{"id":2813,"date":"2024-02-09T19:18:37","date_gmt":"2024-02-09T19:18:37","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/?post_type=chapter&#038;p=2813"},"modified":"2024-02-09T19:18:41","modified_gmt":"2024-02-09T19:18:41","slug":"4-1-square-roots","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/4-1-square-roots\/","title":{"raw":"4.1 Square Roots","rendered":"4.1 Square Roots"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Define square roots<\/li>\r\n \t<li>Calculate the square root of a perfect square<\/li>\r\n \t<li>Simplify a square root that contains perfect square factors<\/li>\r\n \t<li>Estimate square roots using a calculator<\/li>\r\n \t<li>Round a decimal number<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key words<\/h3>\r\n<ul>\r\n \t<li><strong>Perfect square<\/strong>:\u00a0any whole number that has been squared<\/li>\r\n \t<li><strong>Square root<\/strong>:\u00a0a number that when multiplied by itself gives the original number<\/li>\r\n \t<li><strong>Radical<\/strong>: the symbol for square root<\/li>\r\n \t<li><strong>Principal square root<\/strong>:\u00a0the positive square root<\/li>\r\n \t<li><strong>Radicand<\/strong>: the number under a radical sign<\/li>\r\n<\/ul>\r\n<\/div>\r\nWhen we are trying to find the square root of a number (say,\u00a0[latex]25[\/latex]), we are trying to find a number that when multiplied by itself gives that original number. In the case of\u00a0[latex]25[\/latex], we find that [latex]5\\cdot5=25[\/latex], so\u00a0[latex]5[\/latex] is the square root of [latex]25[\/latex].\r\n\r\nThe square root is the inverse of the square (exponent of 2), much like multiplication is the inverse of division. A\u00a0<em><strong>perfect square<\/strong><\/em> is any whole number that has been squared. Consequently, all perfect squares have <em><strong>square roots<\/strong><\/em> that are whole numbers.\r\n\r\nThe symbol for square root is the\u00a0<em><strong>radical sign<\/strong><\/em> [latex]\\sqrt{\\hphantom{5}}[\/latex]. So [latex]\\sqrt{25}=5[\/latex]. The number under the radical sign is called the <em><strong>radicand<\/strong><\/em>. Because we are \"unsquaring\" a number when we find a square root, the radicand must be positive. This is because there is no real number that multiplies by itself to result in a negative number.\r\n<div class=\"textbox examples\">\r\n<h3>Examples<\/h3>\r\nFind the square root of the following numbers:\r\n<ol>\r\n \t<li>[latex]36[\/latex]<\/li>\r\n \t<li>[latex]81[\/latex]<\/li>\r\n \t<li>[latex]-49[\/latex]<\/li>\r\n \t<li>[latex]0[\/latex]<\/li>\r\n<\/ol>\r\nSolution\r\n<ol>\r\n \t<li>[latex]\\sqrt{36}=6[\/latex], since [latex]6^2=36[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{81}=9[\/latex], since [latex]9^2=81[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{-49}[\/latex] is undefined, since there is no real number that squares to -49.<\/li>\r\n \t<li>[latex]\\sqrt{0}=0[\/latex], since [latex]0^2=0[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFind the square root of the following numbers:\r\n<ol>\r\n \t<li>[latex]49[\/latex]<\/li>\r\n \t<li>[latex]64[\/latex]<\/li>\r\n \t<li>[latex]1[\/latex]<\/li>\r\n \t<li>[latex]-25[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"hjm854\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm854\"]\r\n<ol>\r\n \t<li>[latex]\\sqrt{49}=7[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{64}=8[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{1}=1[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{-25}[\/latex] is undefined in the set of real numbers<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<span style=\"font-size: 1rem; text-align: initial;\">Consider [latex] \\sqrt{25}[\/latex] again. If we think about it, there is another number that, when multiplied by itself, also results in\u00a0[latex]25[\/latex]. That number is [latex]\u22125[\/latex].<\/span>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}5\\cdot 5=25\\\\-5\\cdot -5=25\\end{array}[\/latex]<\/p>\r\nBy definition, the square root symbol, [latex]\\sqrt{\\hphantom{5}}[\/latex], always means to find the positive root called the <em><strong>principal root<\/strong><\/em>. So while [latex]5\\cdot5[\/latex] and [latex]\u22125\\cdot\u22125[\/latex] both equal\u00a0[latex]25[\/latex], only\u00a0[latex]5[\/latex] is the principal root. If we want to find the negative square root, we must place a negative sign infant of the radical: [latex]-\\sqrt{25}=-5[\/latex]. Zero is special because it has only one square root:\u00a0 [latex]\\sqrt{0}=0[\/latex]).\r\n\r\nThe notation that we use to express a square root for any real number, [latex]a[\/latex], is as follows:\r\n<div class=\"textbox shaded\">\r\n<h3>Square Root<\/h3>\r\nThe symbol for the square root is called a <strong>radical symbol.<\/strong>\u00a0For any non-negative, real number [latex]a[\/latex], the square root of <em>a<\/em> is written as [latex]\\sqrt{a}[\/latex].\r\n\r\nThe number that is written under the radical symbol is called the <strong>radicand<\/strong>.\r\n\r\nBy definition, the square root symbol, [latex]\\sqrt{\\hphantom{5}}[\/latex] always means to find the nonnegative\u00a0root, called the <strong>principal root<\/strong>.\r\n\r\n[latex]\\sqrt{-a}[\/latex] is undefined in the set of real numbers.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSimplify\u00a0the following square roots:\r\n<ol>\r\n \t<li>[latex]\\sqrt{16}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{9}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{-9}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{5^2}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"24327\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"24327\"]\r\n<ol>\r\n \t<li>[latex]\\sqrt{16}=4[\/latex].<\/li>\r\n \t<li>[latex]\\sqrt{9}=3[\/latex].<\/li>\r\n \t<li>[latex]\\sqrt{-9}[\/latex] is undefined in the set of real numbers.<\/li>\r\n \t<li>[latex]\\sqrt{5^2}=\\sqrt{25}=5[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe last problem in the previous example shows us an important relationship between squares and square roots.\r\n<div class=\"textbox shaded\">\r\n<h3>The square root of a perfect square<\/h3>\r\nFor any nonnegative real number, [latex]a[\/latex], \u00a0[latex]\\sqrt{a^2}=\\big |\\,a\\,\\big |[\/latex].\r\n\r\n<\/div>\r\nRemember that\u00a0\u00a0[latex]\\big |\\,a\\,\\big |[\/latex] represents the distance of the number [latex]a[\/latex] from zero. \u00a0[latex]\\big |\\,a\\,\\big | \\ge 0[\/latex]. Consequently, [latex]\\sqrt{6^2}=\\big |\\,6\\,\\big |=6[\/latex] and\u00a0[latex]\\sqrt{(-6)^2}=\\big |\\,-6\\,\\big |=6[\/latex].\r\n\r\nIn the video that follows, we simplify\u00a0more square roots using the fact that [latex]\\sqrt{a^2}=\\big |\\,a\\,\\big |[\/latex] means finding the principal square root.\r\n\r\nhttps:\/\/youtu.be\/B3riJsl7uZM\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]228[\/ohm_question]\r\n\r\n<\/div>\r\n<h3>Finding Square Roots Using Factoring<\/h3>\r\nWe have previously considered the square root of a fraction. For example, [latex]\\sqrt\\frac{4}{9}=\\frac{2}{3}[\/latex] because [latex]\\frac{2}{3}\\cdot\\frac{2}{3}=\\frac{4}{9}[\/latex]. We also discovered the <em><strong>Quotient Rule for Square Roots<\/strong><\/em> and were able to simply the square root of a fraction by taking the square root of the numerator and denominator separately. For example,\u00a0 [latex]\\sqrt\\frac{4}{9}=\\frac{\\sqrt{4}}{\\sqrt{9}}=\\frac{2}{3}[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>The QUOTIENT Rule for Square Roots<\/h3>\r\nFor any nonnegative, real number [latex]a[\/latex] and positive real number [latex]b[\/latex], \u00a0[latex]\\sqrt{\\frac{a}{b}}=\\frac{\\sqrt{a}}{\\sqrt{b}}[\/latex].