{"id":1054,"date":"2016-02-16T00:17:51","date_gmt":"2016-02-16T00:17:51","guid":{"rendered":"https:\/\/courses.candelalearning.com\/nrocarithmetic\/?post_type=chapter&p=1054"},"modified":"2026-02-01T07:52:27","modified_gmt":"2026-02-01T07:52:27","slug":"7-1-1-roots","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/7-1-1-roots\/","title":{"raw":"7.1: Introduction to Radicals","rendered":"7.1: Introduction to Radicals"},"content":{"raw":"
\r\n

section 7.1 Learning Objectives<\/h3>\r\n7.1: Introduction to Radicals<\/strong>\r\n
    \r\n \t
  • Find the square roots of a perfect square<\/li>\r\n \t
  • Simplify a square root written with the radical symbol\r\n
      \r\n \t
    • When the radicand is a perfect square<\/li>\r\n \t
    • When the radicand is not a perfect square<\/li>\r\n \t
    • When the root has a coefficient<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t
    • Approximate a square root<\/li>\r\n \t
    • Simplify other roots<\/li>\r\n<\/ul>\r\n<\/div>\r\n \r\n

      Find the square roots of a perfect square<\/h2>\r\nWe know how to square a number:\r\n\r\n[latex]5^2=25[\/latex] and [latex]\\left(-5\\right)^2=25[\/latex]\r\n\r\nTaking a square root is the opposite of squaring so we can make these statements:\r\n
        \r\n \t
      • 5 is the nonngeative square root of 25<\/li>\r\n \t
      • -5 is the negative square root of 25<\/li>\r\n<\/ul>\r\n
        \r\n

        Example 1<\/h3>\r\nFind the square roots of the following numbers:\r\n
          \r\n \t
        1. 36<\/li>\r\n \t
        2. 81<\/li>\r\n \t
        3. -49<\/li>\r\n \t
        4. 0<\/li>\r\n<\/ol>\r\n \r\n
            \r\n \t
          1. We want to find a number whose square is 36. [latex]6^2=36[\/latex] therefore, \u00a0the nonnegative square root of 36 is 6 and the negative square root of 36 is -6<\/li>\r\n \t
          2. We want to find a number whose square is 81. [latex]9^2=81[\/latex] therefore, \u00a0the nonnegative square root of 81 is 9 and the negative square root of 81 is -9<\/li>\r\n \t
          3. We want to find a number whose square is -49. When you square a real number, the result is always positive. Stop and think about that for a second.\u00a0A negative number times itself is positive, and a positive number times itself is positive. \u00a0Therefore, -49 does not have square roots.\u00a0 There are no real number solutions to this question.<\/li>\r\n \t
          4. We want to find a number whose square is 0. [latex]0^2=0[\/latex] therefore, \u00a0the nonnegative square root of 0 is 0. \u00a0We do not assign 0 a sign, so it has only one square root, and that is 0.<\/li>\r\n<\/ol>\r\n<\/div>\r\n \r\n

            Simplify a square root - when the radicand is a perfect square<\/h2>\r\nThe notation that we use to express a square root for any real number, a, is as follows:\r\n
            \r\n

            Writing a Square Root<\/h3>\r\nThe symbol for the square root is called a radical symbol.<\/strong>\u00a0For a real number, a<\/em> the square root of a<\/em> is written as [latex]\\sqrt{a}[\/latex]\r\n\r\nThe number that is written under the radical symbol is called the 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 principal root<\/strong>.\r\n\r\n[latex]\\sqrt{-a}[\/latex] is not defined, therefore [latex]\\sqrt{a}[\/latex] is defined for [latex]a\\geq 0[\/latex]\r\n\r\n<\/div>\r\nLet's do an example similar to\u00a0the example from above, this time using square root notation. \u00a0Note that using the square root notation means that you are only finding the principal root - the nonnegative root.\r\n
            \r\n

            Example 2<\/h3>\r\nSimplify\u00a0the following square roots:\r\n
              \r\n \t
            1. [latex]\\sqrt{16}[\/latex]<\/li>\r\n \t
            2. [latex]\\sqrt{9}[\/latex]<\/li>\r\n \t
            3. [latex]\\sqrt{-9}[\/latex]<\/li>\r\n \t
            4. [latex]-\\sqrt{25}[\/latex]<\/li>\r\n \t
            5. [latex]\\sqrt{5^2}[\/latex]<\/li>\r\n \t
            6. [latex]\\sqrt{0.01}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"614386\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"614386\"]\r\n
                \r\n \t
              1. [latex]\\sqrt{16}[\/latex]. We are looking for a number whose square is 16, so\u00a0[latex]\\sqrt{16}=4[\/latex]. We only write the nonnegative root because that is how the root symbol is defined.<\/li>\r\n \t
              2. [latex]\\sqrt{9}[\/latex]. We are looking for a number whose square is 9, so [latex]\\sqrt{9}=3[\/latex].\u00a0We only write the nonnegative root because that is how the root symbol is defined.<\/li>\r\n \t
              3. [latex]\\sqrt{-9}[\/latex]. We are looking for a number whose square is -9. \u00a0There are no real numbers whose square is -9, so this radical is not a real number.<\/li>\r\n \t
              4. [latex]-\\sqrt{25}[\/latex].\u00a0Notice that the minus sign is outside of the radical.\u00a0 That minus sign is asking us to find the opposite of [latex]\\sqrt{25}[\/latex].\u00a0 Since [latex]\\sqrt{25}=5[\/latex], the opposite would be [latex]-5[\/latex]. Therefore, [latex]-\\sqrt{25}=-5[\/latex].<\/li>\r\n \t
              5. [latex]\\sqrt{5^2}[\/latex]. We are looking for a number whose square is [latex]5^2[\/latex]. \u00a0We already have the number whose square is [latex]5^2[\/latex], it's 5!<\/li>\r\n \t
              6. Compared to the other examples, it may be less obvious that [latex]0.01[\/latex] is a perfect square.\u00a0 However, [latex](0.1)(0.1)=0.01[\/latex]. This can also be written as [latex](0.1)^2=0.01[\/latex]. Therefore, [latex]\\sqrt{0.01}=0.1[\/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, and we can summarize it as follows:\r\n
                \r\n

