{"id":1438,"date":"2016-10-24T22:52:20","date_gmt":"2016-10-24T22:52:20","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=1438"},"modified":"2017-04-10T20:27:59","modified_gmt":"2017-04-10T20:27:59","slug":"completing-the-square-and-the-quadratic-formula","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/completing-the-square-and-the-quadratic-formula\/","title":{"raw":"Completing the Square and the Quadratic Formula","rendered":"Completing the Square and the Quadratic Formula"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Complete the square to solve a quadratic equation<\/li>\r\n \t<li>Use the quadratic formula to solve a quadratic equation<\/li>\r\n \t<li>Use the discriminant to determine the number and type of solutions to a quadratic equation<\/li>\r\n<\/ul>\r\n<\/div>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\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 a method for solving a <strong>quadratic equation<\/strong> known as <strong>completing the square<\/strong>. Using this method, 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, <em>a<\/em>, must equal 1. If it does not, then divide the entire equation by <em>a<\/em>. Then, we can use the following procedures to solve a quadratic equation by completing the square.\r\n\r\nWe will use the example [latex]{x}^{2}+4x+1=0[\/latex] to illustrate each step.\r\n<ol>\r\n \t<li>Given a quadratic equation that cannot be factored, and with [latex]a=1[\/latex], first add or subtract the constant term to the right sign of the equal sign.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}+4x=-1[\/latex]<\/div><\/li>\r\n \t<li>Multiply the <em>b <\/em>term by [latex]\\frac{1}{2}[\/latex] and square it.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}\\left(4\\right)=2\\hfill \\\\ {2}^{2}=4\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>Add [latex]{\\left(\\frac{1}{2}b\\right)}^{2}[\/latex] to both sides of the equal sign and simplify the right side. We have\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+4x+4=-1+4\\hfill \\\\ {x}^{2}+4x+4=3\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>The left side of the equation can now be factored as a perfect square.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+4x+4=3\\hfill \\\\ {\\left(x+2\\right)}^{2}=3\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>Use the square root property and solve.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\sqrt{{\\left(x+2\\right)}^{2}}=\\pm \\sqrt{3}\\hfill \\\\ x+2=\\pm \\sqrt{3}\\hfill \\\\ x=-2\\pm \\sqrt{3}\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>The solutions are [latex]x=-2+\\sqrt{3}[\/latex], [latex]x=-2-\\sqrt{3}[\/latex].<\/li>\r\n<\/ol>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Quadratic by Completing the Square<\/h3>\r\nSolve the quadratic equation by completing the square: [latex]{x}^{2}-3x - 5=0[\/latex].\r\n\r\n[reveal-answer q=\"676921\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"676921\"]\r\nFirst, move the constant term to the right side of the equal sign.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}-3x=5[\/latex]<\/div>\r\nThen, take [latex]\\frac{1}{2}[\/latex] of the <em>b <\/em>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>[latex]{\\left(x-\\frac{3}{2}\\right)}^{2}=\\frac{29}{4}[\/latex]<\/div>\r\nUse the square root property and solve.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\sqrt{{\\left(x-\\frac{3}{2}\\right)}^{2}}\\hfill&amp;=\\pm \\sqrt{\\frac{29}{4}}\\hfill \\\\ \\left(x-\\frac{3}{2}\\right)\\hfill&amp;=\\pm \\frac{\\sqrt{29}}{2}\\hfill \\\\ x\\hfill&amp;=\\frac{3}{2}\\pm \\frac{\\sqrt{29}}{2}\\hfill \\end{array}[\/latex]<\/div>\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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve by completing the square: [latex]{x}^{2}-6x=13[\/latex].\r\n\r\n[reveal-answer q=\"222291\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"222291\"]\r\n\r\n[latex]x=3\\pm \\sqrt{22}[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=1384&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=79619&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Using the 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\nWe 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<div class=\"textbox\">\r\n<h3>A General Note: The 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 <em>a<\/em>, <em>b<\/em>, and <em>c<\/em> are real numbers and [latex]a\\ne 0[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a quadratic equation, solve it using the quadratic formula<\/h3>\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<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example : Solve A Quadratic Equation Using the Quadratic Formula<\/h3>\r\nSolve the quadratic equation: [latex]{x}^{2}+5x+1=0[\/latex].