{"id":659,"date":"2024-04-23T22:52:41","date_gmt":"2024-04-23T22:52:41","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=659"},"modified":"2024-11-20T00:53:24","modified_gmt":"2024-11-20T00:53:24","slug":"radicals-and-rational-exponents-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/radicals-and-rational-exponents-learn-it-2\/","title":{"raw":"Radicals and Rational Exponents: Learn It 2","rendered":"Radicals and Rational Exponents: Learn It 2"},"content":{"raw":"<h2>Simplifying Square Roots and Expressing Them in Lowest Terms<\/h2>\r\nTo <strong>simplify a square root<\/strong> means that we rewrite the square root as a rational number times the square root of a number that has no perfect square factors. The act of changing a square root into such a form is simplifying the square root. Before discussing how to simplify a square root, we need to introduce a rule about square roots.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>the product rule for square roots<\/h3>\r\nThe square root of a product of numbers equals the product of the square roots of those number.\r\n\r\nGiven that [latex]a[\/latex] and [latex]b[\/latex] are nonnegative real numbers,\r\n\r\n<center>[latex]\\sqrt{a \\times {b}}=\\sqrt{a} \\times \\sqrt{b}[\/latex]<\/center><\/div>\r\n<\/section>Using this formula, we can factor an integer inside a square root into a perfect square times another integer. Then the square root can be applied to the perfect square, leaving an integer times the square root of another integer. If the number remaining under the square root has no perfect square factors, then we\u2019ve simplified the irrational number into lowest terms.\r\n\r\n<section class=\"textbox proTip\">A perfect square is an integer that can be expressed as the square of another integer. For example, [latex]16[\/latex], [latex]25[\/latex], and [latex]36[\/latex] are perfect squares because they are [latex]4^2[\/latex], [latex]5^2[\/latex], and [latex]6^2[\/latex], respectively.<\/section><section class=\"textbox questionHelp\"><strong>How to: Simplify square roots into lowest terms when [latex]n[\/latex] is an integer<\/strong>\r\n<ul>\r\n \t<li><strong>Step 1:<\/strong> Determine the largest perfect square factor of [latex]n[\/latex], which we denote [latex]a^2[\/latex].<\/li>\r\n \t<li><strong>Step 2:<\/strong> Factor [latex]n[\/latex] into [latex]a^2\u00d7b[\/latex].<\/li>\r\n \t<li><strong>Step 3:<\/strong> Apply [latex]\\sqrt{a^2 \\times b} =\\sqrt{a^2} \\times \\sqrt{b}[\/latex].<\/li>\r\n \t<li><strong>Step 4:<\/strong> Write [latex]\\sqrt{n}[\/latex] in its simplified form, [latex]a\\sqrt{b}[\/latex].<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\">To simplify the given radical expressions, we'll break down the numbers into their [pb_glossary id=\"662\"]prime factors[\/pb_glossary] and simplify the radicals accordingly, while also considering the powers of the variables. Here are the steps:First, factor [latex]300[\/latex] into its prime factors:\r\n<p style=\"text-align: center;\">[latex]300 = 2^2 \\cdot 3 \\cdot 5^2[\/latex]<\/p>\r\nNow, extract the square roots of the perfect squares:\r\n\r\n[latex]\\begin{align} \\sqrt{300} &amp;= \\sqrt{2^2 \\cdot 3 \\cdot 5^2} &amp;&amp; \\text{Factor the number into prime factors.} \\\\ &amp;= \\sqrt{2^2} \\cdot \\sqrt{3} \\cdot \\sqrt{5^2} &amp;&amp; \\text{Separate each factor under its own square root.} \\\\ &amp;= 2 \\cdot \\sqrt{3} \\cdot 5 &amp;&amp; \\text{Simplify the square roots of perfect squares.} \\\\ &amp;= 10\\sqrt{3} &amp;&amp; \\text{Multiply the results to get the simplified form.} \\end{align}[\/latex]\r\n\r\n<\/section><section class=\"textbox example\">Simplify [latex]\\sqrt{162{a}^{5}{b}^{4}}[\/latex].[reveal-answer q=\"30444\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"30444\"]<span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">\\begin{align} \\sqrt{162a^5b^4} &amp;= \\sqrt{2 \\cdot 3^4 \\cdot a^5 \\cdot b^4} &amp;&amp; \\text{Factor the number into prime factors and express variables.} \\\\ &amp;= \\sqrt{2 \\cdot (3^2)^2 \\cdot a^4 \\cdot a \\cdot (b^2)^2} &amp;&amp; \\text{Break down the expression to show squares for clarity.} \\\\ &amp;= \\sqrt{2} \\cdot \\sqrt{(3^2)^2} \\cdot \\sqrt{a^4} \\cdot \\sqrt{a} \\cdot \\sqrt{(b^2)^2} &amp;&amp; \\text{Separate each factor under its own square root.} \\\\ &amp;= \\sqrt{2} \\cdot 3^2 \\cdot a^2 \\cdot \\sqrt{a} \\cdot b^2 &amp;&amp; \\text{Simplify the square roots of perfect squares.} \\\\ &amp;= 9a^2b^2 \\cdot \\sqrt{2a} &amp;&amp; \\text{Combine the constants and simplify further to finalize.