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Exponents and radicals
                                               1     =    1    ´   a    b +  =  a   b +

                                                             b -
                Similarly, for    1    , the rationalizing factor is  a-  =  1  a  ´ b  a  a  b +  =  a  b +  b+  .  ab-
                                a    b -   a    b -  a     b+    ab-
                Thus,                                           .



                Therefore,                      .
          FOR ONLINE READING ONLY
     Mathematics for Secondary Schools  factors:  a Rationalizing factor  a  a ´  Reason
           The following table gives some options and reasons for choosing the rationalising


              Denominator

                                                                           a =
              a
                                                                                )
                                                                   a
                                                                              a =
              a
                                                                                )
                                                                     )(
                                                                    a
                                                                                   ab-
              a +  ab   a - b     a  a +  ab   a - b       ( (  a - b )(  a + b a = ab-
                                                              a + b
                                                                         a - b
           Example 5�36
                                                                          1          1        5    3 +
           Rationalize  the denominator in each of              1       5  1  3 -  =  5  5  3 -  3 +  ´  5  3 +
           the following expressions:                  (b)    5  1   =   5  1   ´  5  5  3 +  5  3 +
                                                                                         3 +
                 3                        1                        3 -  =  1  3 -  ´ = 1  5 =  ´  5  3 +
           (a)                    (b)                       5     3 -  5    3 -  5  3 +(  5  3 +  3 -  )  5  3 +
                                                                                 5
                  5                     5    3 -                    = 5  = (  3 -  5  5 + 3  3 -  5  3 +
                                                                                   3 +
                                                                                3 -
                                                                           5
                                                                                    5
                                                                       (
                                                                     =
                    5                   2    3+                             (  5− 3 )( = ) 5  5 + 3 3 + 5  3 +    )  3 +
           (c)                    (d)                                     5    3 - = )  5  5 3 3 +
                  5 +  3                 2   5 -                          5    3 + (  5  -  3 -    )  5  3 +
                   6                                                          =  5  5 3 3 +  5  3 +
                                                                            -
           (e)                                                       =          =  5   3 +
                                                                                      2
                                                                          -
                  72−                                                    5 3 =     5 3
                                                                               3 +
                                                                          5
                                                                                    -
           Solution                                                  = =  5  1  2  3 +  5  3 +
           (a)   5 is a denominator and a single           Therefore,   5  2  =     2    .

                                                                            3 -
                radical expression.
                Multiply both numerator and            (c)      5          5       5    3 -
                denominator by  5 .                           5    3 +  =  5  3 +  ´  5  3 -
                            3
                      3  3  = = ´ ´ 5  5                                  5  (  5    )
                             3
                         5 5  5 5  5 5                                =            3 -
                                                                              -
                           35                                                53
                            35
                         = =                                            5-   15
                            5  5                                      =
                        3  3 35                                            2
                              35
             Therefore,  \ \ = =  .
                         5 5  5  5                         Therefore,     5   =  5−  15  .
                                                                       5 +  3      2
                                                   110
                                                                            Student's Book Form Two
                                                                                          11/10/2024   20:12:37
     MATHEMATIC F2 v5.indd   110                                                          11/10/2024   20:12:37
     MATHEMATIC F2 v5.indd   110
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