\r\n\r\n<\/div>\r\nA similar rule applies to products. After all, division is just multiplication by the reciprocal, so the quotient rule could be considered a product rule of the reciprocal:\u00a0[latex]\\sqrt{\\frac{a}{b}}=\\sqrt{a\\cdot\\frac{1}{b}}[\/latex] and [latex]\\frac{\\sqrt{a}}{\\sqrt{b}}=\\sqrt{a}\\cdot\\frac{1}{\\sqrt{b}}[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>The Product Rule for Square Roots<\/h3>\r\nFor any nonnegative, real numbers\u00a0[latex]a[\/latex] and [latex]b[\/latex], \u00a0[latex]\\sqrt{a\\cdot{b}}=\\sqrt{a}\\cdot\\sqrt{b}[\/latex].\r\n\r\n<\/div>\r\nThis is useful if we are working with a number whose square we do not know right away. \u00a0We can use factoring and the <em><strong>product rule for square roots<\/strong><\/em> to find square roots such as [latex]\\sqrt{144}[\/latex], or\u00a0\u00a0[latex]\\sqrt{225}[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify\u00a0[latex] \\sqrt{144}[\/latex].\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"620082\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"620082\"]\r\n\r\nDetermine the prime factors of [latex]144[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{144}\\\\\\\\\\sqrt{2\\cdot 72}\\\\\\\\\\sqrt{2\\cdot 2\\cdot 36}\\\\\\\\\\sqrt{2\\cdot 2\\cdot 2\\cdot 18}\\\\\\\\\\sqrt{2\\cdot 2\\cdot 2\\cdot 2\\cdot 9}\\\\\\\\\\sqrt{2\\cdot 2\\cdot 2\\cdot 2\\cdot 3\\cdot 3}\\end{array}[\/latex]<\/p>\r\nBecause we are finding a square root, we regroup these factors into squares.\r\n<p style=\"text-align: center;\">[latex]\\sqrt{2^2\\cdot 2^2\\cdot3^2}[\/latex]<\/p>\r\nNow we can use the product rule for square roots and the square root of a square idea to finish finding the square root.\r\n\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{2^2\\cdot 2^2\\cdot3^2}\\\\\\\\=\\sqrt{2^2}\\cdot\\sqrt{2^2}\\cdot\\sqrt{3^2}\\\\\\\\=2\\cdot 2 \\cdot 3\\\\\\\\=12\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\sqrt{144}=12[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSimplify [latex]\\sqrt{225}[\/latex]\r\n\r\n[reveal-answer q=\"686109\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"686109\"]\r\n\r\nFirst, factor [latex]225[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{225}\\\\\\\\=\\sqrt{5\\cdot45}\\\\\\\\=\\sqrt{5\\cdot5\\cdot9}\\\\\\\\=\\sqrt{5\\cdot5\\cdot3\\cdot3}\\end{array}[\/latex]<\/p>\r\nBecause we are finding a square root, we regroup these factors into squares. Finish simplifying with the product rule for roots, and the square of a square idea.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{5^2\\cdot3^2}\\\\\\\\=\\sqrt{5^2}\\cdot\\sqrt{3^2}\\\\\\\\=5\\cdot3=15\\end{array}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left;\">Answer<\/h4>\r\n[latex]\\sqrt{225}=15[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n\r\n<img class=\" wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"49\" height=\"43\" \/>Caution! \u00a0The square root of a product rule applies\u00a0when you have multiplication ONLY under the square root. You cannot apply the rule to sums or differences:\r\n<p style=\"text-align: center;\">[latex]\\sqrt{a+b}\\ne\\sqrt{a}+\\sqrt{b}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Prove this to yourself with some real numbers: let [latex]a = 64[\/latex] and [latex]b = 36[\/latex], then use the order of operations to simplify each expression.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{64+36}=\\sqrt{100}=10\\\\\\\\\\sqrt{64}+\\sqrt{36}=8+6=14\\\\\\\\10\\ne14\\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\nLet's look at some more examples of expressions with square roots.\u00a0 Pay particular attention to how number 3 is evaluated.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSimplify:\r\n<ol>\r\n \t<li>[latex]\\sqrt{100}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{16}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{25+144}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{49}-\\sqrt{81}[\/latex]<\/li>\r\n \t<li>[latex] -\\sqrt{81}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{-9}[\/latex]<\/li>\r\n<\/ol>\r\n&nbsp;\r\n\r\n[reveal-answer q=\"419579\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"419579\"]\r\n<ol>\r\n \t<li>[latex]\\sqrt{100}=10[\/latex] because [latex]{10}^{2}=100[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{16}=4[\/latex] because [latex]{4}^{2}=16[\/latex]<\/li>\r\n \t<li>Square roots act as grouping symbols in the order of operations, so addition and subtraction must be performed first when they occur under a radical. [latex]\\sqrt{25+144}=\\sqrt{169}=13[\/latex] because [latex]{13}^{2}=169[\/latex].<\/li>\r\n \t<li>This problem is similar to the last one, but this time subtraction should occur after evaluating the root. Stop and think about why these two problems are different. [latex]\\sqrt{49}-\\sqrt{81}=7 - 9=-2[\/latex] because [latex]{7}^{2}=49[\/latex] and [latex]{9}^{2}=81[\/latex]<\/li>\r\n \t<li>\r\n<p style=\"text-align: left;\">The negative in front means to take the opposite of the value after you simplify the radical. [latex] -\\sqrt{81}\\\\-\\sqrt{9\\cdot 9}[\/latex] The square root of\u00a0[latex]81[\/latex] is\u00a0[latex]9.[\/latex] Then, take the opposite of\u00a0[latex]9[\/latex] to get [latex]\u22129[\/latex]<\/p>\r\n<\/li>\r\n \t<li>We are looking for a number that when it is squared returns [latex]-9[\/latex]. We can try [latex](-3)^2[\/latex], but that will give a positive result, and [latex]3^2[\/latex] will also give a positive result. This leads to an important fact - \u00a0you cannot find the square root of a negative number in the system of real numbers.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe last example we showed reminds us of an important characteristic of square roots. We can only take the square root of values that are non-negative.\r\n<div class=\"textbox shaded\">\r\n<h3>SQUARE ROOT OF A NEGATIVE NUMBER<\/h3>\r\nThe square root of a negative number is undefined in the set of real numbers.\r\n\r\nThere is no real number that when squared gives a negative number.\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nSimplify:\r\n<ol>\r\n \t<li>[latex]\\sqrt{49}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{196}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{169-144}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{81}-\\sqrt{36}[\/latex]<\/li>\r\n \t<li>[latex] -\\sqrt{100}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{-4}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"557281\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"557281\"]\r\n<ol>\r\n \t<li>[latex]\\sqrt{49}=7[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{196}=14[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{169-144}=5[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{36}-\\sqrt{81}=-3[\/latex]<\/li>\r\n \t<li>[latex] -\\sqrt{100}=-10[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{-4}[\/latex] is undefined in the set of real numbers<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\nIn the following video, we present more examples of how to find a principal square root.