{"id":6280,"date":"2024-01-23T15:26:34","date_gmt":"2024-01-23T09:56:34","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=6280"},"modified":"2024-01-23T15:26:39","modified_gmt":"2024-01-23T09:56:39","slug":"surds-questions-with-detailed-solutions","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/surds-questions-with-detailed-solutions\/","title":{"rendered":"Surds Questions with Detailed Solutions"},"content":{"rendered":"\n

This article is about surds questions and follow the link surds in math<\/a> if you want to know about surds. Here in this article, you can find surds problem-solving questions. <\/p>\n\n\n\n

Surd Easy Questions<\/h2>\n\n\n\n

Simplify the following<\/strong><\/p>\n\n\n\n

    \n
  1. $\\sqrt{21} \\times \\sqrt{7}$<\/li>\n\n\n\n
  2. $\\sqrt{180} $<\/li>\n\n\n\n
  3. $\\sqrt{14} \\times \\sqrt 6$<\/li>\n\n\n\n
  4. $8 \\sqrt 6 \\times 4 \\sqrt 6$<\/li>\n\n\n\n
  5. $\\sqrt [3]{ab^2c^2} \\times \\sqrt [3]{a^2bc}$<\/li>\n\n\n\n
  6. $\\sqrt [3] {32} \\times \\sqrt [3] {250}$<\/li>\n\n\n\n
  7. $(\\sqrt{2}+\\sqrt 3)(\\sqrt{2}-\\sqrt 3)$<\/li>\n\n\n\n
  8. $(\\sqrt {10} – 2)(1 + \\sqrt {10})$<\/li>\n\n\n\n
  9. $\\frac {2}{1 – \\sqrt 3}$<\/li>\n\n\n\n
  10. $\\sqrt{252}-\\sqrt {112}$<\/li>\n\n\n\n
  11. $6 \\sqrt 3+3 \\sqrt 3$<\/li>\n\n\n\n
  12. $\\frac {2\\sqrt 7}{\\sqrt 11}$<\/li>\n\n\n\n
  13. $ 6 \\sqrt {27}-4 \\sqrt 3$<\/li>\n\n\n\n
  14. $( 1 + \\sqrt 2)^2$<\/li>\n\n\n\n
  15. $\\sqrt {2^4} + \\sqrt [3] {64} + \\sqrt [4] {2^8}$<\/li>\n<\/ol>\n\n\n\n
    Answers<\/summary>
    \n

    (1) $\\sqrt{21} \\times \\sqrt{7}= \\sqrt { 3 \\times 7 \\times 7} = 7 \\sqrt 3$
    (2) $\\sqrt{180} = \\sqrt {2^2 \\times 3^2 \\times 5} = 6 \\sqrt 5$
    (3) $\\sqrt{14} \\times \\sqrt 6 = \\sqrt {2 \\times 7} \\times \\sqrt {2 \\times 3} = 2 \\sqrt {21}$
    (4) $8 \\sqrt 6 \\times 4 \\sqrt 6 = 32 \\times 6 = 192$
    (5) $\\sqrt [3]{ab^2c^2} \\times \\sqrt [3]{a^2bc} = \\sqrt [3] {a^3b^3c^3} = abc$
    (6) $\\sqrt [3] {32} \\times \\sqrt [3] {250}= \\sqrt [3] { 32 \\times 250} = 20$
    (7) $(\\sqrt{2}+\\sqrt 3)(\\sqrt{2}-\\sqrt 3) = 2 -3 =-1$
    (8) $(\\sqrt {10} – 2)(2 + \\sqrt {10})= 10 -4 = 6$
    (9) $\\frac {2}{1 – \\sqrt 3} = \\frac {2}{1 – \\sqrt 3} \\times \\frac {1 + \\sqrt 3}{1 + \\sqrt 3} = -1 – \\sqrt 3 $
    (10) $\\sqrt{252}-\\sqrt {112} = 6 \\sqrt 7 – 4\\sqrt 7 = 2 \\sqrt 7$
    (11) $6 \\sqrt 3+3 \\sqrt 3= 9 \\sqrt 3$
    (12) $\\frac {2\\sqrt 7}{\\sqrt 11}= \\frac {2 \\sqrt {77}}{11}$
    (13) $ 6 \\sqrt {27}-4 \\sqrt 3= 18 \\sqrt 3 – 4 \\sqrt 3 = 14 \\sqrt 3$
    (14) $( 1 + \\sqrt 2)^2= 1 +2 + 2 \\sqrt 2 = 3 + 2 \\sqrt 2$
    (15) $\\sqrt {2^4} + \\sqrt [3] {64} + \\sqrt [4] {2^8} =2^2 + 4 + 2^2 = 12 $<\/p>\n<\/div><\/details><\/div>\n\n\n\n