                \u00a0The square root of a square<\/h3>\r\nFor a nonnegative real number, a, [latex]\\sqrt{a^2}=a[\/latex]\r\n\r\n<\/div>\r\nIn the video that follows, we simplify\u00a0more square roots using the fact that\u00a0\u00a0[latex]\\sqrt{a^2}=a[\/latex] means finding the principal square root.\r\n\r\nhttps:\/\/youtu.be\/B3riJsl7uZM\r\n\r\nWhat if you are working with a number whose square you do not know right away? \u00a0We can use factoring and the product rule for square roots to find square roots such as [latex]\\sqrt{144}[\/latex], or\u00a0\u00a0[latex]\\sqrt{225}[\/latex].\r\n
                \r\n

                The Product Rule for Square Roots<\/h3>\r\nGiven that a and b are nonnegative real numbers, [latex]\\sqrt{a\\cdot{b}}=\\sqrt{a}\\cdot\\sqrt{b}[\/latex]\r\n\r\n<\/div>\r\nIn the examples that follow we will bring together these ideas\u00a0to simplify\u00a0square roots of numbers that are not obvious at first glance:\r\n
                  \r\n \t
                • square root of a square<\/li>\r\n \t
                • the product rule for square roots<\/li>\r\n \t
                • factoring<\/li>\r\n<\/ul>\r\n
                  \r\n

                  Example 3<\/h3>\r\nSimplify\u00a0[latex] \\sqrt{144}[\/latex]\r\n\r\n[reveal-answer q=\"620082\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"620082\"]\r\n\r\nYou might recognize immediately that the answer is 12, since [latex]12^2=144[\/latex]. If not, we look at another approach, where we will use the prime factorization to assist us. This approach will be especially important when the radicand is not a perfect square.\r\n\r\nDetermine the prime factors of 144.\r\n

                  [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 group these factors into pairs, which we can then write as squares.\r\n

                  [latex]\\sqrt{{\\color{blue}{2\\cdot 2}}\\cdot {\\color{red}{2\\cdot 2}}\\cdot {\\color{green} {3\\cdot 3}}}[\/latex]<\/p>\r\n

                  [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 \r\n

                  [latex]\\begin{array}{c}\\sqrt{2^2\\cdot 2^2\\cdot3^2}\\\\\\\\=\\sqrt{2^2}\\cdot\\sqrt{2^2}\\cdot\\sqrt{3^2}\\\\\\\\=2\\cdot2\\cdot3\\\\\\\\=12\\end{array}[\/latex]<\/p>\r\n\r\n

                  Answer<\/h4>\r\n[latex] \\sqrt{144}=12[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n
                  \r\n

                  Example 4<\/h3>\r\nSimplify [latex]\\sqrt{225}[\/latex]\r\n[reveal-answer q=\"686109\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"686109\"]\r\n\r\nFirst, factor 225:\r\n

                  [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 look for pairs of factors, which we could then write as squares. Finish simplifying with the product rule for roots, and the square of a square idea.\r\n

                  [latex]\\begin{array}{c}\\sqrt{{\\color{blue}{5\\cdot 5}}\\cdot {\\color{red}{3\\cdot 3}}}\\\\\\\\=\\sqrt{5^2\\cdot3^2}\\\\\\\\=\\sqrt{5^2}\\cdot\\sqrt{3^2}\\\\\\\\=5\\cdot3=15\\end{array}[\/latex]<\/p>\r\n\r\n

                  Answer<\/h4>\r\n[latex]\\sqrt{225}=15[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n
                  \r\n\r\n\"CautionCaution! \u00a0The square root of a product rule applies\u00a0when you have multiplication ONLY under the square root. You cannot apply the rule to sums:\r\n

                  [latex]\\sqrt{a+b}\\ne\\sqrt{a}+\\sqrt{b}[\/latex]<\/p>\r\n

                  Prove this to yourself with some real numbers: let a = 64 and b = 36, then use the order of operations to simplify each expression.<\/p>\r\n

                  [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\n

                  Simplify a square root - when the radicand is not a perfect square<\/h2>\r\nSo far, you have seen examples that are perfect squares. That is, each is a number whose square root is an integer. But many radical expressions are not perfect squares. Some of these radicals can still be simplified by finding perfect square factors. The example below illustrates how to factor the radicand, looking for pairs of factors that can be expressed as a square.\r\n
                  \r\n