\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/2cWAkmJoaDQ\r\n\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nDoes [latex]\\sqrt{25}=\\pm 5[\/latex]? Write your ideas and a sentence to defend them in the box below before you look at the answer.\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"101071\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"101071\"]\r\n\r\n<em>No. Although both<\/em> [latex]{5}^{2}[\/latex] <em>and<\/em> [latex]{\\left(-5\\right)}^{2}[\/latex] <em>are<\/em> [latex]25[\/latex], <em>the radical symbol implies only a nonnegative root, the principal square root. The principal square root of 25 is<\/em> [latex]\\sqrt{25}=5[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Simplifying Square Roots<\/h3>\r\nSo far, we have seen examples that are perfect squares. That is, each is a number whose square root is a whole number. But many radical expressions are not perfect squares. Some of these radicals can still be simplified by finding <em><strong>perfect square factors<\/strong><\/em>. The example below illustrates how to factor the radicand, looking for pairs of factors that can be expressed as a square.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex] \\sqrt{63}[\/latex]\r\n\r\n[reveal-answer q=\"908978\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"908978\"]Factor [latex]63[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\sqrt{3\\cdot3\\cdot7}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Regroup factors into squares<\/p>\r\n<p style=\"text-align: center;\">[latex]\\sqrt{3^2\\cdot7}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Finish simplifying with the product rule for roots, and the square root of a perfect square idea.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}&amp;\\sqrt{3^2\\cdot7}\\\\\\\\&amp;=\\sqrt{3^2}\\cdot\\sqrt{7}\\\\\\\\&amp;=3\\sqrt{7}\\end{aligned}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Since 7 is prime and we can't write it as a square, it will have to stay under the radical sign. As a matter of convention, we write the constant, [latex]3[\/latex], in front of the radical. \u00a0This helps the reader know that the \u00a0[latex]3[\/latex] is not under the radical anymore.<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\sqrt{63}=3\\sqrt{7}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe final answer [latex] 3\\sqrt{7}[\/latex] may look a bit odd, but it is in simplified form. We read this as \u201cthree radical seven\u201d or \u201cthree times the square root of seven.\u201d\r\n\r\nIn the next example, we make use of the common squares we know, instead of using prime factors. It helps to have the squares of the numbers between [latex]0[\/latex] and [latex]10[\/latex] fresh in your mind to make simplifying radicals faster.\r\n<ul>\r\n \t<li style=\"text-align: left;\">[latex]0^2=0[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]2^2=4[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]3^2=9[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]4^2=16[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]5^2=25[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]6^2=36[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]7^2=49[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]8^2=64[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]9^2=81[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]10^2=100[\/latex]<\/li>\r\n<\/ul>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex] \\sqrt{2,000}[\/latex]\r\n\r\n[reveal-answer q=\"932245\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"932245\"]Factor [latex]2,000[\/latex] to find perfect squares.\r\n\r\n[latex] \\begin{array}{r}\\sqrt{100\\cdot 20}\\\\\\\\=\\sqrt{100\\cdot 4\\cdot 5}\\end{array}[\/latex]\r\n\r\n[latex]100=10^2,4=2^2[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}\\sqrt{100\\cdot 4\\cdot 5}\\\\\\\\=\\sqrt{10^2\\cdot 2^2\\cdot 5}\\\\\\\\=\\sqrt{10^2}\\cdot \\sqrt{2^2}\\cdot \\sqrt{5}\\\\\\\\=10\\cdot 2\\cdot \\sqrt{5}\\end{array}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nMultiply.\r\n<p style=\"text-align: center;\">[latex] 20\\cdot \\sqrt{5}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\sqrt{2,000}=20\\sqrt{5}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn this last video, we show examples of simplifying radicals that are not perfect squares.\r\n\r\nhttps:\/\/youtu.be\/oRd7aBCsmfU\r\n<div class=\"textbox examples\">\r\n<h3>Examples<\/h3>\r\nSimplify the square roots:\r\n<ol>\r\n \t<li>[latex]\\sqrt{44}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{54}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{700}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"HM0656\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"HM0656\"]\r\n<ol>\r\n \t<li>[latex]\\sqrt{44}=\\sqrt{4\\cdot 11}=\\sqrt{4}\\sqrt{11}=2\\sqrt{11}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{54}=\\sqrt{9\\cdot 6}=\\sqrt{9}\\cdot\\sqrt{6}=3\\sqrt{6}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{700}=\\sqrt{100\\cdot 7}=\\sqrt{100}\\cdot\\sqrt{7}=10\\sqrt{7}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nSimplify the square roots:\r\n<ol>\r\n \t<li>[latex]\\sqrt{20}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{75}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{500}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"HM1164\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"HM1164\"]\r\n<ol>\r\n \t<li>[latex]\\sqrt{20}=\\sqrt{4\\cdot 5}=\\sqrt{4}\\cdot\\sqrt{5}=2\\sqrt{5}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{75}=\\sqrt{25}\\cdot 3\\sqrt{25}\\cdot\\sqrt{3}=5\\sqrt{3}[\/latex]<\/li>\r\n \t<li>[latex]\\sqrt{500}=\\sqrt{100\\cdot 5}=\\sqrt{100}\\cdot\\sqrt{5}=10\\sqrt{5}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Decimal Approximations<\/h2>\r\nAll of the radicals we have simplified have resulted in whole number answers or have natural numbers left under the radical after simplifying. These are all <em><strong>exact values<\/strong><\/em> of square roots. However, it may not be necessary or convenient to have an exact answer. If we were to ask a carpenter to cut a board [latex]7\\sqrt{3}[\/latex] inches long, they would tell us to take a hike. Consequently, to get an approximation we can use the radical button on a calculator.\r\n\r\nHow we enter a square root on a calculator depends on the calculator. But it is usually the\u00a0[latex]\\sqrt{\\hphantom{5}}[\/latex] button followed by [latex]3[\/latex] or\u00a0[latex]3[\/latex] followed by the\u00a0[latex]\\sqrt{\\hphantom{5}}[\/latex] button.\r\n\r\nIn the free online Desmos calculator shown here, it is\u00a0the\u00a0[latex]\\sqrt{\\hphantom{5}}[\/latex] button followed by [latex]3[\/latex]. We can also calculate\u00a0[latex]7\\sqrt{3}[\/latex] by using the 7 button, the times button, the square root button and the three button. The results are shown as\u00a0[latex]\\sqrt{3}=1.732050808[\/latex] and\u00a0[latex]7\\sqrt{3}=12.12435565[\/latex]. Both of these are <em><strong>approximations<\/strong><\/em>. The exact values are are<em><strong> irrational numbers<\/strong><\/em>: decimals that never end and never repeat. We can round these decimals to whatever place value is asked for or makes sense. In the case of the board that needs to be cut to\u00a0[latex]7\\sqrt{3}[\/latex] inches, rounding to the tenths place is sufficient:\u00a0[latex]7\\sqrt{3}=12.1[\/latex].