GCSE Mathematics – Basics of numberSLIDE NUMBER 1 May 2019

GCSE Mathematics – Basics of number
SLIDE NUMBER 1
May 2019
© VIDLEARN® 2019
Claire Roberts
1
Session Objectives
The purpose of the session is to:
Understand notation, vocabulary, positive and negative integers
and symbols
Calculate using formal written methods of the four operations
including brackets, powers, roots and reciprocals
Recognise and apply positive integer powers and associated real
roots (square, cube and higher) powers of 2, 3, 4, 5
Describe estimation of calculations and apply the concept to
round numbers and measures
Define factors and multiples and use Prime Factor
Decomposition to identify HCF and LCM
Define the product rule for counting (combinations)
Perform a range of calculations using fractions, decimals and

percentages
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CONSIDER…
At this point you should consider the list of session objectives
and ask yourself:
How many of the session objectives am I confident with
Could I explain these objectives in relation to teaching and
learning
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Session Objectives
and symbols

percentages
SLIDE NUMBER 4
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Key vocabulary
In mathematics, the natural numbers are those used for counting
and ordering. In common language, words used for counting
are cardinal numbers and words used for ordering are ordinal
numbers.
The symbol for the set of all natural numbers is shown
An integer is a number that can be written without a fractional
component. The set of integers consists of the positive and
negative natural numbers and zero.
21, 4, 0, and −2048 are integers, but 9.75, 5 ¹⁄₂ , and √2 are not.
The symbol for the set of all integers is shown
(originating from the German word zahlen)
Basics of number
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Key vocabulary
The type of number we normally use, such as 1, 15.82, −0.1,
3/4, etc, are called real numbers. Positive or negative, large or
small, whole numbers or decimal numbers are all real numbers,
so called because they are not imaginary numbers.
The symbol for the set of all real numbers is shown
An imaginary number is one that when squared gives a negative
result. When we square a real number we always get a positive,
or zero, result. For example, 2×2=4, and (-2)×(-2)=4 as well.
So how can we square a number and get a negative result?
Because we "imagine" that we can. The "unit" for imaginary
numbers (the same as "1" for real numbers) is √(-1), and its
symbol is i or j.
Basics of number
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Key vocabulary
A rational number is any number that can be expressed as the
quotient or fraction of two integers, with a numerator and a
non-zero denominator .
Since may be equal to 1, every integer is a rational number.
The set of all rational numbers is shown
(originating from the Italian word quoziente)
Irrational numbers are all the real numbers which are not
rational numbers.
The most famous irrational number is , sometimes called
Pythagoras’ constant. Other examples include , , , etc
Basics of number
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Positive and Negative Numbers
In mathematics, direction from zero is indicated by use of
positive and negative signs.
Basics of number
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May 2019
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-4-3-2-101234

Negative numbers
Positive numbers
Origin
To denote that a number is negative we use a minus sign in
front of
the number, and, on occasion, a plus sign is used to denote a
positive number but this is commonly assumed.
-9 denotes “negative nine”; +9 or 9 denotes “positive nine”
8
Greater than, less than, equal to
In mathematics, arrows are used to denote whether a number is
greater than, less than or equal to another number.
6 > 3 means that “6 is greater than 3”
-6 < 3 means that “negative 6 is less than 3”
a < b means that “a is less than or equal to b”
b > a means that “b is greater than or equal to a”
It is important to remember that
“ = ” means “equal to”
and not “here’s my next step” or “ the answer is”
Basics of number
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Review of main ideas from above:
We will use real numbers, rational and irrational numbers but
imaginary numbers are only introduced in the Further
Mathematics A-level specification, but here’s something to
consider
Consider this incorrect use of the = symbol
(3 + 4) x 6 + 7 = 7 = 42 = 49
So now pause the recording to consider, once you have done
this play the recording.
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CONSIDER…
10
Could be interpreted mathematically as
“i eight all of pi” or “I ate all of the pie”
(3 + 4) x 6 + 7 = 7 = 42 = 49
The beginning and end of this are true, (3 + 4) x 6 + 7 = 49

But,7≠ 42 ≠ 49 We will look at the correct order of
operations in the next part
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CONSIDER…
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Session Objectives
and symbols
percentages
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Order of operations
Calculations should be completed in a specific order
remembered by the acronym BODMAS
Basics of number
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rackets
B
O
D
M
A
S
rder (power)
ivision
ultiplication
ddition
ubtraction
These operations are interchangeable
These operations are interchangeable
13

BODMAS
First, let’s consider the calculation from earlier
(3 + 4) x 6 + 7
Basics of number
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= 7 x 6 + 7
= 42 + 7
= 49
If we incorporate some more elements of BODMAS, 13 + 12 ÷
(9 - 7)2
= 13 + 12 ÷ (2)2
= 13 + 12 ÷ 4
= 16
= 13 + 3
14
Columnar Addition
If we need to evaluate 2746 + 578, we use the columnar
addition method
Basics of number
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2746

578
3
+
2
1
4
1
1
3
It is important to ensure that the digits are kept in clear place
value columns and the addition sign is included.
Answer: 3324
15
Columnar Subtraction
If we need to evaluate 2746 - 578, we use the columnar
subtraction method
Basics of number
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2746
578
1
-
6
8
2

value columns and the subtraction sign is included.
Answer: 2168
3
1
6
1
16
Long Multiplication
If we need to evaluate 274 x 57, we use the columnar
multiplication method
Basics of number
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8
57
274
19
1
x
5
2
2
0
7
13
0
3

8
6
15
1
1
Answer: 15618
value columns, the multiplication sign is included and you
appreciate that in the second line you are multiplying by a
multiple of 10, hence the zero.
17
Short division
If we need to evaluate 284 ÷ 5, we use the short division
method
Basics of number
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284
5
6
5
3
.
0
8
.
4

If you have a remainder at the end you can introduce a decimal
point and as many zeros as are needed. Alternatively you could
give the answer as a mixed number by putting the remainder as
the numerator and the divisor as the denominator,
Answer: 56.8
i.e.
18
Long division
If we need to evaluate 585 ÷ 15, we use the long division
method
Basics of number
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585
15
9
3
In long division, the remainder is found and shown underneath
as shown
Answer: 39
45
13
5
135
0

19
Consider these questions:
17 + (3 x 6) ÷ 32
672 + 384
965 – 739
361 x 39
391 ÷ 17
So now pause the recording to try these calculations , once you
have done this play the recording
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CONSIDER…
20
Consider these questions:
17 + (3 x 6) ÷ 3219
672 + 3841056
965 – 739226
361 x 3914079
391 ÷ 1723
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CONSIDER…
21
Session Objectives
and symbols
percentages
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Powers
Powers are used when a number is multiplied by itself a number
of times.
For example,
3 x 3 x 3 x 3 = 34 = 81
They can also be used for reciprocals and roots.
and
The rules of indices (or powers) will be covered in more detail
in Session 2 – Indices, roots and surds
Basics of number
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Common Powers and Roots
Basics of number
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Powers are very powerful in mathematics
Modern scientific calculators can find any power of any number
– do you know how to do this on yours?
https://www.youtube.com/results?search_query=how+to+use+a+
scientific+calculator+for+powers
So now pause the recording to consider the use of your
calculator, once you have done this play the recording
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CONSIDER…
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Session Objectives
and symbols

percentages
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Rounding
Numbers can be rounded to a given number of decimal places or
to a given number of significant figures.
The basic rules are the same for both:
Find the digit that will be the new final digit
Look at the digit that follows it
If it is less than 5, the final digit remains unchanged
If it is 5 or more, then the final digit must be increased by one.
If the final digit is 9, then the number before is increased by
one, and so on.
See the next slide for examples
Basics of number
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Rounding to a given number of decimal places
Rounding 34.6456 to one decimal place (1dp) will give 34.6
Rounding 67.6563 to 2dp will give 67.66
Rounding 8.0998 to 3dp will give 8.100
Note on example 3 that, despite the zeros having no value, they
are required to meet the degree of accuracy
Basics of number
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Rounding to a given number of significant figures
Rounding 34.6456 to one significant figure (1sf) will give 30
Rounding 67.6563 to 2sf will give 68
Rounding 8.0998 to 3sf will give 8.10
Rounding 0.030456 to 3sf will give 0.0305