\r\n\r\n[reveal-answer q=\"641400\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"641400\"]\r\nIdentify 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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the quadratic equation using the quadratic formula: [latex]9{x}^{2}+3x - 2=0[\/latex].\r\n\r\n[reveal-answer q=\"232269\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"232269\"][latex]x=-\\frac{2}{3}[\/latex], [latex]x=\\frac{1}{3}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Discriminant<\/h2>\r\nThe <strong>quadratic formula<\/strong> not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the <strong>discriminant<\/strong>, or the expression under the radical, [latex]{b}^{2}-4ac[\/latex]. The discriminant tells us whether the solutions are real numbers or complex numbers (which we will learn about in more depth later in the course), and how many solutions of each type to expect. The table below\u00a0relates the value of the discriminant to the solutions of a quadratic equation.\r\n<table summary=\"A table with 5 rows and 2 columns. The entries in the first row are: Value of Discriminant and Results. The entries in the second row are: b squared minus four times a times c equals zero and One rational solution (double solution). The entries in the third row are: b squared minus four times a times c is greater than zero, perfect square and Two rational solutions. The entries in the fourth row are: b squared minus four times a times c is greater than zero, not a perfect square and Two irrational solutions. The entries in the fifth row are: b squared minus four times a times c is less than zero and Two complex solutions.\">\r\n<thead>\r\n<tr>\r\n<th>Value of Discriminant<\/th>\r\n<th>Results<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]{b}^{2}-4ac=0[\/latex]<\/td>\r\n<td>One rational solution (double solution)<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]{b}^{2}-4ac&gt;0[\/latex], perfect square<\/td>\r\n<td>Two rational solutions<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]{b}^{2}-4ac&gt;0[\/latex], not a perfect square<\/td>\r\n<td>Two irrational solutions<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]{b}^{2}-4ac&lt;0[\/latex]<\/td>\r\n<td>Two complex solutions<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Discriminant<\/h3>\r\nFor [latex]a{x}^{2}+bx+c=0[\/latex], where [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are real numbers, the <strong>discriminant<\/strong> is the expression under the radical in the quadratic formula: [latex]{b}^{2}-4ac[\/latex]. It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation<\/h3>\r\nUse the discriminant to find the nature of the solutions to the following quadratic equations:\r\n<ol>\r\n \t<li>[latex]{x}^{2}+4x+4=0[\/latex]<\/li>\r\n \t<li>[latex]8{x}^{2}+14x+3=0[\/latex]<\/li>\r\n \t<li>[latex]3{x}^{2}-5x - 2=0[\/latex]<\/li>\r\n \t<li>[latex]3{x}^{2}-10x+15=0[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"229118\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"229118\"]\r\nCalculate the discriminant [latex]{b}^{2}-4ac[\/latex] for each equation and state the expected type of solutions.\r\n<ol>\r\n \t<li>[latex]{x}^{2}+4x+4=0[\/latex][latex]{b}^{2}-4ac={\\left(4\\right)}^{2}-4\\left(1\\right)\\left(4\\right)=0[\/latex]. There will be one rational double solution.<\/li>\r\n \t<li>[latex]8{x}^{2}+14x+3=0[\/latex][latex]{b}^{2}-4ac={\\left(14\\right)}^{2}-4\\left(8\\right)\\left(3\\right)=100[\/latex]. As [latex]100[\/latex] is a perfect square, there will be two rational solutions.<\/li>\r\n \t<li>[latex]3{x}^{2}-5x - 2=0[\/latex][latex]{b}^{2}-4ac={\\left(-5\\right)}^{2}-4\\left(3\\right)\\left(-2\\right)=49[\/latex]. As [latex]49[\/latex] is a perfect square, there will be two rational solutions.<\/li>\r\n \t<li>[latex]3{x}^{2}-10x+15=0[\/latex][latex]{b}^{2}-4ac={\\left(-10\\right)}^{2}-4\\left(3\\right)\\left(15\\right)=-80[\/latex]. There will be two complex solutions.