                  Example 5<\/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 63\r\n

                  [latex] \\sqrt{7\\cdot {\\color{red}{3\\cdot3}}}[\/latex]<\/p>\r\n

                  Regroup factors into squares<\/p>\r\n

                  [latex] \\sqrt{7\\cdot3^2}[\/latex]<\/p>\r\n

                  Finish simplifying with the product rule for roots, and the square of a square idea.<\/p>\r\n

                  [latex]\\sqrt{7\\cdot3^2}\\\\\\\\=\\sqrt{7}\\cdot\\sqrt{3^2}\\\\\\\\=\\sqrt{7}\\cdot3[\/latex]<\/p>\r\n

                  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, 3, in front of the radical. \u00a0This helps the reader know that the 3 is not under the radical anymore.<\/p>\r\n

                  [latex] 3\\cdot \\sqrt{7}[\/latex]<\/p>\r\n\r\n

                  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. You can read this as \u201cthree radical seven\u201d or \u201cthree times the square root of seven.\u201d\r\n\r\nLet us look at another example, where we will introduce a small shortcut.\r\n
                  \r\n

                  Example 6<\/h3>\r\nSimplify: [latex]\\sqrt{300}[\/latex]\r\n\r\n[reveal-answer q=\"735257\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"735257\"]\r\n\r\nFind the prime factorization of 300.\r\n

                  [latex]\\sqrt{2\\cdot 2 \\cdot 3 \\cdot 5\\cdot 5}[\/latex]<\/p>\r\nIdentify pairs of identical factors.\r\n

                  [latex]\\sqrt{{\\color{blue}{2\\cdot 2}}\\cdot 3 \\cdot {\\color{red}{5\\cdot 5}}}[\/latex]<\/p>\r\nOptionally, we can rewrite each pair using exponents.\r\n

                  [latex]\\sqrt{2^2\\cdot 3\\cdot 5^2}[\/latex]<\/p>\r\nAs a shortcut, factors that appear in pairs under the radical can be brought outside the radical. (Note, at this point, we will omit rewriting the problem with separate radicals; but be aware, the product rule is hiding behind the scenes, and the reason this works.)\r\n

                  [latex]2\\cdot 5\\sqrt{3}[\/latex]<\/p>\r\nMultiply the outside factors to simplify.\r\n

                  [latex]10\\sqrt{3}[\/latex]<\/p>\r\n\r\n

                  Answer<\/h4>\r\n[latex]\\sqrt{300}=10\\sqrt{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, notice there is already a number in front of the radical. The [latex]7[\/latex] is being multiplied by [latex]\\sqrt{24}[\/latex]. This means that [latex]7[\/latex] will be multiplied by any factors that come out of the radical as we simplify.\r\n
                  \r\n

                  Example 7<\/h3>\r\nSimplify: [latex]7\\sqrt{24}[\/latex]\r\n\r\n[reveal-answer q=\"263512\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"263512\"]\r\n\r\nFind the prime factorization of 24.\r\n

                  [latex]7\\sqrt{2\\cdot 2\\cdot 2\\cdot 3}[\/latex]<\/p>\r\nIdentify pairs and rewrite using exponents if you choose.\r\n

                  [latex]7\\sqrt{{\\color{red}{2\\cdot 2}}\\cdot 2 \\cdot 3}[\/latex]<\/p>\r\n

                  [latex]=7\\sqrt{2^2\\cdot 2\\cdot 3}[\/latex]<\/p>\r\nBring any factors that occur in pairs outside the radical.\r\n

                  [latex]7\\cdot 2\\sqrt{2\\cdot 3}[\/latex]<\/p>\r\nMultiply the numbers outside and those inside to simplify.\r\n

                  [latex]14\\sqrt{6}[\/latex]<\/p>\r\n\r\n

                  Answer<\/h4>\r\n[latex]7\\sqrt{24}=14\\sqrt{6}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n\r\n \r\n\r\n[caption id=\"attachment_4884\" align=\"aligncenter\" width=\"455\"]\"Picture Shortcut This Way[\/caption]\r\n\r\nIn the next example, we introduce another shortcut by making use of the common squares we know, instead of using prime factors. It helps to have the squares of the numbers between 0 and 10 fresh in your mind to make simplifying radicals faster.\r\n
                    \r\n \t
                  • [latex]0^2=0[\/latex]<\/li>\r\n \t
                  • [latex]1^2=1[\/latex]<\/li>\r\n \t
                  • [latex]2^2=4[\/latex]<\/li>\r\n \t
                  • [latex]3^2=9[\/latex]<\/li>\r\n \t
                  • [latex]4^2=16[\/latex]<\/li>\r\n \t
                  • [latex]5^2=25[\/latex]<\/li>\r\n \t
                  • [latex]6^2=36[\/latex]<\/li>\r\n \t
                  • [latex]7^2=49[\/latex]<\/li>\r\n \t
                  • [latex]8^2=64[\/latex]<\/li>\r\n \t
                  • [latex]9^2=81[\/latex]<\/li>\r\n \t
                  • [latex]10^2=100[\/latex]<\/li>\r\n<\/ul>\r\n
                    \r\n

                    Example 8<\/h3>\r\nSimplify. [latex] \\sqrt{2000}[\/latex]\r\n\r\n[reveal-answer q=\"932245\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"932245\"]\r\n\r\nInstead of finding the prime factorization, we search for perfect squares that divide evenly into [latex]2000[\/latex]. An obvious one may be [latex]100[\/latex] as [latex]2000=100\\cdot 20[\/latex]. But then, there is also a perfect square that divides [latex]20[\/latex], namely [latex]4[\/latex]. So, instead of using prime factors, we will write [latex]2000[\/latex] as [latex]100\\cdot 4\\cdot 5[\/latex]. We then apply the product rule and take the square roots of the perfect squares we uncovered.\r\n

