{"id":1758,"date":"2019-08-08T17:06:43","date_gmt":"2019-08-08T21:06:43","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=back-matter&p=1758"},"modified":"2019-08-08T17:59:16","modified_gmt":"2019-08-08T21:59:16","slug":"answer-key-10-2","status":"publish","type":"back-matter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/back-matter\/answer-key-10-2\/","title":{"raw":"Answer Key 10.2","rendered":"Answer Key 10.2"},"content":{"raw":"[latexpage]\r\n
    \r\n \t
  1. \\(\\begin{array}{rrl}\r\n\\\\ \\\\\r\n\\sqrt{x^2}&=&\\sqrt{75} \\\\\r\nx&=&\\pm \\sqrt{25\\cdot 3} \\\\\r\nx&=&\\pm 5\\sqrt{3}\r\n\\end{array}\\)<\/li>\r\n \t
  2. \\(\\begin{array}{rrl}\r\n\\\\\r\n\\sqrt[3]{x^3}&=&\\sqrt[3]{-8} \\\\\r\nx&=&-2\r\n\\end{array}\\)<\/li>\r\n \t
  3. \\(\\begin{array}{rrrrl}\r\n\\\\ \\\\ \\\\ \\\\\r\nx^2&+&5&=&13 \\\\\r\n&-&5&&-5 \\\\\r\n\\midrule\r\n&&\\sqrt{x^2}&=&\\sqrt{8} \\\\\r\n&&x&=&\\pm \\sqrt{4\\cdot 2} \\\\\r\n&&x&=&\\pm 2\\sqrt{2}\r\n\\end{array}\\)<\/li>\r\n \t
  4. \\(\\begin{array}{rrrrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n4x^3&-&2&=&106 \\\\\r\n&+&2&&+2 \\\\\r\n\\midrule\r\n&&\\dfrac{4x^3}{4}&=&\\dfrac{108}{4} \\\\ \\\\\r\n&&x^3&=&27 \\\\\r\n&&\\sqrt[3]{x^3}&=&\\sqrt[3]{27} \\\\\r\n&&x&=&3\r\n\\end{array}\\)<\/li>\r\n \t
  5. \\(\\begin{array}{rrrrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n3x^2&+&1&=&73 \\\\\r\n&-&1&&-1 \\\\\r\n\\midrule\r\n&&\\dfrac{3x^2}{3}&=&\\dfrac{72}{3} \\\\ \\\\\r\n&&x^2&=&24 \\\\\r\n&&\\sqrt{x^2}&=&\\pm \\sqrt{24} \\\\\r\n&&x&=&\\pm \\sqrt{4\\cdot 6} \\\\\r\n&&x&=&\\pm 2\\sqrt{6}\r\n\\end{array}\\)<\/li>\r\n \t
  6. \\(\\sqrt{(x-4)^2}=\\sqrt{49}\\)\r\n\r\n\\(\\begin{array}{rrrrrrr}\r\nx&-&4&=&\\pm 7 && \\\\\r\n&&x&=&4 & \\pm & 7\u00a0 \\\\\r\n&&x&=&11, & -3&\r\n\\end{array}\\)<\/li>\r\n \t
  7. \\(\\sqrt[5]{(x+2)^5}=\\sqrt[5]{-3^5}\\)\r\n\r\n\\(\\begin{array}{rrrrr}\r\nx&+&2&=&-3 \\\\\r\n&-&2&&-2 \\\\\r\n\\midrule\r\n&&x&=&-5\r\n\\end{array}\\)<\/li>\r\n \t
  8. \\(\\sqrt[4]{(5x+1)^4}=\\pm \\sqrt[4]{2^4}\\)\r\n\r\n\\(\\begin{array}{rrrrrrr}\r\n5x&+&1&=&\\pm &2& \\\\\r\n&-&1&&-&1& \\\\\r\n\\midrule\r\n&&5x&=&-1&\\pm &2 \\\\ \\\\\r\n&&x&=&-\\dfrac{3}{5}&\\text{or}&\\dfrac{1}{5}\r\n\\end{array}\\)<\/li>\r\n \t
  9. \\(\\begin{array}{rrrrrrr}\r\n\\\\ \\\\ \\\\\r\n(2x&+&5)^3&-&6&=&21 \\\\\r\n&&&+&6&&+6 \\\\\r\n\\midrule\r\n&&(2x&+&5)^3&=&27 \\\\\r\n\\end{array}\\)\r\n\r\n\\(\\sqrt[3]{(2x+5)^3}=\\sqrt[3]{27}\\)\r\n\r\n\\(\\begin{array}{rrrrr}\r\n2x&+&5&=&3 \\\\\r\n&-&5&&-5 \\\\\r\n\\midrule\r\n&&2x&=&-2 \\\\\r\n&&x&=&-1\r\n\\end{array}\\)<\/li>\r\n \t