Notes
Example 1 – the zero is included as a place value holder
Example 3 – the zero is include to meet the degree of accuracy
Example 4 – the first significant figure is the 3, preceding zeros
are not counted but subsequent zeros are
Basics of number
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Estimation
It is useful to be able to estimate the value of a calculation,
even if you have access to a calculator.
Consider the following, quite complex calculation,
Basics of number
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The exact answer is 20.2 (3sf)
If you were to accidentally type 193 instead of 19.3,
the calculator would give the answer 2020 (3sf) which you

would know was wrong if you had estimated first
30
Round the following numbers to the degree of accuracy given in
brackets:
6.7083 (2dp)
3091.87 (3sf)
0.004001 (2sf)
For the following calculation, first find an estimate then use
your calculator to find the exact answer to 3sf
So now pause the recording to consider the questions, once you
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CONSIDER…
31
6.7083 (2dp)= 6.71
3091.87 (3sf)= 3090
0.004001 (2sf)= 0.0040

Exact answer = 1.65 (3sf)
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CONSIDER…
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Session Objectives
and symbols
percentages
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Factors and Multiples
Factors are numbers which can be multiplied together to give
the original number. They are usually found in pairs, except in
the case of square numbers which always have an odd number of
factors.
For example,
Factors of 24 are 1 and 24, 2 and 12, 3 and 8, 4 and 6
Factors of 16 are 1 and 16, 2 and 8, 4 (and 4)
Multiples are numbers that are generated by multiplying by
integers. Basically they are the numbers found in the times
tables.
For example,
The first 10 multiples of 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18 and
20
The first 5 multiples of 100 are 100, 200, 300, 400 and 500
Basics of number
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Prime numbers
Prime numbers are special numbers which have exactly two
factors – itself and one.
A common misconception is that 1 is a prime number, but it
only has one factor.
The Sieve of Eratosthenes is a simple, ancient algorithm for
finding all prime numbers up to any given limit.
It is completed by marking the multiples
of each prime, starting with the
first prime number, 2, on a number grid.
The numbers left unmarked are
the prime numbers.
Basics of number
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Prime Factor Decomposition
Any integer can be written as a product of its prime factors
which can be found by Prime Factor Decomposition
Let’s write 420 as a product of its prime factors,
Basics of number
SLIDE NUMBER 36
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420

2
2
210
105
35
3
5
7
Answer: 2 x 2 x 3 x 5 x 7
or 22 x 3 x 5 x 7
36
Highest Common Factor (HCF) and Lowest Common Multiple
(LCM)
If we need to find the HCF or LCM of two or more integers we
can use prime factor decomposition.
Let’s find the HCF and LCM of 420 and 66
Basics of number
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66
2
3
33
11
Answer: 2 x 3 x 11
If we now write both numbers as their products one above the
other,

420 = 2 x 2 x 3 x 5 x 7
66 = 2 x 3 x 11
HCF = 2 x 3 = 6
LCM = 2 x 2 x 3 x 5 x 7 x 11 = 4620
37
This youtube video is a quirky explanation of the Sieve of
Eratosthenes https://www.youtube.com/watch?v=V08g_lkKj6Q
Find the HCF and LCM of these pairs of numbers using prime
factor decomposition:
1. 24 and 60
2. 48 and 72
So now pause the recording to watch the youtube video and to
try the technique, once you have done this play the recording
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CONSIDER…
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Find the HCF and LCM of these pairs of numbers using prime

factor decomposition:
1. 24 and 60
24 = 2 x 2 x 2 x 3
60 = 2 x 2 x 3 x 5
HCF = 2 x 2 x 3 = 12; LCM = 2 x 2 x 2 x 3 x 5 = 120
2. 48 and 72
48 = 2 x 2 x 2 x 2 x 3
72 = 2 x 2 x 2 x 3 x 3
HCF = 2 x 2 x 2 x 3 = 24; LCM = 2 x 2 x 2 x 2 x 3 x 3 = 144
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CONSIDER…
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Session Objectives
and symbols

percentages
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The Product Rule for Counting
To find the total number of outcomes for two or more events,
we can use the product rule for counting in which we multiply
the number of outcomes for each event together.
For example, if we wanted to know the total number of
combinations on a menu with 5 starters, 3 main courses and 4
desserts we could use the product rule,
5 x 3 x 4 = 60 different meal combinations
Basics of number
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This will be explained again in more detail in Session 18 –
Probability

41
Session Objectives
and symbols
percentages
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Simplifying Fractions

To simplify a fraction you need to find the HCF of both the
numerator and denominator and divide both by this factor. Once
a fraction has been simplified in this way it is said to be in its
lowest or simplest terms.
For example, to simplify ,we find that the HCF of the
numerator and denominator is 9, so dividing both by 9 will give
Basics of number
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Adding Fractions
To add a pair of fractions with different denominators you need
to:
Find the LCM of both the two denominators
Multiply the numerator by the same factor as the denominator
would be multiplied by to make the LCM; do this to both
fractions
Add the numerators of the two fractions, leaving the
denominator unchanged as the LCM
Simplify the new fraction if possible, or convert to a mixed
number if the fraction is improper.
For example,
Basics of number
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44
Subtracting Fractions
To subtract a fraction from another with a different denominator
you need to:
Find the LCM of both the two denominators
Multiply the numerator by the same factor as the denominator
would be multiplied by to make the LCM; do this to both
fractions
Subtract the second numerator from the first, leaving the
denominator unchanged as the LCM
For example,
Basics of number
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Adding or Subtracting Mixed Numbers
Convert the mixed numbers to improper fractions and then
follow the process as previously described.
For example,
Basics of number
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The same method is used when subtracting mixed numbers
46
Multiplying Fractions and Mixed Numbers
To multiply a pair of fractions you need to:
Multiply the two numerators to give the new numerator
Multiply the two denominators to give the new denominator

For example,
If you need to multiply a pair of mixed numbers you need to
first convert to improper fractions, then follow the method
above.
Basics of number
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Dividing Fractions and Mixed Numbers
To divide a pair of fractions you need to:
Convert the fractions so that they have common denominators
Divide the first numerator by the second numerator
For example,
If you need to divide a pair of mixed numbers you need to first
convert to improper fractions, then follow the method above.
Basics of number
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Converting Fractions to Decimals,
and vice versa
A fraction is another way of representing a division,
so means 3 ÷ 4.
Hence, to convert a fraction to a decimal, simply divide the
numerator by the denominator, meaning that
If you need to write a decimal as a fraction, you need to
consider the meaning of the decimal. For instance, 0.45 means 4
tenths and 5 hundredths, or 45 hundredths,
therefore
Basics of number
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Writing Recurring Decimals as Fractions
A recurring decimal is one that is non-terminating (unlike 0.45)
and has a sequence of digits that repeat infinitely.
Recurring decimals are indicated using small dots over the start

and finish of the repeating sequence, for example,
To convert to a fraction,
Let
So
and
Hence
Basics of number
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Converting Percentages to Fractions to Decimals, and vice versa
A percentage is simply a proportion given as an amount out of
100
Hence, to convert a percentage to a fraction, simply write the
percentage value as the numerator with 100 as the denominator,
and then to write as a decimal simply divide.
For example, and
If you need to write a decimal as a percentage, firstly consider
how many hundredths it is then write as a fraction and then a
percentage.
For example, and
Basics of number
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Converting Percentages to Fractions, and vice versa
Write the percentage value as the numerator with 100 as the
denominator, and then simplify if possible.
For example, and
If you need to write a fraction as a percentage,
divide the numerator by the denominator to give the fraction as
a decimal
then convert the decimal as previously discussed
For example,
and
Basics of number
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Finding Fractions and Percentages of amounts
To find a fraction of an amount ,divide by the denominator then

multiply the result by the numerator.
For example, to find of 64,
first find of 64 = 64 ÷ 8 = 8, then multiply by 7, 8 x 7 = 56
Hence of 64 = 56
If you need to find a percentage of an amount, firstly consider
the percentage as a fraction, then follow the same method
For example, to find 6% of £500, ( of 500)
first find of 500 = 500 ÷ 100 = 5, then multiply by 6, 5 x 6 = 30
Hence 6% of £500 = £30
Basics of number
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Have a go at the following questions:
1. 2. 3. 4.
5. Convert 0.76 to a fraction in its simplest terms
6. Convert to a percentage
7. Write the recurring decimal as a fraction
8. Find 4% of £250
So now pause the recording to consider the questions, once you
SLIDE NUMBER 54
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CONSIDER…

54
Have a go at the following questions:
1. 2.
3. 4.
5. Convert 0.76 to a fraction in its simplest terms
6. Convert to a percentage = 62.5%
7. Write the recurring decimal as a fraction
8. Find 4% of £250 = £10
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CONSIDER…
55
Session Objectives
and symbols

percentages
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56
SLIDE NUMBER 57
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CONSIDER…
End of Presentation
At this point it would be advisable to go back over the
presentation. Ensure that you are fully able to deal accurately
and effectively with each session objective.
You should supplement the content of this session with suitable
reading, research and discussion with others.
Claire Roberts
SLIDE NUMBER 58
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© VIDLEARN® 2019
GCSE Mathematics – Basics of number
End of presentation
58
© 2018 Cengage Learning®. May not be scanned, copied or
duplicated, or posted to a publicly accessible website, in whole
or in part.
© 2018 Cengage Learning.® May not be scanned, copied or
or in part.
‹#›
10.
1
10
Channels of Distribution
or in part.
10. 2
or in part.
‹#›
10.

