{"id":6814,"date":"2020-10-09T21:52:26","date_gmt":"2020-10-09T21:52:26","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6814"},"modified":"2026-02-01T08:04:11","modified_gmt":"2026-02-01T08:04:11","slug":"7-5-applications-with-radicals","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/7-5-applications-with-radicals\/","title":{"raw":"7.5: Further Exploration with Radicals","rendered":"7.5: Further Exploration with Radicals"},"content":{"raw":"
\r\n

section 7.5 Learning Objectives<\/h3>\r\n7.5: Further Exploration with Radicals<\/strong>\r\n
    \r\n \t
  • Use the Pythagorean Theorem to solve applications involving right triangles<\/li>\r\n \t
  • Find the domain of a radical function<\/li>\r\n<\/ul>\r\n<\/div>\r\nThis section will discuss applications which use square roots, in particular the Pythagorean <\/span>Theorem.\u00a0<\/span>As always, the following steps will help to translate and solve the problem.<\/span>\r\n
    \r\n\r\n1.\u00a0<\/span>Read\u00a0<\/span>through the entire problem<\/span><\/strong>\r\n\r\n2.\u00a0<\/span>Organize\u00a0<\/span>the information (a picture ma<\/span>y be useful)<\/span><\/strong>\r\n\r\n3.\u00a0<\/span>Write the\u00a0<\/span>equation<\/span><\/strong>\r\n\r\n4.\u00a0<\/span>Solve\u00a0<\/span>the equation<\/span><\/strong>\r\n\r\n5.\u00a0<\/span>Check\u00a0<\/span>the answer<\/span><\/strong>\r\n\r\n<\/div>\r\n

    Pythagorean Theorem<\/h2>\r\nIf we are given the measurements of two sides of a right triangle (A right triangle is one where one of the angles is 90 degrees; indicated with a square in the angle), we can find the <\/span>measurement of the third side by using the Pythagorean Theorem. The Pythago<\/span>rean Theorem <\/span>states that the square of hypotenuse is equal to the sum of the squares of the other two sides. <\/span>\r\n\r\nThe <\/span>Pythagorean Theorem formula is:<\/span><\/strong>\r\n

    [latex]a^2+b^2=c^2[\/latex]\u00a0 \u00a0<\/strong>where c is the hypotenuse; a and b are the legs.\u00a0<\/strong><\/p>\r\n\"Right\r\n\r\nThe video below will walk you through the proof and a few examples of using The Pythagorean Theorem, including an application example.\r\n\r\n[embed]https:\/\/youtu.be\/Nwp0p-loCZg[\/embed]\r\n

    \r\n

    Example 1<\/h3>\r\nFind the length of the hypotenuse of the right triangle pictured below:\r\n\r\n\"right\r\n\r\n[reveal-answer q=\"839905\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"839905\"]\r\n

    Use the Pythagorean Theorem<\/p>\r\n

    [latex]c^2=a^2+b^2[\/latex]<\/p>\r\nSubstitute the values into the formula. Note, both given sides are the legs, a and b.\r\n

    [latex]c^2=7^2+24^2[\/latex]<\/p>\r\nSimplify the squares\r\n

    [latex]c^2=49+576[\/latex]<\/p>\r\nAdd\r\n

    [latex]c^2=625[\/latex]<\/p>\r\nTake the square root of each side\r\n

    [latex]\\sqrt{c^2}=\\sqrt{625}[\/latex]<\/p>\r\n

    [latex]c=25[\/latex]<\/p>\r\nSolution:\u00a0<\/strong>\r\n\r\nThe hypotenuse of the triangle is 25 units long.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    \r\n

    Example 2<\/h3>\r\nFind the missing length of the leg of the right triangle pictured below:\r\n\r\n\"right\r\n\r\n[reveal-answer q=\"613728\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"613728\"]\r\n

