{"id":1978,"date":"2021-12-02T19:40:17","date_gmt":"2021-12-03T00:40:17","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/answer-key-10-2\/"},"modified":"2022-11-02T10:38:52","modified_gmt":"2022-11-02T14:38:52","slug":"answer-key-10-2","status":"publish","type":"back-matter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/back-matter\/answer-key-10-2\/","title":{"raw":"Answer Key 10.2","rendered":"Answer Key 10.2"},"content":{"raw":"
    \n \t
  1. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\n\\sqrt{x^2}&=&\\sqrt{75} \\\\\nx&=&\\pm \\sqrt{25\\cdot 3} \\\\\nx&=&\\pm 5\\sqrt{3}\n\\end{array}[\/latex]<\/li>\n \t
  2. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrl}\n\\sqrt[3]{x^3}&=&\\sqrt[3]{-8} \\\\\nx&=&-2\n\\end{array}[\/latex]<\/li>\n \t
  3. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrrrl}\nx^2&+&5&=&13 \\\\\n&-&5&&-5 \\\\\n\\hline\n&&\\sqrt{x^2}&=&\\sqrt{8} \\\\\n&&x&=&\\pm \\sqrt{4\\cdot 2} \\\\\n&&x&=&\\pm 2\\sqrt{2}\n\\end{array}[\/latex]<\/li>\n \t
  4. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrrrl}\n4x^3&-&2&=&106 \\\\\n&+&2&&+2 \\\\\n\\hline\n&&\\dfrac{4x^3}{4}&=&\\dfrac{108}{4} \\\\ \\\\\n&&x^3&=&27 \\\\\n&&\\sqrt[3]{x^3}&=&\\sqrt[3]{27} \\\\\n&&x&=&3\n\\end{array}[\/latex]<\/li>\n \t
  5. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrrrl}\n3x^2&+&1&=&73 \\\\\n&-&1&&-1 \\\\\n\\hline\n&&\\dfrac{3x^2}{3}&=&\\dfrac{72}{3} \\\\ \\\\\n&&x^2&=&24 \\\\\n&&\\sqrt{x^2}&=&\\pm \\sqrt{24} \\\\\n&&x&=&\\pm \\sqrt{4\\cdot 6} \\\\\n&&x&=&\\pm 2\\sqrt{6}\n\\end{array}[\/latex]<\/li>\n \t
  6. [latex]\\sqrt{(x-4)^2}=\\sqrt{49}[\/latex]\n[latex]\\begin{array}{rrrrrrr}\nx&-&4&=&\\pm 7 && \\\\\n&&x&=&4 & \\pm & 7 \\\\\n&&x&=&11, & -3&\n\\end{array}[\/latex]<\/li>\n \t
  7. [latex]\\sqrt[5]{(x+2)^5}=\\sqrt[5]{-3^5}[\/latex]\n[latex]\\begin{array}{rrrrr}\nx&+&2&=&-3 \\\\\n&-&2&&-2 \\\\\n\\hline\n&&x&=&-5\n\\end{array}[\/latex]<\/li>\n \t
  8. [latex]\\sqrt[4]{(5x+1)^4}=\\pm \\sqrt[4]{2^4}[\/latex]\n[latex]\\begin{array}{rrrrrrr}\n5x&+&1&=&\\pm &2& \\\\\n&-&1&&-&1& \\\\\n\\hline\n&&5x&=&-1&\\pm &2 \\\\ \\\\\n&&x&=&-\\dfrac{3}{5}&\\text{or}&\\dfrac{1}{5}\n\\end{array}[\/latex]<\/li>\n \t
  9. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrrrrrr}\n(2x&+&5)^3&-&6&=&21 \\\\\n&&&+&6&&+6 \\\\\n\\hline\n&&(2x&+&5)^3&=&27 \\\\\n\\end{array}[\/latex]\n[latex]\\sqrt[3]{(2x+5)^3}=\\sqrt[3]{27}[\/latex]\n[latex]\\begin{array}{rrrrr}2x&+&5&=&3 \\\\\n&-&5&&-5 \\\\\n\\hline\n&&2x&=&-2 \\\\\n&&x&=&-1\n\\end{array}[\/latex]<\/li>\n \t
