{"id":1009,"date":"2025-06-20T17:28:32","date_gmt":"2025-06-20T17:28:32","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1009"},"modified":"2025-08-28T13:38:50","modified_gmt":"2025-08-28T13:38:50","slug":"polar-coordinates-and-conic-sections-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/polar-coordinates-and-conic-sections-background-youll-need-1\/","title":{"raw":"Polar Coordinates and Conic Sections: Background You'll Need 1","rendered":"Polar Coordinates and Conic Sections: Background You&#8217;ll Need 1"},"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 by completing the square.<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Completing the Square<\/h2>\r\nNot all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use other methods for solving a quadratic equation. One method of solving quadratic equation is known as <strong>completing the square<\/strong>.\r\n\r\nUsing this process, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, [latex]a[\/latex],\u00a0must equal [latex]1[\/latex]. If it does not, then divide the entire equation by [latex]a[\/latex]. Then, we can use the following procedures to solve a quadratic equation by completing the square.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>completing the square<\/h3>\r\nThe goal of <strong>completing the square<\/strong> is to transform a quadratic equation of the form [latex]ax^2+bx+c = 0[\/latex] into a perfect square trinomial:\r\n<p style=\"text-align: center;\">[latex](x-h)^2 = k[\/latex],<\/p>\r\nwhich can easily be solved by taking square roots.\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to: Solve a quadratic equation by completing the square<\/strong>\r\n<ol>\r\n \t<li>Rearrange your equation so that it is in standard form: [latex]ax^2 + bx + c = 0[\/latex]. Divide by [latex]a[\/latex] if [latex]a \\neq 1[\/latex].<\/li>\r\n \t<li>Isolate the constant term by moving it to the right side of the equation.<\/li>\r\n \t<li>Add [latex](\\frac{b}{2})^2[\/latex] on both sides of the equation.<\/li>\r\n \t<li>Form the perfect square trinomial.<\/li>\r\n \t<li>Solve for [latex]x[\/latex] using the square root property.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox recall\" aria-label=\"Recall\">Perfect Square Trinomials:\r\n<ul>\r\n \t<li>[latex]a^2 + 2ab + b^2 = (a + b)^2[\/latex]<\/li>\r\n \t<li>[latex]a^2 - 2ab + b^2 = (a - b)^2[\/latex]<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">We 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<\/section><section class=\"textbox recall\">Remember that we are permitted, by the properties of equality, to add, subtract, multiply, or divide the same amount to both sides of an equation. Doing so won't change the value of the equation but it will enable us to isolate the variable on one side (that is, to solve the equation for the variable).The square root property gives us another operation we can do to both sides of an equation, taking the square root. We just have to remember when taking the square root (or any even root, as we'll see later), to consider both the positive and negative possibilities of the constant.<\/section><section class=\"textbox example\" aria-label=\"Example\">Solve the quadratic equation by completing the square:<center>[latex]{x}^{2}-3x - 5=0[\/latex]<\/center>[reveal-answer q=\"349481\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"349481\"]First, move the constant term to the right side of the equal sign by adding [latex]5[\/latex] to both sides of the equation.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}-3x=5[\/latex]<\/div>\r\nThen, take [latex]\\frac{1}{2}[\/latex] of the [latex]b[\/latex] term and square it.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}\\left(-3\\right)=-\\frac{3}{2}\\hfill \\\\ {\\left(-\\frac{3}{2}\\right)}^{2}=\\frac{9}{4}\\hfill \\end{array}[\/latex]<\/div>\r\nAdd the result to both sides of the equal sign.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }{x}^{2}-3x+{\\left(-\\frac{3}{2}\\right)}^{2}=5+{\\left(-\\frac{3}{2}\\right)}^{2}\\hfill \\\\ {x}^{2}-3x+\\frac{9}{4}=5+\\frac{9}{4}\\hfill \\end{array}[\/latex]<\/div>\r\nFactor the left side as a perfect square and simplify the right side.\r\n<div style=\"text-align: center;\">[latex]{\\left(x-\\frac{3}{2}\\right)}^{2}=\\frac{29}{4}[\/latex]<\/div>\r\nUse the square root property and solve.