    Use the Pythagorean Theorem<\/p>\r\n

    [latex]a^2+b^2=c^2[\/latex]<\/p>\r\nSubstitute the values into the formula. Note, this time sides labeled 13 and 5 aren't both legs of the triangle like in the last example. The side labeled 13 is the hypotenuse so must replace the [latex]c[\/latex] in the formula.\r\n

    [latex]a^2+5^2=13^2[\/latex]<\/p>\r\nSimplify the squares\r\n

    [latex]a^2+25=169[\/latex]<\/p>\r\nSubtract 25 from both sides of the equation\r\n

    [latex]\\begin{array}{r}a^2+25=169\\\\\\underline{\\,\\,\\,\\,-25\\,\\,\\,-25}\\\\a^2=144\\end{array}[\/latex]<\/p>\r\nTake the square root of each side\r\n

    [latex]\\sqrt{a^2}=\\sqrt{144}[\/latex]<\/p>\r\n

    [latex]a=12[\/latex]<\/p>\r\nSolution:\u00a0<\/strong>\r\n\r\nThe vertical leg of this triangle is 12 units long.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe Pythagorean Theorem can be used in a variety of real-world applications. The next two examples show some simple applications.\r\n

    \r\n

    Example 3<\/h3>\r\nGiven a rectangular picture frame whose opening is 12 inches by 10 inches, how long is the diagonal from <\/span>one corner of the opening to the other? Give both an exact answer and an approximation to the nearest hundredth.\u00a0<\/span>\r\n\r\n[reveal-answer q=\"22263\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"22263\"]\r\n\r\n\"picture\r\n

    Use the Pythagorean Theorem<\/p>\r\n

    [latex]c^2=a^2+b^2[\/latex]<\/p>\r\nSubstitute the values into the formula. Note, both given sides are the legs, a and b.\r\n

    [latex]c^2=12^2+10^2[\/latex]<\/p>\r\nSimplify the squares\r\n

    [latex]c^2=144+100[\/latex]<\/p>\r\nAdd\r\n

    [latex]c^2=244[\/latex]<\/p>\r\nTake the square root of each side\r\n

    [latex]\\sqrt{c^2}=\\sqrt{244}[\/latex]<\/p>\r\nSimplify square root to list answer in exact form\r\n

    [latex]c=2\\sqrt{61}[\/latex]<\/p>\r\nType root in calculator to find approximation for the answer\r\n

    [latex]c\u224815.6204...[\/latex]<\/p>\r\nRound to the nearest hundredth\r\n

    [latex]c\u224815.62[\/latex]<\/p>\r\nSolution:\u00a0<\/strong>\r\n\r\nThe diagonal from one corner of the opening to the other is of length\u00a0[latex]2\\sqrt{61}[\/latex] inches which is approximately 15.62 inches long.\r\n\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    \r\n

    Example 4<\/h3>\r\nA <\/span>6<\/span>-<\/span>foot\u00a0<\/span>ladder is placed against a wall. If the base of the <\/span>ladder is 3 feet from the\u00a0<\/span>wall, how high <\/span>up the wall will the ladder reach? Give both an exact answer and an approximation to the nearest hundredth.\u00a0<\/span>\r\n\r\n[reveal-answer q=\"993096\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"993096\"]\r\n\r\n\"image\r\n

    Since the triangle formed has a right angle we can use the Pythagorean Theorem<\/p>\r\n

    [latex]a^2+b^2=c^2[\/latex]<\/p>\r\nSubstitute the values into the formula. Note, this time sides labeled 3 and 6 aren't both legs of the triangle like in the last example. The side 6 ft ladder is the hypotenuse of the triangle so must replace the [latex]c[\/latex] in the formula.\r\n

    [latex]a^2+3^2=6^2[\/latex]<\/p>\r\nSimplify the squares\r\n

    [latex]a^2+9=36[\/latex]<\/p>\r\nSubtract 9 from both sides of the equation\r\n

    [latex]\\begin{array}{r}a^2+9=36\\\\\\underline{\\,\\,\\,\\,-9\\,\\,\\,-9}\\\\a^2=27\\end{array}[\/latex]<\/p>\r\nTake the square root of each side\r\n