  10. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrrrrrr}\n(2x&+&1)^2&+&3&=&21 \\\\\n&&&-&3&&-3 \\\\\n\\hline\n&&(2x&+&1)^2&=&18\n\\end{array}[\/latex]\n[latex]\\sqrt{{(2x+1)}^{2}}=\\sqrt{18} \\Rightarrow\\sqrt{9\\cdot 2}\\Rightarrow\\pm 3\\sqrt{2}[\/latex]\n[latex]\\begin{array}{rrrrl}\n2x&+&1&=&\\pm 3\\sqrt{2} \\\\\n&-&1&&-1 \\\\\n\\hline\n&&\\dfrac{2x}{2}&=&\\dfrac{-1\\pm 3\\sqrt{2}}{2} \\\\ \\\\\n&&x&=&\\dfrac{-1\\pm 3\\sqrt{2}}{2}\n\\end{array}[\/latex]<\/li>\n \t
  11. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrrrl}\n(x&-&1)^{\\frac{2}{3}}&=&2^4 \\\\\n(x&-&1)^{\\frac{2}{3}\\cdot \\frac{3}{2}}&=&2^{4\\cdot \\frac{3}{2}} \\\\\nx&-&1&=&\\pm 2^6 \\\\\n&+&1&&+1 \\\\\n\\hline\n&&x&=&1 \\pm 2^6 \\\\\n&&x&=&65\\text{ or }-63\n\\end{array}[\/latex]<\/li>\n \t
  12. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrrrl}\n(x&-&1)^{\\frac{3}{2}}&=&2^3 \\\\\n(x&-&1)^{\\frac{3}{2}\\cdot \\frac{2}{3}}&=&2^{3\\cdot \\frac{2}{3}} \\\\\nx&-&1&=&2^2 \\\\\n&+&1&=&+1 \\\\\n\\hline\n&&x&=&5\n\\end{array}[\/latex]<\/li>\n \t
  13. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrlrl}\n(2&-&\\phantom{-}x)^{\\frac{3}{2}}&=&\\phantom{-}3^3 \\\\\n(2&-&\\phantom{-}x)^{\\frac{3}{2}\\cdot \\frac{2}{3}}&=&\\phantom{-}3^{3\\cdot \\frac{2}{3}} \\\\\n2&-&\\phantom{-}x&=&\\phantom{-}3^2 \\\\\n-2&&&&-2 \\\\\n\\hline\n&&-x&=&\\phantom{-}7 \\\\\n&&\\phantom{-}x&=&-7\n\\end{array}[\/latex]<\/li>\n \t
  14. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrlrl}\n(2x&+&3)^{\\frac{4}{3}}&=&2^4 \\\\\n(2x&+&3)^{\\frac{4}{3}\\cdot \\frac{3}{4}}&=&2^{4\\cdot \\frac{3}{4}} \\\\\n2x&+&3&=&\\pm 2^3 \\\\\n&-&3&&-3 \\\\\n\\hline\n&&2x&=&5 \\\\\n&&2x&=&-11 \\\\ \\\\\n&&x&=&\\dfrac{5}{2}, -\\dfrac{11}{2}\n\\end{array}[\/latex]<\/li>\n \t
  15. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrlrl}\n(2x&-&3)^{\\frac{2}{3}}&=&2^2 \\\\\n(2x&-&3)^{\\frac{2}{3}\\cdot \\frac{3}{2}}&=&2^{2\\cdot \\frac{3}{2}} \\\\\n2x&-&3&=&\\pm 2^3 \\\\\n&+&3&&+3 \\\\\n\\hline\n&&2x&=&11 \\\\\n&&2x&=&-5 \\\\ \\\\\n&&x&=&\\dfrac{11}{2}, -\\dfrac{5}{2}\n\\end{array}[\/latex]<\/li>\n \t
  16. [latex]\\phantom{a}[\/latex]\n[latex]\\begin{array}[t]{rrlrl}\n(3x&-&2)^{\\frac{4}{5}}&=&2^4 \\\\\n(3x&-&2)^{\\frac{4}{5}\\cdot \\frac{5}{4}}&=&2^{4\\cdot \\frac{5}{4}} \\\\\n3x&-&2&=&\\pm 2^5 \\\\\n&+&2&&+2 \\\\\n\\hline\n&&\\dfrac{3x}{3}&=&\\dfrac{34}{3} \\\\ \\\\\n&&\\dfrac{3x}{3}&=&\\dfrac{-30}{3} \\\\ \\\\\n&&x&=&\\dfrac{34}{3}, -10\n\\end{array}[\/latex]<\/li>\n<\/ol>","rendered":"
      \n
    1. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} \\sqrt{x^2}&=&\\sqrt{75} \\\\ x&=&\\pm \\sqrt{25\\cdot 3} \\\\ x&=&\\pm 5\\sqrt{3} \\end{array}[\/latex]<\/li>\n
    2. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrl} \\sqrt[3]{x^3}&=&\\sqrt[3]{-8} \\\\ x&=&-2 \\end{array}[\/latex]<\/li>\n