\r\n\r\n<img class=\"aligncenter wp-image-1713 size-large\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5676\/2021\/06\/14005147\/Desmos-square-roots-1024x594.png\" alt=\"Desmos calculator image\" width=\"1024\" height=\"594\" \/><img \/>\r\n\r\nRounding a decimal to a certain place value requires us to identify where in the decimal number that place value lies, then looking at the digit immediately after it to decide whether to leave the place value digit alone or to increase it by one.\r\n<div class=\"textbox shaded\">\r\n<h3>Round a decimal.<\/h3>\r\n<ol id=\"eip-id1168466730764\" class=\"stepwise\">\r\n \t<li>Locate the given place value and mark it with an arrow.<\/li>\r\n \t<li>Underline the digit to the right of the given place value.<\/li>\r\n \t<li>Is this digit greater than or equal to [latex]5?[\/latex]\r\n<ul id=\"fs-id1367462\">\r\n \t<li>Yes - add [latex]1[\/latex] to the digit in the given place value.<\/li>\r\n \t<li>No - do not change the digit in the given place value<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Rewrite the number, removing all digits to the right of the given place value.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nRound [latex]18.379[\/latex] to the nearest\r\n<ol>\r\n \t<li style=\"list-style-type: none;\">\r\n<ol>\r\n \t<li>tenth<\/li>\r\n \t<li>whole number<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"414963\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"414963\"]\r\n\r\nSolution\r\n<table id=\"eip-id1168467479198\" class=\"unnumbered unstyled\" summary=\"The number 18.379 is shown. The first line says, \">\r\n<tbody>\r\n<tr>\r\n<td>1. Round [latex]18.379[\/latex] to the nearest tenth.<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]18.379[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Locate the tenths place and mark it with an arrow.<\/td>\r\n<td><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221520\/CNX_BMath_Figure_05_01_022_img-02.png\" alt=\".\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Underline the digit to the right of the tenths digit.<\/td>\r\n<td><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221521\/CNX_BMath_Figure_05_01_022_img-03.png\" alt=\".\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Because [latex]7[\/latex] is greater than or equal to [latex]5[\/latex], add [latex]1[\/latex] to the [latex]3[\/latex].<\/td>\r\n<td><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221523\/CNX_BMath_Figure_05_01_022_img-04.png\" alt=\".\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rewrite the number, deleting all digits to the right of the tenths place.<\/td>\r\n<td>[latex]18.4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>So, [latex]18.379[\/latex] rounded to the nearest tenth is [latex]18.4[\/latex].<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<table id=\"eip-id1168466047205\" class=\"unnumbered unstyled\" summary=\"The number 18.379 is shown. The first line says, \">\r\n<tbody>\r\n<tr>\r\n<td>2. Round [latex]18.379[\/latex] to the nearest whole number.<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]18.379[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Locate the ones place and mark it with an arrow.<\/td>\r\n<td><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221526\/CNX_BMath_Figure_05_01_023_img-02.png\" alt=\".\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Underline the digit to the right of the ones place.<\/td>\r\n<td><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221527\/CNX_BMath_Figure_05_01_023_img-03.png\" alt=\".\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Since [latex]3[\/latex] is not greater than or equal to [latex]5[\/latex], do not add [latex]1[\/latex] to the [latex]8[\/latex].<\/td>\r\n<td><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221528\/CNX_BMath_Figure_05_01_023_img-04.png\" alt=\".\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rewrite the number, deleting all digits to the right of the ones place.<\/td>\r\n<td>[latex]18[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>So [latex]18.379[\/latex] rounded to the nearest whole number is [latex]18[\/latex].<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]146244[\/ohm_question]\r\n\r\n<\/div>\r\nWatch the following video to see an example of how to round a number to several different place values.\r\n\r\nhttps:\/\/youtu.be\/qu4Y9DGqXlk\r\n<div class=\"textbox examples\">\r\n<h3>Examples<\/h3>\r\nUse a calculator to find the following square roots:\r\n\r\nSimplify the square roots:\r\n<ol>\r\n \t<li>[latex]\\sqrt{17}[\/latex]\u00a0 \u00a0 Round to 4 decimal places.<\/li>\r\n \t<li>[latex]\\sqrt{68}[\/latex]\u00a0 \u00a0 Round to 2 decimal places.<\/li>\r\n \t<li>[latex]\\sqrt{547}[\/latex]\u00a0 \u00a0 Round to 3 decimal places.<\/li>\r\n \t<li>[latex]\\sqrt{-75}[\/latex]\u00a0 \u00a0 Round to 2 decimal places.<\/li>\r\n<\/ol>\r\n<h4>Solution<\/h4>\r\n<ol>\r\n \t<li>[latex]\\sqrt{17} = 4.123056... 4.1231[\/latex]<\/li>\r\n \t<li><span style=\"font-size: 1rem; orphans: 1; text-align: initial; widows: 2;\">[latex]\\sqrt{68}=8.246211...\\approx 8.25[\/latex]<\/span><\/li>\r\n \t<li><span style=\"font-size: 1rem; orphans: 1; text-align: initial; widows: 2;\">[latex]\\sqrt{547}=23.388031...\\approx 23.388[\/latex]<\/span><\/li>\r\n \t<li>[latex]\\sqrt{-75}[\/latex] is undefined in the set of real numbers. Calculator gives an error message.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nUse a calculator to find the following square roots:\r\n<ol>\r\n \t<li>[latex]\\sqrt{2}[\/latex]. Round to 3 decimal places.<\/li>\r\n \t<li><span style=\"font-size: 1rem; orphans: 1; text-align: initial; widows: 2;\">[latex]\\sqrt{90}[\/latex]. Round to 2 decimal places.<\/span><\/li>\r\n \t<li><span style=\"font-size: 1rem; orphans: 1; text-align: initial; widows: 2;\">[latex]\\sqrt{436}[\/latex]. Round to 3 decimal places.<\/span><\/li>\r\n<\/ol>\r\n[reveal-answer q=\"759987\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"759987\"]\r\n<ol>\r\n \t<li>[latex]\\sqrt{2}\\approx 1.141[\/latex]. Round to 3 decimal places.<\/li>\r\n \t<li>[latex]\\sqrt{90}\\approx 9.49[\/latex]. Round to 2 decimal places.<\/li>\r\n \t<li>[latex]\\sqrt{436}\\approx 20.881[\/latex]. Round to 3 decimal places.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Summary<\/h2>\r\nThe square root of a number is the number which, when multiplied by itself, gives the original number. Principal square roots are always positive and the square root of [latex]0[\/latex] is [latex]0[\/latex]. You can only take the square root of values that are nonnegative. The square root of a perfect square will be a whole number. Other square roots can be simplified by identifying factors that are perfect squares and taking their square root.\u00a0 Square roots that are not perfect can also be estimated by using a calculator and rounding to an appropriate place value.