{"id":1801,"date":"2019-08-13T16:22:38","date_gmt":"2019-08-13T20:22:38","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=back-matter&p=1801"},"modified":"2019-08-13T16:59:05","modified_gmt":"2019-08-13T20:59:05","slug":"answer-key-10-8","status":"publish","type":"back-matter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/back-matter\/answer-key-10-8\/","title":{"raw":"Answer Key 10.8","rendered":"Answer Key 10.8"},"content":{"raw":"[latexpage]\r\n
    \r\n \t
  1. \\(x=5, x=2\\)\r\n\\(x-5=0, x-2=0\\)\r\n\r\n\\(0=(x-5)(x-2)\\)\r\n\\(0=x^2-7x+10\\)<\/li>\r\n \t
  2. \\(x=3, x=6\\)\r\n\\(x-3=0, x-6=0\\)\r\n\r\n\\((x-3)(x-6)=0\\)\r\n\\(0=x^2-9x+18\\)<\/li>\r\n \t
  3. \\(x=20, x=2\\)\r\n\\(x-20=0, x-2=0\\)\r\n\r\n\\(0=(x-20)(x-2)\\)\r\n\\(0=x^2-22x+40\\)<\/li>\r\n \t
  4. \\(x=13, x=1\\)\r\n\\(x-13=0, x-1=0\\)\r\n\r\n\\(0=(x-13)(x-1)\\)\r\n\\(0=x^2-14x+13\\)<\/li>\r\n \t
  5. \\(x=4, x=4\\)\r\n\\(x-4=0, x-4=0\\)\r\n\r\n\\(0=(x-4)(x-4)\\)\r\n\\(0=x^2-8x+16\\)<\/li>\r\n \t
  6. \\(x=0, x=9\\)\r\n\\(x-9=0, x\\)\r\n\r\n\\(x(x-9)=0\\)\r\n\\(0=x^2-9x\\)<\/li>\r\n \t
  7. \\(x=\\dfrac{3}{4}, x=\\dfrac{1}{4} \\\\ \\)\r\n\\(x-\\dfrac{3}{4}=0, x-\\dfrac{1}{4}=0 \\)\r\n\r\n\\(0=\\left(x-\\dfrac{3}{4}\\right)\\left(x-\\dfrac{1}{4}\\right) \\\\ \\)\r\n\\(0=x^2-x+\\dfrac{3}{16}\\)<\/li>\r\n \t
  8. \\(x=\\dfrac{5}{8}, x=\\dfrac{5}{7} \\\\ \\)\r\n\\(x-\\dfrac{5}{8}=0, x-\\dfrac{5}{7}=0\\)\r\n\r\n\\(0=\\left(x-\\dfrac{5}{8}\\right)\\left(x-\\dfrac{5}{7}\\right) \\\\ \\)\r\n\\(0=x^2-\\dfrac{75}{56}x+\\dfrac{25}{56}\\)<\/li>\r\n \t
  9. \\(x=\\dfrac{1}{2}, x=\\dfrac{1}{3} \\\\ \\)\r\n\\(x-\\dfrac{1}{2}=0, x-\\dfrac{1}{3}=0\\)\r\n\r\n\\(0=\\left(x-\\dfrac{1}{2}\\right)\\left(x-\\dfrac{1}{3}\\right) \\\\ \\)\r\n\\(0=x^2-\\dfrac{5}{6}x+\\dfrac{1}{6}\\)<\/li>\r\n \t
  10. \\(x=\\dfrac{1}{2}, x=\\dfrac{2}{3} \\\\ \\)\r\n\\(x-\\dfrac{1}{2}=0, x-\\dfrac{2}{3}=0\\)\r\n\r\n\\(0=\\left(x-\\dfrac{1}{2}\\right)\\left(x-\\dfrac{2}{3}\\right) \\\\ \\)\r\n\\(0=x^2-\\dfrac{7}{6}x+\\dfrac{1}{3}\\)<\/li>\r\n \t
  11. \\(x=5, x=-5\\)\r\n\\(x-5=0, x+5=0\\)\r\n\r\n\\(0=(x-5)(x+5)\\)\r\n\\(0=x^2-25\\)<\/li>\r\n \t