  10. \\(\\begin{array}{rrrrrrr}\r\n\\\\ \\\\ \\\\\r\n(2x&+&1)^2&+&3&=&21 \\\\\r\n&&&-&3&&-3 \\\\\r\n\\midrule\r\n&&(2x&+&1)^2&=&18\r\n\\end{array}\\)\r\n\r\n\\(\\sqrt{(2x+1)^2}&=&\\sqrt{18} \\Rightarrow \\sqrt{9\\cdot 2}\\Rightarrow \\pm 3\\sqrt{2}\\)\r\n\r\n\\(\\begin{array}{rrrrl}\r\n2x&+&1&=&\\pm 3\\sqrt{2} \\\\\r\n&-&1&&-1 \\\\\r\n\\midrule\r\n&&\\dfrac{2x}{2}&=&\\dfrac{-1\\pm 3\\sqrt{2}}{2} \\\\ \\\\\r\n&&x&=&\\dfrac{-1\\pm 3\\sqrt{2}}{2}\r\n\\end{array}\\)<\/li>\r\n \t
  11. \\(\\begin{array}{rrrrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(x&-&1)^{\\frac{2}{3}}&=&2^4 \\\\\r\n(x&-&1)^{\\frac{2}{3}\\cdot \\frac{3}{2}}&=&2^{4\\cdot \\frac{3}{2}} \\\\\r\nx&-&1&=&\\pm 2^6 \\\\\r\n&+&1&&+1 \\\\\r\n\\midrule\r\n&&x&=&1 \\pm 2^6 \\\\\r\n&&x&=&65\\text{ or }-63\r\n\\end{array}\\)<\/li>\r\n \t
  12. \\(\\begin{array}{rrrrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\\r\n(x&-&1)^{\\frac{3}{2}}&=&2^3 \\\\\r\n(x&-&1)^{\\frac{3}{2}\\cdot \\frac{2}{3}}&=&2^{3\\cdot \\frac{2}{3}} \\\\\r\nx&-&1&=&2^2 \\\\\r\n&+&1&=&+1 \\\\\r\n\\midrule\r\n&&x&=&5\r\n\\end{array}\\)<\/li>\r\n \t
  13. \\(\\begin{array}{rrlrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(2&-&\\phantom{-}x)^{\\frac{3}{2}}&=&\\phantom{-}3^3 \\\\\r\n(2&-&\\phantom{-}x)^{\\frac{3}{2}\\cdot \\frac{2}{3}}&=&\\phantom{-}3^{3\\cdot \\frac{2}{3}} \\\\\r\n2&-&\\phantom{-}x&=&\\phantom{-}3^2 \\\\\r\n-2&&&&-2 \\\\\r\n\\midrule\r\n&&-x&=&\\phantom{-}7 \\\\\r\n&&\\phantom{-}x&=&-7\r\n\\end{array}\\)<\/li>\r\n \t
  14. \\(\\begin{array}{rrlrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(2x&+&3)^{\\frac{4}{3}}&=&2^4 \\\\\r\n(2x&+&3)^{\\frac{4}{3}\\cdot \\frac{3}{4}}&=&2^{4\\cdot \\frac{3}{4}} \\\\\r\n2x&+&3&=&\\pm 2^3 \\\\\r\n&-&3&&-3 \\\\\r\n\\midrule\r\n&&2x&=&5 \\\\\r\n&&2x&=&-11 \\\\ \\\\\r\n&&x&=&\\dfrac{5}{2}, -\\dfrac{11}{2}\r\n\\end{array}\\)<\/li>\r\n \t
  15. \\(\\begin{array}{rrlrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(2x&-&3)^{\\frac{2}{3}}&=&2^2 \\\\\r\n(2x&-&3)^{\\frac{2}{3}\\cdot \\frac{3}{2}}&=&2^{2\\cdot \\frac{3}{2}} \\\\\r\n2x&-&3&=&\\pm 2^3 \\\\\r\n&+&3&&+3 \\\\\r\n\\midrule\r\n&&2x&=&11 \\\\\r\n&&2x&=&-5 \\\\ \\\\\r\n&&x&=&\\dfrac{11}{2}, -\\dfrac{5}{2}\r\n\\end{array}\\)<\/li>\r\n \t
  16. \\(\\begin{array}{rrlrl}\r\n\\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\ \\\\\r\n(3x&-&2)^{\\frac{4}{5}}&=&2^4 \\\\\r\n(3x&-&2)^{\\frac{4}{5}\\cdot \\frac{5}{4}}&=&2^{4\\cdot \\frac{5}{4}} \\\\\r\n3x&-&2&=&\\pm 2^5 \\\\\r\n&+&2&&+2 \\\\\r\n\\midrule\r\n&&\\dfrac{3x}{3}&=&\\dfrac{34}{3} \\\\ \\\\\r\n&&\\dfrac{3x}{3}&=&\\dfrac{-30}{3} \\\\ \\\\\r\n&&x&=&\\dfrac{34}{3}, -10\r\n\\end{array}\\)<\/li>\r\n<\/ol>","rendered":"
      \n
    1. \"\begin{array}{rrl}<\/li>\n
    2. \"\begin{array}{rrl}<\/li>\n
    3. \"\begin{array}{rrrrl}<\/li>\n
    4. \"\begin{array}{rrrrl}<\/li>\n
    5. \"\begin{array}{rrrrl}<\/li>\n
    6. \"\sqrt{(x-4)^2}=\sqrt{49}\"\n