    True and False statement<\/strong><\/p>\n\n\n\n

      \n
    1. $\\sqrt {1} + \\sqrt {2} = \\sqrt {1+2}$<\/li>\n\n\n\n
    2. $\\sqrt {3} = \\sqrt {-1 \\times -3}= \\sqrt {-1} . \\sqrt {-3}$<\/li>\n\n\n\n
    3. $\\sqrt [3] {8} \\div \\sqrt {4}$ is a rational number<\/li>\n\n\n\n
    4. $(1 + \\sqrt 5) ( 1- \\sqrt 5)$ is a irrational number<\/li>\n<\/ol>\n\n\n\n
      Answers<\/summary>
      \n

      True and False statement
      (1) $\\sqrt {1} + \\sqrt {2} = \\sqrt {1+2}$ : False
      (2) $\\sqrt {3} = \\sqrt {-1 \\times -3}= \\sqrt {-1} . \\sqrt {-3}$ :False , we cannot have negative number inside the square root
      (3) $\\sqrt [3] {8} \\div \\sqrt {4}$ is a rational number : $2 \\div 2 =1$, Hence Rational number. So True
      (4) $(1 + \\sqrt 5) ( 1- \\sqrt 5) $ is a irrational number : $ 1 – 5 = -4, Hence Rational number, So False<\/p>\n<\/div><\/details><\/div>\n\n\n\n

      Moderate Surds Questions<\/h2>\n\n\n\n

      Simplify the following<\/strong><\/p>\n\n\n\n

        \n
      1. $\\frac {1+ \\sqrt 5}{1 – \\sqrt 5} + \\frac {1 – \\sqrt 5}{1 + \\sqrt 5}$<\/li>\n\n\n\n
      2. $\\frac {1}{1 + \\sqrt 2} + \\frac {1}{\\sqrt 2+ \\sqrt 3 } + \\frac {1}{\\sqrt 3+ \\sqrt 4 } + \\frac {1}{\\sqrt 4+ \\sqrt 5 } +\u2026.+ \\frac {1}{\\sqrt 8+ \\sqrt 9 }$<\/li>\n\n\n\n
      3. $\\sqrt {31 + 8 \\sqrt {15}}$<\/li>\n\n\n\n
      4. $(\\sqrt {72} + \\sqrt {108})(\\sqrt {\\frac {1}{3}} – \\sqrt {\\frac {2}{9}})$<\/li>\n\n\n\n
      5. $ \\frac {7 \\sqrt 3}{\\sqrt {10} + \\sqrt 3} – \\frac {2 \\sqrt 5}{\\sqrt 6 + \\sqrt 5} – \\frac {3 \\sqrt 2}{\\sqrt {15} + 3 \\sqrt 2}$<\/li>\n<\/ol>\n\n\n\n
        Answers<\/summary>
        \n

        (1) $\\frac {1+ \\sqrt 5}{1 – \\sqrt 5} + \\frac {1 – \\sqrt 5}{1 + \\sqrt 5}$
        $= \\frac { (1+ \\sqrt 5)^ + (1- \\sqrt 5)^2 }{1 -5} =-3$<\/p>\n

