Download 1. a × a = a 2. a ÷ a = a

Algebra Refresher
Workshop: Algebra Refresher
Topics Covered:
Rules of indices
Factorisation of Quadratics
Completing the Square
The rules of indices
The rules of indices are used to manipulate algebraic and numerical
expressions. These rules can only be applied to terms that have the same
base.
1.
am × an = am+n
Examples:
34 × 32 = 36
y6 × y10 = y16
4xy × 2x2y3 = 8x3y4
2. am ÷ an = am – n
am
= a m −n
an
Examples:
57
= 54
3
5
a 25
= a19
6
a
20x 3 y 10
= 5xy4
2 6
4x y
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1
Algebra Refresher
3. (am)n = amn
Examples:
(85)3 = 815
This rule (rule 3), can also be used to manipulate expressions such as:
(a m b n ) k = a m k b n k
For example,
(a 4 b 3 ) 3 = a12b9
(2x4y3)5 = 25x20y15 = 32x20y15
Other points to note:
Any number (or base) raised to the power of zero is always 1, e.g.
20=1. More generally, a0 = 1.
Theory
Clearly
am
= 1 . However, using rule 2 above suggests that
am
am
= a m −m = a 0 .
m
a
Therefore a0 is equal to 1.
Any number (or base) raised to the power of 1 is itself, e.g. 21 = 2.
More generally, a1 = a.
Negative Powers:
a-m = 1/am
Theory:
Let’s look at
am
a m +n
 am
1
1
 = m n = n  , clearly this is the same as n , using
a 
a
 a a
rule 2, this can also be re-written as: am-(m+n) = a-n.
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Algebra Refresher
Therefore, when a number is raised to a negative power, e.g. a-m, it can
be rewritten as:
1
, i.e.
am
a-m = 1/am
Example: 4-2 =
1
42
1
Roots: A basic root that you should know is x 2 = x .
In general,
Positive integer
1
n
x =n x
(where n is a natural number)
Taking this further,
( ) = ( x) .
m
1
m
m
n
x n = xn
Examples:
1
42 = 4 = 2
2
( ) = ( 27 ) = (3) = 9
1
27 3 = 27 3
2
3
2
2
Other examples:
1. (x3y4z3)(x2yz3) = x5y5z6
2.
15x 3 y 4 5x 5
=
y
3x −2 y 5
3.
x 4 y5 y3z3 x 4 y8z3
× 2 =
= x 2y8z2
2
z
x
x z
4.
a10 b 5 a 2b 3 a10 b 5
c6
a10 b 5 c 6
÷ 6 =
× 2 3 = 2 3 = a 8b 2c 5
c
c
c
a b
a b c
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Algebra Refresher
Questions:
Simplify the following:
2.
3x6 × 5x9
3.
(a2bc2)(ab2c)
4.
53
52
5.
6x 2 y 3
3xy 2
6.
(3x2y3)4
7.
(x-2y5)3
8.
64 3
9.
16 2
11.
a 9 b 4 b 3 c 12
× 2
c5
a
12.
x5 y2 x3z2
÷ -1
z3
y
1.
a-2 × a6
10.
2 3
-1 2
3x y (2x y z)
1
3
Solutions:
1. a4
2. 15x15
3. a3b3c3
4. 51 = 5
5. 2xy
6. 81x8y12
7. x-6y15
8. 64 3 =
3
( ) = (4)
1
9. 16 2 = 16 2
3
3
= 64
1
3
64 = 4
10. 3x2y3(2x-1y2z) = 6xy5z
a 9 b 4 b 3 c 12 a 9 b 7 c 12
11.
× 2 =
= a7b7c7
5
5 2
c
a
c a
x5 y2 x3z2 x5 y2
y -1
x5y x2y
12.
÷ -1 =
× 3 2 = 3 5 = 5
z3
y
z3
x z