  12. \\(x=1, x=-1\\)\r\n\\(x-1=0, x+1=0\\)\r\n\r\n\\(0=(x-1)(x+1)\\)\r\n\\(0=x^2-1\\)<\/li>\r\n \t
  13. \\(x=\\dfrac{1}{5}, x=-\\dfrac{1}{5} \\\\ \\)\r\n\\(x-\\dfrac{1}{5}=0, x+\\dfrac{1}{5}=0\\)\r\n\r\n\\(0=(x-\\dfrac{1}{5})(x+\\dfrac{1}{5}) \\\\ \\)\r\n\\(0=x^2-\\dfrac{1}{25}\\)<\/li>\r\n \t
  14. \\(x=\\sqrt{7}, x=-\\sqrt{7}\\)\r\n\\(x-\\sqrt{7}=0, x+\\sqrt{7}=0\\)\r\n\r\n\\(0=(x-\\sqrt{7})(x+\\sqrt{7})\\)\r\n\\(0=x^2-7\\)<\/li>\r\n \t
  15. \\(x=\\sqrt{11}, x=-\\sqrt{11}\\)\r\n\\(x-\\sqrt{11}=0, x+\\sqrt{11}=0\\)\r\n\r\n\\(0=(x-\\sqrt{11})(x+\\sqrt{11})\\)\r\n\\(0=x^2-11\\)<\/li>\r\n \t
  16. \\(x=2\\sqrt{3}, x=-2\\sqrt{3}\\)\r\n\\(x-2\\sqrt{3}=0, x+2\\sqrt{3}=0\\)\r\n\r\n\\(0=(x-2\\sqrt{3})(x+2\\sqrt{3})\\)\r\n\\(0=x^2-12\\)<\/li>\r\n \t
  17. \\(x=3, x=5, x=8\\)\r\n\\((x-3)=0, (x-5)=0, (x-8)=0\\)\r\n\r\n\\(\\begin{array}{rrcrcrrrr}\r\n(x&-&3)(x&-&5)(x&-&8)&=&0 \\\\\r\n(x^2&-&8x&+&15)(x&-&8)&=&0 \\\\\r\nx^3&-&8x^2&+&15x&&&& \\\\\r\n&-&8x^2&+&64x&-&120&=&0 \\\\\r\n\\midrule\r\nx^3&-&16x^2&+&79x&-&120&=&0\r\n\\end{array}\\)<\/li>\r\n \t
  18. \\(x=-4, x=0, x=4\\)\r\n\\(x+4=0, x, x-4=0\\)\r\n\r\n\\(x(x+4)(x-4)=0\\)\r\n\\(x(x^2-16)=0\\)\r\n\\(x^3-16x=0\\)<\/li>\r\n \t
  19. \\(x=-9, x+6=0, x=-2\\)\r\n\\(x+9=0, x+6=0, x+2=0\\)\r\n\r\n\\(\\begin{array}{rrcrcrrrr}\r\n(x&+&9)(x&+&6)(x&+&2)&=&0 \\\\\r\n(x^2&+&15x&+&54)(x&+&2)&=&0 \\\\\r\nx^3&+&15x^2&+&54x&&&& \\\\\r\n&+&2x^2&+&30x&+&108&=&0 \\\\\r\n\\midrule\r\nx^3&+&17x^2&+&84x&+&108&=&0\r\n\\end{array}\\)<\/li>\r\n \t
  20. \\(x=-1, x=1, x=5\\)\r\n\\(x+1=0, x-1=0, x-5=0\\)\r\n\r\n\\((x+1)(x-1)(x-5)=0\\)\r\n\\((x^2-1)(x-5)=0\\)\r\n\\(x^3-5x^2-x+5=0\\)<\/li>\r\n \t
  21. \\(x=-2, x=2, x=5, x=-5\\)\r\n\\(x+2=0, x-2=0, x-5=0, x+5=0\\)\r\n\r\n\\((x+2)(x-2)(x-5)(x+5)=0\\)\r\n\\((x^2-4)(x^2-25)=0\\)\r\n\\(x^4-29x^2+100=0\\)<\/li>\r\n \t
  22. \\(x=2\\sqrt{3}, x=-2\\sqrt{3}, x=\\sqrt{5}, x=-\\sqrt{5}\\)\r\n\\(x-2\\sqrt{3}=0, x+2\\sqrt{3}=0, x-\\sqrt{5}=0, x+\\sqrt{5}=0\\)\r\n\r\n\\((x-2\\sqrt{3})(x+2\\sqrt{3})(x-\\sqrt{5})(x+\\sqrt{5})=0\\)\r\n\\((x^2-12)(x^2-5)=0\\)\r\n\\(x^4-17x^2+60=0\\)<\/li>\r\n<\/ol>","rendered":"
      \n
    1. \"x=5, x=2\"
      \n\"x-5=0, x-2=0\"<\/p>\n

