A sheet of metal is 20 m,it's length
is  1m  greater than its width.
    What  is  its length  and what is
its width ?
    Let  x = the  width of  the metal
therefore  x + 1 = the length.
    Therefore the area equals
          x(x + 1) = 20
           x + x  = 20
    How  do  we  solve  this  type of
equation ?
^z
First bring the 20 over to  the (LHS)
(left hand side) side :
     x + x -20 = 0
Now  using the  rules for factorising
quadratic   equations  as   shown  in
another program factorise the expres-
sion.
   -20 is the guide number.
    The  factors  of  20  which  when
added together give 1 (  the coeffic-
ient of the x term  )  are  5 and -4.
^z
Rewriting  the   expression  so  that
these  are the coefficients of  the x
term gives us . . .
    x + 5x -4x -20 = 0
    Now  factorising this  expression
gives us:
    x(x + 5) -4(x + 5) = 0
        (x - 4)(x + 5) = 0
Therefore x - 4 and x + 5 are the two
factors of the expression ,which when
multiplied together give us 0.
^z
When two numbers , factors or expres-
sions  multiply together to give us 0
one or both of them must equal 0.
    Therefore  one  or  both  of  the
factors x - 4  or x + 5 must equal 0.
    x - 4 = 0       x =  4
    x + 5 = 0       x = -5
    Here common  sense  tells  you to
disregard  the value  -5  because you
cannot have a negative length theref-
ore  the width  of the  sheet x  must
equal 4m and the length x + 1 = 5m.
^z
1
-1
-1
rewrite
^t
Lets go through another example.

x -2x + 1 = 0

1: The guide number equals blank

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2:The  factors of this  guide number
which when  added together  give the
coefficient  of the x term are blank
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and blank

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We will blank   the expression using
these factors as the coefficients of
the x term.
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-x -x + 1 = 0
factorise
(x - 1)
factors
0
^t
Giving us   x blank

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Next we blank     this expression to
give us

^t
(x - 1) blank    = 0

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These are the blank   of the
expression.

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One or  both of  these factors  must
equal blank in order for their
product to equal 0
^z
therefore x - 1 must equal 0

Note: Here  we  have  the   unusual
situation where both factors of the
expression are the same.

x = 1

We can prove that this value makes
the expression true by substituting
it for x in the original expression.

   x -2x + 1 = 0
   1  -2  + 1 = 0
^z
12
coefficient
4
3
(x + 4)
(x + 3)
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Solve the equation  x + 7x + 12 = 0

x + 7x + 12 = 0

The guide number = blank

^t
The  factors of 12  which when added
together  give 7  the blank       of
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the x term are blank
^t
and blank

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The factors of the equation therefore
are blank   and
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    blank
^z
(x + 4)(x + 3) = 0
-4
-3
^t


The equation can be written as
blank


^t
Therefore the values of x which make
it true are blank

^t
and         blank

^z
42
-6x -7x
2x(x - 3) -7(x - 3)
(2x -7)(x - 3) = 0
3.5
3
^t
Solve 2x - 13x = -21

      2x - 13x + 21 = 0
The Guide number = blank
^t
Rewriting the equation as
2x blank   + 21 = 0

^t
Factorising to give
blank               = 0
^t
Which simplifies to give
blank
^t

Therefore the values of x which make
the equation true are blank
^t
and blank
^z
-8
4x - 2x
x(x + 4)  -2(x + 4) = 0
(x + 4)(x - 2) = 0
-4
2
^t
Solve x + 2x - 8 = 0

The Guide number = blank

^t
Rewriting the equation as
x + blank   -8 = 0

^t
Factorising to give
blank
^t
Which simplifies to give
blank

^t
Therefore the values of x which make
the equation true are blank and
^t
blank
^z
-50
50x - x
x(x + 50)  -(x + 50) = 0
(x - 1)(x + 50) = 0
1
-50
^t
Solve x + 49x - 50 = 0

The Guide number = blank

^t
Rewriting the equation as
x + blank   -50 = 0

^t
Factorising to give
blank
^t
Which simplifies to give
blank

^t
Therefore the values of x which make
the equation true are blank and
^t
blank
^z
-28
-7x + 4x
x(x - 7) + 4(x - 7)
(x + 4)(x - 7)
-4
7
^t
Solve x - 3x - 28 = 0

The Guide number = blank

^t
Rewriting the equation as
x blank    -28 = 0

^t
Factorising to give
blank
^t
Which simplifies to give
blank

^t
Therefore the values of x which make
the equation true are blank and
^t
blank
^z
Factorisation  is  one method of sol-
ving  quadratic  equations however it
involves  a certain  amount  of trial
and  error  and  the factors  may not
always be whole numbers.
    eg    x -1.5x = 2.5
    { x = -1 and x = 2.5 }
    It would be nice if we could find
a  formula  to generate  the solution
set of an equation.
{ The  solution set is the two values
of x  which  make  the  equation true
^z
these  values  are  also  called  the
roots of the equation }
    Such a  formulae  exists  and  we
will now derive it.
    The proof involves a mathematical
technique  which we  will not go into
here  called  completing the  square.

