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Solve quadratic equations<\/span><\/li>\r\n<\/ul>\r\n<\/section>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.\r\n\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\nThe 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: [latex]{x}^{2}+x - 6=0[\/latex].[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[latex]x^2+x - 6 = (x+2)(x-3)[\/latex]\r\n\r\n<strong>Step 2: Apply the Zero-Product Property to find [latex]x[\/latex].<\/strong>\r\n\r\n[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]\r\n\r\n<\/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\">With the equation in standard form, let\u2019s review the grouping procedures:\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: [latex]4x^2 + 15x + 9 = 0[\/latex].[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\" \/> Graph of a parabola[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]318848[\/ohm_question]<\/section><section>\r\n<h2>Quadratic Formula<\/h2>\r\nAnother 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 connectIt\" aria-label=\"Connect It\">We can derive the quadratic formula by <strong>completing the square<\/strong>. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by [latex]-1[\/latex] and obtain a positive <em>a<\/em>. Given [latex]a{x}^{2}+bx+c=0[\/latex], [latex]a\\ne 0[\/latex], we will complete the square as follows:\r\n<ol>\r\n \t<li>First, move the constant term to the right side of the equal sign:\r\n<div style=\"text-align: center;\">[latex]a{x}^{2}+bx=-c[\/latex]<\/div><\/li>\r\n \t<li>As we want the leading coefficient to equal 1, divide through by <em>a<\/em>:\r\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x=-\\frac{c}{a}[\/latex]<\/div><\/li>\r\n \t<li>Then, find [latex]\\frac{1}{2}[\/latex] of the middle term, and add [latex]{\\left(\\frac{1}{2}\\frac{b}{a}\\right)}^{2}=\\frac{{b}^{2}}{4{a}^{2}}[\/latex] to both sides of the equal sign:\r\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x+\\frac{{b}^{2}}{4{a}^{2}}=\\frac{{b}^{2}}{4{a}^{2}}-\\frac{c}{a}[\/latex]<\/div><\/li>\r\n \t<li>Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:\r\n<div style=\"text-align: center;\">[latex]{\\left(x+\\frac{b}{2a}\\right)}^{2}=\\frac{{b}^{2}-4ac}{4{a}^{2}}[\/latex]<\/div><\/li>\r\n \t<li>Now, use the square root property, which gives\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x+\\frac{b}{2a}=\\pm \\sqrt{\\frac{{b}^{2}-4ac}{4{a}^{2}}}\\hfill \\\\ x+\\frac{b}{2a}=\\frac{\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>Finally, add [latex]-\\frac{b}{2a}[\/latex] to both sides of the equation and combine the terms on the right side. Thus,\r\n<div style=\"text-align: center;\">[latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/div><\/li>\r\n<\/ol>\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]318849[\/ohm_question]<\/section><\/div>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Solve quadratic equations<\/span><\/li>\n<\/ul>\n<\/section>\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<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>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: [latex]{x}^{2}+x - 6=0[\/latex].<\/p>\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>[latex]x^2+x - 6 = (x+2)(x-3)[\/latex]<\/p>\n<p><strong>Step 2: Apply the Zero-Product Property to find [latex]x[\/latex].<\/strong><\/p>\n<p>[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]<\/p>\n<\/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\">With the equation in standard form, let\u2019s review the grouping procedures:<\/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: [latex]4x^2 + 15x + 9 = 0[\/latex].<\/p>\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\">Graph of a parabola<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm318848\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=318848&theme=lumen&iframe_resize_id=ohm318848&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section>\n<h2>Quadratic Formula<\/h2>\n<p>Another 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 connectIt\" aria-label=\"Connect It\">We can derive the quadratic formula by <strong>completing the square<\/strong>. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by [latex]-1[\/latex] and obtain a positive <em>a<\/em>. Given [latex]a{x}^{2}+bx+c=0[\/latex], [latex]a\\ne 0[\/latex], we will complete the square as follows:<\/p>\n<ol>\n<li>First, move the constant term to the right side of the equal sign:\n<div style=\"text-align: center;\">[latex]a{x}^{2}+bx=-c[\/latex]<\/div>\n<\/li>\n<li>As we want the leading coefficient to equal 1, divide through by <em>a<\/em>:\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x=-\\frac{c}{a}[\/latex]<\/div>\n<\/li>\n<li>Then, find [latex]\\frac{1}{2}[\/latex] of the middle term, and add [latex]{\\left(\\frac{1}{2}\\frac{b}{a}\\right)}^{2}=\\frac{{b}^{2}}{4{a}^{2}}[\/latex] to both sides of the equal sign:\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x+\\frac{{b}^{2}}{4{a}^{2}}=\\frac{{b}^{2}}{4{a}^{2}}-\\frac{c}{a}[\/latex]<\/div>\n<\/li>\n<li>Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:\n<div style=\"text-align: center;\">[latex]{\\left(x+\\frac{b}{2a}\\right)}^{2}=\\frac{{b}^{2}-4ac}{4{a}^{2}}[\/latex]<\/div>\n<\/li>\n<li>Now, use the square root property, which gives\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x+\\frac{b}{2a}=\\pm \\sqrt{\\frac{{b}^{2}-4ac}{4{a}^{2}}}\\hfill \\\\ x+\\frac{b}{2a}=\\frac{\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>Finally, add [latex]-\\frac{b}{2a}[\/latex] to both sides of the equation and combine the terms on the right side. Thus,\n<div style=\"text-align: center;\">[latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/div>\n<\/li>\n<\/ol>\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=\"ohm318849\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=318849&theme=lumen&iframe_resize_id=ohm318849&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/div>\n<\/section>\n","protected":false},"author":67,"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":506,"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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2582"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/67"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2582\/revisions"}],"predecessor-version":[{"id":5321,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2582\/revisions\/5321"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/506"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2582\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=2582"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=2582"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=2582"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=2582"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}