        (2) $\\frac {1}{1 + \\sqrt 2} + \\frac {1}{\\sqrt 2+ \\sqrt 3 } + \\frac {1}{\\sqrt 3+ \\sqrt 4 } + \\frac {1}{\\sqrt 4+ \\sqrt 5 } +\u2026.+ \\frac {1}{\\sqrt 8+ \\sqrt 9 }$
        $=\\frac {1}{1 + \\sqrt 2} \\frac {1 – \\sqrt 2}{1 – \\sqrt 2} + \\frac {1}{\\sqrt 2+ \\sqrt 3 } \\frac {\\sqrt 2 – \\sqrt 3}{\\sqrt 2 – \\sqrt 3} + \\frac {1}{\\sqrt 3+ \\sqrt 4 } \\frac {\\sqrt 3 – \\sqrt 4}{\\sqrt 3 – \\sqrt 4}+\u2026.+ \\frac {1}{\\sqrt 8+ \\sqrt 9 }\\frac {\\sqrt 8 – \\sqrt 9}{\\sqrt 8 – \\sqrt 9}$
        $= -1(1- \\sqrt 2) -1(\\sqrt 2 – \\sqrt 3) – 1( \\sqrt 3 – \\sqrt 4) + \u2026.-1(\\sqrt 8 – \\sqrt 9)$
        $= \\sqrt 9 -1 = 2$

        (3) $\\sqrt {31 + 8 \\sqrt {15}}$
        $=\\sqrt { 4^2 + 15 + 2 \\times 4 \\sqrt {15}} = \\sqrt {(4+ \\sqrt {15})^2} = 4 + \\sqrt {15}$

        (4) $(\\sqrt {72} + \\sqrt {108})(\\sqrt {\\frac {1}{3}} – \\sqrt {\\frac {2}{9}})$
        $=\\sqrt {24} -\\sqrt {16} + \\sqrt {36} -\\sqrt {24}$
        $=6 -4 = 2$

        (5) $\\frac {7 \\sqrt 3}{\\sqrt {10} + \\sqrt 3} – \\frac {2 \\sqrt 5}{\\sqrt 6 + \\sqrt 5} – \\frac {3 \\sqrt 2}{\\sqrt {15} + 3 \\sqrt 2}$
        $=\\frac {7 \\sqrt 3( \\sqrt {10} – \\sqrt 3)}{7} – \\frac { 2 \\sqrt 5(\\sqrt 6 – \\sqrt 5)}{1} – \\frac {3 \\sqrt 2(\\sqrt {15} – 3\\sqrt 2}{-3}$
        $=\\frac { 7 \\sqrt {30} -21}{7} – \\frac {2 \\sqrt {30} -10}{1} – \\frac {3 \\sqrt {15} -18}{-3}$
        $=1$<\/p>\n<\/div><\/details><\/div>\n\n\n\n

          \n
        1. Find the value of a and b : $\\frac {3 + \\sqrt 7}{ 3 – 4 \\sqrt 7} = a + b \\sqrt 7$<\/li>\n<\/ol>\n\n\n\n
          Answers<\/summary>
          \n

          LHS
          $\\frac {3 + \\sqrt 7}{ 3 – 4 \\sqrt 7}$
          $= \\frac {3 + \\sqrt 7}{ 3 – 4 \\sqrt 7} \\frac {3 + 4 \\sqrt 7}{ 3 + 4 \\sqrt 7}$
          $= \\frac { 9 + 12 \\sqrt 7 + 3 \\sqrt 7 + 28}{9 – 112}$
          $=\\frac {37 + 15 \\sqrt 7}{-103}$
          $= -\\frac {37}{103} – \\frac {15}{103} \\sqrt 7$
          Comparing
          $a= -\\frac {37}{103}$ and $b=- \\frac {15}{103}$<\/p>\n<\/div><\/details><\/div>\n\n\n\n