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom6\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=35145&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Complete the square to solve a quadratic equation<\/li>\n<li>Use the quadratic formula to solve a quadratic equation<\/li>\n<li>Use the discriminant to determine the number and type of solutions to a quadratic equation<\/li>\n<\/ul>\n<\/div>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><br \/>\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 a method for solving a <strong>quadratic equation<\/strong> known as <strong>completing the square<\/strong>. Using this method, 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, <em>a<\/em>, must equal 1. If it does not, then divide the entire equation by <em>a<\/em>. Then, we can use the following procedures to solve a quadratic equation by completing the square.<\/p>\n<p>We will use the example [latex]{x}^{2}+4x+1=0[\/latex] to illustrate each step.<\/p>\n<ol>\n<li>Given a quadratic equation that cannot be factored, and with [latex]a=1[\/latex], first add or subtract the constant term to the right sign of the equal sign.\n<div style=\"text-align: center;\">[latex]{x}^{2}+4x=-1[\/latex]<\/div>\n<\/li>\n<li>Multiply the <em>b <\/em>term by [latex]\\frac{1}{2}[\/latex] and square it.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}\\left(4\\right)=2\\hfill \\\\ {2}^{2}=4\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>Add [latex]{\\left(\\frac{1}{2}b\\right)}^{2}[\/latex] to both sides of the equal sign and simplify the right side. We have\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+4x+4=-1+4\\hfill \\\\ {x}^{2}+4x+4=3\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>The left side of the equation can now be factored as a perfect square.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{x}^{2}+4x+4=3\\hfill \\\\ {\\left(x+2\\right)}^{2}=3\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>Use the square root property and solve.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\sqrt{{\\left(x+2\\right)}^{2}}=\\pm \\sqrt{3}\\hfill \\\\ x+2=\\pm \\sqrt{3}\\hfill \\\\ x=-2\\pm \\sqrt{3}\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>The solutions are [latex]x=-2+\\sqrt{3}[\/latex], [latex]x=-2-\\sqrt{3}[\/latex].<\/li>\n<\/ol>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Quadratic by Completing the Square<\/h3>\n<p>Solve the quadratic equation by completing the square: [latex]{x}^{2}-3x - 5=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q676921\">Solution<\/span><\/p>\n<div id=\"q676921\" class=\"hidden-answer\" style=\"display: none\">\nFirst, move the constant term to the right side of the equal sign.<\/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 <em>b <\/em>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>[latex]{\\left(x-\\frac{3}{2}\\right)}^{2}=\\frac{29}{4}[\/latex]<\/div>\n<p>Use the square root property and solve.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\sqrt{{\\left(x-\\frac{3}{2}\\right)}^{2}}\\hfill&=\\pm \\sqrt{\\frac{29}{4}}\\hfill \\\\ \\left(x-\\frac{3}{2}\\right)\\hfill&=\\pm \\frac{\\sqrt{29}}{2}\\hfill \\\\ x\\hfill&=\\frac{3}{2}\\pm \\frac{\\sqrt{29}}{2}\\hfill \\end{array}[\/latex]<\/div>\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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve by completing the square: [latex]{x}^{2}-6x=13[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q222291\">Solution<\/span><\/p>\n<div id=\"q222291\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=3\\pm \\sqrt{22}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=1384&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=79619&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Using the 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<p>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<div class=\"textbox\">\n<h3>A General Note: The 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 <em>a<\/em>, <em>b<\/em>, and <em>c<\/em> are real numbers and [latex]a\\ne 0[\/latex].<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a quadratic equation, solve it using the quadratic formula<\/h3>\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<\/div>\n<div class=\"textbox exercises\">\n<h3>Example : Solve A Quadratic Equation Using the Quadratic Formula<\/h3>\n<p>Solve the quadratic equation: [latex]{x}^{2}+5x+1=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q641400\">Solution<\/span><\/p>\n<div id=\"q641400\" class=\"hidden-answer\" style=\"display: none\">\nIdentify 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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the quadratic equation using the quadratic formula: [latex]9{x}^{2}+3x - 2=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q232269\">Solution<\/span><\/p>\n<div id=\"q232269\" class=\"hidden-answer\" style=\"display: none\">[latex]x=-\\frac{2}{3}[\/latex], [latex]x=\\frac{1}{3}[\/latex]<\/div>\n<\/div>\n<\/div>\n<h2>The Discriminant<\/h2>\n<p>The <strong>quadratic formula<\/strong> not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the <strong>discriminant<\/strong>, or the expression under the radical, [latex]{b}^{2}-4ac[\/latex]. The discriminant tells us whether the solutions are real numbers or complex numbers (which we will learn about in more depth later in the course), and how many solutions of each type to expect. The table below\u00a0relates the value of the discriminant to the solutions of a quadratic equation.