2
Marketing Framework
or in part.
‹#›
10.
Discussion Questions #1
Apple introduced the Apple retail store in 2001 when it had less
than 3% of the computer market—prior to its introduction of the
iPod. Previously, Apple computers were sold through local
computer retailers.
What do you think prompted the idea for Apple’s new retail
strategy?
What were the risks associated with this strategy?
or in part.
‹#›
10.
Place
The market realigns discrepancies between buyers and sellers
Sellers have large quantities; Buyers want a few
Breaking bulk
Making goods available in smaller batches

or in part.
‹#›
10.
Distribution Channels
Distribution channel
A network of firms that are interconnected in their quest to
provide sellers a means of infusing the marketplace with their
goods, and buyers a means of purchasing those goods
The goal is to do this efficiently and profitably
Channel members include
Manufacturers, wholesalers, retailers, consumers, etc.
or in part.
‹#›
10.
Functions of a Channel
Activities that are
Customer-oriented (e.g., ordering)
Product-oriented (e.g., storage)
Marketing-centric (e.g., promotion)
Logistics
Coordinating flow of goods, services, and information
throughout channel
or in part.

‹#›
10.
Channel Tension
All channel functions must be done by someone, the question is
…
What is the most effective and efficient way to distribute the
product?
Tension in channels can be created by each channel member
Does member provide more benefit than cost?
The make-or-buy decision
or in part.
‹#›
10.
Channel Questions #1
Which of these is more efficient? Why?
or in part.
‹#›
10.
Channels and Supply Chains
Supply chain
Upstream partners
Channel members
Downstream partners

or in part.
‹#›
10.
Channel Questions #2
Who is in Amazon’s supply chain?
Who is in Pixar’s channel?
How is Dell’s distribution different from the others?
or in part.
‹#›
10.
The What, Why, & How of Channels
The “what” of channels
Network of suppliers and providers
The “why” of channels
Effectiveness and efficiency
The “how” of channels
Designing effective and efficient channels
or in part.
‹#›
10.

How to Design Channels
(slide 1 of 4)
Intensive distribution: widely distributed
Drugstores, supermarkets, discount stores, convenience stores,
etc.
Usually for simple, inexpensive, easily transported products
e.g., Snack food, shampoo, newspapers
Pull strategy: promote directly to end consumers to pull through
channel
or in part.
‹#›
10.
(slide 2 of 4)
Selective distribution: limited distribution
Usually for complex and/or expensive products that require
assistance
e.g., Most cars, computers, appliances
Push strategy: promote to distribution partners to push goods to
consumer
Manufacturer has more control due to fewer relationships to
manage
or in part.
‹#›
10.

(slide 3 of 4)
Exclusive distribution: extremely selective
e.g., Ferrari and Rolex
Manufacturers have the most control
May become monopolistic
or in part.
‹#›
10.
(slide 4 of 4)
How much distribution?
Design needs to be consistent with other marketing elements
Wide distribution
Usually goes with heavy promotion, lower prices, and average
or lower-quality products
Exclusive distribution
Usually goes with less promotion, higher prices, and higher-
quality products
or in part.
‹#›
10.
Push vs. Pull Strategies
(slide 1 of 2)
Push strategy
Incentives are offered to distribution partners to push products
through the channel

Pull strategy
Incentives are offered to consumers to pull products through the
channel
or in part.
‹#›
10.
Push vs. Pull Strategies
(slide 2 of 2)
or in part.
‹#›
10.
Power and Conflict in Channels
Conflict arises in distribution channels
Some conflict can be healthy
Some conflict can end a partnership
Power
Power is usually defined by size
Power can be used to win conflict
Exerting power over distribution partners can lead to resentment
and lack of cooperation
or in part.
‹#›

10.
Transaction Cost Analysis (TCA)
Model that considers channel members’ production costs &
governance costs
Goal is to minimize both costs
Production cost
Cost of producing/bringing product to market
Governance cost
Cost involved with relational issues incurred by coordinati ng
enterprise and controlling one’s partners
or in part.
‹#›
10.
Transaction Value Analysis
Perspective that emphasizes the benefits a company brings to its
partners
Goes beyond cost reductions
Uses human relationship terms
Communication enhances trust
Trust is the willingness and ability to deliver on promises
or in part.
‹#›
10.

Ways to Resolve Conflict
Communicate
Exchange personnel
Sponsor joint research projects
Mediation
Negotiate through a third party that determines the two parties’
utility functions
Arbitration
The third party makes a binding decision for the two parties
or in part.
‹#›
10.
Revenue Sharing
(slide 1 of 2)
Double marginalization: the problem
The manufacturer wants a markup
The retailer wants a second markup
or in part.
‹#›
10.
Revenue Sharing
(slide 2 of 2)
Double marginalization: solutions

or in part.
‹#›
10.
Discussion Question #2
Why wouldn’t the manufacturer just avoid the double
marginalization problem entirely and go directly to the
consumer?
or in part.
‹#›
10.
Integration
(slide 1 of 3)
All functions within a channel need to be completed
Revisit make-or-buy decision
Make: complete a function yourself
Buy: outsource a function
or in part.
‹#›
10.
Integration
(slide 2 of 3)
Vertical integration
Moving backward or forward in a channel
Forward integration
Moving forward in a distribution channel
e.g., Manufacturer opens its own retail stores

Backward integration
Moving backward in a distribution channel
e.g., Manufacturer controls raw materials or retailer sets up
private label
or in part.
‹#›
10.
Integration
(slide 3 of 3)
Private label
Type of backward integration
Advantages
Gives retailer negotiating power with manufacturers
Offers significant margins
Helps differentiate retailer from other retailers
e.g., Great Value oatmeal is only at Walmart
or in part.
‹#›
10.
Global Channels
Channels can be complicated
or in part.

‹#›
10.
How might Anheuser-Busch engage in forward integration?
How might Google engage in backward integration?
or in part.
‹#›
10.
Retailing & Retail Classifications
Retailers have been gaining power and momentum over the past
10–20 years
Retailers are classified by ownership, level of service, and
product assortment
Management’s level of ownership
Independent retailers
Branded store chains
Franchises
or in part.
‹#›
10.
Retailing Classifications
Level of service provided
Usually related to price points
Product assortment carried
Specialty: carry depth not much breadth

e.g., Toy stores
General merchandise: carry breadth but not much depth
e.g., Department stores
or in part.
‹#›
10.
Retailing Employees
Retail employees are important
Connect the retailer and its customers
Retailers should hire selectively, train well, and pay fairly
Dissatisfied employees can lead to dissatisfied customers and
employee turnover
Employee turnover leads to new associates who cause further
customer dissatisfaction
or in part.
‹#›
10.
Retailing Operations
Retailing is a service
Retailers should flowchart operations
Front-stage: elements customers see
Backstage: elements customers do not see
Must be run efficiently to support front-stage
The goal is to create effective and efficient processes
Self-service is a way to streamline

or in part.
‹#›
10.
Retailing Location
Location is important
Determine appropriate success factors for your specific
business; analyze locations to pick ideal sites
e.g., Population densities, income and social class distributions,
median ages, household composition
or in part.
‹#›
10.
Retailing Growth Strategies
Provide additional services
Target additional segments
Open multiple stores
Expand internationally
e.g., Exporting, joint ventures, direct foreign investment, and
license agreements
Global outsourcing
e.g., India & technology, China & manufacturing
or in part.
‹#›