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Define square roots<\/li>\n<li>Calculate the square root of a perfect square<\/li>\n<li>Simplify a square root that contains perfect square factors<\/li>\n<li>Estimate square roots using a calculator<\/li>\n<li>Round a decimal number<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key words<\/h3>\n<ul>\n<li><strong>Perfect square<\/strong>:\u00a0any whole number that has been squared<\/li>\n<li><strong>Square root<\/strong>:\u00a0a number that when multiplied by itself gives the original number<\/li>\n<li><strong>Radical<\/strong>: the symbol for square root<\/li>\n<li><strong>Principal square root<\/strong>:\u00a0the positive square root<\/li>\n<li><strong>Radicand<\/strong>: the number under a radical sign<\/li>\n<\/ul>\n<\/div>\n<p>When we are trying to find the square root of a number (say,\u00a0[latex]25[\/latex]), we are trying to find a number that when multiplied by itself gives that original number. In the case of\u00a0[latex]25[\/latex], we find that [latex]5\\cdot5=25[\/latex], so\u00a0[latex]5[\/latex] is the square root of [latex]25[\/latex].<\/p>\n<p>The square root is the inverse of the square (exponent of 2), much like multiplication is the inverse of division. A\u00a0<em><strong>perfect square<\/strong><\/em> is any whole number that has been squared. Consequently, all perfect squares have <em><strong>square roots<\/strong><\/em> that are whole numbers.<\/p>\n<p>The symbol for square root is the\u00a0<em><strong>radical sign<\/strong><\/em> [latex]\\sqrt{\\hphantom{5}}[\/latex]. So [latex]\\sqrt{25}=5[\/latex]. The number under the radical sign is called the <em><strong>radicand<\/strong><\/em>. Because we are &#8220;unsquaring&#8221; a number when we find a square root, the radicand must be positive. This is because there is no real number that multiplies by itself to result in a negative number.<\/p>\n<div class=\"textbox examples\">\n<h3>Examples<\/h3>\n<p>Find the square root of the following numbers:<\/p>\n<ol>\n<li>[latex]36[\/latex]<\/li>\n<li>[latex]81[\/latex]<\/li>\n<li>[latex]-49[\/latex]<\/li>\n<li>[latex]0[\/latex]<\/li>\n<\/ol>\n<p>Solution<\/p>\n<ol>\n<li>[latex]\\sqrt{36}=6[\/latex], since [latex]6^2=36[\/latex]<\/li>\n<li>[latex]\\sqrt{81}=9[\/latex], since [latex]9^2=81[\/latex]<\/li>\n<li>[latex]\\sqrt{-49}[\/latex] is undefined, since there is no real number that squares to -49.<\/li>\n<li>[latex]\\sqrt{0}=0[\/latex], since [latex]0^2=0[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Find the square root of the following numbers:<\/p>\n<ol>\n<li>[latex]49[\/latex]<\/li>\n<li>[latex]64[\/latex]<\/li>\n<li>[latex]1[\/latex]<\/li>\n<li>[latex]-25[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm854\">Show Answer<\/span><\/p>\n<div id=\"qhjm854\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\sqrt{49}=7[\/latex]<\/li>\n<li>[latex]\\sqrt{64}=8[\/latex]<\/li>\n<li>[latex]\\sqrt{1}=1[\/latex]<\/li>\n<li>[latex]\\sqrt{-25}[\/latex] is undefined in the set of real numbers<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"font-size: 1rem; text-align: initial;\">Consider [latex]\\sqrt{25}[\/latex] again. If we think about it, there is another number that, when multiplied by itself, also results in\u00a0[latex]25[\/latex]. That number is [latex]\u22125[\/latex].<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}5\\cdot 5=25\\\\-5\\cdot -5=25\\end{array}[\/latex]<\/p>\n<p>By definition, the square root symbol, [latex]\\sqrt{\\hphantom{5}}[\/latex], always means to find the positive root called the <em><strong>principal root<\/strong><\/em>. So while [latex]5\\cdot5[\/latex] and [latex]\u22125\\cdot\u22125[\/latex] both equal\u00a0[latex]25[\/latex], only\u00a0[latex]5[\/latex] is the principal root. If we want to find the negative square root, we must place a negative sign infant of the radical: [latex]-\\sqrt{25}=-5[\/latex]. Zero is special because it has only one square root:\u00a0 [latex]\\sqrt{0}=0[\/latex]).<\/p>\n<p>The notation that we use to express a square root for any real number, [latex]a[\/latex], is as follows:<\/p>\n<div class=\"textbox shaded\">\n<h3>Square Root<\/h3>\n<p>The symbol for the square root is called a <strong>radical symbol.<\/strong>\u00a0For any non-negative, real number [latex]a[\/latex], the square root of <em>a<\/em> is written as [latex]\\sqrt{a}[\/latex].<\/p>\n<p>The number that is written under the radical symbol is called the <strong>radicand<\/strong>.<\/p>\n<p>By definition, the square root symbol, [latex]\\sqrt{\\hphantom{5}}[\/latex] always means to find the nonnegative\u00a0root, called the <strong>principal root<\/strong>.<\/p>\n<p>[latex]\\sqrt{-a}[\/latex] is undefined in the set of real numbers.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Simplify\u00a0the following square roots:<\/p>\n<ol>\n<li>[latex]\\sqrt{16}[\/latex]<\/li>\n<li>[latex]\\sqrt{9}[\/latex]<\/li>\n<li>[latex]\\sqrt{-9}[\/latex]<\/li>\n<li>[latex]\\sqrt{5^2}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q24327\">Show Answer<\/span><\/p>\n<div id=\"q24327\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\sqrt{16}=4[\/latex].<\/li>\n<li>[latex]\\sqrt{9}=3[\/latex].<\/li>\n<li>[latex]\\sqrt{-9}[\/latex] is undefined in the set of real numbers.<\/li>\n<li>[latex]\\sqrt{5^2}=\\sqrt{25}=5[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>The last problem in the previous example shows us an important relationship between squares and square roots.<\/p>\n<div class=\"textbox shaded\">\n<h3>The square root of a perfect square<\/h3>\n<p>For any nonnegative real number, [latex]a[\/latex], \u00a0[latex]\\sqrt{a^2}=\\big |\\,a\\,\\big |[\/latex].<\/p>\n<\/div>\n<p>Remember that\u00a0\u00a0[latex]\\big |\\,a\\,\\big |[\/latex] represents the distance of the number [latex]a[\/latex] from zero. \u00a0[latex]\\big |\\,a\\,\\big | \\ge 0[\/latex]. Consequently, [latex]\\sqrt{6^2}=\\big |\\,6\\,\\big |=6[\/latex] and\u00a0[latex]\\sqrt{(-6)^2}=\\big |\\,-6\\,\\big |=6[\/latex].<\/p>\n<p>In the video that follows, we simplify\u00a0more square roots using the fact that [latex]\\sqrt{a^2}=\\big |\\,a\\,\\big |[\/latex] means finding the principal square root.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Simplify Square Roots (Perfect Square Radicands)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/B3riJsl7uZM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm228\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=228&theme=oea&iframe_resize_id=ohm228&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h3>Finding Square Roots Using Factoring<\/h3>\n<p>We have previously considered the square root of a fraction. For example, [latex]\\sqrt\\frac{4}{9}=\\frac{2}{3}[\/latex] because [latex]\\frac{2}{3}\\cdot\\frac{2}{3}=\\frac{4}{9}[\/latex]. We also discovered the <em><strong>Quotient Rule for Square Roots<\/strong><\/em> and were able to simply the square root of a fraction by taking the square root of the numerator and denominator separately. For example,\u00a0 [latex]\\sqrt\\frac{4}{9}=\\frac{\\sqrt{4}}{\\sqrt{9}}=\\frac{2}{3}[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>The QUOTIENT Rule for Square Roots<\/h3>\n<p>For any nonnegative, real number [latex]a[\/latex] and positive real number [latex]b[\/latex], \u00a0[latex]\\sqrt{\\frac{a}{b}}=\\frac{\\sqrt{a}}{\\sqrt{b}}[\/latex].<\/p>\n<\/div>\n<p>A similar rule applies to products. After all, division is just multiplication by the reciprocal, so the quotient rule could be considered a product rule of the reciprocal:\u00a0[latex]\\sqrt{\\frac{a}{b}}=\\sqrt{a\\cdot\\frac{1}{b}}[\/latex] and [latex]\\frac{\\sqrt{a}}{\\sqrt{b}}=\\sqrt{a}\\cdot\\frac{1}{\\sqrt{b}}[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>The Product Rule for Square Roots<\/h3>\n<p>For any nonnegative, real numbers\u00a0[latex]a[\/latex] and [latex]b[\/latex], \u00a0[latex]\\sqrt{a\\cdot{b}}=\\sqrt{a}\\cdot\\sqrt{b}[\/latex].