{"id":1432,"date":"2021-12-02T19:37:59","date_gmt":"2021-12-03T00:37:59","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/9-5-rationalizing-denominators\/"},"modified":"2023-08-31T18:13:46","modified_gmt":"2023-08-31T22:13:46","slug":"rationalizing-denominators","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/rationalizing-denominators\/","title":{"raw":"9.5 Rationalizing Denominators","rendered":"9.5 Rationalizing Denominators"},"content":{"raw":"It is considered non-conventional to have a radical in the denominator. When this happens, generally the numerator and denominator are multiplied by the same factors to remove the radical denominator. The problems in the previous section dealt with removing a monomial radical. In this section, the previous strategy is expanded to include binomial radicals.\r\n
\r\n

Example 9.5.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nRationalize [latex]\\dfrac{\\sqrt{3}-9}{2\\sqrt{6}}[\/latex].\r\n\r\nTo rationalize the denominator, multiply out the [latex]\\sqrt{6}[\/latex].\r\n\r\nThis will look like:\r\n

[latex]\\dfrac{(\\sqrt{3}-9)(\\sqrt{6})}{2\\sqrt{6}(\\sqrt{6})}[\/latex]<\/p>\r\nMultiplying the [latex]\\sqrt{6}[\/latex] throughout yields:\r\n

[latex]\\dfrac{(\\sqrt{3})(\\sqrt{6})-(9)(\\sqrt{6})}{2\\sqrt{36}}[\/latex]<\/p>\r\nWhich reduces to:\r\n

[latex]\\dfrac{3\\sqrt{2}-9\\sqrt{6}}{2\\cdot 6}[\/latex]<\/p>\r\nAnd simplifies to:\r\n

[latex]\\dfrac{\\sqrt{2}-3\\sqrt{6}}{4}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nPlease note that, in reducing the numerator and denominator by the factor 3, reduce each term in the numerator by 3.\r\n\r\nQuite often, there will be a denominator binomial that contains radicals. For these problems, it is easiest to use a feature from the sum and difference of squares: [latex]a^2 - b^2 = (a + b)(a - b)[\/latex].\r\n\r\n[latex](a + b)(a - b)[\/latex] are termed conjugates of each other. They are identical binomials, except that their signs are opposite. When encountering radical binomials, simply multiply by the conjugates to square out the radical.\r\n

\r\n

Example 9.5.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSquare out the radical of the binomial [latex](\\sqrt{3} - \\sqrt{5})[\/latex] using its conjugate.\r\n\r\nThe conjugate of [latex](\\sqrt{3} - \\sqrt{5})[\/latex] is [latex](\\sqrt{3} + \\sqrt{5})[\/latex].\r\n\r\nWhen multiplied, these conjugates yield [latex](\\sqrt{3}-\\sqrt{5})(\\sqrt{3}+\\sqrt{5})[\/latex] or [latex](\\sqrt{3})^2-(\\sqrt{5})^2[\/latex].\r\n\r\nThis yields 3 \u2212 5 = \u22122.\r\n\r\n<\/div>\r\n<\/div>\r\nWhen encountering a radicalized binomial denominator, the best solution is to multiply both the numerator and denominator by the conjugate of the denominator.\r\n
\r\n

Example 9.5.3<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nRationalize the denominator of [latex]\\dfrac{\\sqrt{6}}{\\sqrt{6}+\\sqrt{13}}[\/latex].\r\n\r\nMultiplying the numerator and denominator by the denominator's conjugate yields:\r\n

[latex]\\dfrac{\\sqrt{6}}{\\sqrt{6}+\\sqrt{13}} \\cdot \\dfrac{(\\sqrt{6}-\\sqrt{13})}{(\\sqrt{6}-\\sqrt{13})}[\/latex]<\/p>\r\nWhen multiplied out, this yields:\r\n

[latex]\\dfrac{(\\sqrt{6})^2-\\sqrt{6}\\sqrt{13}}{(\\sqrt{6})^2-(\\sqrt{13})^2}[\/latex]<\/p>\r\nWhich reduces to:\r\n

[latex]\\dfrac{6-\\sqrt{78}}{6-13}[\/latex]<\/p>\r\nOr:\r\n

[latex]\\dfrac{6-\\sqrt{78}}{-7}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