x z
z
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Algebra Refresher
Factorisation of Quadratics
A quadratic equation takes the form
ax2 + bx + c = 0
where a, b and c are numbers and a ≠ 0.
Quadratics can be factorised into two brackets (
)(
)
(however, not all quadratics can be factorised in this way).
→ Factorising quadratics when a = 1
x2 + bx + c = 0
the first term in each bracket is x: (x
)(x
)
The two missing spaces are numbers that when multiplied gives c and when
added gives b.
Once these two values have been found (for now lets call them d and e), then
the solutions to the quadratic can be identified by setting the term in each
bracket to zero, for example:
x2 + bx + c = 0
(x + d)(x + e) = 0
(x + d) = 0
and
x = -d
(x + e) = 0
x = -e
Therefore the solutions to the quadratic equation x2 + bx + c are x = -d and
x = -e.
(To check your answers are correct you can substitute these values into your
original equation to see if the value zero is obtained.)
Examples: Find the solutions to the following quadratics by factorising
1. x2 + 4x + 3 = 0
(x + 3)(x + 1) = 0
∴x = -3 & x = -1
Check:
Substitute x = -3 and x = -1
into the equation x2 + 4x + 3
x = -3: (-3)2 + 4(-3) + 3
= 9 – 12 + 3 = 0 x = -1: (-1)2 + 4(-1) + 3
=1–4+3=0
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Algebra Refresher
2. x2 – 4x + 3 = 0
(x – 3)(x – 1) = 0
∴x = 3 & x = 1
3. x2 + 8x – 9 = 0
(x + 9)(x – 1) = 0
∴x = -9 & x = 1
4. x2 – 8x – 9 = 0
(x – 9)(x + 1) = 0
∴x = 9 & x = -1
Questions (factorising quadratics; a = 1):
Find the solutions to the following quadratics by factorising:
1. x2 + 8x + 12 = 0
2. x2 – 5x + 6 = 0
3. x2 + 2x – 15 = 0
4. x2 – x – 12 = 0
Solutions (factorising quadratics; a = 1):
1. (x + 6)(x + 2) = 0
∴x = -6 & x = -2
2. (x – 3)(x – 2) = 0
∴x = 3 & x = 2
3. (x + 5)(x – 3) = 0
∴x = -5 & x = 3
4. (x – 4)(x + 3) = 0
∴x = 4 & x = -3
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Algebra Refresher
→ Factorising quadratics when a > 1
ax2 + bx + c = 0
To factorise quadratics in this form you can use:
1. Trial & Error (which with practice you’ll find quite quick and easy).
2. Algebraic method.
Trial & Error method
2x2 + 3x – 2 = 0
(2x – 1)(x + 2) = 0
Hence,
(2x – 1) = 0
&
(x + 2) = 0
x=½
x = -2
Algebraic method
6x2 + 7x – 3 = 0
Look for two numbers which multiply to give -18 (coefficient of x2 and constant
term multiplied together: 6 × -3 = -18) and add to give 7 (coefficient of x term).
Possibilities: 9 × -2 = -18
9 + (-2) = 7
Therefore the two numbers are 9 and -2.
Use these two numbers to write 7x as 9x – 2x, then