<\/p>\n<\/div>\n<p>This is useful if we are working with a number whose square we do not know right away. \u00a0We can use factoring and the <em><strong>product rule for square roots<\/strong><\/em> to find square roots such as [latex]\\sqrt{144}[\/latex], or\u00a0\u00a0[latex]\\sqrt{225}[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify\u00a0[latex]\\sqrt{144}[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q620082\">Show Solution<\/span><\/p>\n<div id=\"q620082\" class=\"hidden-answer\" style=\"display: none\">\n<p>Determine the prime factors of [latex]144[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{144}\\\\\\\\\\sqrt{2\\cdot 72}\\\\\\\\\\sqrt{2\\cdot 2\\cdot 36}\\\\\\\\\\sqrt{2\\cdot 2\\cdot 2\\cdot 18}\\\\\\\\\\sqrt{2\\cdot 2\\cdot 2\\cdot 2\\cdot 9}\\\\\\\\\\sqrt{2\\cdot 2\\cdot 2\\cdot 2\\cdot 3\\cdot 3}\\end{array}[\/latex]<\/p>\n<p>Because we are finding a square root, we regroup these factors into squares.<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{2^2\\cdot 2^2\\cdot3^2}[\/latex]<\/p>\n<p>Now we can use the product rule for square roots and the square root of a square idea to finish finding the square root.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{2^2\\cdot 2^2\\cdot3^2}\\\\\\\\=\\sqrt{2^2}\\cdot\\sqrt{2^2}\\cdot\\sqrt{3^2}\\\\\\\\=2\\cdot 2 \\cdot 3\\\\\\\\=12\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\sqrt{144}=12[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Simplify [latex]\\sqrt{225}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q686109\">Show Solution<\/span><\/p>\n<div id=\"q686109\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, factor [latex]225[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{225}\\\\\\\\=\\sqrt{5\\cdot45}\\\\\\\\=\\sqrt{5\\cdot5\\cdot9}\\\\\\\\=\\sqrt{5\\cdot5\\cdot3\\cdot3}\\end{array}[\/latex]<\/p>\n<p>Because we are finding a square root, we regroup these factors into squares. Finish simplifying with the product rule for roots, and the square of a square idea.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{5^2\\cdot3^2}\\\\\\\\=\\sqrt{5^2}\\cdot\\sqrt{3^2}\\\\\\\\=5\\cdot3=15\\end{array}[\/latex]<\/p>\n<h4 style=\"text-align: left;\">Answer<\/h4>\n<p>[latex]\\sqrt{225}=15[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"49\" height=\"43\" \/>Caution! \u00a0The square root of a product rule applies\u00a0when you have multiplication ONLY under the square root. You cannot apply the rule to sums or differences:<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{a+b}\\ne\\sqrt{a}+\\sqrt{b}[\/latex]<\/p>\n<p style=\"text-align: left;\">Prove this to yourself with some real numbers: let [latex]a = 64[\/latex] and [latex]b = 36[\/latex], then use the order of operations to simplify each expression.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\sqrt{64+36}=\\sqrt{100}=10\\\\\\\\\\sqrt{64}+\\sqrt{36}=8+6=14\\\\\\\\10\\ne14\\end{array}[\/latex]<\/p>\n<\/div>\n<p>Let&#8217;s look at some more examples of expressions with square roots.\u00a0 Pay particular attention to how number 3 is evaluated.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Simplify:<\/p>\n<ol>\n<li>[latex]\\sqrt{100}[\/latex]<\/li>\n<li>[latex]\\sqrt{16}[\/latex]<\/li>\n<li>[latex]\\sqrt{25+144}[\/latex]<\/li>\n<li>[latex]\\sqrt{49}-\\sqrt{81}[\/latex]<\/li>\n<li>[latex]-\\sqrt{81}[\/latex]<\/li>\n<li>[latex]\\sqrt{-9}[\/latex]<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q419579\">Show Solution<\/span><\/p>\n<div id=\"q419579\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\sqrt{100}=10[\/latex] because [latex]{10}^{2}=100[\/latex]<\/li>\n<li>[latex]\\sqrt{16}=4[\/latex] because [latex]{4}^{2}=16[\/latex]<\/li>\n<li>Square roots act as grouping symbols in the order of operations, so addition and subtraction must be performed first when they occur under a radical. [latex]\\sqrt{25+144}=\\sqrt{169}=13[\/latex] because [latex]{13}^{2}=169[\/latex].<\/li>\n<li>This problem is similar to the last one, but this time subtraction should occur after evaluating the root. Stop and think about why these two problems are different. [latex]\\sqrt{49}-\\sqrt{81}=7 - 9=-2[\/latex] because [latex]{7}^{2}=49[\/latex] and [latex]{9}^{2}=81[\/latex]<\/li>\n<li>\n<p style=\"text-align: left;\">The negative in front means to take the opposite of the value after you simplify the radical. [latex]-\\sqrt{81}\\\\-\\sqrt{9\\cdot 9}[\/latex] The square root of\u00a0[latex]81[\/latex] is\u00a0[latex]9.[\/latex] Then, take the opposite of\u00a0[latex]9[\/latex] to get [latex]\u22129[\/latex]<\/p>\n<\/li>\n<li>We are looking for a number that when it is squared returns [latex]-9[\/latex]. We can try [latex](-3)^2[\/latex], but that will give a positive result, and [latex]3^2[\/latex] will also give a positive result. This leads to an important fact &#8211; \u00a0you cannot find the square root of a negative number in the system of real numbers.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>The last example we showed reminds us of an important characteristic of square roots. We can only take the square root of values that are non-negative.<\/p>\n<div class=\"textbox shaded\">\n<h3>SQUARE ROOT OF A NEGATIVE NUMBER<\/h3>\n<p>The square root of a negative number is undefined in the set of real numbers.<\/p>\n<p>There is no real number that when squared gives a negative number.<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Simplify:<\/p>\n<ol>\n<li>[latex]\\sqrt{49}[\/latex]<\/li>\n<li>[latex]\\sqrt{196}[\/latex]<\/li>\n<li>[latex]\\sqrt{169-144}[\/latex]<\/li>\n<li>[latex]\\sqrt{81}-\\sqrt{36}[\/latex]<\/li>\n<li>[latex]-\\sqrt{100}[\/latex]<\/li>\n<li>[latex]\\sqrt{-4}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q557281\">Show Answer<\/span><\/p>\n<div id=\"q557281\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\sqrt{49}=7[\/latex]<\/li>\n<li>[latex]\\sqrt{196}=14[\/latex]<\/li>\n<li>[latex]\\sqrt{169-144}=5[\/latex]<\/li>\n<li>[latex]\\sqrt{36}-\\sqrt{81}=-3[\/latex]<\/li>\n<li>[latex]-\\sqrt{100}=-10[\/latex]<\/li>\n<li>[latex]\\sqrt{-4}[\/latex] is undefined in the set of real numbers<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In the following video, we present more examples of how to find a principal square root.<\/p>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Simplify a Variety of Square Expressions (Simplify Perfectly)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/2cWAkmJoaDQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Does [latex]\\sqrt{25}=\\pm 5[\/latex]? Write your ideas and a sentence to defend them in the box below before you look at the answer.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q101071\">Show Solution<\/span><\/p>\n<div id=\"q101071\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>No. Although both<\/em> [latex]{5}^{2}[\/latex] <em>and<\/em> [latex]{\\left(-5\\right)}^{2}[\/latex] <em>are<\/em> [latex]25[\/latex], <em>the radical symbol implies only a nonnegative root, the principal square root. The principal square root of 25 is<\/em> [latex]\\sqrt{25}=5[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Simplifying Square Roots<\/h3>\n<p>So far, we have seen examples that are perfect squares. That is, each is a number whose square root is a whole number. But many radical expressions are not perfect squares. Some of these radicals can still be simplified by finding <em><strong>perfect square factors<\/strong><\/em>. The example below illustrates how to factor the radicand, looking for pairs of factors that can be expressed as a square.