{"id":416,"date":"2019-07-15T22:44:51","date_gmt":"2019-07-15T22:44:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/chapter\/complete-the-square\/"},"modified":"2019-07-15T22:44:51","modified_gmt":"2019-07-15T22:44:51","slug":"complete-the-square","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/chapter\/complete-the-square\/","title":{"raw":"Complete the Square","rendered":"Complete the Square"},"content":{"raw":"\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcome<\/h3>\n<ul>\n \t<li>Complete the square to create a perfect square trinomial<\/li>\n<\/ul>\n<\/div>\nWe have seen the technique of completing the square before when solving quadratic equations. Using this method, we added or subtracted terms to both sides of the equation until we had a perfect square trinomial on one side of the equal sign.&nbsp; Here is an example demonstrated earlier in the text.\n\nWe will use the example [latex]{x}^{2}+4x+1=0[\/latex] to illustrate each step. 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.\n<ol>\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.\n<div style=\"text-align: center\">[latex]{x}^{2}+4x=-1[\/latex]<\/div><\/li>\n \t<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><\/li>\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\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>\n \t<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><\/li>\n \t<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><\/li>\n \t<li>The solutions are [latex]x=-2+\\sqrt{3}[\/latex], [latex]x=-2-\\sqrt{3}[\/latex].<\/li>\n<\/ol>\nWe will see when studying conic sections that the method of completing the square comes in handy when rewriting the equation of a conic section given in general form. In preparation to use the method in this manner, it will be good to practice it first.\n<div class=\"textbox\">\n<h3>How To: use the method of complete the square to write a perfect square trinomial from an expression.<\/h3>\n<ol>\n \t<li>Given an expression of the form [latex]a\\left(x^2+bx\\right)[\/latex], add [latex]\\left(\\dfrac{b}{2}\\right)^2[\/latex] inside the parentheses.<\/li>\n \t<li>Then subtract [latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex] to counteract the change you made to the expression.<\/li>\n \t<li>If completing the square on one side of an equation, you may either subtract the value of&nbsp;[latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex] from that side, or add it to the other to maintain equality.<\/li>\n \t<li>&nbsp;Then factor the perfect square trinomial you created inside the original parentheses.<\/li>\n<\/ol>\n<h3>The resulting form will look like this:<\/h3>\nGiven\n<p style=\"text-align: center\">[latex]\\qquad a\\left(x^2+bx\\right)[\/latex]<\/p>\nadd [latex]\\left(b\/2\\right)^2[\/latex] inside the parentheses and subtract&nbsp;[latex]a\\left(b\/2\\right)^2[\/latex] to counteract the change you made to the expression\n<p style=\"text-align: center\">[latex]=a\\left(x^2+bx+ \\left(\\dfrac{b}{2}\\right)^2\\right)-a\\left(\\dfrac{b}{2}\\right)^2[\/latex]<\/p>\nthen factor the resulting perfect square trinomial\n<p style=\"text-align: center\">[latex]=a\\left(x+ \\dfrac{b}{2}\\right)^2-a\\left(\\dfrac{b}{2}\\right)^2[\/latex].<\/p>\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example : Create a perfect square trinomial using the method of complete the square<\/h3>\nComplete the square on: [latex]3\\left(x^2 - 10x\\right)[\/latex].\n\n[reveal-answer q=\"107002\"]Show Solution[\/reveal-answer]\n\n[hidden-answer a=\"107002\"]\n\n&nbsp;\n\nAdd [latex]\\left(\\dfrac{b}{2}\\right)^2[\/latex] inside the parentheses and subtract [latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex]\n<p style=\"text-align: center\">[latex]3\\left(x^2 - 10x+25\\right)-3\\cdot25[\/latex]<\/p>\n&nbsp;\n\nFactor the perfect square trinomial and simplify\n<p style=\"text-align: center\">[latex]3\\left(x -5\\right)^2-75[\/latex].<\/p>\n.\n<h4><span style=\"text-decoration: underline\">Analysis<\/span><\/h4>\nThe resulting expression is equivalent to the original expression. To test this, substitute a small value for [latex]x[\/latex], say [latex]x=3[\/latex].\n\n[latex]3\\left(3^2-10\\cdot3\\right) = \\quad -63 \\quad = 3(3-5)^2-75[\/latex]. True.\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n[ohm_question]15002[\/ohm_question]\n\n<\/div>\n&nbsp;\n","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcome<\/h3>\n<ul>\n<li>Complete the square to create a perfect square trinomial<\/li>\n<\/ul>\n<\/div>\n<p>We have seen the technique of completing the square before when solving quadratic equations. Using this method, we added or subtracted terms to both sides of the equation until we had a perfect square trinomial on one side of the equal sign.