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{{\\left(x-\\frac{3}{2}\\right)}^{2}}=\\pm \\sqrt{\\frac{29}{4}}\\hfill \\\\ \\left(x-\\frac{3}{2}\\right)=\\pm \\frac{\\sqrt{29}}{2}\\hfill \\\\ x=\\frac{3}{2}\\pm \\frac{\\sqrt{29}}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nThe solutions are [latex]x=\\frac{3}{2}+\\frac{\\sqrt{29}}{2}[\/latex], [latex]x=\\frac{3}{2}-\\frac{\\sqrt{29}}{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">Solve the quadratic equation using completing the square:<center>[latex]3x^2-6x-9=0[\/latex]<\/center>[reveal-answer q=\"479550\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"479550\"][latex]\\begin{align*} \\text{Divide by the coefficient of } x^2: &amp; \\quad \\frac{3x^2 - 6x - 9}{3} = 0 \\\\ &amp; \\quad x^2 - 2x - 3 = 0 \\\\ \\text{Isolate the constant term:} &amp; \\quad x^2 - 2x = 3 \\\\ \\text{Add } \\left(\\frac{-2}{2}\\right)^2 = (-1)^2 = 1 \\text{ on both sides}: &amp; \\quad x^2 - 2x + 1 = 3 +1 \\\\ \\text{Form the perfect square trinomial:} &amp; \\quad (x - 1)^2 = 3 +1 \\\\ \\text{Simplify and solve:} &amp; \\quad (x - 1)^2 = 4 \\\\ &amp; \\quad x - 1 = \\pm 2 \\\\ &amp; \\quad x = 1 \\pm 2 \\\\ &amp; \\quad x = 1 + 2 = 3 \\text{ and } x = 1 - 2 = -1 \\end{align*}[\/latex][\/hidden-answer]<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">Note that when solving a quadratic by completing the square, a negative value will sometimes arise under the square root symbol. Later, we'll see that this value can be represented by a complex number (as shown in the video help for the problem below). We may also treat this type of solution as <em>unreal<\/em>, stating that no real solutions exist for this equation, by writing <em>DNE<\/em>.<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18967[\/ohm2_question]<\/section><section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18965[\/ohm2_question]<\/section><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Solve quadratic equations by completing the square.<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Completing the Square<\/h2>\n<p>Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use other methods for solving a quadratic equation. One method of solving quadratic equation is known as <strong>completing the square<\/strong>.<\/p>\n<p>Using this process, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, [latex]a[\/latex],\u00a0must equal [latex]1[\/latex]. If it does not, then divide the entire equation by [latex]a[\/latex]. Then, we can use the following procedures to solve a quadratic equation by completing the square.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>completing the square<\/h3>\n<p>The goal of <strong>completing the square<\/strong> is to transform a quadratic equation of the form [latex]ax^2+bx+c = 0[\/latex] into a perfect square trinomial:<\/p>\n<p style=\"text-align: center;\">[latex](x-h)^2 = k[\/latex],<\/p>\n<p>which can easily be solved by taking square roots.<\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to: Solve a quadratic equation by completing the square<\/strong><\/p>\n<ol>\n<li>Rearrange your equation so that it is in standard form: [latex]ax^2 + bx + c = 0[\/latex]. Divide by [latex]a[\/latex] if [latex]a \\neq 1[\/latex].<\/li>\n<li>Isolate the constant term by moving it to the right side of the equation.<\/li>\n<li>Add [latex](\\frac{b}{2})^2[\/latex] on both sides of the equation.<\/li>\n<li>Form the perfect square trinomial.<\/li>\n<li>Solve for [latex]x[\/latex] using the square root property.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox recall\" aria-label=\"Recall\">Perfect Square Trinomials:<\/p>\n<ul>\n<li>[latex]a^2 + 2ab + b^2 = (a + b)^2[\/latex]<\/li>\n<li>[latex]a^2 - 2ab + b^2 = (a - b)^2[\/latex]<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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<\/section>\n<section class=\"textbox recall\">Remember that we are permitted, by the properties of equality, to add, subtract, multiply, or divide the same amount to both sides of an equation. Doing so won&#8217;t change the value of the equation but it will enable us to isolate the variable on one side (that is, to solve the equation for the variable).The square root property gives us another operation we can do to both sides of an equation, taking the square root. We just have to remember when taking the square root (or any even root, as we&#8217;ll see later), to consider both the positive and negative possibilities of the constant.