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\sqrt{63}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q908978\">Show Solution<\/span><\/p>\n<div id=\"q908978\" class=\"hidden-answer\" style=\"display: none\">Factor [latex]63[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{3\\cdot3\\cdot7}[\/latex]<\/p>\n<p style=\"text-align: left;\">Regroup factors into squares<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{3^2\\cdot7}[\/latex]<\/p>\n<p style=\"text-align: left;\">Finish simplifying with the product rule for roots, and the square root of a perfect square idea.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}&\\sqrt{3^2\\cdot7}\\\\\\\\&=\\sqrt{3^2}\\cdot\\sqrt{7}\\\\\\\\&=3\\sqrt{7}\\end{aligned}[\/latex]<\/p>\n<p style=\"text-align: left;\">Since 7 is prime and we can&#8217;t write it as a square, it will have to stay under the radical sign. As a matter of convention, we write the constant, [latex]3[\/latex], in front of the radical. \u00a0This helps the reader know that the \u00a0[latex]3[\/latex] is not under the radical anymore.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\sqrt{63}=3\\sqrt{7}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The final answer [latex]3\\sqrt{7}[\/latex] may look a bit odd, but it is in simplified form. We read this as \u201cthree radical seven\u201d or \u201cthree times the square root of seven.\u201d<\/p>\n<p>In the next example, we make use of the common squares we know, instead of using prime factors. It helps to have the squares of the numbers between [latex]0[\/latex] and [latex]10[\/latex] fresh in your mind to make simplifying radicals faster.<\/p>\n<ul>\n<li style=\"text-align: left;\">[latex]0^2=0[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]2^2=4[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]3^2=9[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]4^2=16[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]5^2=25[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]6^2=36[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]7^2=49[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]8^2=64[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]9^2=81[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]10^2=100[\/latex]<\/li>\n<\/ul>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\sqrt{2,000}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q932245\">Show Solution<\/span><\/p>\n<div id=\"q932245\" class=\"hidden-answer\" style=\"display: none\">Factor [latex]2,000[\/latex] to find perfect squares.<\/p>\n<p>[latex]\\begin{array}{r}\\sqrt{100\\cdot 20}\\\\\\\\=\\sqrt{100\\cdot 4\\cdot 5}\\end{array}[\/latex]<\/p>\n<p>[latex]100=10^2,4=2^2[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}\\sqrt{100\\cdot 4\\cdot 5}\\\\\\\\=\\sqrt{10^2\\cdot 2^2\\cdot 5}\\\\\\\\=\\sqrt{10^2}\\cdot \\sqrt{2^2}\\cdot \\sqrt{5}\\\\\\\\=10\\cdot 2\\cdot \\sqrt{5}\\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Multiply.<\/p>\n<p style=\"text-align: center;\">[latex]20\\cdot \\sqrt{5}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\sqrt{2,000}=20\\sqrt{5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In this last video, we show examples of simplifying radicals that are not perfect squares.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Simplify Square Roots (Not Perfect Square Radicands)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/oRd7aBCsmfU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox examples\">\n<h3>Examples<\/h3>\n<p>Simplify the square roots:<\/p>\n<ol>\n<li>[latex]\\sqrt{44}[\/latex]<\/li>\n<li>[latex]\\sqrt{54}[\/latex]<\/li>\n<li>[latex]\\sqrt{700}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qHM0656\">Show Answer<\/span><\/p>\n<div id=\"qHM0656\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\sqrt{44}=\\sqrt{4\\cdot 11}=\\sqrt{4}\\sqrt{11}=2\\sqrt{11}[\/latex]<\/li>\n<li>[latex]\\sqrt{54}=\\sqrt{9\\cdot 6}=\\sqrt{9}\\cdot\\sqrt{6}=3\\sqrt{6}[\/latex]<\/li>\n<li>[latex]\\sqrt{700}=\\sqrt{100\\cdot 7}=\\sqrt{100}\\cdot\\sqrt{7}=10\\sqrt{7}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Simplify the square roots:<\/p>\n<ol>\n<li>[latex]\\sqrt{20}[\/latex]<\/li>\n<li>[latex]\\sqrt{75}[\/latex]<\/li>\n<li>[latex]\\sqrt{500}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qHM1164\">Show Answer<\/span><\/p>\n<div id=\"qHM1164\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\sqrt{20}=\\sqrt{4\\cdot 5}=\\sqrt{4}\\cdot\\sqrt{5}=2\\sqrt{5}[\/latex]<\/li>\n<li>[latex]\\sqrt{75}=\\sqrt{25}\\cdot 3\\sqrt{25}\\cdot\\sqrt{3}=5\\sqrt{3}[\/latex]<\/li>\n<li>[latex]\\sqrt{500}=\\sqrt{100\\cdot 5}=\\sqrt{100}\\cdot\\sqrt{5}=10\\sqrt{5}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>Decimal Approximations<\/h2>\n<p>All of the radicals we have simplified have resulted in whole number answers or have natural numbers left under the radical after simplifying. These are all <em><strong>exact values<\/strong><\/em> of square roots. However, it may not be necessary or convenient to have an exact answer. If we were to ask a carpenter to cut a board [latex]7\\sqrt{3}[\/latex] inches long, they would tell us to take a hike. Consequently, to get an approximation we can use the radical button on a calculator.<\/p>\n<p>How we enter a square root on a calculator depends on the calculator. But it is usually the\u00a0[latex]\\sqrt{\\hphantom{5}}[\/latex] button followed by [latex]3[\/latex] or\u00a0[latex]3[\/latex] followed by the\u00a0[latex]\\sqrt{\\hphantom{5}}[\/latex] button.<\/p>\n<p>In the free online Desmos calculator shown here, it is\u00a0the\u00a0[latex]\\sqrt{\\hphantom{5}}[\/latex] button followed by [latex]3[\/latex]. We can also calculate\u00a0[latex]7\\sqrt{3}[\/latex] by using the 7 button, the times button, the square root button and the three button. The results are shown as\u00a0[latex]\\sqrt{3}=1.732050808[\/latex] and\u00a0[latex]7\\sqrt{3}=12.12435565[\/latex]. Both of these are <em><strong>approximations<\/strong><\/em>. The exact values are are<em><strong> irrational numbers<\/strong><\/em>: decimals that never end and never repeat. We can round these decimals to whatever place value is asked for or makes sense. In the case of the board that needs to be cut to\u00a0[latex]7\\sqrt{3}[\/latex] inches, rounding to the tenths place is sufficient:\u00a0[latex]7\\sqrt{3}=12.1[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1713 size-large\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5676\/2021\/06\/14005147\/Desmos-square-roots-1024x594.png\" alt=\"Desmos calculator image\" width=\"1024\" height=\"594\" \/><img decoding=\"async\" src=\"src\" alt=\"image\" \/><\/p>\n<p>Rounding a decimal to a certain place value requires us to identify where in the decimal number that place value lies, then looking at the digit immediately after it to decide whether to leave the place value digit alone or to increase it by one.