      \"0=(x-5)(x-2)\"
      \n\"0=x^2-7x+10\"<\/li>\n

    2. \"x=3, x=6\"
      \n\"x-3=0, x-6=0\"<\/p>\n

      \"(x-3)(x-6)=0\"
      \n\"0=x^2-9x+18\"<\/li>\n

    3. \"x=20, x=2\"
      \n\"x-20=0, x-2=0\"<\/p>\n

      \"0=(x-20)(x-2)\"
      \n\"0=x^2-22x+40\"<\/li>\n

    4. \"x=13, x=1\"
      \n\"x-13=0, x-1=0\"<\/p>\n

      \"0=(x-13)(x-1)\"
      \n\"0=x^2-14x+13\"<\/li>\n

    5. \"x=4, x=4\"
      \n\"x-4=0, x-4=0\"<\/p>\n

      \"0=(x-4)(x-4)\"
      \n\"0=x^2-8x+16\"<\/li>\n

    6. \"x=0, x=9\"
      \n\"x-9=0, x\"<\/p>\n

      \"x(x-9)=0\"
      \n\"0=x^2-9x\"<\/li>\n

    7. \"x=\dfrac{3}{4}, x=\dfrac{1}{4} \\\"
      \n\"x-\dfrac{3}{4}=0, x-\dfrac{1}{4}=0\"<\/p>\n

      \"0=\left(x-\dfrac{3}{4}\right)\left(x-\dfrac{1}{4}\right) \\\"
      \n\"0=x^2-x+\dfrac{3}{16}\"<\/li>\n

    8. \"x=\dfrac{5}{8}, x=\dfrac{5}{7} \\\"
      \n\"x-\dfrac{5}{8}=0, x-\dfrac{5}{7}=0\"<\/p>\n

      \"0=\left(x-\dfrac{5}{8}\right)\left(x-\dfrac{5}{7}\right) \\\"
      \n\"0=x^2-\dfrac{75}{56}x+\dfrac{25}{56}\"<\/li>\n

    9. \"x=\dfrac{1}{2}, x=\dfrac{1}{3} \\\"
      \n\"x-\dfrac{1}{2}=0, x-\dfrac{1}{3}=0\"<\/p>\n

      \"0=\left(x-\dfrac{1}{2}\right)\left(x-\dfrac{1}{3}\right) \\\"
      \n\"0=x^2-\dfrac{5}{6}x+\dfrac{1}{6}\"<\/li>\n

    10. \"x=\dfrac{1}{2}, x=\dfrac{2}{3} \\\"
      \n\"x-\dfrac{1}{2}=0, x-\dfrac{2}{3}=0\"<\/p>\n

      \"0=\left(x-\dfrac{1}{2}\right)\left(x-\dfrac{2}{3}\right) \\\"
      \n\"0=x^2-\dfrac{7}{6}x+\dfrac{1}{3}\"<\/li>\n