10.
Franchising
Unique format of multisite expansion
Company can retain some control without complete ownership
or capital expenditure
Benefits
Franchisor: receives capital, scales of economy, committed
people, less risk, can focus on core functions
Franchisee: well-known brand and some market awareness,
supplier relationships, templates for training, central support
or in part.
‹#›
10.
Types of Franchising
Product franchising
Supplier authorizes a distributor in a territory to carry its
products, use its brand name, benefit from its advertising, etc.
e.g., Ford dealers, Coca-Cola bottlers
Business format franchising
Company offers a proven system to conduct business, marketing
support, brand name, advertising, etc., to the franchisee
e.g., McDonald’s, Holiday Inn
or in part.
‹#›
10.
E-Commerce
The Internet is an important channel

Online retail sales are about $180 billion, growing about 10% a
year
Still only 11% of total retail sales
Customers are younger and more affluent
Customer characteristics are changing to match customers in
general markets
United States dominates but not by much
or in part.
‹#›
10.
Asian Internet Penetration Percentages
or in part.
‹#›
10.
Discussion Question #4
How do you see the future for the distribution of entertainment
programs?
or in part.
‹#›
10.
Catalog Sales

Top 10 catalogers are B2B companies
e.g., Dell, Staples, etc.
80 of the top 100 catalogers continue to see sales growth
Internet is well-suited for a search while catalogs still dominate
browsing
Catalogs often complement not compete with Internet
or in part.
‹#›
10.
Sales Force
(slide 1 of 3)
Utilized extensively with a push strategy
Important with undifferentiated products
Issues
How many?
How to compensate them?
Usually salary plus bonuses
Tie compensation to performance evaluation
Sales force evaluation factors
e.g., Sales, time with clients, expertise
or in part.
‹#›
10.
Sales Force
(slide 2 of 3)
Sales force size
Estimate workload

100,000 stores
12 visits each per year for 30 minutes
50 weeks per year × 40 hours a week = 2,000 hours
500 of these hours will be spent on travel and administrative
duties
(100,000 accounts × 12 visits per year × 0.5 hour)/1,500 hours =
400 salespeople
or in part.
‹#›
10.
Sales Force
(slide 3 of 3)
B2B customers’ biggest complaints about salespeople
The salesperson isn’t following my company’s buying process
They don’t listen to my needs
They didn’t bother to follow up
or in part.
‹#›
10.
What criteria would you utilize to evaluate a car salesperson?
How would you tie compensation to this evaluation?
or in part.
‹#›

10.
Integrated Marketing Channels
As the number of channels proliferates, increasing care must be
taken to coordinate and integrate across them
Companies must understand customer behavior in order to
design effective distribution channels and to allocate resources
across channel options
Know your customer!
or in part.
‹#›
10.
Managerial Recap
(slide 1 of 2)
Distribution channels are the link from the manufacturer to the
customer
Numerous thoughtful decisions must be made in designing
channels
or in part.
‹#›
10.
Managerial Recap
(slide 2 of 2)
Channel entities are independent yet interdependent
organizations; thus, conflicts may arise
Conflicts are best addressed by employing good communication

and trust, revenue sharing, or greater vertical integration
or in part.
‹#›
10.
GCSE Mathematics – Algebra
SLIDE NUMBER 1
February 2019
© VIDLEARN® 2019
Joe Hammond
1
Session Objectives
Define key vocabulary and notation used in algebra
Describe simplification by collecting like terms
Demonstrate expanding linear and quadratic expressions
Practice factorising linear and quadratic expressions

Illustrate the process of rearranging formulae
Interpret functions and find inverse functions
Identify composite functions through worked examples
SLIDE NUMBER 2
February 2019
© VIDLEARN® 2019
2
CONSIDER…
and ask yourself:
learning
SLIDE NUMBER 3
February 2019
© VIDLEARN® 2019
3
Session Objectives
SLIDE NUMBER 4
February 2019
© VIDLEARN® 2019

4
A new language…
“the part of mathematics in which letters and symbols are used
to represent numbers and quantities in formulae, equations and
expressions”
A formal efficient way of solving problems
SLIDE NUMBER 5
February 2019
© VIDLEARN® 2019
Algebra
5
For example
2 + = 10

2 + x = 10
SLIDE NUMBER 6
February 2019
© VIDLEARN® 2019
Algebra
?
6
Variable: The letters x, y, z, a, b, c…
Term: x, 2x, 5y, y, -y, -3x2, x ...
3
Expression: x + 3, t - 4 , x2 + 4x - 3…
2
Equation/formulae: y = 2x, C = 3.4t ...
SLIDE NUMBER 7
February 2019
© VIDLEARN® 2019
Algebra
Vocabulary
7

2 x x = 2x
-3 x x = -3x
1 x x = x
-3 x -x = 3x
-x x -4 = 4x
-2x x 4y = -8xy
SLIDE NUMBER 8
February 2019
© VIDLEARN® 2019
Algebra
Multiplying and dividing with algebra
x ÷ 3 = x
3
10x ÷ 2 = 10y = 5y
2
x ÷ y = x
y
8
Pause here
What are “like terms”?
3x4y-3y-y23y4y34y2
Like terms share the same variable with the same power/index
SLIDE NUMBER 9

February 2019
© VIDLEARN® 2019
CONSIDER…
9
Pause here
What are “like terms”?
3x
SLIDE NUMBER 10
February 2019
© VIDLEARN® 2019
CONSIDER…
4y-3y3y
-y24y2
4y3
10
Session Objectives

SLIDE NUMBER 11
February 2019
© VIDLEARN® 2019
11
x + 3x + 4x =
y + y + y =
-y + y + y =
2t + 4t - 10t =
4a - 6a + 2b - 4b =
x - 4x2 - 2x + 2x2 =
SLIDE NUMBER 12
February 2019
© VIDLEARN® 2019
Algebra
Collecting like terms
8x
3y
y
-4t

-2a -2b
-x - 2x2
12
4a - 2b - 3a + 4b =
4a - 3a = a
- 2b + 4b = 2b
SLIDE NUMBER 13
February 2019
© VIDLEARN® 2019
Algebra
Modelling in the classroom
x2y + 2x2y - 3xy2 =
x2y + 2x2y = 3x2y
- 3xy2 = - 3xy2
a + 2b
3x2y - 3xy2
13
Pause here
-y + 4y =
3a - 5a + 7b - 3b =
-y2 + 4y2 =
d + 3d - t - 2d + 3t =

4y3 - 10y3 + 2y2 - 2y =
33q100 + 27q99 - 27q100 =
3p-1 - 7p - 3p2 =
SLIDE NUMBER 14
February 2019
© VIDLEARN® 2019
CONSIDER…
14
Pause here
-y + 4y =
3a - 5a + 7b - 3b =
-y2 + 4y2 =
d + 3d - t - 2d + 3t =
4y3 - 10y3 + 2y2 - 2y =
33q100 + 27q99 - 27q100 =
3p-1 - 7p - 3p2 =
SLIDE NUMBER 15
February 2019
© VIDLEARN® 2019
CONSIDER…
3y
- 2a + 4b
3y2
2d + 2t
- 6y3 + 2y2 - 2y
6q100 + 27q99
3p-1 - 7p - 3p2

15
Session Objectives
SLIDE NUMBER 16
February 2019
© VIDLEARN® 2019
16
3 x 54
3 x 50 = 150
3 x 4 = 12
162
SLIDE NUMBER 17
February 2019

© VIDLEARN® 2019
Algebra
Expanding brackets
3 x 54
3(50 + 4) = 150 + 12
= 162
17
SLIDE NUMBER 18
February 2019
© VIDLEARN® 2019
Algebra
Expanding brackets
3(x + 8)
-4(x - 8)
6(1 - 2y)
6(1 - 2y) =
-5(-2 - x)
-5(-2 - x) =
3(x + 8) =
3x + 24
-4(x - 8) =
-4x + 32
6 - 12y
10 + 5x

18
SLIDE NUMBER 19
February 2019
© VIDLEARN® 2019
Algebra
Expanding and simplifying
4(2 + x) - 3(x - 2)
4(2 + x) - 3(x - 2)
8 + 4x - 3x + 6
14 + x or x + 14
19
SLIDE NUMBER 20
February 2019
© VIDLEARN® 2019
Algebra
Expanding and simplifying
3(x2 - 2) - 7(2 - x2)
3(x2 - 2) - 7(2 - x2)
3x2 - 6 - 14 + 7x2

10x2 - 20 or - 20 + 10x2
20
SLIDE NUMBER 21
February 2019
© VIDLEARN® 2019
Algebra
Expanding double brackets
3 x 54
3(50 + 4) = 150 + 12
= 162
23 x 54
(20 + 3)(50 + 4) = 1242
20 x 50 = 1000
20 x 4 = 80
3 x 50 = 150
3 x 4 = 12
21
SLIDE NUMBER 22