} \\end{align}[\/hidden-answer]<\/span><\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18752[\/ohm2_question]<\/section><section class=\"textbox proTip\">For the variable [latex]x[\/latex], [latex]\\sqrt{x^2} = |x|[\/latex] , but why is that?When you square any values, the result is always non-negative, meaning it's either positive or zero. Then, when you take the square root of this non-negative squared value, you get back the original number without its sign\u2014just its size or magnitude. Thus, taking the square root of [latex]x^2[\/latex] always yields the absolute value of [latex]x[\/latex] ensuring that we consider [latex]x[\/latex] in its non-negative form.<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18753[\/ohm2_question]<\/section>Given the product of multiple radical expressions, we can use the product rule to combine them into one radical expression and then simplify as we did above.\r\n\r\n<section class=\"textbox example\">Simplify the following radical expression.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"text-align: left;\">[latex]\\sqrt{12}\\cdot \\sqrt{3}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{50x}\\cdot \\sqrt{2x}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"13428714\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"13428714\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\begin{align}\\sqrt{12}\\cdot \\sqrt{3} &amp; = \\sqrt{12\\cdot 3} &amp;&amp; \\text{Express the product as a single radical expression}. \\\\ &amp;= \\sqrt{36} &amp;&amp; \\text{Simplify}. \\\\ &amp;= 6 \\end{align}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{align*} \\sqrt{50x} \\cdot \\sqrt{2x} &amp;= \\sqrt{50x \\cdot 2x} &amp; \\text{Multiply under the radicals} \\\\ &amp;= \\sqrt{100x^2} &amp; \\text{Simplify the product inside the radical} \\\\ &amp;= 10|x| &amp; \\text{Take the square root of \\(100\\) and \\(x^2\\)} \\end{align*}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18754[\/ohm2_question]<\/section>","rendered":"<h2>Simplifying Square Roots and Expressing Them in Lowest Terms<\/h2>\n<p>To <strong>simplify a square root<\/strong> means that we rewrite the square root as a rational number times the square root of a number that has no perfect square factors. The act of changing a square root into such a form is simplifying the square root. Before discussing how to simplify a square root, we need to introduce a rule about square roots.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>the product rule for square roots<\/h3>\n<p>The square root of a product of numbers equals the product of the square roots of those number.<\/p>\n<p>Given that [latex]a[\/latex] and [latex]b[\/latex] are nonnegative real numbers,<\/p>\n<div style=\"text-align: center;\">[latex]\\sqrt{a \\times {b}}=\\sqrt{a} \\times \\sqrt{b}[\/latex]<\/div>\n<\/div>\n<\/section>\n<p>Using this formula, we can factor an integer inside a square root into a perfect square times another integer. Then the square root can be applied to the perfect square, leaving an integer times the square root of another integer. If the number remaining under the square root has no perfect square factors, then we\u2019ve simplified the irrational number into lowest terms.<\/p>\n<section class=\"textbox proTip\">A perfect square is an integer that can be expressed as the square of another integer. For example, [latex]16[\/latex], [latex]25[\/latex], and [latex]36[\/latex] are perfect squares because they are [latex]4^2[\/latex], [latex]5^2[\/latex], and [latex]6^2[\/latex], respectively.<\/section>\n<section class=\"textbox questionHelp\"><strong>How to: Simplify square roots into lowest terms when [latex]n[\/latex] is an integer<\/strong><\/p>\n<ul>\n<li><strong>Step 1:<\/strong> Determine the largest perfect square factor of [latex]n[\/latex], which we denote [latex]a^2[\/latex].<\/li>\n<li><strong>Step 2:<\/strong> Factor [latex]n[\/latex] into [latex]a^2\u00d7b[\/latex].<\/li>\n<li><strong>Step 3:<\/strong> Apply [latex]\\sqrt{a^2 \\times b} =\\sqrt{a^2} \\times \\sqrt{b}[\/latex].<\/li>\n<li><strong>Step 4:<\/strong> Write [latex]\\sqrt{n}[\/latex] in its simplified form, [latex]a\\sqrt{b}[\/latex].<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\">To simplify the given radical expressions, we&#8217;ll break down the numbers into their <a class=\"glossary-term\" aria-haspopup=\"dialog\" aria-describedby=\"definition\" href=\"#term_659_662\">prime factors<\/a> and simplify the radicals accordingly, while also considering the powers of the variables. Here are the steps:First, factor [latex]300[\/latex] into its prime factors:<\/p>\n<p style=\"text-align: center;\">[latex]300 = 2^2 \\cdot 3 \\cdot 5^2[\/latex]<\/p>\n<p>Now, extract the square roots of the perfect squares:<\/p>\n<p>[latex]\\begin{align} \\sqrt{300} &= \\sqrt{2^2 \\cdot 3 \\cdot 5^2} && \\text{Factor the number into prime factors.} \\\\ &= \\sqrt{2^2} \\cdot \\sqrt{3} \\cdot \\sqrt{5^2} && \\text{Separate each factor under its own square root.} \\\\ &= 2 \\cdot \\sqrt{3} \\cdot 5 && \\text{Simplify the square roots of perfect squares.