      \"\begin{array}{rrrrrrr}<\/li>\n

    7. \"\sqrt[5]{(x+2)^5}=\sqrt[5]{-3^5}\"\n

      \"\begin{array}{rrrrr}<\/li>\n

    8. \"\sqrt[4]{(5x+1)^4}=\pm \sqrt[4]{2^4}\"\n

      \"\begin{array}{rrrrrrr}<\/li>\n

    9. \"\begin{array}{rrrrrrr}\n

      \"\sqrt[3]{(2x+5)^3}=\sqrt[3]{27}\"<\/p>\n

      \"\begin{array}{rrrrr}<\/li>\n

    10. \"\begin{array}{rrrrrrr}\n

      \"\sqrt{(2x+1)^2}&=&\sqrt{18} \Rightarrow \sqrt{9\cdot 2}\Rightarrow \pm 3\sqrt{2}\"<\/p>\n

      \"\begin{array}{rrrrl}<\/li>\n

    11. \"\begin{array}{rrrrl}<\/li>\n
    12. \"\begin{array}{rrrrl}<\/li>\n
    13. \"\begin{array}{rrlrl}<\/li>\n
    14. \"\begin{array}{rrlrl}<\/li>\n
    15. \"\begin{array}{rrlrl}<\/li>\n
    16. \"\begin{array}{rrlrl}<\/li>\n<\/ol>\n","protected":false},"author":727,"menu_order":89,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"back-matter-type":[],"contributor":[],"license":[],"class_list":["post-1758","back-matter","type-back-matter","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1758","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/types\/back-matter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/users\/727"}],"version-history":[{"count":3,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1758\/revisions"}],"predecessor-version":[{"id":1761,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1758\/revisions\/1761"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1758\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=1758"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter-type?post=1758"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=1758"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=1758"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}