Take the general quadratic equation :
         ax + bx + c = 0
^z
         ax + bx + c = 0

         x + b/a x = -c/a

x + b/a x + (b/2a) = -c/a + b/4a

     (x + b/2a) = b/4a - c/a
                 = (b - 4ac)/4a
                      _________
       x + b/2a  =   b - 4ac
                     
                         2a
^z
                    _________
        x   = -b   b - 4ac
                   
              2a       2a

The  roots  of  ax + bx + c = 0  are
                    _________
        x   = -b   b - 4ac
               
                     2a
^z
Lets try an example using this
formulae : x + 13x -27 = 0

     a = 1   b = 13   c = -27
                    _________
      x     = -b   b - 4ac
               
                     2a
                    ____________
      x     = -13  169 - 4(27)
               
                      2
^z
                     ___________
      x     = -13   169 - 108
               
                      2
                     ____
      x     = -13   61
               
                   2
             ____                ____
x =  -6.5 +  61  , x =  -6.5 -  61
                             
               2                   2
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Heres another one :
           2x + 7x -15 = 0
                    
           a     b    c
                    _________
      x     = -b   b - 4ac
               
                     2a
                    _______________
      x     = -7   49 - 4(2)(-15)
               
                        4
^z
                    ___________
      x     = -7   49 + 120
               
                      4
                    ____
      x     = -7   169
               
                   4

x =  -7/4 + 13/4  , x =  -7/4 - 13/4

x =  1           , x =  -5
^z
1
9
-2
- 4ac
^t
Now try one yourself x + 9x - 2 = 0

The a , b  and  c values are
a = blank
^t
b = blank
^t
c = blank

^t
Complete the general formulae to
give :
                      _________
        x     = -b   b blank
                 
                       2a
^z
9
81 + 8
^t
Inserting the a ,b ,c values into
the formulae gives us

                  ______________
 x     = -blank  81 - 4(1)(-2)
          
                      2
^t
Giving us
               _______
 x     = -9   blank
          
                2

Which simplifies to give
^z
89
-9/2
-9/2
^t


                   ______
     x     = -9   blank
            
                 2


^t
Therefore
     x =  blank +  89/2      or

^t
     x =  blank -  89/2
^z
0
1
0
-3
^t
Solve  x = 3  using the
formula.

In its usual form the equation can
be written as x -3 = blank

^t

The a , b  and  c values are
a = blank
^t
b = blank
^t
c = blank

^z
4(1)(-3)
2
12
^t
inserting the a ,b ,c values into
the formulae gives us

                   _____________
 x     =      0   0 - blank
              
^t
                     blank
^t
Giving us
                   ____
 x     =      0   blank
              
                  2

which can be written as
^z
12
4
2
3
3
^t
               _____
 x     =  0   blank
          
               2

^t
               ___     ___
 x     =  0   blank( 3 )
          
                 2
^t
                    ___
 x     =  0  blank 3
          
                2
^t
Therefore              x =  blank
^t
                       x = -blank
^z
2
-9
4
81 - 32
4
^t
Solve  2x - 9x + 4 = 0 using the
formula.

The a , b  and  c values are
a = blank
^t
b = blank
^t
c = blank

^t
Inserting the a ,b ,c values into
the formulae gives us

                   __________
 x     =      9  blank
              
^t
                    blank
^z
49
7
4
1/2
^t
Giving us
                   ____
 x     =      9   blank
              
                  4

^t

 x     =      9  blank
              
                4

^t
Therefore              x = blank
^t
                       x = blank
^z
-1
1
3
-1
-2
^t
Solve   -x + x + 3 = 0   using the
formula.

The a , b  and  c values are
a = blank
^t
b = blank
^t
c = blank

^t
Inserting the a ,b ,c values into
the formulae gives us

                        __________
 x     =      blank    1 + 12
              
^t
                     blank
^z
13
13/2
13/2
^t
Giving us

                   ____
 x     =     -1   blank
              
                  -2

^t

Therefore            x =  - blank
^t
                     x =  + blank
^z
7
9
-2
49 + 56
14
^t
Solve  7x + 9x - 2 = 0 using the
formula.

The a , b  and  c values are
a = blank
^t
b = blank
^t
c = blank

^t
Inserting the a ,b ,c values into
the formulae gives us

                     __________
 x     =      -9    blank
              
^t
                     blank
^z
105
9/14
9/14
^t
Giving us

                    ____
 x     =      -9   blank
              
                  14

^t

Therefore       x = blank + 105/14

^t
                x = blank - 105/14
^z
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