6x2 + 7x – 3 = 0
6x2 + 9x – 2x – 3 = 0
3x(2x + 3) – (2x + 3) = 0
(2x + 3)(3x – 1) = 0
The two solutions to this quadratic equation can now be found.
(2x + 3) = 0
x= -
(3x – 1) = 0
3
2
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x=
1
3
7
Algebra Refresher
Example 1: Factorise the quadratic 4x2 + 6x + 2 = 0 and find the solutions.
4x2 + 6x + 2 = 0
2(2x2 + 3x + 1) = 0
The quadratic that needs to be factorised is 2x2 + 3x + 1
Trial & error method
(2x + 1)(x + 1)
4x2 + 6x + 2 = 0
Therefore,
⇒ 2(2x2 + 3x + 1) = 0
⇒ 2(2x + 1)(x + 1) = 0
To find the solutions of the quadratic, we can divide both sides by 2 to leave:
(2x + 1)(x + 1) = 0
Hence,
(2x + 1) = 0
⇒ x= -
and
1
2
(x + 1) = 0
⇒ x = -1
Algebraic Method
Find two numbers that when multiplied gives 2 and when added gives 3.
Possibilities: 2 × 1 = 2
2+1=3
So the two numbers are 2 and 1.
Write 3x as 2x + x
2x2 + 3x + 1 = 0
2x2 + 2x + x + 1 = 0
2x(x + 1) + (x + 1) = 0
(x + 1)(2x + 1) = 0
Therefore
4x2 + 6x + 2 = 0
2(2x2 + 3x + 1) = 0
2(x + 1)(2x + 1) = 0
To find the solutions of the quadratic, we can divide both sides by 2 to leave:
(2x + 1)(x + 1) = 0
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Algebra Refresher
Hence,
(2x + 1) = 0
⇒ x= -
and
1
2
(x + 1) = 0
⇒ x = -1
Example 2: Factorise the quadratic 2x2 + 11x + 12 = 0 and find the solutions.
Trial & error method
(2x + 3)(x + 4)
2x2 + 11x + 12 = 0
Therefore,
⇒ (2x + 3)(x + 4) = 0
To find the solutions of the quadratic, put each term equal to zero and
rearrange:
(2x + 3)(x + 4) = 0
Hence,
(2x + 3) = 0
⇒ x= -
and
3
2
(x + 4) = 0
⇒ x = -4
Algebraic Method
Find two numbers that when multiplied gives 24 and when added gives 11.
Possibilities: 8 × 3 = 24
8 + 3 = 11
So the two numbers are 8 and 3.
Write 11x as 8x + 3x
2x2 + 11x + 12 = 0
2x2 + 8x + 3x + 12 = 0
2x(x + 4) + 3(x + 4) = 0
(x + 4)(2x + 3) = 0
Therefore
2x2 + 11x + 12 = 0
(x + 4)(2x + 3) = 0
To find the solutions of the quadratic, put each term equal to zero and
rearrange:
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Algebra Refresher
(x + 4)(2x + 3) = 0
Hence,
(x + 4) = 0
and
(2x + 3) = 0
⇒ x = -4
⇒ x= -
3
2
Questions (factorising quadratics; a>1):
Factorise the following quadratics either by trial and error or by using the
algebraic method and hence find the solutions.
1. 2x2 + 5x + 3 = 0
2. 3x2 – 17x + 20 = 0
3. 2x2 + 3x – 5 = 0
4. 10x2 + 14x – 12 = 0
Solutions (factorising quadratics; a>1): There is no practical way to show on
paper how to apply the trial and error method but your answers will be the
same no matter which method you use.