            \n
          1. If $x = 2 + \\sqrt 3$
            Find the value
            (a) $x + \\frac {1}{x}$
            (b) $x^2 + \\frac {1}{x^2}$<\/li>\n<\/ol>\n\n\n\n
            Answer<\/summary>
            \n

            (a) $ 2 + \\sqrt 3 + \\frac {1}{2 + \\sqrt 3}$
            $=2 + \\sqrt 3 -(2- \\sqrt 3)$
            $=2 \\sqrt 3 $
            (b)
            $x^2 + \\frac {1}{x^2}$
            $=(x + \\frac {1}{x})^2 – 2$
            $=(2 \\sqrt 3)^2 -2$
            $=12 -2=10$<\/p>\n<\/div><\/details><\/div>\n\n\n\n

              \n
            1. Rationalize the Denominator of $\\frac {1}{\\sqrt 3 – \\sqrt 2 -1 }$<\/li>\n<\/ol>\n\n\n\n
              Answer<\/summary>
              \n

              $=\\frac {1}{\\sqrt 3 – 1 – \\sqrt 2 }$
              $=\\frac {1}{\\sqrt 3 – 1 – \\sqrt 2 } \\frac {\\sqrt 3 – 1 + \\sqrt 2}{\\sqrt 3 – 1 + \\sqrt 2 }$
              $= \\frac {\\sqrt 3 – 1 + \\sqrt 2}{ (\\sqrt 3 -1)^2 – 2}$
              $=\\frac {\\sqrt 3 – 1 + \\sqrt 2}{4 -2 \\sqrt 3 -2}$
              $=\\frac {\\sqrt 3 – 1 + \\sqrt 2}{2 -2 \\sqrt 3 }$
              $=\\frac {\\sqrt 3 – 1 + \\sqrt 2}{2 -2 \\sqrt 3 } \\frac {2 +2 \\sqrt 3 }{2 +2 \\sqrt 3 } $
              $=\\frac {(\\sqrt 3 – 1 + \\sqrt 2)(2 +2 \\sqrt 3)}{4 – 12}$
              $=\\frac{ 2 \\sqrt 3 + 6 -2- 2\\sqrt 3 + 2\\sqrt 2 + 2 \\sqrt 6}{-8}$
              $=-\\frac {4 + 2\\sqrt 2 + 2 \\sqrt 6}{8}$
              $= – \\frac { 2 + \\sqrt 2 + \\sqrt 6}{4}$<\/p>\n<\/div><\/details><\/div>\n\n\n\n

                \n
              1. Convert into pure surds $2 \\sqrt {5 \\sqrt 7}$<\/li>\n<\/ol>\n\n\n\n
                Answer<\/summary>
                \n

                $2 \\sqrt {5 \\sqrt 7} = \\sqrt {2^2 \\times 5 \\times \\sqrt 7} = \\sqrt {\\sqrt {20^2 \\times 7}}= \\sqrt {\\sqrt {2800}}= \\sqrt [4] {2800}$<\/p>\n<\/div><\/details><\/div>\n\n\n\n

                  \n
                1. Which of the following has greatest value $\\sqrt [3] {7}$, $\\sqrt [6] {15}$, $\\sqrt [4] {10}$<\/li>\n<\/ol>\n\n\n\n
                  Answer<\/summary>
                  \n

                  We need to convert into equal surd power to compare,
                  LCM 3,6 and 4 is 12
                  $\\sqrt [3] {7} = \\sqrt [12] {7^4} = \\sqrt [12] { 2401}$
                  $\\sqrt [6] {15}=\\sqrt [12] {15^2} = \\sqrt [12] { 225}$
                  $\\sqrt [4] {10}=\\sqrt [12] {10^3} = \\sqrt [12] { 1000}$<\/p>\n

                  So greatest is $\\sqrt [3] {7}$<\/p>\n<\/div><\/details><\/div>\n\n\n\n

                  Tough Surd Questions<\/h2>\n\n\n\n

                  (1) <\/strong>Simplify the expression
                  $\\frac {\\sqrt 3}{ \\sqrt {19 + 8 \\sqrt 3} – \\sqrt {19 – 8 \\sqrt 3}}$<\/p>\n\n\n\n