    [latex]\\sqrt{a^2}=\\sqrt{27}[\/latex]<\/p>\r\nSimplify square root to list answer in exact form\r\n

    [latex]a=3\\sqrt{3}[\/latex]<\/p>\r\nType root in calculator to find approximation for the answer\r\n

    [latex]a\u22485.1961...[\/latex]<\/p>\r\nRound to the nearest hundredth\r\n

    [latex]a\u22485.20[\/latex]<\/p>\r\nSolution:\u00a0<\/strong>\r\n\r\nThe 6-foot ladder will reach\u00a0[latex]3\\sqrt{3}[\/latex] ft. up the wall, which is approximately 5.20 ft.\r\n\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe next example includes a radical in one of the lengths. See if you can apply the skills of this chapter to this problem.\r\n

    \r\n

    Example 5<\/h3>\r\nA right triangle\u2019s hypotenuse is 8 meters, and one leg is [latex]4\\sqrt{3}[\/latex]\u00a0<\/span><\/span>meters.\u00a0 Find the length of the other leg.\u00a0\u00a0<\/span><\/span>\r\n\r\n[reveal-answer q=\"650980\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"650980\"]\r\n

    Use the Pythagorean Theorem<\/p>\r\n

    [latex]a^2+b^2=c^2[\/latex]<\/p>\r\nSubstitute [latex]4\\sqrt{3}[\/latex]<\/span><\/span>\u00a0for either a<\/em> or b<\/em>; Substitute 8 for c<\/em>.<\/span><\/span><\/span><\/span><\/span>\r\n

    [latex]a^2+(4\\sqrt{3})^2=8^2[\/latex]<\/span><\/span><\/p>\r\nSimplify the squares\r\n

    [latex]a^2+(16\\cdot3)=64[\/latex]<\/p>\r\n

    [latex]a^2+48=64[\/latex]<\/p>\r\nSubtract 48 from both sides of the equation\r\n

    [latex]\\begin{array}{r}a^2+48=64\\\\\\underline{\\,\\,\\,\\,-48\\,\\,\\,-48}\\\\a^2=16\\end{array}[\/latex]<\/p>\r\nTake the square root of each side\r\n

    [latex]\\sqrt{a^2}=\\sqrt{16}[\/latex]<\/p>\r\n

    [latex]a=4[\/latex]<\/p>\r\nSolution:\u00a0<\/strong>\r\n\r\nThe length of the other leg is 4 meters long.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAs mentioned earlier, the Pythagorean Theorem is used in many applications. In fact, it has played a critical role in real-world applications for centuries! If you are interested in exploring another real-world example, celestial navigation and the mathematics behind it, watch this video for fun.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=XWLZKmPU17M\r\n

    Domain of Radical Functions<\/h2>\r\nIn module 3, we introduced the domain of a function as the set of input values that produce valid outputs.\u00a0 In this module, we have been exploring radical expressions.\u00a0 In the next example, let us examine a radical function and determine if any domain issues arise.\r\n
    \r\n

    Example 6<\/h3>\r\nConsider the radical function [latex]f(x)=\\sqrt{3x-2}[\/latex]\r\n

    A.\u00a0 Compute [latex]f(1)[\/latex], [latex]f(6)[\/latex], [latex]f(7)[\/latex], [latex]f\\left(\\frac{2}{3}\\right)[\/latex], and [latex]f(0)[\/latex]. Round to three decimal places if needed.<\/p>\r\n

    B.\u00a0 Determine if each of the [latex]x[\/latex]-values from part A is in the domain of the function.<\/p>\r\n[reveal-answer q=\"587337\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"587337\"]\r\n

    A.\u00a0 As with any function, we will substitute each value into the function for [latex]x[\/latex].<\/p>\r\n