February 2019
© VIDLEARN® 2019
Algebra
(x + 2)(x + 3)
(x + 2)(x + 3)
= x2 + 3x + 2x + 6
= x2 + 5x + 6
x x x = x2
x x 3 = 3x
2 x x = 2x
2 x 3 = 6
22
SLIDE NUMBER 23
February 2019
© VIDLEARN® 2019
Algebra
(x + 7)(x - 2)
(x + 7)(x - 2)
= x2 + -2x + 7x - 14
= x2 + 5x - 14
x x x = x2
x x -2 = -2x
7 x x = 7x
7 x -2 = -14

23
SLIDE NUMBER 24
February 2019
© VIDLEARN® 2019
Algebra
(3x + 1)(2x + 2)
(3x + 1)(2x + 2)
= 6x2 + 6x + 2x + 2
= 6x2 + 8x + 2
3x x 2x = 6x2
3x x 2 = 6x
1 x 2x = 2x
1 x 2 = 2
24
Pause here
Expand and simplify the following
-3(x + 4)
2(10 + x) + 4(x - 3)
-(x2 + 7)
(x - 4)(x + 6)
(2y + 4)(8 - y)

(3x - 5)(10 + 2x)
SLIDE NUMBER 25
February 2019
© VIDLEARN® 2019
CONSIDER…
25
Pause here
Expand and simplify the following
-3(x + 4)
2(10 + x) + 4(x - 3)
-(x2 + 7)
(x - 4)(x + 6)
(2y + 4)(8 - y)
(3x - 5)(10 + 2x)
SLIDE NUMBER 26
February 2019
© VIDLEARN® 2019
CONSIDER…
-3x - 12
8 + 6x
-x2 - 7
x2 + 2x - 24
-2y2 + 12y + 32
6x2 + 5x - 50

26
Session Objectives
SLIDE NUMBER 27
February 2019
© VIDLEARN® 2019
27
The process of writing an expression as a product or
combination of factors
Expanding: 3(x + 9) = 3x + 27
Factorising: 3x + 27 = 3(x + 9)
SLIDE NUMBER 28
February 2019
© VIDLEARN® 2019

Algebra
Factorising
28
2x + 10 = 2(x + 5)
5x + 20 = 5(x + 4)
7y - 28 = 7(y - 4)
SLIDE NUMBER 29
February 2019
© VIDLEARN® 2019
Algebra
Some quick factorising examples
7(y - 4) = 7y - 28
29
4x + 18 = 4( __ + __ )
4 isn’t a factor of 18
4x + 18 = 2( __ + __ )

Because 2 is the highest common factor
SLIDE NUMBER 30
February 2019
© VIDLEARN® 2019
Algebra
Trickier factorising examples
= 2(2x + 9)
30
8y - 60 = 8( __ + __ )
8 isn’t a factor of -60
8y - 60 = 4( __ + __ )
SLIDE NUMBER 31
February 2019
© VIDLEARN® 2019
Algebra
Trickier factorising examples
= 4(2y - 15)
31

8y + 20 = 8( __ + __ )
8 isn’t a factor of 20
8y + 20 = 4( __ + __ )
SLIDE NUMBER 32
February 2019
© VIDLEARN® 2019
Algebra
Two more examples
= 4(2y + 5)
32
12y2 + 54y = 12( __ + __ )
6 is the HCF of 12 and 54
y is the HCF of y2 and y
12y2 + 54y = 6y( __ + __ )
SLIDE NUMBER 33
February 2019
© VIDLEARN® 2019
Algebra
Two more examples
= 6y(2y + 9)

33
Pause here
Fully factorise
30x2 + 42x
6(5x2 + 7x)b. 2x(15x + 21)
c. 6x(5x + 7)d. 6x(5x + 7x)
SLIDE NUMBER 34
February 2019
© VIDLEARN® 2019
CONSIDER…
34
Pause here
Fully factorise
6x+ 30
14y + 42
10x2 + 42x
12xy + 28x
4xy2 - 8x2y
x2yz2 - xy + xy2

SLIDE NUMBER 35
February 2019
© VIDLEARN® 2019
CONSIDER…
35
Pause here
Fully factorise
6x+ 30
14y + 42
10x2 + 42x
12xy + 28x
4xy2 - 8x2y
x2yz2 - xy + xy2
SLIDE NUMBER 36
February 2019
© VIDLEARN® 2019
CONSIDER…
6(x+ 5)
14(y + 3)
2x(5x + 21)
4x(3y + 7)
4xy(y - 2x)
xy(xz2 - 1 + y)

36
SLIDE NUMBER 37
February 2019
© VIDLEARN® 2019
Algebra
Factorising quadratic expressions
(x + 2)(x + 3)
(x + 2)(x + 3)
= x2 + 3x + 2x + 6
= x2 + 5x + 6
x2 + 5x + 6 = (x + 2)(x + 3)
37
SLIDE NUMBER 38
February 2019
© VIDLEARN® 2019
Algebra
Factorising quadratic expressions
(x + 2)(x + 4) = x2 + 6x + 8
(x + 2)(x + __ ) = x2 + 7x + 10

(x + __ )(x - __ ) = x2 + 2x - 3
(x - __ )(x - __ ) = x2 - 8x + 15
5
3
1
5
3
38
SLIDE NUMBER 39
February 2019
© VIDLEARN® 2019
Algebra
Examples
x2 + 4x + 3 = (x __ )(x __ )
x2 + 2x - 3 = (x __ )(x __ )
x2 - 4x + 3 = (x __ )(x __ )
x2 - 2x - 3 = (x __ )(x __ )
+ 1 + 3
+ 3 - 1
3 - 1
3 + 1

39
SLIDE NUMBER 40
February 2019
© VIDLEARN® 2019
Algebra
Trickier examples
Sometimes we might have something like
2x2 + 9x + 4
= (2x __ )(x __ )
Numbers will multiply to make 4
*
2, 2: (2x + 2)(x + 2) = 2x2 + 6x + 4
4, 1: (2x + 4)(x + 1) = 2x2 + 6x + 4
1, 4: (2x + 1)(x + 4) = 2x2 + 9x + 4
40
SLIDE NUMBER 41
February 2019
© VIDLEARN® 2019
Algebra

A second example
3x2 + x - 14
= (3x __ )(x __ )
Numbers will multiply to make -14
*
-1, 14
14, -1
1, -14
-14, 1
-2, 7
7, -2
2, -7
-7, 2
-1, 42
14, -3
1, -42
-14, 3
-2, 21
7, -6
2, -21
-7, 6
*
(3x + 7)(x - 2)
41
Pause here
Fully factorise
x2 + 10x + 24
x2 - 10x + 24
x2 + 2x - 24

x2 - 2x - 24
2x2 + 11x + 5
3x2 + x - 10
SLIDE NUMBER 42
February 2019
© VIDLEARN® 2019
CONSIDER…
42
Pause here
Fully factorise
x2 + 10x + 24
x2 - 10x + 24
x2 + 2x - 24
x2 - 2x - 24
2x2 + 11x + 5
3x2 + x - 10
SLIDE NUMBER 43
February 2019
© VIDLEARN® 2019
CONSIDER…
(x + 4)(x + 6)
(x - 4)(x - 6)
(x - 4)(x + 6)
(x + 4)(x - 6)
(2x + 1)(x + 5)

(3x - 5)(x + 2)
43
Session Objectives
SLIDE NUMBER 44
February 2019
© VIDLEARN® 2019
Define key vocabulary and notation used in algebra.
Describe simplification by collecting like terms.
Demonstrate expanding linear and quadratic expressions.
Practice factorising linear and quadratic expressions.
Illustrate the process of rearranging formulae.
Interpret functions and find inverse functions.
Identify composite functions through worked examples.
44
SLIDE NUMBER 45
February 2019
© VIDLEARN® 2019
Algebra
The basic idea...
“ = ” is equal to

5 + 7 = 12
5 + 7 - 7 = 12 - 7 - 7
5 = 12 - 7
45
SLIDE NUMBER 46
February 2019
© VIDLEARN® 2019
Algebra
The basic idea...
“ = ” is equal to
x + y = z
x + y - y = z - y - y
x = z - y
This is called making x the subject
46