                    [latex] \\sqrt{2000}=\\sqrt{100\\cdot 4 \\cdot 5}[\/latex]<\/p>\r\n

                    [latex]=\\sqrt{100}\\cdot \\sqrt{4}\\cdot \\sqrt{5}[\/latex]<\/p>\r\n

                    [latex]=10\\cdot 2\\cdot \\sqrt{5}[\/latex]<\/p>\r\n

                    [latex]= 20\\cdot \\sqrt{5}[\/latex]<\/p>\r\nAs we did in the previous two examples, you might choose to skip the step of formally splitting the problem into separate radicals. Also, this process is shortest if you find the largest perfect square that divides the given number. In this case, that was actually [latex]400[\/latex]. See if you can verify that this would produce the same answer.\r\n

                    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

                    Approximate a square root<\/h2>\r\nWhen the radicand is not a perfect square, sometimes it is beneficial to approximate the root rather than simplify it.\u00a0 The square roots of radicands that are not perfect squares are irrational numbers.\u00a0 This means they cannot be written exactly in fraction or decimal form.\u00a0 However, if we use what we know about perfect squares, it isn\u2019t too difficult to determine which two consecutive whole numbers the square root would be between.\r\n
                    \r\n

                    Example 9<\/h3>\r\n[latex]\\sqrt{29}[\/latex] is between which consecutive whole numbers?\r\n\r\n[reveal-answer q=\"621777\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"621777\"]\r\n\r\nBelow is a list of perfect squares and their square roots. Notice that [latex]29[\/latex] is between the perfect squares [latex]25[\/latex] and [latex]36[\/latex].\r\n

                    \"Table<\/p>\r\nTherefore, we can write\r\n

                    [latex]\\sqrt{25}<\\sqrt{29}<\\sqrt{36}[\/latex]<\/p>\r\nSimplifying the square roots on the left and right, we get\r\n

                    [latex]5<\\sqrt{29}<6[\/latex]<\/p>\r\nSo, we know that the [latex]\\sqrt{29}[\/latex] is between [latex]5[\/latex] and [latex]6[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nA square root whose radicand is not a perfect square can be approximated using a calculator.\u00a0 Look for a calculator key that looks like [latex]\\sqrt{x}[\/latex] or simply [latex]\\sqrt{\\hspace{.08in}}[\/latex] . Depending on the brand of the calculator, either the radicand is entered first and then the square root key is pressed, or the square root key is pressed first and then the radicand is entered.\r\n

                    \r\n

                    Example 10<\/h3>\r\nUse a calculator to approximate [latex]\\sqrt{29}[\/latex] to two decimal places.\r\n\r\n[reveal-answer q=\"995790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"995790\"]\r\n\r\nTrying putting this in your calculator. It should initially give you several digits after the decimal, which we can then round to two decimal places (nearest hundredth).\r\n

                    [latex]\\sqrt{29}\\approx 5.385164807[\/latex]<\/p>\r\n

                    [latex]\\sqrt{29}\\approx 5.39 [\/latex]<\/p>\r\nNotice that this agrees with our conclusion earlier when we determined that [latex]\\sqrt{29}[\/latex] would be between [latex]5[\/latex] and [latex]6[\/latex].\r\n\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

                    \r\n

                    Example 11<\/h3>\r\n[latex]\\sqrt{118}[\/latex] is between which two consecutive whole numbers? Approximate [latex]\\sqrt{118}[\/latex] to two decimal places.\r\n\r\n[reveal-answer q=\"12106\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"12106\"]\r\n\r\n[latex]118[\/latex] is between the perfect squares [latex]100[\/latex] and [latex]121[\/latex].\r\n\r\nTherefore, we can write\r\n

                    [latex]\\sqrt{100}<\\sqrt{118}<\\sqrt{121}[\/latex]<\/p>\r\nSimplifying the square roots on the left and right, we get\r\n

                    [latex]10<\\sqrt{118}<11[\/latex]<\/p>\r\nWe now know that [latex]\\sqrt{118}[\/latex] is between [latex]10[\/latex] and [latex]11[\/latex].\r\n\r\nUsing the calculator,\r\n

                    [latex]\\sqrt{118}\\approx 10.86278049[\/latex]<\/p>\r\n

                    [latex]\\sqrt{118}\\approx 10.86[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n<\/div>\r\n \r\n

