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

^q

1

-1

-1

rewrite

^t

Lets go through another example.



x} -2x + 1 = 0



1: The guide number equals blank



^t

2:The  factors of this  guide number

which when  added together  give the

coefficient  of the x term are blank

^t

and blank



^t

We will blank   the expression using

these factors as the coefficients of

the x term.

^z

^q

-x-x+1=0

factorise

(x-1)

factors

0

^t

Giving us   x} blank



^t

Next we blank     this expression to

give us



^t

(x - 1) blank    = 0



^t

These are the blank   of the

expression.



^t

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

^q

12

coefficient

4

3

(x+4)

(x+3)

^t

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

^t

the x term are blank

^t

and blank



^t

The factors of the equation therefore

are blank   and

^t

    blank

^z

^q

(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

^q

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

^q

-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

^q

-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

^q

-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

^z

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

^q

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

^q

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

^q

89

-9/2

-9/2

^t





                   ______

     x     = -9   blank

            -------------

                 2





^t

Therefore

     x =  blank +  89/2      or



^t

     x =  blank -  89/2

^z

^q

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

^q

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

^q

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

^q

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

^q

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

^q

-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

^q

13

13/2

13/2

^t

Giving us



                   ____

 x     =     -1   blank

              ---------

                  -2



^t



Therefore            x = + - blank

^t

                     x = + + blank

^z

^q

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

^q

105

9/14

9/14

^t

Giving us



                    ____

 x     =      -9   blank

              ----------

                  14



^t



Therefore     x = blank + 105/14



^t

              x = blank - 105/14



^t