SLIDE NUMBER 47
February 2019
© VIDLEARN® 2019
Algebra
Make x the subject
x + 4 = y
x + 4 - 4= y - 4 - 4
x = y - 4
x - a = b
x - a + a= b + a + a
x = b + a
3 - x = y
3 = y + x + x
3 - y = x - y
2x = y
x = y ÷ 2
2

47
SLIDE NUMBER 48
February 2019
© VIDLEARN® 2019
Algebra
Make x the subject
4x + b = 2
4x = 2 - b- b
x = 2 - b÷ 4
4
48
SLIDE NUMBER 49
February 2019
© VIDLEARN® 2019
Algebra
Make x the subject
4(x + b) = 2
4x = 2 - 4b- 4b
x = 2 - 4b÷ 4

4
49
SLIDE NUMBER 50
February 2019
© VIDLEARN® 2019
Algebra
Make x the subject
3x2 + a = b
3x2 = b - a- a
x2 = b - a ÷ 3
3
x = b - asquare root
3
50
Pause here
Make x the subject
m + 3x = ax + d
3x = ax + d - m
3x - ax = d - m

x(3 - a) = d - m
x = d - m
3 - a
SLIDE NUMBER 51
February 2019
© VIDLEARN® 2019
CONSIDER…
51
SLIDE NUMBER 52
February 2019
© VIDLEARN® 2019
Algebra
Make x the subject
y + x = y(x + 2)
y + x= yx + 2yexpand
x = yx + 2y - y- y
x - yx = 2y - y- yx
x(1 - y) = 2y - yfactorise
x = 2y - 2÷ (1 - y)
1 - y
52

Pause here
SLIDE NUMBER 53
February 2019
© VIDLEARN® 2019
CONSIDER…
Make x the subject
3x + 5 = t
4(x - a) = b
y = 1 x + p
2
4. r = 4x2
5. a(x + b) = 2x + a
53
Pause here
SLIDE NUMBER 54
February 2019
© VIDLEARN® 2019
CONSIDER…
t - 5 = x

3
b + 4a = x
4
2(y - p) = x
r = x
4
a - ab = x
a - 2
Make x the subject
3x + 5 = t
4(x - a) = b
y = 1 x + p
2
4. r = 4x2
5. a(x + b) = 2x + a
54
Session Objectives
.
SLIDE NUMBER 55
February 2019
© VIDLEARN® 2019
Define key vocabulary and notation used in algebra.
Describe simplification by collecting like terms.

Demonstrate expanding linear and quadratic expressions.
Practice factorising linear and quadratic expressions.
Illustrate the process of rearranging formulae.
Interpret functions and find inverse functions.
Identify composite functions through worked examples.
55
SLIDE NUMBER 56
February 2019
© VIDLEARN® 2019
Algebra
Functions
x 4
xy
56
SLIDE NUMBER 57
February 2019
© VIDLEARN® 2019
Algebra
Functions

x4
x 4
y = 4x
f(x) = 4x
f(2) = 4 x 2 = 8
f(-3) = 4 x -3 = -12
57
SLIDE NUMBER 58
February 2019
© VIDLEARN® 2019
Algebra
Functions
f(x) = x - 7
Find; f(13) =
f(100) =
f(-2) =
f(a) =
13 - 7 = 6
100 - 7 = 93
-2 - 7 = -9
a - 7

58
Pause here
y = 2x + 3
y = x - 3
2
SLIDE NUMBER 59
February 2019
© VIDLEARN® 2019
CONSIDER…
y = 2x + 3
y - 3= 2x
y - 3 = x
2
59
SLIDE NUMBER 60
February 2019

© VIDLEARN® 2019
Algebra
Inverse functions
y = 2x + 3
y = x - 3
2
y = 2x + 3
y - 3= 2x
y - 3 = x
2
x - 3 = y
2
y = x - 3 or f-1(x) = x - 3
2 2
60
SLIDE NUMBER 61
February 2019
© VIDLEARN® 2019
Algebra
Inverse functions

x 4
xy
61
SLIDE NUMBER 62
February 2019
© VIDLEARN® 2019
Algebra
Finding the inverse function
f(x) = x2 - 5
y = x2 - 5
y + 5= x2
y + 5 = x
x + 5 = y
f-1(x) =x + 5
62

Pause here
Find the inverse functions for the following:
SLIDE NUMBER 63
February 2019
© VIDLEARN® 2019
CONSIDER…
63
Pause here
Find the inverse functions for the following:
SLIDE NUMBER 64
February 2019
© VIDLEARN® 2019
CONSIDER…
f -1(x) = x - 2
f-1(x) = x + 2
4
f -1(x) = x
f -1(x) = 3
x
f -1(x) = 1 - 2x
x

64
SLIDE NUMBER 65
February 2019
© VIDLEARN® 2019
Algebra
Finding the inverse function
f (x) = 1
x + 2
f -1(x) = 1 - 2x
x
y = 1
x + 2
y(x + 2)= 1
yx + 2y = 1
yx = 1 - 2y
x = 1 - 2y
y
65
Pause here

SLIDE NUMBER 66
February 2019
© VIDLEARN® 2019
CONSIDER…
www.mathspad.co.uk
66
Session Objectives
SLIDE NUMBER 67
February 2019
© VIDLEARN® 2019
Define key vocabulary and notation for algebra
67
SLIDE NUMBER 68
February 2019

© VIDLEARN® 2019
Algebra
Composite functions
f(x) = x - 7
Find; f(13) =
f(100) =
f(-2) =
f(a) =
13 - 7 = 6
100 - 7 = 93
-2 - 7 = -9
a - 7
68
SLIDE NUMBER 69
February 2019
© VIDLEARN® 2019
Algebra
Composite functions
f(x) = x - 7
Find; f(2x) =
f(x - 4) =
f(x + 4) =
f(3x + 1) =

2x - 7
x - 4 - 7 = x - 11
x + 4 - 7 = x - 3
3x + 1 - 7 = 3x - 6
69
SLIDE NUMBER 70
February 2019
© VIDLEARN® 2019
Algebra
Composite functions
f(x) = 3x + 2
Find; f(2x) =
f(x - 4) =
f(x + 4) =
f(3x + 1) =
3(2x) + 2 = 6x + 2
3(x - 4) + 2 = 3x - 10
3(x + 4) + 2 = 3x + 14
3(3x + 1) + 2 = 9x + 5
70

SLIDE NUMBER 71
February 2019
© VIDLEARN® 2019
Algebra
An example
Given f(x) = 2x + 3 and g(x) = 4x
Find:
fg(x)gf(x)
= f(g(x))
= 2(4x) + 3
= 8x + 3
= g(f(x))
= 4(2x + 3)
= 8x + 12
71
SLIDE NUMBER 72
February 2019
© VIDLEARN® 2019
Algebra
An example
Given f(x) = x2 + x and g(x) = -2x
Find:
fg(x)gf(x)
= f(g(x))

= (-2x)2 + (-2x)
= 4x2 - 2x
= g(f(x))
= -2(x2 + x)
= -2x2 - 2x
72
SLIDE NUMBER 73
February 2019
© VIDLEARN® 2019
Algebra
An example
Given g(x) = 3 + x and h(x) = x2 + 2
Find:
gh(x)hg(x)
= g(h(x))
= 3 + (x2 + 2)
= 5 + x2
= h(g(x))
= (3 + x)2 + 2
= 9 + 6x + x2 + 2
= x2 + 6x + 11
73

Pause here
Given that g(x) = (x + 2)(x + 4), h(x) = x2 - 1, j(x) = x + 3,
show that hj(x) = g(x)
SLIDE NUMBER 74
February 2019
© VIDLEARN® 2019
CONSIDER…
www.mathspad.co.uk
74
Pause here
Given that g(x) = (x + 2)(x + 4), h(x) = x2 - 1, j(x) = x + 3,
show that hj(x) = g(x)
SLIDE NUMBER 75
February 2019
© VIDLEARN® 2019
CONSIDER…
hj(x)= h(j(x))
= (x + 3)2 - 1
= x2 + 6x + 9 - 1
= x2 + 6x + 8
= (x + 2)(x + 4)
= g(x)

www.mathspad.co.uk
75
SLIDE NUMBER 76
February 2019
© VIDLEARN® 2019
CONSIDER…
Pause here
f(x) = x2 - 4x + 2, g(x) = 3x - 7, find fg(x)
g(x) = -6x + 5 and h(x) = -9x - 11, find gh(x)
f(x) = 2x - 5 and g(x) = 5x2 - 3, find gf(x)
f(x) = -2x + 9 and g(x) = -4x2 + 5x - 3, find fg(x)
f(x) = x - 3 and g(x) = 4x2 - 3x - 9, find gf(x)
www.mathspad.co.uk
76
SLIDE NUMBER 77
February 2019
© VIDLEARN® 2019
CONSIDER…
Answers
f(x) = x2 - 4x + 2, g(x) = 3x - 7, find fg(x)
fg(x) = 9x2 - 54x + 79
www.mathspad.co.uk
77