                    Simplify other roots<\/h2>\r\n[caption id=\"attachment_5110\" align=\"aligncenter\" width=\"289\"]\"Rubik's Rubik's Cube[\/caption]\r\n\r\nWhile square roots are probably the most common radical, you can also find the third root, the fifth root, the 10th\u00a0root, or really any other n<\/i>th root of a number. Just as the square root is a number that, when squared, gives the radicand, the cube root<\/strong> is a number that, when cubed, gives the radicand.\r\n\r\nFind the cube roots of the following numbers:\r\n
                      \r\n \t
                    1. 27<\/li>\r\n \t
                    2. 8<\/li>\r\n \t
                    3. -8<\/li>\r\n \t
                    4. 0<\/li>\r\n<\/ol>\r\n
                        \r\n \t
                      1. We want to find a number whose cube is 27. \u00a0[latex]3\\cdot9=27[\/latex] and [latex]9=3^2[\/latex], so [latex]3\\cdot3\\cdot3=3^3=27[\/latex]<\/li>\r\n \t
                      2. We want to find a number whose cube is 8. [latex]2\\cdot2\\cdot2=8[\/latex] the cube root of 8 is 2.<\/li>\r\n \t
                      3. We want to find a number whose cube is -8. We know 2 is the cube root of 8, so maybe we can try -2. [latex]-2\\cdot{-2}\\cdot{-2}=-8[\/latex], so the cube root of -8 is -2. This is different from square roots because multiplying three negative numbers together results in a negative number.<\/li>\r\n \t
                      4. We want to find a number whose cube is 0. [latex]0\\cdot0\\cdot0[\/latex], no matter how many times you multiply [latex]0[\/latex] by itself, you will always get\u00a0[latex]0[\/latex].<\/li>\r\n<\/ol>\r\nThe cube root of a number is written with a small number 3, called the index<\/strong>, just outside and above the radical symbol. For example, the cube root of x\u00a0<\/em>looks like [latex] \\sqrt[3]{{x}}[\/latex]. This little 3 distinguishes cube roots from square roots which are written without a small number outside and above the radical symbol.\r\n
                        \r\n\r\n\"Caution\"Caution! Be careful to distinguish between [latex] \\sqrt[3]{x}[\/latex], the cube root of x<\/i>, and [latex] 3\\sqrt{x}[\/latex], three times<\/i> the square<\/i> root of x<\/i>. They may look similar at first, but they lead you to much different expressions!\r\n\r\n<\/div>\r\n
                        \r\n

                        Example 12<\/h3>\r\nSimplify each of the following:\r\n
                          \r\n \t
                        1. [latex]\\sqrt[3]{27}[\/latex]<\/li>\r\n \t
                        2. [latex]-\\sqrt[3]{27}[\/latex]<\/li>\r\n \t
                        3. [latex]\\sqrt[3]{-27}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"203559\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"203559\"]\r\n\r\nIn each case, we look for a number that, when cubed, results in the number under the radical.\r\n
                            \r\n \t
                          1. [latex]\\sqrt[3]{27}[\/latex]. We see that [latex]3\\cdot 3\\cdot 3=27[\/latex], or equivalently, [latex]3^{3}=27[\/latex]. Therefore, [latex]\\sqrt[3]{27}=3[\/latex].<\/li>\r\n \t
                          2. [latex]-\\sqrt[3]{27}[\/latex]. The minus sign in front of the radical is asking us to find the opposite of [latex]\\sqrt[3]{27}[\/latex]. Based on the answer above, the result would then be [latex]-3[\/latex]. Therefore, [latex]-\\sqrt[3]{27}=-3[\/latex].<\/li>\r\n \t
                          3. [latex]\\sqrt[3]{-27}[\/latex]. With square roots, the answer was \"Not a Real Number\" if the radicand was negative since no real number times itself will give a negative result. However, this is not the case with cubes! Since [latex](-3)(-3)(-3)=-27[\/latex], or equivalently, [latex](-3)^{3}=-27[\/latex], it follows that [latex]\\sqrt[3]{-27}=-3[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n\r\nLike we did with square roots, we can also use factoring to simplify cube roots such as [latex] \\sqrt[3]{125}[\/latex]. You can read this as \u201cthe third root of 125\u201d or \u201cthe cube root of 125.\u201d To simplify this expression, look for a number that, when multiplied by itself three times (for a total of three identical factors), equals 125. Let\u2019s factor 125 and find that number.\r\n
                            \r\n

                            Example 13<\/h3>\r\nSimplify. [latex] \\sqrt[3]{125}[\/latex]\r\n\r\n[reveal-answer q=\"517592\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"517592\"]\r\n\r\nWe first search for the prime factorization of 125. Since 125 ends in 5, so you know that 5 is a factor. Expand 125 into [latex]5\\cdot25[\/latex].\r\n

                            [latex] \\sqrt[3]{5\\cdot 25}[\/latex]<\/p>\r\nFactor 25 into 5 and 5.\r\n

                            [latex] \\sqrt[3]{5\\cdot 5\\cdot 5}[\/latex]<\/p>\r\nThe factors are [latex]5\\cdot5\\cdot5[\/latex], or [latex]5^{3}[\/latex].\r\n

                            [latex] \\sqrt[3]{{{5}^{3}}}[\/latex]<\/p>\r\nBecause of the index of 3, we are now looking for factors that appear in groups of three, so that they can be written with an exponent of 3. Then,\r\n

                            [latex]\\sqrt[3]{5^3}=5[\/latex]<\/p>\r\n\r\n

                            Answer<\/h4>\r\n[latex] \\sqrt[3]{125}=5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe prime factors of 125 are [latex]5\\cdot5\\cdot5[\/latex], which can be rewritten as [latex]5^{3}[\/latex]. The cube root of a cubed number is the number itself, so [latex] \\sqrt[3]{{{5}^{3}}}=5[\/latex]. You have found the cube root, the three identical factors that when multiplied together give 125. 125 is known as a perfect cube<\/strong> because its cube root is an integer.\r\n\r\nThe next example shows how to simplify a cube root when the radical is not a perfect cube.\r\n
                            \r\n