    11. \"x=5, x=-5\"
      \n\"x-5=0, x+5=0\"<\/p>\n

      \"0=(x-5)(x+5)\"
      \n\"0=x^2-25\"<\/li>\n

    12. \"x=1, x=-1\"
      \n\"x-1=0, x+1=0\"<\/p>\n

      \"0=(x-1)(x+1)\"
      \n\"0=x^2-1\"<\/li>\n

    13. \"x=\dfrac{1}{5}, x=-\dfrac{1}{5} \\\"
      \n\"x-\dfrac{1}{5}=0, x+\dfrac{1}{5}=0\"<\/p>\n

      \"0=(x-\dfrac{1}{5})(x+\dfrac{1}{5}) \\\"
      \n\"0=x^2-\dfrac{1}{25}\"<\/li>\n

    14. \"x=\sqrt{7}, x=-\sqrt{7}\"
      \n\"x-\sqrt{7}=0, x+\sqrt{7}=0\"<\/p>\n

      \"0=(x-\sqrt{7})(x+\sqrt{7})\"
      \n\"0=x^2-7\"<\/li>\n

    15. \"x=\sqrt{11}, x=-\sqrt{11}\"
      \n\"x-\sqrt{11}=0, x+\sqrt{11}=0\"<\/p>\n

      \"0=(x-\sqrt{11})(x+\sqrt{11})\"
      \n\"0=x^2-11\"<\/li>\n

    16. \"x=2\sqrt{3}, x=-2\sqrt{3}\"
      \n\"x-2\sqrt{3}=0, x+2\sqrt{3}=0\"<\/p>\n

      \"0=(x-2\sqrt{3})(x+2\sqrt{3})\"
      \n\"0=x^2-12\"<\/li>\n

    17. \"x=3, x=5, x=8\"
      \n\"(x-3)=0, (x-5)=0, (x-8)=0\"<\/p>\n

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

    18. \"x=-4, x=0, x=4\"
      \n\"x+4=0, x, x-4=0\"<\/p>\n

      \"x(x+4)(x-4)=0\"
      \n\"x(x^2-16)=0\"
      \n\"x^3-16x=0\"<\/li>\n

    19. \"x=-9, x+6=0, x=-2\"
      \n\"x+9=0, x+6=0, x+2=0\"<\/p>\n

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

    20. \"x=-1, x=1, x=5\"
      \n\"x+1=0, x-1=0, x-5=0\"<\/p>\n

      \"(x+1)(x-1)(x-5)=0\"
      \n\"(x^2-1)(x-5)=0\"
      \n\"x^3-5x^2-x+5=0\"<\/li>\n

    21. \"x=-2, x=2, x=5, x=-5\"
      \n\"x+2=0, x-2=0, x-5=0, x+5=0\"<\/p>\n

      \"(x+2)(x-2)(x-5)(x+5)=0\"
      \n\"(x^2-4)(x^2-25)=0\"
      \n\"x^4-29x^2+100=0\"<\/li>\n

    22. \"x=2\sqrt{3}, x=-2\sqrt{3}, x=\sqrt{5}, x=-\sqrt{5}\"
      \n\"x-2\sqrt{3}=0, x+2\sqrt{3}=0, x-\sqrt{5}=0, x+\sqrt{5}=0\"<\/p>\n

      \"(x-2\sqrt{3})(x+2\sqrt{3})(x-\sqrt{5})(x+\sqrt{5})=0\"
      \n\"(x^2-12)(x^2-5)=0\"
      \n\"x^4-17x^2+60=0\"<\/li>\n<\/ol>\n","protected":false},"author":727,"menu_order":95,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"back-matter-type":[],"contributor":[],"license":[],"class_list":["post-1801","back-matter","type-back-matter","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1801","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\/1801\/revisions"}],"predecessor-version":[{"id":1804,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1801\/revisions\/1804"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter\/1801\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=1801"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/back-matter-type?post=1801"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=1801"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=1801"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}