    [latex]f(1)=\\sqrt{3(1)-2}=\\sqrt{3-2}=\\sqrt{1}=1[\/latex]<\/p>\r\n

    [latex]f(6)=\\sqrt{3(6)-2}=\\sqrt{18-2}=\\sqrt{16}=4[\/latex]<\/p>\r\n

    [latex]f(7)=\\sqrt{3(7)-2}=7\\sqrt{21-2}=\\sqrt{19}\\approx 4.359[\/latex]<\/p>\r\n

    [latex]f\\left(\\frac{2}{3}\\right)=\\sqrt{3\\left(\\frac{2}{3}\\right)-2}=\\sqrt{2-2}=\\sqrt{0}=0[\/latex]<\/p>\r\n

    [latex]f(0)=\\sqrt{3(0)-2}=\\sqrt{0-2}=\\sqrt{-2}\\hspace{.2in}[\/latex] Not a real number<\/p>\r\n

    B.\u00a0 The first four inputs produced valid results, but the last input of [latex]0[\/latex] did not, as we are reminded that we cannot currently take the even root of a negative number.\u00a0 We conclude that<\/p>\r\n\r\n

      \r\n \t
    • [latex]1, 6, 7, \\frac{2}{3}[\/latex] are in the domain<\/li>\r\n \t
    • [latex]0[\/latex] is not in the domain<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe previous example shows us that any input that results in a negative radicand is excluded from the domain. In other words, we must always ensure that the radicand is nonnegative ([latex]\\geq 0[\/latex]).\r\n
      \r\n\r\nFinding the domain of a radical function with an even index<\/strong>\r\n
        \r\n \t
      • Set the radicand (the expression under the radical) to be [latex]\\geq 0[\/latex].<\/li>\r\n \t
      • Solve the corresponding inequality for the variable.<\/li>\r\n \t
      • Write your answer in the appropriate notation.<\/li>\r\n<\/ul>\r\n<\/div>\r\n
        So now, let us find the domain of the radical function given in the previous example.<\/div>\r\n
        \r\n
        \r\n

        Example 7<\/h3>\r\nFind the domain of the radical function [latex]f(x)=\\sqrt{3x-2}[\/latex]. Give your answer in set-builder notation and interval notation.\r\n\r\n[reveal-answer q=\"619800\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"619800\"]\r\n\r\nWe need to determine when the radicand, [latex]3x-2[\/latex], is nonnegative. We can do this by setting it to be [latex]\\geq 0[\/latex] and then solving.\r\n

        [latex]3x-2\\geq 0[\/latex]<\/p>\r\n

        [latex]\\underline{\\hspace{.25in}+\\hspace{.05in} 2 \\hspace{.05in}+2}[\/latex]<\/p>\r\n

        [latex]3x\\geq 2[\/latex]<\/p>\r\n

        [latex]\\frac{3x}{3}\\geq \\frac{2}{3}[\/latex]<\/p>\r\n

        [latex]x\\geq \\frac{2}{3}[\/latex]<\/p>\r\nIn set-builder notation, we simply write this as\u00a0[latex]\\left\\{x|x\\geq \\frac{2}{3}\\right\\}[\/latex].\r\n\r\nOn a number line, this corresponds to all numbers to the right of [latex]\\frac{2}{3}[\/latex], including [latex]\\frac{2}{3}[\/latex] itself. So, the interval notation is [latex]\\left[ \\frac{2}{3},\\infty\\right)[\/latex].\r\n