                  Answer<\/summary>
                  \n

                  $\\frac {\\sqrt 3}{ \\sqrt {19 + 8 \\sqrt 3} – \\sqrt {19 – 8 \\sqrt 3}}$
                  $=\\frac {\\sqrt 3}{ \\sqrt {19 + 8 \\sqrt 3} – \\sqrt {19 – 8 \\sqrt 3}} \\frac {\\sqrt {19 + 8 \\sqrt 3} + \\sqrt {19 – 8 \\sqrt 3}}{\\sqrt {19 + 8 \\sqrt 3} + \\sqrt {19 – 8 \\sqrt 3}}$
                  $= \\frac {\\sqrt 3 (\\sqrt {19 + 8 \\sqrt 3} + \\sqrt {19 – 8 \\sqrt 3})}{19 + 8 \\sqrt 3 – 19 + 8 \\sqrt 3}$
                  $=\\frac {\\sqrt {19 + 8 \\sqrt 3} + \\sqrt {19 – 8 \\sqrt 3}}{16}$
                  Now $19 + 8 \\sqrt 3= 4^2 + (\\sqrt 3)^2 + 8 \\sqrt 3= (4 + \\sqrt 3)^2$
                  Therefore
                  $=\\frac {\\sqrt {19 + 8 \\sqrt 3} + \\sqrt {19 – 8 \\sqrt 3}}{16}$
                  $= \\frac {\\sqrt {(4 + \\sqrt 3)^2} + \\sqrt {(4 – \\sqrt 3)^2}}{16}$
                  $=\\frac {4+ \\sqrt 3 + 4 – \\sqrt 3}{16} = \\frac {1}{2}$<\/p>\n<\/div><\/details><\/div>\n\n\n\n

                  (2)<\/strong> if $y= \\frac { \\sqrt {a+b} + \\sqrt {a-b}}{ \\sqrt {a+b} – \\sqrt {a-b}}$
                  Find the value of $by^2 – 2ay + b$<\/p>\n\n\n\n

                  Answer<\/summary>
                  \n

                  $y= \\frac { \\sqrt {a+b} + \\sqrt {a-b}}{ \\sqrt {a+b} – \\sqrt {a-b}}$
                  $y= \\frac { \\sqrt {a+b} + \\sqrt {a-b}}{ \\sqrt {a+b} – \\sqrt {a-b}} \\times \\frac { \\sqrt {a+b} + \\sqrt {a-b}}{ \\sqrt {a+b} + \\sqrt {a-b}}$
                  $y= \\frac { a+b +a-b + 2 \\sqrt {a^2 -b^2}}{a+b -a +b}$
                  $y=\\frac { 2a + 2 \\sqrt {a^2 -b^2}}{2b}$
                  $y=\\frac {a + \\sqrt {a^2 -b^2}}{b}$<\/p>\n

                  $by^2 – 2ay + b$
                  $= b [\\frac {a + \\sqrt {a^2 -b^2}}{b}]^2 – 2a [\\frac {a + \\sqrt {a^2 -b^2}}{b}] + b$
                  $=\\frac {a^2 + a^2 -b^2 + 2a \\sqrt {a^2 -b^2}}{b} – 2a [\\frac {a + \\sqrt {a^2 -b^2}}{b}] + b$
                  $=\\frac {a^2 + a^2 -b^2 + 2a \\sqrt {a^2 -b^2} -2 a^2 -2a \\sqrt {a^2 -b^2} + b^2}{b}$
                  $=0$<\/p>\n<\/div><\/details><\/div>\n\n\n\n

                  (3)<\/strong> if $x = 3 + 2 \\sqrt 2$
                  Then find the value of $( \\sqrt x – \\frac {1}{\\sqrt x})$<\/p>\n\n\n\n