} \\\\ &= 10\\sqrt{3} && \\text{Multiply the results to get the simplified form.} \\end{align}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">Simplify [latex]\\sqrt{162{a}^{5}{b}^{4}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q30444\">Show Answer<\/button><\/p>\n<div id=\"q30444\" class=\"hidden-answer\" style=\"display: none\"><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">\\begin{align} \\sqrt{162a^5b^4} &amp;= \\sqrt{2 \\cdot 3^4 \\cdot a^5 \\cdot b^4} &amp;&amp; \\text{Factor the number into prime factors and express variables.} \\\\ &amp;= \\sqrt{2 \\cdot (3^2)^2 \\cdot a^4 \\cdot a \\cdot (b^2)^2} &amp;&amp; \\text{Break down the expression to show squares for clarity.} \\\\ &amp;= \\sqrt{2} \\cdot \\sqrt{(3^2)^2} \\cdot \\sqrt{a^4} \\cdot \\sqrt{a} \\cdot \\sqrt{(b^2)^2} &amp;&amp; \\text{Separate each factor under its own square root.} \\\\ &amp;= \\sqrt{2} \\cdot 3^2 \\cdot a^2 \\cdot \\sqrt{a} \\cdot b^2 &amp;&amp; \\text{Simplify the square roots of perfect squares.} \\\\ &amp;= 9a^2b^2 \\cdot \\sqrt{2a} &amp;&amp; \\text{Combine the constants and simplify further to finalize.} \\end{align}<\/div>\n<\/div>\n<p><\/span><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18752\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18752&theme=lumen&iframe_resize_id=ohm18752&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox proTip\">For the variable [latex]x[\/latex], [latex]\\sqrt{x^2} = |x|[\/latex] , but why is that?When you square any values, the result is always non-negative, meaning it&#8217;s either positive or zero. Then, when you take the square root of this non-negative squared value, you get back the original number without its sign\u2014just its size or magnitude. Thus, taking the square root of [latex]x^2[\/latex] always yields the absolute value of [latex]x[\/latex] ensuring that we consider [latex]x[\/latex] in its non-negative form.<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18753\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18753&theme=lumen&iframe_resize_id=ohm18753&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<p>Given the product of multiple radical expressions, we can use the product rule to combine them into one radical expression and then simplify as we did above.<\/p>\n<section class=\"textbox example\">Simplify the following radical expression.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"text-align: left;\">[latex]\\sqrt{12}\\cdot \\sqrt{3}[\/latex]<\/li>\n<li>[latex]\\sqrt{50x}\\cdot \\sqrt{2x}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q13428714\">Show Solution<\/button><\/p>\n<div id=\"q13428714\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\begin{align}\\sqrt{12}\\cdot \\sqrt{3} & = \\sqrt{12\\cdot 3} && \\text{Express the product as a single radical expression}. \\\\ &= \\sqrt{36} && \\text{Simplify}. \\\\ &= 6 \\end{align}[\/latex]<\/li>\n<li>[latex]\\begin{align*} \\sqrt{50x} \\cdot \\sqrt{2x} &= \\sqrt{50x \\cdot 2x} & \\text{Multiply under the radicals} \\\\ &= \\sqrt{100x^2} & \\text{Simplify the product inside the radical} \\\\ &= 10|x| & \\text{Take the square root of \\(100\\) and \\(x^2\\)} \\end{align*}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18754\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18754&theme=lumen&iframe_resize_id=ohm18754&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<div class=\"glossary\"><span class=\"screen-reader-text\" id=\"definition\">definition<\/span><template id=\"term_659_662\"><div class=\"glossary__definition\" role=\"dialog\" data-id=\"term_659_662\"><div tabindex=\"-1\"><p>A prime factor is a factor of a number that is a prime number itself.<br \/>\nPrime numbers are natural numbers greater than 1 that have no positive divisors other than 1 and themselves.<\/p>\n<\/div><button><span aria-hidden=\"true\">&times;<\/span><span class=\"screen-reader-text\">Close definition<\/span><\/button><\/div><\/template><\/div>","protected":false},"author":12,"menu_order":23,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":32,"module-header":"learn_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\/659"}],"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":28,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/659\/revisions"}],"predecessor-version":[{"id":6219,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/659\/revisions\/6219"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/32"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/659\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=659"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=659"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=659"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=659"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}