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Additional Mathematics  SPM  Chapter  2 Quadratic Functions

                      2.1     Quadratic Equations and Inequalities



                    1.  The general  form of the quadratic     (b)  Formula method
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                      equation is ax  + bx + c = 0 where a,            –b ±
                                  2
                                                                           b
                                                                               4
                                                                                 ac

                                                                              –
                                                                             2
                      b and c are constants.                       x =      2a
                    2.  Method  for  solving  quadratic      3.  If α and β are the roots of a quadratic
                      equations:                               equation then, x  – (α + β)x + αβ = 0,   Form 4
                                                                             2
                      (a)  Method of completing the            where α + β is the sum of the roots
                          square                               and αβ is the product of the roots.
                          Make sure the coefficient of x  is
                                                    2
                          1 before we start completing the
                          square.
                    4.  Method to determine the range for a quadratic inequality:
                      (a)  Graph sketching
                          For a quadratic equation in the form of (x – a)(x – b) = 0, where a  b:
                          If (x – a)(x – b)  0, then x  a or x > b.
                          If (x – a)(x – b)  0, then a  x  b.
                      (b)  Number line
                      Select any  integer  x    a  Select any integer   x  Select any  integer  x    b
                      to test whether the value  that  satisfies  a <  x <  b   to test whether the value
                      of (x – a)(x – b) is positive  to test whether the value of   of (x – a)(x – b) is positive
                      or negative.            (x – a)(x – b) is positive or  or negative.
                                              negative.
                                             a                       b                        x
                                x  a                a  x  b                 x  b
                      (c)  Table
                                                           Range of values of x
                                                       x  a    a  x  b   x  b
                                          (x – a)        –          +         +
                                          (x – b)        –          –         +
                                       (x – a)(x – b)    +          –         +



                  Example 1                                  Solution
                   Solve the following quadratic equations by   (a)    x  – 8x + 4 = 0
                                                                    2
                   using completing the square method.                x  – 8x = –4
                                                                       2
                   (a)  x  – 8x + 4 = 0                         2        2         2
                       2
                                                                                    –8
                                                                        –8
                   (b)  –2x  – 7x + 5 = 0                      x  – 8x +   2   = –4 +   2
                         2
                                                                x  – 8x + (–4)  = –4 + (–4) 2
                                                                 2
                                                                           2
                                                                      (x – 4)  = 12
                                                                           2
                                                                        x – 4 = ±12
                                                               x = 0.5359  or  x = 7.4641
                                                       15


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