1. 2x2 + 5x + 3 = 0
Factorises as: (2x + 3)(x + 1)
Solution to the quadratic are: x = − 32 and x = -1
Algebraic method: find two numbers that when multiplied gives 6 and
when added gives 5.
Two numbers are 3 & 2
Replace 5x by 3x + 2x:
2x2 + 5x + 3 = 0
2x2 + 3x + 2x + 3 = 0
x(2x + 3) + (2x + 3) = 0
(2x + 3)(x + 1) = 0
Roots are: (2x + 3) = 0
⇒ x= -
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(x + 1) = 0
3
2
⇒ x = -1
10
Algebra Refresher
2. 3x2 – 17x + 20 = 0
Factorises as: (x – 4)(3x – 5)
Solution to the quadratic are: x = 4 and x =
5
3
Algebraic method: find two numbers that when multiplied gives 60 and
when added gives -17.
Two numbers are -12 & -5
Replace -17x by -12x – 5x:
3x2 – 17x + 20 = 0
3x2 – 12x – 5x + 20 = 0
3x(x – 4) – 5(x – 4) = 0
(x – 4)(3x – 5) = 0
Roots are: (x – 4) = 0
⇒ x=4
(3x – 5) = 0
⇒ x=
5
3
3. 2x2 + 3x – 5 = 0
Factorises as: (x – 1)(2x + 5)
Solution to the quadratic are: x = 1 and x = − 52
Algebraic method: find two numbers that when multiplied gives -10 and
when added gives 3.
Two numbers are -2 & 5
Replace 3x by -2x + 5x:
2x2 + 3x – 5 = 0
2x2 – 2x + 5x – 5 = 0
2x(x – 1) + 5(x – 1) = 0
(x – 1)(2x + 5) = 0
Roots are: (x – 1) = 0
⇒ x=1
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(2x + 5) = 0
⇒ x= -
5
2
11
Algebra Refresher
4. 10x2 + 14x – 12 = 0
5x2 + 7x – 6 = 0
Factorises as: (x + 1)(5x – 3)
Solution to the quadratic are: x = -1 and x = 35
Algebraic method: find two numbers that when multiplied gives -30 and
when added gives 7.
Two numbers are 10 & -3
Replace 7x by 10x – 3x:
5x2 + 7x – 6 = 0
5x2 + 10x – 3x – 6 = 0
5x(x + 2) – 3(x + 1) = 0
(x + 2)(5x – 3) = 0
Roots are: (x + 1) = 0
⇒ x = -1
(5x – 3) = 0
⇒ x= -
3
5
Differences of squares
You may get expressions such as
x2 – 4
which can be factorised into two brackets as:
(x + 2)(x – 2)
Expressions that can be factorised in this way are known as difference of
squares.
(Each term, x2 and the constant term, is a ‘square’ number)
The numbers in each bracket are the same the only thing that is different is
the sign.
Example 1: x2 – 16 = (x + 4)(x – 4)
Example 2: 4x2 – 25 = (2x – 5)(2x + 5)
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Algebra Refresher
Questions (difference of squares):
Factorise:
1. x2 – 1
2. 9x2 – 16
Solutions (difference of squares):
1. (x + 1)(x – 1)
2. (3x + 4)(3x – 4)
Completing the Square
Completing the square is another procedure used to factorise quadratic
expressions. The procedure is to transform the quadratic ax2 + bx + c into the
form: (
)2 + constant
When a = 1
x2 + bx + c
= (x2 + bx) + c
2
b