SLIDE NUMBER 78
February 2019
© VIDLEARN® 2019
CONSIDER…
Answers
g(x) = -6x + 5 and h(x) = -9x - 11, find gh(x)
gh(x) = 54x + 71
www.mathspad.co.uk
78
SLIDE NUMBER 79
February 2019
© VIDLEARN® 2019
CONSIDER…
Answers
f(x) = 2x - 5 and g(x) = 5x2 - 3, find gf(x)
gf(x) = 10x - 28
www.mathspad.co.uk
79
SLIDE NUMBER 80
February 2019

© VIDLEARN® 2019
CONSIDER…
Answers
f(x) = -2x + 9 and g(x) = -4x2 + 5x - 3, find fg(x)
fg(x) = 8x2 - 10x + 15
www.mathspad.co.uk
80
SLIDE NUMBER 81
February 2019
© VIDLEARN® 2019
CONSIDER…
Answers
f(x) = x - 3 and g(x) = 4x2 - 3x - 9, find gf(x)
gf(x) = 4x2 - 27x + 36
www.mathspad.co.uk
81
Session Objectives
SLIDE NUMBER 82
February 2019

© VIDLEARN® 2019
Define key vocabulary and notation for algebra
82
SLIDE NUMBER 83
February 2019
© VIDLEARN® 2019
CONSIDER…
End of Presentation
SLIDE NUMBER 84
February 2019
© VIDLEARN® 2019

Joe Hammond
End of presentation
84
GCSE Mathematics – Quadratic Equations
SLIDE NUMBER 1
May 2019
© VIDLEARN® 2019
Rebecca Wigfull
Insert a relevant picture here to fill the whole space
1
Session Objectives
The purpose of the session is to be able to:
Recall the key characteristics of linear and quadratic graphs.
Factorise a quadratic equation including the difference of two

squares.
Apply factorising to solving a quadratic equation.
Solve quadratic equations by completing the square.
Solve quadratic equations by using the quadratic formula.
Solve quadratic inequalities.
Solve quadratic equations using an iterative method.
SLIDE NUMBER 2
May 2019
© VIDLEARN® 2019
2
CONSIDER…
and ask yourself:
learning
SLIDE NUMBER 3
May 2019
© VIDLEARN® 2019
3
Session Objectives
SLIDE NUMBER 4

May 2019
© VIDLEARN® 2019
squares.
4
Linear Graphs
Quadratic Equations
SLIDE NUMBER 5
May 2019
© VIDLEARN® 2019
y = mx + c
5
Negative curve

Positive curve
Negative curve
Positive curve
Quadratic Graphs
Quadratic Equations
SLIDE NUMBER 6
May 2019
© VIDLEARN® 2019
y = ax2 +bx + c
6
Key Characteristics of a Quadratic Graph
Quadratic Equations
SLIDE NUMBER 7
May 2019
© VIDLEARN® 2019
Root
Root
(-2,0)
(2,0)
Turning point
(0, -4)

Positive curve
7
Quadratic Equations
SLIDE NUMBER 8
May 2019
© VIDLEARN® 2019
Root
Root
(-4,0)
(2,0)
Turning point
(-1, 9)
Negative curve
Y - intercept
(0, 8)
8

Quadratic Equations
SLIDE NUMBER 9
May 2019
© VIDLEARN® 2019
Root
Turning point
(-1, 0)
Positive curve
y - intercept
(0, 1)
Repeated root
9
Quadratic Equations
SLIDE NUMBER 10
May 2019
© VIDLEARN® 2019
Turning point
(3, 2)
Positive curve
y - intercept
(0, 11)
Turning point
(-1, -1)
Negative curve

(0, -2)
Y - intercept
No Roots
No Roots
10
For the following 3 graphs list the key characteristic of each
one.
SLIDE NUMBER 11
May 2019
© VIDLEARN® 2019
CONSIDER…
The green graph is a positive quadratic curve.
It has no roots.
The turning point is (1,2).
The intercept on the y-axis is (0,3).
The blue graph is a positive quadratic curve.
It’s roots are at (-3,0) and (3,0).
The turning point is (0,-9).
The intercept on the y-axis is (0,-9).
The red graph is a negative quadratic curve.
It has a repeated root where x = 2.
The turning point is (2,0).
The intercept on the y-axis is (0, -4).
11

Session Objectives
SLIDE NUMBER 12
May 2019
© VIDLEARN® 2019
squares.
12
Factorising Quadratics
Quadratic Equations
SLIDE NUMBER 13
May 2019
© VIDLEARN® 2019
If (x + a)(x + b) is expanded we get
(x + a)(x + b) = x2 + ax + bx + ab
= x2 + (a + b)x + ab

13
Quadratic Equations
SLIDE NUMBER 14
May 2019
© VIDLEARN® 2019QuadraticFactorisedRule appliedx2 + 5x +
6 (x + 3)(x + 2)A positive number in each bracket.x2 - 5x + 6(x
- 3)(x - 2)A negative number in each bracket.x2 - x - 6(x - 3)(x
+ 2)A negative and a positive number in each bracket. x2 + x - 6
(x + 3)(x - 2)A negative and a positive number in each bracket.
14
Quadratic Equations
SLIDE NUMBER 15
May 2019
© VIDLEARN® 2019
x2 + 6x + 8
x2 + 6x + 8
(x )(x )
(x + )(x + )
x2 + 6x + 8abtotal181+8 = 9242+4 = 6
(x + a )(x + b )

(x + 2 )(x + 4 )
(x + 2 )(x + 4 )
= x2 + 4x + 2x + 8
= x2 + 6x + 8
15
Quadratic Equations
SLIDE NUMBER 16
May 2019
© VIDLEARN® 2019
x2 - 7x + 12
x2 - 7x + 12
(x )(x )
(x - )(x - )
x2 - 7x + 12abtotal1121+12 = 13262+6 = 8343+4=7
(x - a )(x - b )
(x - 3 )(x - 4 )
(x - 3 )(x - 4 )
= x2 - 4x - 3x + 12
= x2 - 7x + 12

16
Quadratic Equations
SLIDE NUMBER 17
May 2019
© VIDLEARN® 2019
x2 + 6x - 16
x2 + 6x - 16
(x )(x )
(x + )(x - )abdifference11615286440
(x + 8 )(x - 2 )
(x + 8 )(x - 2 )
= x2 - 2x + 8x - 16
= x2 + 6x - 16
17
Quadratic Equations
SLIDE NUMBER 18
May 2019
© VIDLEARN® 2019
x2 - x - 20
x2 - x - 20

(x )(x )
(x + )(x - )abdifference120192108451
(x + 4 )(x - 5 )
(x + 4)(x - 5 )
= x2 - 5x + 4x - 20
= x2 - x - 20
18
Quadratic Equations
SLIDE NUMBER 19
May 2019
© VIDLEARN® 2019
x2 - 36
x2 - 36
(x )(x )
(x + )(x - )
(x + 6 )(x - 6 )
(x + 6)(x - 6 )
= x2 - 6x + 6x - 36
= x2 - 36
The difference of two squares
19

Quadratic Equations
SLIDE NUMBER 20
May 2019
© VIDLEARN® 2019
9x2 - 36
9x2 - 36
( )( )
( + )( - )
(3x + 6 )(3x - 6 )
(3x + 6)(3x - 6 )
= 9x2 - 18x + 18x - 36
= 9x2 - 36
The difference of two squares
20
Factorising Quadratics – single bracket
Quadratic Equations
SLIDE NUMBER 21
May 2019
© VIDLEARN® 2019
x2 - 23x
x(x )
= x2 - 23x
x(x - 23 )
x(x - 23 )
2x2 + 14x

2x(x )
= 2x2 + 14x
2x(x + 7 )
2x(x + 7 )
21
– where the coefficient of x2 is not 1
Quadratic Equations
SLIDE NUMBER 22
May 2019
© VIDLEARN® 2019
Can you do an initial factorisation to make the coefficient of x2
1?
5x2 - 20
5(x2 - 4)
5(x + 2)(x - 2)
Difference of two squares.
22
– where the coefficient of x2 is not 1 (prime)
Quadratic Equations
SLIDE NUMBER 23