Questions<\/h1>\r\nRationalize the following radical fractions.\r\n
    \r\n \t
  1. [latex]\\dfrac{4+2\\sqrt{3}}{\\sqrt{3}}[\/latex]<\/li>\r\n \t
  2. [latex]\\dfrac{-4+\\sqrt{3}}{4\\sqrt{3}}[\/latex]<\/li>\r\n \t
  3. [latex]\\dfrac{4+2\\sqrt{3}}{5\\sqrt{6}}[\/latex]<\/li>\r\n \t
  4. [latex]\\dfrac{2\\sqrt{3}-2}{2\\sqrt{3}}[\/latex]<\/li>\r\n \t
  5. [latex]\\dfrac{2-5\\sqrt{5}}{4\\sqrt{3}}[\/latex]<\/li>\r\n \t
  6. [latex]\\dfrac{\\sqrt{5}+4}{4\\sqrt{5}}[\/latex]<\/li>\r\n \t
  7. [latex]\\dfrac{\\sqrt{2}-3\\sqrt{3}}{\\sqrt{3}}[\/latex]<\/li>\r\n \t
  8. [latex]\\dfrac{\\sqrt{5}-\\sqrt{2}}{3\\sqrt{6}}[\/latex]<\/li>\r\n \t
  9. [latex]\\dfrac{5}{3\\sqrt{5}+\\sqrt{2}}[\/latex]<\/li>\r\n \t
  10. [latex]\\dfrac{5}{\\sqrt{3}+4\\sqrt{5}}[\/latex]<\/li>\r\n \t
  11. [latex]\\dfrac{2}{5+\\sqrt{2}}[\/latex]<\/li>\r\n \t
  12. [latex]\\dfrac{5}{2\\sqrt{3}-\\sqrt{2}}[\/latex]<\/li>\r\n \t
  13. [latex]\\dfrac{3}{4-\\sqrt{3}}[\/latex]<\/li>\r\n \t
  14. [latex]\\dfrac{4}{\\sqrt{2}-2}[\/latex]<\/li>\r\n \t
  15. [latex]\\dfrac{4}{3+\\sqrt{5}}[\/latex]<\/li>\r\n \t
  16. [latex]\\dfrac{2}{\\sqrt{5}+2\\sqrt{3}}[\/latex]<\/li>\r\n \t
  17. [latex]\\dfrac{-3+2\\sqrt{3}}{\\sqrt{3}+2}[\/latex]<\/li>\r\n \t
  18. [latex]\\dfrac{4+\\sqrt{5}}{2+2\\sqrt{5}}[\/latex]<\/li>\r\n \t
  19. [latex]\\dfrac{2-\\sqrt{3}}{1+\\sqrt{2}}[\/latex]<\/li>\r\n \t
  20. [latex]\\dfrac{-1+\\sqrt{3}}{\\sqrt{3}-1}[\/latex]<\/li>\r\n<\/ol>\r\nAnswer Key 9.5<\/a>","rendered":"

    It is considered non-conventional to have a radical in the denominator. When this happens, generally the numerator and denominator are multiplied by the same factors to remove the radical denominator. The problems in the previous section dealt with removing a monomial radical. In this section, the previous strategy is expanded to include binomial radicals.<\/p>\n

    \n
    \n

    Example 9.5.1<\/p>\n<\/header>\n

    \n

    Rationalize [latex]\\dfrac{\\sqrt{3}-9}{2\\sqrt{6}}[\/latex].<\/p>\n

    To rationalize the denominator, multiply out the [latex]\\sqrt{6}[\/latex].<\/p>\n

    This will look like:<\/p>\n

    [latex]\\dfrac{(\\sqrt{3}-9)(\\sqrt{6})}{2\\sqrt{6}(\\sqrt{6})}[\/latex]<\/p>\n

    Multiplying the [latex]\\sqrt{6}[\/latex] throughout yields:<\/p>\n

    [latex]\\dfrac{(\\sqrt{3})(\\sqrt{6})-(9)(\\sqrt{6})}{2\\sqrt{36}}[\/latex]<\/p>\n

    Which reduces to:<\/p>\n

    [latex]\\dfrac{3\\sqrt{2}-9\\sqrt{6}}{2\\cdot 6}[\/latex]<\/p>\n

    And simplifies to:<\/p>\n

    [latex]\\dfrac{\\sqrt{2}-3\\sqrt{6}}{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n

    Please note that, in reducing the numerator and denominator by the factor 3, reduce each term in the numerator by 3.<\/p>\n

    Quite often, there will be a denominator binomial that contains radicals. For these problems, it is easiest to use a feature from the sum and difference of squares: [latex]a^2 - b^2 = (a + b)(a - b)[\/latex].<\/p>\n

    [latex](a + b)(a - b)[\/latex] are termed conjugates of each other. They are identical binomials, except that their signs are opposite. When encountering radical binomials, simply multiply by the conjugates to square out the radical.<\/p>\n

    \n
    \n

    Example 9.5.2<\/p>\n<\/header>\n

    \n

    Square out the radical of the binomial [latex](\\sqrt{3} - \\sqrt{5})[\/latex] using its conjugate.<\/p>\n

    The conjugate of [latex](\\sqrt{3} - \\sqrt{5})[\/latex] is [latex](\\sqrt{3} + \\sqrt{5})[\/latex].<\/p>\n