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Solve the quadratic equation by completing the square:<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}-3x - 5=0[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q349481\">Show Answer<\/button><\/p>\n<div id=\"q349481\" class=\"hidden-answer\" style=\"display: none\">First, move the constant term to the right side of the equal sign by adding [latex]5[\/latex] to both sides of the equation.<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}-3x=5[\/latex]<\/div>\n<p>Then, take [latex]\\frac{1}{2}[\/latex] of the [latex]b[\/latex] term and square it.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}\\left(-3\\right)=-\\frac{3}{2}\\hfill \\\\ {\\left(-\\frac{3}{2}\\right)}^{2}=\\frac{9}{4}\\hfill \\end{array}[\/latex]<\/div>\n<p>Add the result to both sides of the equal sign.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }{x}^{2}-3x+{\\left(-\\frac{3}{2}\\right)}^{2}=5+{\\left(-\\frac{3}{2}\\right)}^{2}\\hfill \\\\ {x}^{2}-3x+\\frac{9}{4}=5+\\frac{9}{4}\\hfill \\end{array}[\/latex]<\/div>\n<p>Factor the left side as a perfect square and simplify the right side.<\/p>\n<div style=\"text-align: center;\">[latex]{\\left(x-\\frac{3}{2}\\right)}^{2}=\\frac{29}{4}[\/latex]<\/div>\n<p>Use the square root property and solve.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{{\\left(x-\\frac{3}{2}\\right)}^{2}}=\\pm \\sqrt{\\frac{29}{4}}\\hfill \\\\ \\left(x-\\frac{3}{2}\\right)=\\pm \\frac{\\sqrt{29}}{2}\\hfill \\\\ x=\\frac{3}{2}\\pm \\frac{\\sqrt{29}}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>The solutions are [latex]x=\\frac{3}{2}+\\frac{\\sqrt{29}}{2}[\/latex], [latex]x=\\frac{3}{2}-\\frac{\\sqrt{29}}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Solve the quadratic equation using completing the square:<\/p>\n<div style=\"text-align: center;\">[latex]3x^2-6x-9=0[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q479550\">Show Answer<\/button><\/p>\n<div id=\"q479550\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{align*} \\text{Divide by the coefficient of } x^2: & \\quad \\frac{3x^2 - 6x - 9}{3} = 0 \\\\ & \\quad x^2 - 2x - 3 = 0 \\\\ \\text{Isolate the constant term:} & \\quad x^2 - 2x = 3 \\\\ \\text{Add } \\left(\\frac{-2}{2}\\right)^2 = (-1)^2 = 1 \\text{ on both sides}: & \\quad x^2 - 2x + 1 = 3 +1 \\\\ \\text{Form the perfect square trinomial:} & \\quad (x - 1)^2 = 3 +1 \\\\ \\text{Simplify and solve:} & \\quad (x - 1)^2 = 4 \\\\ & \\quad x - 1 = \\pm 2 \\\\ & \\quad x = 1 \\pm 2 \\\\ & \\quad x = 1 + 2 = 3 \\text{ and } x = 1 - 2 = -1 \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Note that when solving a quadratic by completing the square, a negative value will sometimes arise under the square root symbol. Later, we&#8217;ll see that this value can be represented by a complex number (as shown in the video help for the problem below). We may also treat this type of solution as <em>unreal<\/em>, stating that no real solutions exist for this equation, by writing <em>DNE<\/em>.<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18967\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18967&theme=lumen&iframe_resize_id=ohm18967&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18965\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18965&theme=lumen&iframe_resize_id=ohm18965&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/section>\n","protected":false},"author":15,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":677,"module-header":"- Select Header -","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\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1009"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":3,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1009\/revisions"}],"predecessor-version":[{"id":2061,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1009\/revisions\/2061"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/677"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1009\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1009"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1009"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1009"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1009"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}