<\/p>\n<table summary=\"A table with 5 rows and 2 columns. The entries in the first row are: Value of Discriminant and Results. The entries in the second row are: b squared minus four times a times c equals zero and One rational solution (double solution). The entries in the third row are: b squared minus four times a times c is greater than zero, perfect square and Two rational solutions. The entries in the fourth row are: b squared minus four times a times c is greater than zero, not a perfect square and Two irrational solutions. The entries in the fifth row are: b squared minus four times a times c is less than zero and Two complex solutions.\">\n<thead>\n<tr>\n<th>Value of Discriminant<\/th>\n<th>Results<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]{b}^{2}-4ac=0[\/latex]<\/td>\n<td>One rational solution (double solution)<\/td>\n<\/tr>\n<tr>\n<td>[latex]{b}^{2}-4ac>0[\/latex], perfect square<\/td>\n<td>Two rational solutions<\/td>\n<\/tr>\n<tr>\n<td>[latex]{b}^{2}-4ac>0[\/latex], not a perfect square<\/td>\n<td>Two irrational solutions<\/td>\n<\/tr>\n<tr>\n<td>[latex]{b}^{2}-4ac<0[\/latex]<\/td>\n<td>Two complex solutions<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox\">\n<h3>A General Note: The Discriminant<\/h3>\n<p>For [latex]a{x}^{2}+bx+c=0[\/latex], where [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are real numbers, the <strong>discriminant<\/strong> is the expression under the radical in the quadratic formula: [latex]{b}^{2}-4ac[\/latex]. It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation<\/h3>\n<p>Use the discriminant to find the nature of the solutions to the following quadratic equations:<\/p>\n<ol>\n<li>[latex]{x}^{2}+4x+4=0[\/latex]<\/li>\n<li>[latex]8{x}^{2}+14x+3=0[\/latex]<\/li>\n<li>[latex]3{x}^{2}-5x - 2=0[\/latex]<\/li>\n<li>[latex]3{x}^{2}-10x+15=0[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q229118\">Solution<\/span><\/p>\n<div id=\"q229118\" class=\"hidden-answer\" style=\"display: none\">\nCalculate the discriminant [latex]{b}^{2}-4ac[\/latex] for each equation and state the expected type of solutions.<\/p>\n<ol>\n<li>[latex]{x}^{2}+4x+4=0[\/latex][latex]{b}^{2}-4ac={\\left(4\\right)}^{2}-4\\left(1\\right)\\left(4\\right)=0[\/latex]. There will be one rational double solution.<\/li>\n<li>[latex]8{x}^{2}+14x+3=0[\/latex][latex]{b}^{2}-4ac={\\left(14\\right)}^{2}-4\\left(8\\right)\\left(3\\right)=100[\/latex]. As [latex]100[\/latex] is a perfect square, there will be two rational solutions.<\/li>\n<li>[latex]3{x}^{2}-5x - 2=0[\/latex][latex]{b}^{2}-4ac={\\left(-5\\right)}^{2}-4\\left(3\\right)\\left(-2\\right)=49[\/latex]. As [latex]49[\/latex] is a perfect square, there will be two rational solutions.<\/li>\n<li>[latex]3{x}^{2}-10x+15=0[\/latex][latex]{b}^{2}-4ac={\\left(-10\\right)}^{2}-4\\left(3\\right)\\left(15\\right)=-80[\/latex]. There will be two complex solutions.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom6\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=35145&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/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-1438\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 1384. <strong>Authored by<\/strong>: WebWork-Rochester. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC- BY + GPL<\/li><li>Question ID 79619. <strong>Authored by<\/strong>: Edward Wicks. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC- BY + GPL<\/li><li>Question ID 35145. <strong>Authored by<\/strong>: Jim Smart. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC- BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: OpenStax College Algebra. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/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":21,"menu_order":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"College Algebra\",\"author\":\"OpenStax College Algebra\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen 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