<\/p>\n<div class=\"textbox shaded\">\n<h3>Round a decimal.<\/h3>\n<ol id=\"eip-id1168466730764\" class=\"stepwise\">\n<li>Locate the given place value and mark it with an arrow.<\/li>\n<li>Underline the digit to the right of the given place value.<\/li>\n<li>Is this digit greater than or equal to [latex]5?[\/latex]\n<ul id=\"fs-id1367462\">\n<li>Yes &#8211; add [latex]1[\/latex] to the digit in the given place value.<\/li>\n<li>No &#8211; do not change the digit in the given place value<\/li>\n<\/ul>\n<\/li>\n<li>Rewrite the number, removing all digits to the right of the given place value.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Round [latex]18.379[\/latex] to the nearest<\/p>\n<ol>\n<li style=\"list-style-type: none;\">\n<ol>\n<li>tenth<\/li>\n<li>whole number<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q414963\">Show Solution<\/span><\/p>\n<div id=\"q414963\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution<\/p>\n<table id=\"eip-id1168467479198\" class=\"unnumbered unstyled\" summary=\"The number 18.379 is shown. The first line says,\">\n<tbody>\n<tr>\n<td>1. Round [latex]18.379[\/latex] to the nearest tenth.<\/td>\n<\/tr>\n<tr>\n<td>[latex]18.379[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Locate the tenths place and mark it with an arrow.<\/td>\n<td><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221520\/CNX_BMath_Figure_05_01_022_img-02.png\" alt=\".\" \/><\/td>\n<\/tr>\n<tr>\n<td>Underline the digit to the right of the tenths digit.<\/td>\n<td><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221521\/CNX_BMath_Figure_05_01_022_img-03.png\" alt=\".\" \/><\/td>\n<\/tr>\n<tr>\n<td>Because [latex]7[\/latex] is greater than or equal to [latex]5[\/latex], add [latex]1[\/latex] to the [latex]3[\/latex].<\/td>\n<td><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221523\/CNX_BMath_Figure_05_01_022_img-04.png\" alt=\".\" \/><\/td>\n<\/tr>\n<tr>\n<td>Rewrite the number, deleting all digits to the right of the tenths place.<\/td>\n<td>[latex]18.4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>So, [latex]18.379[\/latex] rounded to the nearest tenth is [latex]18.4[\/latex].<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table id=\"eip-id1168466047205\" class=\"unnumbered unstyled\" summary=\"The number 18.379 is shown. The first line says,\">\n<tbody>\n<tr>\n<td>2. Round [latex]18.379[\/latex] to the nearest whole number.<\/td>\n<\/tr>\n<tr>\n<td>[latex]18.379[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Locate the ones place and mark it with an arrow.<\/td>\n<td><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221526\/CNX_BMath_Figure_05_01_023_img-02.png\" alt=\".\" \/><\/td>\n<\/tr>\n<tr>\n<td>Underline the digit to the right of the ones place.<\/td>\n<td><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221527\/CNX_BMath_Figure_05_01_023_img-03.png\" alt=\".\" \/><\/td>\n<\/tr>\n<tr>\n<td>Since [latex]3[\/latex] is not greater than or equal to [latex]5[\/latex], do not add [latex]1[\/latex] to the [latex]8[\/latex].<\/td>\n<td><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221528\/CNX_BMath_Figure_05_01_023_img-04.png\" alt=\".\" \/><\/td>\n<\/tr>\n<tr>\n<td>Rewrite the number, deleting all digits to the right of the ones place.<\/td>\n<td>[latex]18[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>So [latex]18.379[\/latex] rounded to the nearest whole number is [latex]18[\/latex].<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146244\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146244&theme=oea&iframe_resize_id=ohm146244&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Watch the following video to see an example of how to round a number to several different place values.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Examples:  Rounding Decimals\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/qu4Y9DGqXlk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox examples\">\n<h3>Examples<\/h3>\n<p>Use a calculator to find the following square roots:<\/p>\n<p>Simplify the square roots:<\/p>\n<ol>\n<li>[latex]\\sqrt{17}[\/latex]\u00a0 \u00a0 Round to 4 decimal places.<\/li>\n<li>[latex]\\sqrt{68}[\/latex]\u00a0 \u00a0 Round to 2 decimal places.<\/li>\n<li>[latex]\\sqrt{547}[\/latex]\u00a0 \u00a0 Round to 3 decimal places.<\/li>\n<li>[latex]\\sqrt{-75}[\/latex]\u00a0 \u00a0 Round to 2 decimal places.<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<ol>\n<li>[latex]\\sqrt{17} = 4.123056... 4.1231[\/latex]<\/li>\n<li><span style=\"font-size: 1rem; orphans: 1; text-align: initial; widows: 2;\">[latex]\\sqrt{68}=8.246211...\\approx 8.25[\/latex]<\/span><\/li>\n<li><span style=\"font-size: 1rem; orphans: 1; text-align: initial; widows: 2;\">[latex]\\sqrt{547}=23.388031...\\approx 23.388[\/latex]<\/span><\/li>\n<li>[latex]\\sqrt{-75}[\/latex] is undefined in the set of real numbers. Calculator gives an error message.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Use a calculator to find the following square roots:<\/p>\n<ol>\n<li>[latex]\\sqrt{2}[\/latex]. Round to 3 decimal places.<\/li>\n<li><span style=\"font-size: 1rem; orphans: 1; text-align: initial; widows: 2;\">[latex]\\sqrt{90}[\/latex]. Round to 2 decimal places.<\/span><\/li>\n<li><span style=\"font-size: 1rem; orphans: 1; text-align: initial; widows: 2;\">[latex]\\sqrt{436}[\/latex]. Round to 3 decimal places.<\/span><\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q759987\">Show Answer<\/span><\/p>\n<div id=\"q759987\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\sqrt{2}\\approx 1.141[\/latex]. Round to 3 decimal places.<\/li>\n<li>[latex]\\sqrt{90}\\approx 9.49[\/latex]. Round to 2 decimal places.<\/li>\n<li>[latex]\\sqrt{436}\\approx 20.881[\/latex]. Round to 3 decimal places.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>Summary<\/h2>\n<p>The square root of a number is the number which, when multiplied by itself, gives the original number. Principal square roots are always positive and the square root of [latex]0[\/latex] is [latex]0[\/latex]. You can only take the square root of values that are nonnegative. The square root of a perfect square will be a whole number. Other square roots can be simplified by identifying factors that are perfect squares and taking their square root.\u00a0 Square roots that are not perfect can also be estimated by using a calculator and rounding to an appropriate place value.<\/p>\n","protected":false},"author":370291,"menu_order":1,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2813","chapter","type-chapter","status-publish","hentry"],"part":2811,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/2813","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/users\/370291"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/2813\/revisions"}],"predecessor-version":[{"id":2815,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/2813\/revisions\/2815"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/parts\/2811"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/2813\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/media?parent=2813"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2813"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/contributor?post=2813"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/license?post=2813"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}