    3. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrrrl} x^2&+&5&=&13 \\\\ &-&5&&-5 \\\\ \\hline &&\\sqrt{x^2}&=&\\sqrt{8} \\\\ &&x&=&\\pm \\sqrt{4\\cdot 2} \\\\ &&x&=&\\pm 2\\sqrt{2} \\end{array}[\/latex]<\/li>\n
    4. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrrrl} 4x^3&-&2&=&106 \\\\ &+&2&&+2 \\\\ \\hline &&\\dfrac{4x^3}{4}&=&\\dfrac{108}{4} \\\\ \\\\ &&x^3&=&27 \\\\ &&\\sqrt[3]{x^3}&=&\\sqrt[3]{27} \\\\ &&x&=&3 \\end{array}[\/latex]<\/li>\n
    5. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrrrl} 3x^2&+&1&=&73 \\\\ &-&1&&-1 \\\\ \\hline &&\\dfrac{3x^2}{3}&=&\\dfrac{72}{3} \\\\ \\\\ &&x^2&=&24 \\\\ &&\\sqrt{x^2}&=&\\pm \\sqrt{24} \\\\ &&x&=&\\pm \\sqrt{4\\cdot 6} \\\\ &&x&=&\\pm 2\\sqrt{6} \\end{array}[\/latex]<\/li>\n
    6. [latex]\\sqrt{(x-4)^2}=\\sqrt{49}[\/latex]
      \n[latex]\\begin{array}{rrrrrrr} x&-&4&=&\\pm 7 && \\\\ &&x&=&4 & \\pm & 7 \\\\ &&x&=&11, & -3& \\end{array}[\/latex]<\/li>\n
    7. [latex]\\sqrt[5]{(x+2)^5}=\\sqrt[5]{-3^5}[\/latex]
      \n[latex]\\begin{array}{rrrrr} x&+&2&=&-3 \\\\ &-&2&&-2 \\\\ \\hline &&x&=&-5 \\end{array}[\/latex]<\/li>\n
    8. [latex]\\sqrt[4]{(5x+1)^4}=\\pm \\sqrt[4]{2^4}[\/latex]
      \n[latex]\\begin{array}{rrrrrrr} 5x&+&1&=&\\pm &2& \\\\ &-&1&&-&1& \\\\ \\hline &&5x&=&-1&\\pm &2 \\\\ \\\\ &&x&=&-\\dfrac{3}{5}&\\text{or}&\\dfrac{1}{5} \\end{array}[\/latex]<\/li>\n
    9. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrrrrrr} (2x&+&5)^3&-&6&=&21 \\\\ &&&+&6&&+6 \\\\ \\hline &&(2x&+&5)^3&=&27 \\\\ \\end{array}[\/latex]
      \n[latex]\\sqrt[3]{(2x+5)^3}=\\sqrt[3]{27}[\/latex]
      \n[latex]\\begin{array}{rrrrr}2x&+&5&=&3 \\\\ &-&5&&-5 \\\\ \\hline &&2x&=&-2 \\\\ &&x&=&-1 \\end{array}[\/latex]<\/li>\n
    10. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrrrrrr} (2x&+&1)^2&+&3&=&21 \\\\ &&&-&3&&-3 \\\\ \\hline &&(2x&+&1)^2&=&18 \\end{array}[\/latex]
      \n[latex]\\sqrt{{(2x+1)}^{2}}=\\sqrt{18} \\Rightarrow\\sqrt{9\\cdot 2}\\Rightarrow\\pm 3\\sqrt{2}[\/latex]
      \n[latex]\\begin{array}{rrrrl} 2x&+&1&=&\\pm 3\\sqrt{2} \\\\ &-&1&&-1 \\\\ \\hline &&\\dfrac{2x}{2}&=&\\dfrac{-1\\pm 3\\sqrt{2}}{2} \\\\ \\\\ &&x&=&\\dfrac{-1\\pm 3\\sqrt{2}}{2} \\end{array}[\/latex]<\/li>\n
    11. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrrrl} (x&-&1)^{\\frac{2}{3}}&=&2^4 \\\\ (x&-&1)^{\\frac{2}{3}\\cdot \\frac{3}{2}}&=&2^{4\\cdot \\frac{3}{2}} \\\\ x&-&1&=&\\pm 2^6 \\\\ &+&1&&+1 \\\\ \\hline &&x&=&1 \\pm 2^6 \\\\ &&x&=&65\\text{ or }-63 \\end{array}[\/latex]<\/li>\n
