{"id":1285,"date":"2024-05-08T00:30:17","date_gmt":"2024-05-08T00:30:17","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=1285"},"modified":"2024-11-25T18:28:34","modified_gmt":"2024-11-25T18:28:34","slug":"quadratic-equations-apply-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/quadratic-equations-apply-it-1\/","title":{"raw":"Quadratic Equations: Apply It 1","rendered":"Quadratic Equations: Apply It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n \t<li>Solve quadratic equations by factoring.<\/li>\r\n \t<li>Solve quadratic equations by square root property.<\/li>\r\n \t<li>Solve quadratic equations by completing the square.<\/li>\r\n \t<li>Solve quadratic equations by using quadratic formula.<\/li>\r\n<\/ul>\r\n<\/section>When it comes to solving quadratic equations, you have learned four powerful methods: factoring, using the square root property, completing the square, and the quadratic formula. Each method has its unique advantages and is suited for different types of quadratic equations. It's important to understand that you are not restricted to using just one method exclusively for all problems. Instead, think of these methods as tools in your <em>mathematical toolkit<\/em>.\r\n\r\nAs you gain experience with these techniques, you'll start to see which ones are the most efficient or straightforward for a given equation. For example, factoring might be the quickest way if the equation simplifies neatly, but the quadratic formula is a reliable all-purpose tool when factoring is complex or unclear.\r\n\r\nThe key is to assess the equation you are working with and choose the method that simplifies your problem-solving process the most. Don\u2019t hesitate to try a different approach if one method seems cumbersome or inadequate. Over time, you'll develop a sense of which method to use based on the specific characteristics of the equation and the context in which you're working.\r\n\r\n<section class=\"textbox example\">Solve [latex]x^2-6x+9 = 0[\/latex] using factoring, completing the square, square root method, and the quadratic formula.[reveal-answer q=\"712326\"]Factoring[\/reveal-answer]\r\n[hidden-answer a=\"712326\"]This equation factors as it is a perfect square: [latex]x^2 - 6x + 9 = (x - 3)(x - 3) = (x - 3)^2[\/latex]Setting the factor equal to zero gives:\r\n<p style=\"text-align: center;\">[latex]x - 3 = 0[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x = 3[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[\/hidden-answer]<\/p>\r\n[reveal-answer q=\"674302\"]Square Root Property[\/reveal-answer]\r\n[hidden-answer a=\"674302\"]\r\n\r\nFirst, observe that the equation is already a perfect square: [latex](x - 3)^2 = 0[\/latex]\r\n\r\nTake the square root of both sides:\r\n<p style=\"text-align: center;\">[latex]x - 3 = \\pm 0 = 0[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x = 3[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"946832\"]Completing the Square[\/reveal-answer]\r\n[hidden-answer a=\"946832\"]\r\n\r\nThe equation is presented as a completed square, but let's illustrate the steps:\r\n\r\n[latex]x^2 - 6x = -9[\/latex]\r\n\r\nAdd [latex](\\frac{-6}{2})^2 = 3^2 = 9[\/latex] to both sides:\r\n\r\n[latex]x^2 - 6x + 9 = -9 + 9[\/latex]\r\n\r\n[latex](x - 3)^2 = 0[\/latex]\r\n\r\nTake the square root of both sides: [latex]x = 3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"476917\"]Quadratic Formula[\/reveal-answer]\r\n[hidden-answer a=\"476917\"]\r\n\r\nApplying the quadratic formula where [latex]a=1, b=-6, c=9[\/latex]:\r\n\r\n[latex]x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}[\/latex]\r\n\r\n[latex]x = \\frac{-(-6) \\pm \\sqrt{(-6)^2 - 4 \\cdot 1 \\cdot 9}}{2 \\cdot 1}[\/latex]\r\n\r\n[latex]x = \\frac{6 \\pm \\sqrt{36 - 36}}{2}[\/latex]\r\n\r\n[latex]x = \\frac{6 \\pm 0}{2}[\/latex]\r\n\r\n[latex]x=3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\nThe solution to the equation by all four methods\u2014factoring, using the square root property, completing the square, and the quadratic formula\u2014is [latex]x=3[\/latex].\r\n\r\nThis demonstrates the versatility and connectivity of different algebraic methods in solving quadratic equations.\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18971[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18972[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18973[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Solve quadratic equations by factoring.<\/li>\n<li>Solve quadratic equations by square root property.<\/li>\n<li>Solve quadratic equations by completing the square.<\/li>\n<li>Solve quadratic equations by using quadratic formula.