|
Completion
of squares
Clement Radcliffe,Contributor
In
this week's lesson we will continue
to review quadratic equations using
both the formula and the factorisation
methods. Let us begin with the solution
to the homework given last week.
Solve x2 + 3x + 1 = 0
x
= ( - b ± vb2 - 4ac )/2a
From
the equation, a = 1, b = 3 and c =
1. Substituting
x
= ( - 3 ± v( 3 )2 - 4 x 1 x
1 )/2
x 1
x
( - 3 ± v9 - 4 )/2
x
= ( - 3 ±v 5 )/2
x
= ( - 3 ± v2.24 )/2
x
= ( - 3 + 2.24 )/2 = ( - 0.76 )/2
= - 0.38
OR
x = ( - 3 - 2.24 )/2 = - 5.24/2 =
- 2.62
Answer:
x = - 0.38 and - 2.62
Solve
2x2 - 6x - 1 = 0
Using
the formula x = ( - b ± vb2
- 4ac )/2a
From
the equation, a = 2, b = - 6 and c
= - 1.
x
= ( - (- 6) ± v(-6)2 - 4 x
2 x -1 )/2
x 2
x
= ( 6 ± v36 + 8 )/4
x
= ( 6 ± v44 )/4
x
= ( 6 ± v6.63 )/4
x
= ( 6 + 6.63 )/4 = 12.63/4 = 3.16
x
= ( 6 - 6.63 )/4 = -0.63/4 = - 0.16
Answer:
x =3.16 and - 0.16
Solve
2x2 - 3x - 4 = 2 - 4x
First
make the right-hand side equal zero:
2x2
- 3x + 4x - 4 - 2 = 0
2x2
+ x - 6 = 0 Factorising
(2x
- 3)(x + 2) = 0
2x
- 3 = 0, that is x = 3⁄2
And
x + 2 = 0, that is x = -2.
Answer:
x = 3⁄2
and -2
Let
us now proceed to use the completion
of squares method to reorganise a
quadratic expression.
The
quadratic expression is changed to
a format which enables you to determine
the following:
(1)
The axis of symmetry
(2)
The maximum or minimum value of the
expression.
Given
the quadratic expression x2
+ bx + c, the aim is to convert the
expression to the form (x + d)2 +
k, where d and k are constants. The
method requires two steps as follows.
- You
are required to CONVERT x2
+ bx to a perfecet square of the
form
(x + d)2. This is based on the following
equation: (x + d)2 =
x2 + 2dx + d2.
Given x2 + 2dx, then
d2, the square of half
the coefficient of x, must be added
to complete the perfect square.
- When
d2 is added to make x2
+ bx a perfect square then d2
is also subtracted to avoid change
in value of the expression. It follows,
therefore, that x2 +
bx +c is converted to x2
+ bx + d2 +c - d2.
EXAMPLE
Express
x2 + 6x + 3 in the form
(x + d )2 + k, where d
and k are constants.
Covert
x2 + 6x + 3 to the form
(x2 + 6x) + 3. As shown
above, 9 is added to x2
+ 6x to make it perfect square. When
9 is added then 9 is also subtracted
to avoid changing the value of the
expression.
(x2
+ 6x) + 3 = (x2 + 6x) +
[6]2 /2 + 3 - [6]2
/2
=
(x2 + 6x + 9) + 3 - 9
=
(x + 3)2 - 6
d
= 3 and k = - 6
Let
us try another example together
EXAMPLE
Express
x2 - 3x - 4 in the form
(x - a )2 - b
x2
- 3x - 4 is expressed in the form
(x2 - 3x ) - 4 Considering
the brackets
(x2
- 3x + 3/22) is a perfect
square.
...
x2 - 3x - 4 = (x2
- 3x + 3/22) - (3/2)2
- 4
=
(x - 3/2)2 - 9⁄4
- 4
=
(x - 3⁄2)2
- 25⁄4
On
your own please attempt the following.
Convert
the following expressions in the form
(x + a) + b
x2
+ 4x + 5
x2
- 3x - 7
In
the exam-type questions, the coefficient
of x2 is usually greater than or less
than one. In this case the initial
step is to factorise to make the coefficient
of x2 one.
EXAMPLE
Express
2x2 - 3x +1 in the form
a(x + h)2 + k where a,
h and k are real numbers.
2x2
- 3x + 1 = 2(x2 - 3⁄2x)
+ 1
Complete
the squares inside the bracket.
...
2(x2 + 3⁄2x
+ (3⁄4)2
- 2x(3⁄4)2
+ 1
NB
(3⁄2)2
is added within the bracket so the
real value is 2x(3⁄4)2
=
2(x2 - 3⁄2x
+ 9⁄16)
- 9⁄8
+ 1
=
2(x - 3⁄4)2
- 1⁄8
You
may expand to show that the value
of the expression remain the same.
a=2,
h= -3⁄4
and k = -17⁄8
We
will complete the lesson with another
example.
EXAMPLE
Express
f(x)= 2x2 - 4x - 13 in
the form f(x) = a (x + h)2
+ k.
2x2
- 4x - 13 = (2x2 - 4x)
- 13
=
2(x2 - 2x) - 13
Completing
the square
=
2(x2 - 2x + 1) - 2 x 1
- 13
=
2(x - 1)2 - 15
f(x)
= 2(x -1)2 - 15
For
homework please attempt the following:
1.
Solve the equation x2 +
4x - 3 = 0
2.
Solve the equation 2x2
+ 5x = 9
3.
Solve the equation 4x2
+ 9x + 10 = 4 - 2x
4.
Express 2x2 + 4x - 7 in
the form a(x + b)2 + c
5.
Express 3x2 - 2x + 1 in
the form a(x + b)2 + C
Clement
Radcliffe is an independent contributor.
Send questions and comments to kerry-ann.hepburn@gleanerjm.com
|