        Answer<\/h4>\r\nSet-Builder Notation:\u00a0[latex]\\left\\{x|x\\geq \\frac{2}{3}\\right\\}[\/latex]\r\n\r\nInterval Notation:\u00a0[latex]\\left[ \\frac{2}{3},\\infty\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIt is worth noting that had the function in the previous example been a fourth root, sixth root, or any even root, the process and final answer would not have changed. However, we saw in the beginning of this module that odd roots do not have the same issue with negative radicands.\u00a0 For example, [latex]\\sqrt[3]{-8}=-2[\/latex]. It follows that odd roots will not have any domain issues.\r\n
        \r\n\r\nDomain of a radical function with an odd index<\/strong>\r\n
          \r\n \t
        • The domain will be\u00a0all real numbers<\/em>. In interval notation, this is [latex](-\\infty,\\infty)[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\nLet us conclude the section with one more example to emphasize the difference between these two cases.\r\n
          \r\n

          Example 8<\/h3>\r\nFind the domain in set-builder notation for each radical function.\u00a0 If the domain is all real numbers, state this.\r\n

          A.\u00a0 [latex]f(x)=\\sqrt{5-x}[\/latex]<\/p>\r\n

          B.\u00a0 [latex]g(x)=\\sqrt[3]{5-x}[\/latex]<\/p>\r\n[reveal-answer q=\"709176\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"709176\"]\r\n

          A.\u00a0 First note that this is an even radical (an index of 2).\u00a0 So, we set the radicand to be [latex]\\geq 0[\/latex].<\/p>\r\n

          [latex]5-x\\geq 0[\/latex]<\/p>\r\n

          [latex]\\underline{\\hspace{-.14in}-5 \\hspace{.35in} -5}[\/latex]<\/p>\r\n

          [latex]-x\\geq -5[\/latex]<\/p>\r\n

          [latex]\\frac{-x}{-1}\\leq \\frac{-5}{-1}[\/latex]<\/p>\r\n

          [latex]x\\leq 5[\/latex]<\/p>\r\n

          So, the domain is [latex]\\{x|x\\leq 5\\}[\/latex].<\/p>\r\n

          B.\u00a0 The function is an odd radical (an index of 3).\u00a0 So, the domain is\u00a0all real numbers<\/em>.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n\r\n<\/div>\r\n

          <\/div>\r\n ","rendered":"
          \n

          section 7.5 Learning Objectives<\/h3>\n

          7.5: Further Exploration with Radicals<\/strong><\/p>\n

            \n
          • Use the Pythagorean Theorem to solve applications involving right triangles<\/li>\n
          • Find the domain of a radical function<\/li>\n<\/ul>\n<\/div>\n

            This section will discuss applications which use square roots, in particular the Pythagorean <\/span>Theorem.\u00a0<\/span>As always, the following steps will help to translate and solve the problem.<\/span><\/p>\n

            \n

            1.\u00a0<\/span>Read\u00a0<\/span>through the entire problem<\/span><\/strong><\/p>\n

            2.\u00a0<\/span>Organize\u00a0<\/span>the information (a picture ma<\/span>y be useful)<\/span><\/strong><\/p>\n

            3.\u00a0<\/span>Write the\u00a0<\/span>equation<\/span><\/strong><\/p>\n

            4.\u00a0<\/span>Solve\u00a0<\/span>the equation<\/span><\/strong><\/p>\n

            5.\u00a0<\/span>Check\u00a0<\/span>the answer<\/span><\/strong><\/p>\n<\/div>\n

            Pythagorean Theorem<\/h2>\n

            If we are given the measurements of two sides of a right triangle (A right triangle is one where one of the angles is 90 degrees; indicated with a square in the angle), we can find the <\/span>measurement of the third side by using the Pythagorean Theorem. The Pythago<\/span>rean Theorem <\/span>states that the square of hypotenuse is equal to the sum of the squares of the other two sides. <\/span><\/p>\n

            The <\/span>Pythagorean Theorem formula is:<\/span><\/strong><\/p>\n

            [latex]a^2+b^2=c^2[\/latex]\u00a0 \u00a0<\/strong>where c is the hypotenuse; a and b are the legs.\u00a0<\/strong><\/p>\n

            \"Right<\/p>\n

            The video below will walk you through the proof and a few examples of using The Pythagorean Theorem, including an application example.<\/p>\n