                            Example 14<\/h3>\r\nSimplify: [latex]\\sqrt[3]{80}[\/latex]\r\n\r\n[reveal-answer q=\"92686\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"92686\"]\r\n\r\nFind the prime factorization of 80.\r\n

                            [latex]\\sqrt[3]{80}=\\sqrt[3]{2\\cdot 2\\cdot 2\\cdot 2\\cdot 5}[\/latex]<\/p>\r\nIdentify groups of three identical factors. Optionally (and formally), we can then write any such group with an exponent of 3.\r\n

                            [latex]=\\sqrt[3]{{\\color{red}{2\\cdot 2\\cdot 2}}\\cdot 2\\cdot 5}[\/latex]<\/p>\r\n

                            [latex]=\\sqrt[3]{2^{3}\\cdot 2\\cdot 5}[\/latex]<\/p>\r\nApplying a similar shortcut to what we used for square roots, any factor that appears in a group of three (or can be written as a perfect cube) can come out of the radical. Any other factors must remain inside the radicand.\r\n

                            [latex]=2\\sqrt[3]{2\\cdot 5}[\/latex]<\/p>\r\nMultiply the factors inside the radicand to simplify.\r\n

                            [latex]=2\\sqrt[3]{10}[\/latex]<\/p>\r\n\r\n

                            Answer<\/h4>\r\n[latex]\\sqrt[3]{80}=2\\sqrt[3]{10}\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow that we have explored cube roots, we can extend these same idea to even higher order roots (those with larger indices). For example, consider the fourth root of 625, written [latex]\\sqrt[4]{625}[\/latex]. This is now asking us to determine \"what number when multiplied by itself four times equals 625,\" or equivalently, \"what number raised to the fourth power is 625.\" We answer this, and some variations of this, in the next example.\r\n
                            \r\n

                            Example 15<\/h3>\r\nSimplify each of the following:\r\n
                              \r\n \t
                            1. [latex]\\sqrt[4]{625}[\/latex]<\/li>\r\n \t
                            2. [latex]-\\sqrt[4]{625}[\/latex]<\/li>\r\n \t
                            3. [latex]\\sqrt[4]{-625}[\/latex]<\/li>\r\n<\/ol>\r\nIn each case, we are looking for a number that, when raised to the fourth power, results in the number under the radical.\r\n
                                \r\n \t
                              1. [latex]\\sqrt[4]{625}[\/latex]. We see that [latex]5\\cdot 5\\cdot 5\\cdot 5=625[\/latex], or [latex]5^{4}=625[\/latex]. Therefore, [latex]\\sqrt[4]625[\/latex]=5.<\/li>\r\n \t
                              2. [latex]-\\sqrt[4]{625}[\/latex]. The minus sign in front of the radical is asking us to find the opposite of [latex]\\sqrt[4]{625}[\/latex]. Based on our answer above, the result would be [latex]-\\sqrt{625}=-5[\/latex].<\/li>\r\n \t
                              3. [latex]\\sqrt[4]{-625}[\/latex]. There is no real number that, when raised to the fourth power, will result in a negative number. Therefore, [latex]\\sqrt[4]{-625}[\/latex] is: Not a Real Number.<\/li>\r\n<\/ol>\r\n<\/div>\r\nThis is a good time to summarize what we have observed about negatives and radicals.\r\n
                                  \r\n \t
                                • If the index is an odd<\/strong> number, a negative radicand will result in a negative answer<\/strong>.<\/li>\r\n \t
                                • If the index is an even<\/strong> number, a negative radicand results in an answer of Not a Real Number<\/strong>.<\/li>\r\n \t
                                • A negative outside the radical always indicates a negative answer, regardless of whether the index is even or odd (provided the radicand is positive).<\/li>\r\n<\/ul>\r\nThe rules we have introduced for simplifying, including the slight shortcuts, can easily be extended to a radical with any index. The next example shows a fifth root.\r\n
                                  \r\n

                                  Example 16<\/h3>\r\nSimplify: [latex]3\\sqrt[5]{64}[\/latex]\r\n\r\n[reveal-answer q=\"98999\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"98999\"]\r\n\r\nFirst note that the 3 in front of the radical is\u00a0not<\/em> an index, but is simply a constant being multiplied by the root (as we saw in Example 7).\r\n\r\nAs with any index, we will begin by finding the prime factorization of the radicand, 64.\r\n

                                  [latex]3\\sqrt[5]{64}=3\\sqrt[5]{2\\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2}[\/latex]<\/p>\r\nBecause of the index of 5, we identify groups of 5 identical factors, which we can then write using an exponent of 5 (or omit this latter step as a shortcut).\r\n

                                  [latex]=3\\sqrt[5]{{\\color{red}{2\\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2}}\\cdot 2}[\/latex]<\/p>\r\n

                                  [latex]=3\\sqrt[5]{2^{5}\\cdot 2}[\/latex]<\/p>\r\nAny factor that appears in a group of 5 (or a perfect fifth power) can be brought outside the radical, while any other factor must remain inside the radical.\r\n

                                  [latex]=3\\cdot 2\\sqrt[5]{2}[\/latex]<\/p>\r\nMultiply the numbers in front of the radical.\r\n

                                  [latex]=6\\sqrt[5]{2}[\/latex]<\/p>\r\nAs a point of caution, notice the index must remain on the radical. Avoid the common error of losing this since, if we drop the index, that would instead mean square root.\r\n