    12. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrrrl} (x&-&1)^{\\frac{3}{2}}&=&2^3 \\\\ (x&-&1)^{\\frac{3}{2}\\cdot \\frac{2}{3}}&=&2^{3\\cdot \\frac{2}{3}} \\\\ x&-&1&=&2^2 \\\\ &+&1&=&+1 \\\\ \\hline &&x&=&5 \\end{array}[\/latex]<\/li>\n
    13. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrlrl} (2&-&\\phantom{-}x)^{\\frac{3}{2}}&=&\\phantom{-}3^3 \\\\ (2&-&\\phantom{-}x)^{\\frac{3}{2}\\cdot \\frac{2}{3}}&=&\\phantom{-}3^{3\\cdot \\frac{2}{3}} \\\\ 2&-&\\phantom{-}x&=&\\phantom{-}3^2 \\\\ -2&&&&-2 \\\\ \\hline &&-x&=&\\phantom{-}7 \\\\ &&\\phantom{-}x&=&-7 \\end{array}[\/latex]<\/li>\n
    14. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrlrl} (2x&+&3)^{\\frac{4}{3}}&=&2^4 \\\\ (2x&+&3)^{\\frac{4}{3}\\cdot \\frac{3}{4}}&=&2^{4\\cdot \\frac{3}{4}} \\\\ 2x&+&3&=&\\pm 2^3 \\\\ &-&3&&-3 \\\\ \\hline &&2x&=&5 \\\\ &&2x&=&-11 \\\\ \\\\ &&x&=&\\dfrac{5}{2}, -\\dfrac{11}{2} \\end{array}[\/latex]<\/li>\n
    15. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrlrl} (2x&-&3)^{\\frac{2}{3}}&=&2^2 \\\\ (2x&-&3)^{\\frac{2}{3}\\cdot \\frac{3}{2}}&=&2^{2\\cdot \\frac{3}{2}} \\\\ 2x&-&3&=&\\pm 2^3 \\\\ &+&3&&+3 \\\\ \\hline &&2x&=&11 \\\\ &&2x&=&-5 \\\\ \\\\ &&x&=&\\dfrac{11}{2}, -\\dfrac{5}{2} \\end{array}[\/latex]<\/li>\n
    16. [latex]\\phantom{a}[\/latex]
      \n[latex]\\begin{array}[t]{rrlrl} (3x&-&2)^{\\frac{4}{5}}&=&2^4 \\\\ (3x&-&2)^{\\frac{4}{5}\\cdot \\frac{5}{4}}&=&2^{4\\cdot \\frac{5}{4}} \\\\ 3x&-&2&=&\\pm 2^5 \\\\ &+&2&&+2 \\\\ \\hline &&\\dfrac{3x}{3}&=&\\dfrac{34}{3} \\\\ \\\\ &&\\dfrac{3x}{3}&=&\\dfrac{-30}{3} \\\\ \\\\ &&x&=&\\dfrac{34}{3}, -10 \\end{array}[\/latex]<\/li>\n<\/ol>\n","protected":false},"author":90,"menu_order":94,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"back-matter-type":[],"contributor":[],"license":[56],"class_list":["post-1978","back-matter","type-back-matter","status-publish","hentry","license-cc-by-nc-sa"],"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/1978","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter"}],"about":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/types\/back-matter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":1,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/1978\/revisions"}],"predecessor-version":[{"id":1979,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/1978\/revisions\/1979"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter\/1978\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=1978"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/back-matter-type?post=1978"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=1978"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=1978"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}