    When multiplied, these conjugates yield [latex](\\sqrt{3}-\\sqrt{5})(\\sqrt{3}+\\sqrt{5})[\/latex] or [latex](\\sqrt{3})^2-(\\sqrt{5})^2[\/latex].<\/p>\n

    This yields 3 \u2212 5 = \u22122.<\/p>\n<\/div>\n<\/div>\n

    When encountering a radicalized binomial denominator, the best solution is to multiply both the numerator and denominator by the conjugate of the denominator.<\/p>\n

    \n
    \n

    Example 9.5.3<\/p>\n<\/header>\n

    \n

    Rationalize the denominator of [latex]\\dfrac{\\sqrt{6}}{\\sqrt{6}+\\sqrt{13}}[\/latex].<\/p>\n

    Multiplying the numerator and denominator by the denominator’s conjugate yields:<\/p>\n

    [latex]\\dfrac{\\sqrt{6}}{\\sqrt{6}+\\sqrt{13}} \\cdot \\dfrac{(\\sqrt{6}-\\sqrt{13})}{(\\sqrt{6}-\\sqrt{13})}[\/latex]<\/p>\n

    When multiplied out, this yields:<\/p>\n

    [latex]\\dfrac{(\\sqrt{6})^2-\\sqrt{6}\\sqrt{13}}{(\\sqrt{6})^2-(\\sqrt{13})^2}[\/latex]<\/p>\n

    Which reduces to:<\/p>\n

    [latex]\\dfrac{6-\\sqrt{78}}{6-13}[\/latex]<\/p>\n

    Or:<\/p>\n

    [latex]\\dfrac{6-\\sqrt{78}}{-7}[\/latex]<\/p>\n<\/div>\n<\/div>\n

    Questions<\/h1>\n

    Rationalize the following radical fractions.<\/p>\n

      \n
    1. [latex]\\dfrac{4+2\\sqrt{3}}{\\sqrt{3}}[\/latex]<\/li>\n
    2. [latex]\\dfrac{-4+\\sqrt{3}}{4\\sqrt{3}}[\/latex]<\/li>\n
    3. [latex]\\dfrac{4+2\\sqrt{3}}{5\\sqrt{6}}[\/latex]<\/li>\n
    4. [latex]\\dfrac{2\\sqrt{3}-2}{2\\sqrt{3}}[\/latex]<\/li>\n
    5. [latex]\\dfrac{2-5\\sqrt{5}}{4\\sqrt{3}}[\/latex]<\/li>\n
    6. [latex]\\dfrac{\\sqrt{5}+4}{4\\sqrt{5}}[\/latex]<\/li>\n
    7. [latex]\\dfrac{\\sqrt{2}-3\\sqrt{3}}{\\sqrt{3}}[\/latex]<\/li>\n
    8. [latex]\\dfrac{\\sqrt{5}-\\sqrt{2}}{3\\sqrt{6}}[\/latex]<\/li>\n
    9. [latex]\\dfrac{5}{3\\sqrt{5}+\\sqrt{2}}[\/latex]<\/li>\n
    10. [latex]\\dfrac{5}{\\sqrt{3}+4\\sqrt{5}}[\/latex]<\/li>\n
    11. [latex]\\dfrac{2}{5+\\sqrt{2}}[\/latex]<\/li>\n
    12. [latex]\\dfrac{5}{2\\sqrt{3}-\\sqrt{2}}[\/latex]<\/li>\n
    13. [latex]\\dfrac{3}{4-\\sqrt{3}}[\/latex]<\/li>\n
    14. [latex]\\dfrac{4}{\\sqrt{2}-2}[\/latex]<\/li>\n
    15. [latex]\\dfrac{4}{3+\\sqrt{5}}[\/latex]<\/li>\n
    16. [latex]\\dfrac{2}{\\sqrt{5}+2\\sqrt{3}}[\/latex]<\/li>\n
    17. [latex]\\dfrac{-3+2\\sqrt{3}}{\\sqrt{3}+2}[\/latex]<\/li>\n
    18. [latex]\\dfrac{4+\\sqrt{5}}{2+2\\sqrt{5}}[\/latex]<\/li>\n
    19. [latex]\\dfrac{2-\\sqrt{3}}{1+\\sqrt{2}}[\/latex]<\/li>\n
    20. [latex]\\dfrac{-1+\\sqrt{3}}{\\sqrt{3}-1}[\/latex]<\/li>\n<\/ol>\n

      Answer Key 9.5<\/a><\/p>\n","protected":false},"author":90,"menu_order":5,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-1432","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":1422,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1432","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":2,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1432\/revisions"}],"predecessor-version":[{"id":2148,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1432\/revisions\/2148"}],"part":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/parts\/1422"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1432\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=1432"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapter-type?post=1432"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=1432"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=1432"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}