                                  Answer<\/span><\/h4>\r\n[latex]3\\sqrt[5]{64}=6\\sqrt[5]{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow that you have practiced simplifying radicals, we introduce two \"Think About It\" problems. In these problems, the radicals contain variables. Although you will not be held accountable for this situation, it is worth looking at to see that the same approach we have taken throughout this section applies to variables as well. You may also want to explore these for more practice with simplifying integers under radicals as well.\r\n
                                  \r\n

                                  think about it 1<\/h3>\r\nSimplify. [latex] \\sqrt[4]{32{{m}^{6}}}[\/latex]\r\n\r\n[reveal-answer q=\"617053\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"617053\"]\r\n\r\nFind the prime factorization of 32 and also break down [latex]m^5[\/latex] to its components.\r\n

                                  [latex]\\sqrt[4]{2\\cdot 2\\cdot 2\\cdot 2\\cdot 2\\cdot m\\cdot m\\cdot m\\cdot m\\cdot m\\cdot m}[\/latex]<\/p>\r\nSince the index is 4, we look for identical factors appearing in groups of 4.\r\n

                                  [latex]\\sqrt[4]{{\\color{blue}{2\\cdot 2\\cdot 2\\cdot 2}}\\cdot 2\\cdot {\\color{red}{m\\cdot m\\cdot m\\cdot m}}\\cdot m\\cdot m}[\/latex]<\/p>\r\n

                                  [latex]=\\sqrt[4]{2^4\\cdot 2\\cdot m^4 \\cdot m\\cdot m}[\/latex]<\/p>\r\nAny factor appearing in a group of 4 (or perfect fourth power) can be brought out of the radical.\r\n

                                  [latex]2m\\cdot \\sqrt[4]{2\\cdot m \\cdot m}[\/latex]<\/p>\r\n

                                  [latex]=2m\\sqrt[4]{2m^2}[\/latex]<\/p>\r\n\r\n

                                  Answer<\/h4>\r\n[latex] \\sqrt[4]{32{{m}^{6}}}=2m\\sqrt[4]{2{{m}^{2}}}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
                                  \r\n

                                  think about it 2<\/h3>\r\nSimplify. [latex] \\sqrt[3]{-27{{x}^{4}}{{y}^{3}}}[\/latex]\r\n\r\n[reveal-answer q=\"670300\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"670300\"]\r\n\r\nBreak down [latex]-27[\/latex] as well as the variables.\r\n

                                  [latex]\\sqrt[3]{(-3)(-3)(-3)\\cdot x \\cdot x\\cdot x\\cdot x\\cdot y \\cdot y \\cdot y}[\/latex]<\/p>\r\nSearch for groups of 3 since it is a cube root. As always, you can then include the formal step of writing as cubes or immediately skip to the next step.\r\n

                                  [latex]\\sqrt[3]{{\\color{blue}{(-3)(-3)(-3)}}\\cdot {\\color{red}{x\\cdot x\\cdot x}}\\cdot x \\cdot {\\color{green}{y\\cdot y \\cdot y}}}[\/latex]<\/p>\r\n

                                  [latex]=\\sqrt[3]{(-3)^3\\cdot x^3\\cdot x\\cdot y^3}[\/latex]<\/p>\r\nAny factor appearing in a group of 3 (or a perfect cube) can come outside of the radical.\r\n

                                  [latex](-3)\\cdot x\\cdot y\\sqrt[3]{x}[\/latex]<\/p>\r\n

                                  [latex]=-3xy\\sqrt[3]{x}[\/latex]<\/p>\r\n\r\n

                                  Answer<\/h4>\r\n[latex] \\sqrt[3]{-27{{x}^{4}}{{y}^{3}}}=-3xy\\sqrt[3]{x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nTo conclude the section, the following video shows more examples of simplifying cube roots (including two with variables).\r\n\r\nhttps:\/\/youtu.be\/9Nh-Ggd2VJo\r\n

                                  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 0 is 0. You can only take the square root of values that are nonnegative. The square root of a perfect square will be an integer. Other square roots can be simplified by identifying factors that are perfect squares and taking their square root.\r\n\r\nA radical expression is a mathematical way of representing the [latex]n[\/latex]th root of a number. Square roots and cube roots are the most common radicals. To simplify radical expressions, look for identical factors that occur in groups of sizes matching the index.\r\n\r\n ","rendered":"
                                  \n

                                  section 7.1 Learning Objectives<\/h3>\n

                                  7.1: Introduction to Radicals<\/strong><\/p>\n

                                    \n
                                  • Find the square roots of a perfect square<\/li>\n
                                  • Simplify a square root written with the radical symbol\n
                                      \n
                                    • When the radicand is a perfect square<\/li>\n
                                    • When the radicand is not a perfect square<\/li>\n
                                    • When the root has a coefficient<\/li>\n<\/ul>\n<\/li>\n
                                    • Approximate a square root<\/li>\n
                                    • Simplify other roots<\/li>\n<\/ul>\n<\/div>\n

                                       <\/p>\n

                                      Find the square roots of a perfect square<\/h2>\n

                                      We know how to square a number:<\/p>\n

                                      [latex]5^2=25[\/latex] and [latex]\\left(-5\\right)^2=25[\/latex]<\/p>\n

                                      Taking a square root is the opposite of squaring so we can make these statements:<\/p>\n

                                        \n
                                      • 5 is the nonngeative square root of 25<\/li>\n
                                      • -5 is the negative square root of 25<\/li>\n<\/ul>\n
                                        \n