b
= x +  + c -  
2

2
2
Example: Complete the square on x2 + 2x + 5
x2 + 2x + 5
= (x2 + 2x) + 5
= (x + 1)2 + 5 – (1)2
Check:
(x + 1)2 + 4
= (x2 + 2x + 1) + 4
= x2 + 2x + 5
= (x + 1)2 + 5 – 1
= (x + 1)2 + 4
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Algebra Refresher
When a > 1
ax2 + bx + c
b
c

= a x 2 + x + 
a
a

2
2

b 
c  b  
= a  x +
 + −  
2a 
a  2a  

2
b 

 b 
= a x +
 + c − a 
2a 

 2a 
2
Example: Complete the square on 2x2 + 4x + 7
2x2 + 4x + 7
7

= 2 x 2 + 2x + 
2

7

2
2
= 2(x + 1) + − (1) 
2


7 

2
= 2(x + 1) + − 1
2 

5

2
= 2(x + 1) + 
2

= 2(x + 1)2 + 5
You can also find the roots (or solutions) to a quadratic using the method of
completing the square.
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Algebra Refresher
When a = 1
2
b

b
x2 + bx + c =  x +  + c -  
2

2
2
The roots occur when the equation equals zero, i.e.
2
2
b

b
x +  + c -   = 0
2

2
b

x + 
2

2
2
b
=   -c
2
2
b
b
x+
= ±   −c
2
2
2
b
b
x = - ±   −c
2
2
Example: Find the roots of the quadratic x2 + 2x – 7 = 0 by using the
method of completing the square
x2 + 2x – 7 = 0
(x + 1)2 – 7 – (1)2 = 0
(x + 1)2 – 7 – 1 = 0
(x + 1)2 – 8 = 0
(x + 1)2 = 8
x+1= ± 8
x = -1 ± 8
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Algebra Refresher
When a > 1
2
b 

 b 
ax2 + bx + c = a x +
 + c − a 
2a 

 2a 
2
The roots occur when the equation equals zero, i.e.
2
2
b 

 b 
a x +
 + c − a  = 0
2a 

 2a 
2
2
b 

 b 
a x +
 = a  − c
2a 
 2a 

2
 b 
a  − c
2
b 

 2a 
x +
 =
a
2a 

2
2
b 
c

 b 
x +
 =  −
2a 
a

 2a 
2
b
c
 b 
x+
= ±   −
2a
a
 2a 
2
b
c
 b 
x=±   −
2a
a
 2a 
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Algebra Refresher
Example: Find the roots of the quadratic 2x2 + 8x – 9 = 0 by using the
method of completing the square.
2x2 + 8x – 9 = 0
9

2 x 2 + 4x −  = 0
2

9

2 x 2 + 4x −  = 0
2

9

2
2
2(x + 2) − − (2)  = 0
2


9


2
2  (x + 2 ) − − 4  = 0
2


17 

2
2  (x + 2 ) −  = 0
2

(x + 2)2 − 17 = 0
2
(x + 2)2 =
17
2
x+2= ±
17
2
x = -2 ±
17
2
Questions (completing the square):
Complete the square on the following:
1. x2 + 8x + 5
2. x2 + 8x – 20
3. 2x2 + 8x + 4
4. 3x2 – 12x – 2
Find the roots of the following quadratics using the method of completing the
square:
5. x2 – 5x + 6 = 0
6. 2x2 – 4x – 7 = 0
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Algebra Refresher
Solutions (completing the square):
1.
3.
x2 + 8x + 5
2.
x2 + 8x – 20
= (x + 4)2 + 5 – (4)2
= (x + 4)2 – 20 – (4)2
= (x + 4)2 + 5 – 16
= (x + 4)2 – 20 – 16
= (x + 4)2 – 11
= (x + 4)2 - 36
2x2 + 8x + 4
4.
3x2 – 12x – 2
= 2(x2 + 4x + 2)
= 3(x 2 − 4x − 32 )
= 2[(x + 2)2 + 2 – (2)2]
[
= 3 [(x − 2)
= 3 [(x − 2)
2
2
= 3 (x − 2) − 32 − (2)
2
= 2[(x + 2) + 2 – 4]
= 2[(x + 2)2 – 2]
= 2(x + 2)2 - 4
2
− 32 − 4
2
− 143
]
]
]
= 3(x – 2)2 – 14
5.
x2 – 5x + 6 = 0
(x − 52 )2 + 6 − ( 52 )2
=0
(x − 52 )2 + 6 − 254
=0
(x − 52 )2 − 14
=0
(x − 52 )2
=
1
4
x − 52 = ±
1
4
x − 52 = ± 12
x=
5
2
± 12
Therefore, x = 52 + 12 = 3 and x = 52 − 12 = 2
6.
2x2 – 4x – 7 = 0
2(x 2 − 2x − 72 ) = 0
[
2[(x − 1)
2[(x − 1)
2
2
]
2 (x − 1) − 72 − (1) = 0
]
2
− 72 − 1 = 0
2
− 92 = 0
]
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Algebra Refresher
(x − 1)2 − 92 = 0
(x − 1)2 = 92
9
2
x–1= ±
x = 1±
9
2
Therefore, x = 1 +
x = 1−
9
2
9
2
= 3.121 (3 decimal places) and
= -1.121 (3 decimal places).
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19