                  Answer<\/summary>
                  \n

                  $x = 3 + 2 \\sqrt 2$
                  $x = 1 + (\\sqrt 2)^2 + 2 \\sqrt 2$
                  $x = (1+ \\sqrt 2)^2$
                  Now
                  $( \\sqrt x – \\frac {1}{\\sqrt x})$
                  $= ( (1+ \\sqrt 2) – \\frac {1}{1 + \\sqrt 2})$
                  $= (1 + \\sqrt 2 +(1- \\sqrt 2))= 2$<\/p>\n<\/div><\/details><\/div>\n\n\n\n

                  (4)<\/strong> if $a = \\frac {1}{7 + 4 \\sqrt 3}$ and $b= \\frac {1}{7 – 4 \\sqrt 3}$
                  Then find the value of $\\frac {1}{a+1} + \\frac {1}{b+1}$<\/p>\n\n\n\n

                  Answer<\/summary>
                  \n

                  $a = \\frac {1}{7 + 4 \\sqrt 3}$
                  $a= \\frac { 7 – 4 \\sqrt 3}{49 – 48}= 7 – 4 \\sqrt 3$
                  $b= \\frac {1}{7 – 4 \\sqrt 3}$
                  $a= \\frac { 7 + 4 \\sqrt 3}{49 – 48}= 7 + 4 \\sqrt 3$

                  $\\frac {1}{a+1} + \\frac {1}{b+1}$
                  $= \\frac {1}{8 – 4 \\sqrt 3} + \\frac {1}{8 + 4 \\sqrt 3}$
                  $ = \\frac { 16 }{64 – 48}=1$<\/p>\n<\/div><\/details><\/div>\n\n\n\n

                  (5)<\/strong> if $y =\\frac {\\sqrt {3}}{2}$,then find the value of
                  $\\frac {\\sqrt {1+y}}{1+ \\sqrt {1+y}} + \\frac {\\sqrt {1-y}}{1+ \\sqrt {1-y}}$<\/p>\n\n\n\n

                  Answer<\/summary>
                  \n

                  $y =\\frac {\\sqrt {3}}{2}$
                  $1 +y = \\frac { 2 + \\sqrt 3}{2} = \\frac { 4 + 2\\sqrt 3}{4} = \\frac {(\\sqrt 3 + 1)^2}{4}$
                  Therefore
                  $\\sqrt {1 + y} = \\frac {\\sqrt 3 + 1}{2}$
                  Similarly
                  $1 -y = \\frac { 2 – \\sqrt 3}{2} = \\frac { 4 – 2\\sqrt 3}{4} = \\frac {(\\sqrt 3 – 1)^2}{4}$
                  Therefore
                  $\\sqrt {1 + y} = \\frac {\\sqrt 3 + 1}{2}$<\/p>\n

                  $\\frac {\\sqrt {1+y}}{1+ \\sqrt {1+y}} + \\frac {\\sqrt {1-y}}{1+ \\sqrt {1-y}}$
                  $=\\frac { (\\sqrt 3 + 1)\/2}{1 + (\\sqrt 3 + 1)\/2} + \\frac { (\\sqrt 3 – 1)\/2}{1 – (\\sqrt 3 -1 )\/2}$
                  $=\\frac {\\sqrt 3 + 1}{3 + \\sqrt 3} + \\frac {\\sqrt 3 – 1}{3 – \\sqrt 3} $
                  $\\frac {1 + \\sqrt 3}{ \\sqrt 3(1 + \\sqrt 3)} + \\frac { \\sqrt 3 – 1}{\\sqrt 3 ( \\sqrt 3 – 1)} $
                  $= \\frac {2}{\\sqrt 3}$<\/p>\n<\/div><\/details><\/div>\n\n\n\n

                  I hope you like these surd questions with detailed solution and it helps in your preparation for the examination<\/p>\n\n\n\n

                  Further References<\/strong><\/p>\n\n\n\n