&nbsp; Here is an example demonstrated earlier in the text.<\/p>\n<p>We will use the example [latex]{x}^{2}+4x+1=0[\/latex] to illustrate each step. 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<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<p>We will see when studying conic sections that the method of completing the square comes in handy when rewriting the equation of a conic section given in general form. In preparation to use the method in this manner, it will be good to practice it first.<\/p>\n<div class=\"textbox\">\n<h3>How To: use the method of complete the square to write a perfect square trinomial from an expression.<\/h3>\n<ol>\n<li>Given an expression of the form [latex]a\\left(x^2+bx\\right)[\/latex], add [latex]\\left(\\dfrac{b}{2}\\right)^2[\/latex] inside the parentheses.<\/li>\n<li>Then subtract [latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex] to counteract the change you made to the expression.<\/li>\n<li>If completing the square on one side of an equation, you may either subtract the value of&nbsp;[latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex] from that side, or add it to the other to maintain equality.<\/li>\n<li>&nbsp;Then factor the perfect square trinomial you created inside the original parentheses.<\/li>\n<\/ol>\n<h3>The resulting form will look like this:<\/h3>\n<p>Given<\/p>\n<p style=\"text-align: center\">[latex]\\qquad a\\left(x^2+bx\\right)[\/latex]<\/p>\n<p>add [latex]\\left(b\/2\\right)^2[\/latex] inside the parentheses and subtract&nbsp;[latex]a\\left(b\/2\\right)^2[\/latex] to counteract the change you made to the expression<\/p>\n<p style=\"text-align: center\">[latex]=a\\left(x^2+bx+ \\left(\\dfrac{b}{2}\\right)^2\\right)-a\\left(\\dfrac{b}{2}\\right)^2[\/latex]<\/p>\n<p>then factor the resulting perfect square trinomial<\/p>\n<p style=\"text-align: center\">[latex]=a\\left(x+ \\dfrac{b}{2}\\right)^2-a\\left(\\dfrac{b}{2}\\right)^2[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example : Create a perfect square trinomial using the method of complete the square<\/h3>\n<p>Complete the square on: [latex]3\\left(x^2 - 10x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q107002\">Show Solution<\/span><\/p>\n<div id=\"q107002\" class=\"hidden-answer\" style=\"display: none\">\n<p>&nbsp;<\/p>\n<p>Add [latex]\\left(\\dfrac{b}{2}\\right)^2[\/latex] inside the parentheses and subtract [latex]a\\left(\\dfrac{b}{2}\\right)^2[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]3\\left(x^2 - 10x+25\\right)-3\\cdot25[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Factor the perfect square trinomial and simplify<\/p>\n<p style=\"text-align: center\">[latex]3\\left(x -5\\right)^2-75[\/latex].<\/p>\n<p>.<\/p>\n<h4><span style=\"text-decoration: underline\">Analysis<\/span><\/h4>\n<p>The resulting expression is equivalent to the original expression. To test this, substitute a small value for [latex]x[\/latex], say [latex]x=3[\/latex].<\/p>\n<p>[latex]3\\left(3^2-10\\cdot3\\right) = \\quad -63 \\quad = 3(3-5)^2-75[\/latex]. True.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm15002\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15002&theme=oea&iframe_resize_id=ohm15002&show_question_numbers\" width=\"100%\" height=\"150\"><\/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-416\">\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 15002. <strong>Authored by<\/strong>: Pieter Rousseau. <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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra.\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\" http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2.\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Question ID 15002\",\"author\":\"Pieter Rousseau\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-416","chapter","type-chapter","status-publish","hentry"],"part":413,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapters\/416","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":0,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapters\/416\/revisions"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/parts\/413"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapters\/416\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/media?parent=416"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/pressbooks\/v2\/chapter-type?post=416"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/contributor?post=416"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/wp-json\/wp\/v2\/license?post=416"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}