Page 22 - CHAPTER 4 (Quadratic equations)
P. 22

CHAPTER 4

                                                                    QUADRATIC EQUATIONS


                                                                   3
              Example  3:  Find  the  roots    of  4x +3x+5=0  by  the  method  of
              completing the square.

                                             2
              Solution: Note that 4x -3x-5=0 is the same as

                                                2
                                         2
                                      3
                                             3
                                 3
                   2
              (2x) +2×(2x)× + ( ) - ( ) +5=0
                                 4    4      4
                                        2
                                      3
              i.e.,            (2x+ ) -     9  +5=0
                                      4     16
                                        2
                                      3
              i.e.,            (2x+ ) +     71  =0
                                      4     16
                                        2
                                      3
              i.e.,            (2x+ ) -     -71  <0
                                      4     6

                             2
                          3
              but (2x+ ) cannot be negative for any real value of x(why?) . So, there
                          4
              is no real value of x satisfying the given equation. Therefore , the given

              equation has real roots.


              Now  you  have  seen  several  examples  of  the  use  of  the  method  of

              completing the square . So, let us give this method in general.

              Consider  the  quadratic  equation       +      −    = 0  (   ≠ 0)  .  dividing
                                                                   2
              throughout by a, we get

                   b     c
               2
              x +    x+ =0
                   a     a
                                                          2
                                                 2
                                                      b
                                                             c
                                              b
              This is the same as (x+ ) - ( ) + =0
                                              2a      2a     a
                           2
                                2
                        b
              i.e., (x+ ) -    b -4ac =0
                        2a      4a 2
              So the roots of the given equation are the same as those of
                      2
                                                               2
                                                                    2
                           2
                                                            b
                   b
              (x+ ) -     b -4ac =0, i.e., those of (x+ ) =        b -4ac
                   2a      4a 2                             2a       4a 2
                  2
              If b -4ac≥0, then by taking the square roots in (1), we get
                             2
                         √
                  b    ± b -4ac
              x+     =
                 2a         2a

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