                                        Example 1<\/h3>\n

                                        Find the square roots of the following numbers:<\/p>\n

                                          \n
                                        1. 36<\/li>\n
                                        2. 81<\/li>\n
                                        3. -49<\/li>\n
                                        4. 0<\/li>\n<\/ol>\n

                                           <\/p>\n

                                            \n
                                          1. We want to find a number whose square is 36. [latex]6^2=36[\/latex] therefore, \u00a0the nonnegative square root of 36 is 6 and the negative square root of 36 is -6<\/li>\n
                                          2. We want to find a number whose square is 81. [latex]9^2=81[\/latex] therefore, \u00a0the nonnegative square root of 81 is 9 and the negative square root of 81 is -9<\/li>\n
                                          3. We want to find a number whose square is -49. When you square a real number, the result is always positive. Stop and think about that for a second.\u00a0A negative number times itself is positive, and a positive number times itself is positive. \u00a0Therefore, -49 does not have square roots.\u00a0 There are no real number solutions to this question.<\/li>\n
                                          4. We want to find a number whose square is 0. [latex]0^2=0[\/latex] therefore, \u00a0the nonnegative square root of 0 is 0. \u00a0We do not assign 0 a sign, so it has only one square root, and that is 0.<\/li>\n<\/ol>\n<\/div>\n

                                             <\/p>\n

                                            Simplify a square root – when the radicand is a perfect square<\/h2>\n

                                            The notation that we use to express a square root for any real number, a, is as follows:<\/p>\n

                                            \n

                                            Writing a Square Root<\/h3>\n

                                            The symbol for the square root is called a radical symbol.<\/strong>\u00a0For a real number, a<\/em> the square root of a<\/em> is written as [latex]\\sqrt{a}[\/latex]<\/p>\n

                                            The number that is written under the radical symbol is called the radicand<\/strong>.<\/p>\n

                                            By definition, the square root symbol, [latex]\\sqrt{\\hphantom{5}}[\/latex] always means to find the nonnegative\u00a0root, called the principal root<\/strong>.<\/p>\n

                                            [latex]\\sqrt{-a}[\/latex] is not defined, therefore [latex]\\sqrt{a}[\/latex] is defined for [latex]a\\geq 0[\/latex]<\/p>\n<\/div>\n

                                            Let’s do an example similar to\u00a0the example from above, this time using square root notation. \u00a0Note that using the square root notation means that you are only finding the principal root – the nonnegative root.<\/p>\n

                                            \n

                                            Example 2<\/h3>\n

                                            Simplify\u00a0the following square roots:<\/p>\n

                                              \n
                                            1. [latex]\\sqrt{16}[\/latex]<\/li>\n
                                            2. [latex]\\sqrt{9}[\/latex]<\/li>\n
                                            3. [latex]\\sqrt{-9}[\/latex]<\/li>\n
                                            4. [latex]-\\sqrt{25}[\/latex]<\/li>\n
                                            5. [latex]\\sqrt{5^2}[\/latex]<\/li>\n
                                            6. [latex]\\sqrt{0.01}[\/latex]<\/li>\n<\/ol>\n
                                              Show Solution<\/span><\/p>\n
                                              \n
                                                \n
                                              1. [latex]\\sqrt{16}[\/latex]. We are looking for a number whose square is 16, so\u00a0[latex]\\sqrt{16}=4[\/latex]. We only write the nonnegative root because that is how the root symbol is defined.<\/li>\n
                                              2. [latex]\\sqrt{9}[\/latex]. We are looking for a number whose square is 9, so [latex]\\sqrt{9}=3[\/latex].\u00a0We only write the nonnegative root because that is how the root symbol is defined.<\/li>\n
                                              3. [latex]\\sqrt{-9}[\/latex]. We are looking for a number whose square is -9. \u00a0There are no real numbers whose square is -9, so this radical is not a real number.<\/li>\n
                                              4. [latex]-\\sqrt{25}[\/latex].\u00a0Notice that the minus sign is outside of the radical.\u00a0 That minus sign is asking us to find the opposite of [latex]\\sqrt{25}[\/latex].\u00a0 Since [latex]\\sqrt{25}=5[\/latex], the opposite would be [latex]-5[\/latex]. Therefore, [latex]-\\sqrt{25}=-5[\/latex].<\/li>\n
                                              5. [latex]\\sqrt{5^2}[\/latex]. We are looking for a number whose square is [latex]5^2[\/latex]. \u00a0We already have the number whose square is [latex]5^2[\/latex], it’s 5!<\/li>\n
                                              6. Compared to the other examples, it may be less obvious that [latex]0.01[\/latex] is a perfect square.\u00a0 However, [latex](0.1)(0.1)=0.01[\/latex]. This can also be written as [latex](0.1)^2=0.01[\/latex]. Therefore, [latex]\\sqrt{0.01}=0.1[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n

                                                The last problem in the previous example shows us an important relationship between squares and square roots, and we can summarize it as follows:<\/p>\n

                                                \n

                                                \u00a0The square root of a square<\/h3>\n

                                                For a nonnegative real number, a, [latex]\\sqrt{a^2}=a[\/latex]<\/p>\n<\/div>\n

                                                In the video that follows, we simplify\u00a0more square roots using the fact that\u00a0\u00a0[latex]\\sqrt{a^2}=a[\/latex] means finding the principal square root.<\/p>\n