<\/li>\n<\/ul>\n<\/section>\n<p>When it comes to solving quadratic equations, you have learned four powerful methods: factoring, using the square root property, completing the square, and the quadratic formula. Each method has its unique advantages and is suited for different types of quadratic equations. It&#8217;s important to understand that you are not restricted to using just one method exclusively for all problems. Instead, think of these methods as tools in your <em>mathematical toolkit<\/em>.<\/p>\n<p>As you gain experience with these techniques, you&#8217;ll start to see which ones are the most efficient or straightforward for a given equation. For example, factoring might be the quickest way if the equation simplifies neatly, but the quadratic formula is a reliable all-purpose tool when factoring is complex or unclear.<\/p>\n<p>The key is to assess the equation you are working with and choose the method that simplifies your problem-solving process the most. Don\u2019t hesitate to try a different approach if one method seems cumbersome or inadequate. Over time, you&#8217;ll develop a sense of which method to use based on the specific characteristics of the equation and the context in which you&#8217;re working.<\/p>\n<section class=\"textbox example\">Solve [latex]x^2-6x+9 = 0[\/latex] using factoring, completing the square, square root method, and the quadratic formula.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q712326\">Factoring<\/button><\/p>\n<div id=\"q712326\" class=\"hidden-answer\" style=\"display: none\">This equation factors as it is a perfect square: [latex]x^2 - 6x + 9 = (x - 3)(x - 3) = (x - 3)^2[\/latex]Setting the factor equal to zero gives:<\/p>\n<p style=\"text-align: center;\">[latex]x - 3 = 0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x = 3[\/latex]<\/p>\n<p style=\"text-align: center;\"><\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q674302\">Square Root Property<\/button><\/p>\n<div id=\"q674302\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, observe that the equation is already a perfect square: [latex](x - 3)^2 = 0[\/latex]<\/p>\n<p>Take the square root of both sides:<\/p>\n<p style=\"text-align: center;\">[latex]x - 3 = \\pm 0 = 0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x = 3[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q946832\">Completing the Square<\/button><\/p>\n<div id=\"q946832\" class=\"hidden-answer\" style=\"display: none\">\n<p>The equation is presented as a completed square, but let&#8217;s illustrate the steps:<\/p>\n<p>[latex]x^2 - 6x = -9[\/latex]<\/p>\n<p>Add [latex](\\frac{-6}{2})^2 = 3^2 = 9[\/latex] to both sides:<\/p>\n<p>[latex]x^2 - 6x + 9 = -9 + 9[\/latex]<\/p>\n<p>[latex](x - 3)^2 = 0[\/latex]<\/p>\n<p>Take the square root of both sides: [latex]x = 3[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q476917\">Quadratic Formula<\/button><\/p>\n<div id=\"q476917\" class=\"hidden-answer\" style=\"display: none\">\n<p>Applying the quadratic formula where [latex]a=1, b=-6, c=9[\/latex]:<\/p>\n<p>[latex]x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}[\/latex]<\/p>\n<p>[latex]x = \\frac{-(-6) \\pm \\sqrt{(-6)^2 - 4 \\cdot 1 \\cdot 9}}{2 \\cdot 1}[\/latex]<\/p>\n<p>[latex]x = \\frac{6 \\pm \\sqrt{36 - 36}}{2}[\/latex]<\/p>\n<p>[latex]x = \\frac{6 \\pm 0}{2}[\/latex]<\/p>\n<p>[latex]x=3[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>The solution to the equation by all four methods\u2014factoring, using the square root property, completing the square, and the quadratic formula\u2014is [latex]x=3[\/latex].<\/p>\n<p>This demonstrates the versatility and connectivity of different algebraic methods in solving quadratic equations.<\/p>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18971\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18971&theme=lumen&iframe_resize_id=ohm18971&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18972\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18972&theme=lumen&iframe_resize_id=ohm18972&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18973\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18973&theme=lumen&iframe_resize_id=ohm18973&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":12,"menu_order":9,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":92,"module-header":"apply_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1285"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":4,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1285\/revisions"}],"predecessor-version":[{"id":3569,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1285\/revisions\/3569"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/92"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1285\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=1285"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1285"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=1285"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=1285"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}