May 2019
© VIDLEARN® 2019
3x2 + 11x + 6
(3x + )(x + )3xTotal16192311329619
(3x + 2 )(x + 3 )
23
– where the coefficient of x2 is not 1 (prime)
- alternative method
Quadratic Equations
SLIDE NUMBER 24
May 2019
© VIDLEARN® 2019
3x2 + 11x + 6
(3x + )(x + )
3 x 6 = 18abtotal118192911369
(3x + 2)(x + 3)

24
- where the coefficient of x2 is not 1 (prime)
Quadratic Equations
SLIDE NUMBER 25
May 2019
© VIDLEARN® 2019
5x2 + 7x - 6
(5x )(x )ab1-6-292-3-133-2-76-11
(5x - 3)(x + 2)ab-1629-2313-327-61-1
25
- where the coefficient of x2 is not 1 (prime)
Quadratic Equations
SLIDE NUMBER 26
May 2019
© VIDLEARN® 2019
5x2 + 7x - 6

(5x )(x )
5 x 6 = 30
(5x - 3)(x + 2)abDifference13029215133107651
26
- where the coefficient of x2 is not 1 (not a prime)
Quadratic Equations
SLIDE NUMBER 27
May 2019
© VIDLEARN® 2019
8x2 - 2x - 15
(2x )(4x )ab1-15-1193-5-375-3-1915-17
ab-115119-3537-5319-151-7
(8x )(x )
27

- where the coefficient of x2 is not 1 ( not a prime)
Quadratic Equations
SLIDE NUMBER 28
May 2019
© VIDLEARN® 2019
8x2 - 2x - 15
(2x )(4x )ab1-15-263-525-31415-158
ab-11526-35-2-53-14-151-58
(2x -3)(4x + 5)
28
where the coefficient of x2 is not 1 ( not a prime)
alternative method
Quadratic Equations
SLIDE NUMBER 29
May 2019
© VIDLEARN® 2019
8x2 - 2x - 15ab1120119260583403743026
ab5241962014815710122
8x15 = 120
8x2 + 10x - 12x - 15
2x(4x + 5) - 3(4x + 5)
(2x-3)(4x + 5)

29
Factorise each of the following questions
SLIDE NUMBER 30
May 2019
© VIDLEARN® 2019
CONSIDER…
x2 - 12x + 35
x2 - 17x + 30
x2 + 9x + 20
x2 + 19x + 18
x2 + 2x - 48
x2 - 1
3x2 - 15x
4x2 - 64
(x-7)(x-5)
(x-15)(x-2)
(x+4)(x+5)
(x+18)(x+1)
(x+8)(x-6)
(x+1)(x-1)
3x(x-5)
(2x+8)(2x-8)
2x2 + 5x +3
5x2 - 38x +21
6x2 +17x + 12
9x2 +9x - 10
(2x+3)(x+1)
(5x-3)(x-7)
(3x+4)(2x+3)
(3x-2)(3x+5)

30
Session Objectives
SLIDE NUMBER 31
May 2019
© VIDLEARN® 2019
squares.
31
Solving a quadratic equation
Quadratic Equations
SLIDE NUMBER 32
May 2019
© VIDLEARN® 2019

Root
Root
(0,-2)
(0, 2)
x2 - 4
(x-2)(x+2)= 0
x = -2 or 2
32
Solving a Quadratic Equation
Quadratic Equations
SLIDE NUMBER 33
May 2019
© VIDLEARN® 2019
Root
Root
(-4,0)
(2, 0)
-(x+4)(x-2)= 0
x = -4 or 2
-x2 - 2x + 8
-(x2 +2x - 8)= 0
33

Solving Quadratic Equations
Quadratic Equations
SLIDE NUMBER 34
May 2019
© VIDLEARN® 2019
x2 + 6x - 16 = 0
(x + 8 )(x - 2 ) = 0
(x + 8 )= 0 and (x - 2 ) = 0
x + 8 = 0
x = -8
and
x - 2 = 0
x = 2
The roots are (2,0) and (-8,0)
34
Solving Quadratic Equations
- where the coefficient of x2 is not 1
Quadratic Equations
SLIDE NUMBER 35
May 2019
© VIDLEARN® 2019
8x2 - 2x – 15 = 0
(2x -3)(4x + 5) = 0
(2x - 3)= 0 and (4x + 5) = 0
2x - 3 = 0

2x = 3
x = 1.5
and
4x + 5 = 0
4x = -5
x = -1.25
The roots are (1.5,0) and (-1.25,0)
35
Solve each of the following questions you previously factorised
SLIDE NUMBER 36
May 2019
© VIDLEARN® 2019
CONSIDER…
x2 - 12x + 35=0
x2 - 17x + 30=0
x2 + 9x + 20=0
x2 + 19x + 18=0
x2 + 2x - 48=0
x2 – 1=0
3x2 - 15x=0
4x2 – 64=0
x = 5 or 7
x = 2 or 15
x = -2 or -15
x = -18 or -1
x = -8 or 6
x = 1 or -1
x = 0 or 5
x = 4 or -4

2x2 + 5x +3=0
5x2 - 38x +21=0
6x2 +17x + 12=0
9x2 +9x - 10=0
x = -1.5 or -1
x = 0.6 or 7
x = -1.3 or -1.5
x = -1.6 or 0.6
.
.
.
36
Session Objectives
SLIDE NUMBER 37
May 2019
© VIDLEARN® 2019
squares.

37
Completing the Square
Quadratic Equations
SLIDE NUMBER 38
May 2019
© VIDLEARN® 2019
Minimum Point
Minimum point (-3, -21)
Minimum value is -21
X2 + 6x - 12
Turning point
38
x2 + 6x – 12 = 0
(x + 3)2 – 9 – 12 = 0
(x + 3)2 – 21 = 0
(x + 3)2 = 21
x + 3 = ± 21
x = ± 21 – 3
x = 1.58 or x = -7.58 (2dp)
Solving using Completing the Square
Quadratic Equations

SLIDE NUMBER 39
May 2019
© VIDLEARN® 2019
Roots (1.58,0) and (-7.58,0)
Half the coefficient of x
Take away the square of half the coefficient of x
Completed square form
Minimum point (-3, -21)
Minimum value is -21
√
√
39
Completing the Square
Quadratic Equations
SLIDE NUMBER 40
May 2019
© VIDLEARN® 2019
Turning point
(-1, 9)
Maximum point
-x2 - 2x + 8

Maximum point (-1, 9)
Maximum value is 9
40
-x2 - 2x + 8 = 0
-(x2 + 2x – 8) = 0
x2 + 2x – 8 = 0
(x + 1)2 – 1 – 8 = 0
(x + 1)2 – 9 = 0
(x + 1)2 = 9
x + 1 = ± 9
x = ± 3 – 1
x = -4 or x = 2
√
Quadratic Equations
SLIDE NUMBER 41
May 2019
© VIDLEARN® 2019
Roots (-4,0) and (2,0)
Completed square form

Maximum point (-1, 9)
Maximum value is 9
41
4x2 - 8x + 1 = 0
4(x2 - 2x + 0.25) = 0
x2 - 2x + 0.25 = 0
(x - 1)2 -1 + 0.25 = 0
(x - 1)2 - 0.75 = 0
(x - 1)2 = 0.75
x - 1 = ± 0.75
x = 1 ± 0.75
x = 1.87 (2dp) or x = 0.13 (2dp)
√
√
Quadratic Equations
SLIDE NUMBER 42
May 2019
© VIDLEARN® 2019

Roots (1.87,0) and (0.13,0)
Minimum point (1, -0.75)
Minimum value is -0.75
42
2x2 = 8x +1 1
2x2 - 8x -11 = 0
2(x2 - 4x – 5.5) = 0
x2 - 4x – 5.5 = 0
(x - 2)2 -4 -5.5 = 0
(x - 2)2 -9.5 = 0
(x - 2)2 = 9.5
x - 2 = ± 9.5
x = 2 ± 9.5
x = 5.08 (2dp) or x = -1.08 (2dp)
√
√
Quadratic Equations
SLIDE NUMBER 43
May 2019
© VIDLEARN® 2019

Minimum point (2, -9.5)
Minimum value is -9.5
43
Use the completing the square to find the turning point and to
solve the following quadratics. (Give answers to 2dp where
necessary)
SLIDE NUMBER 44
May 2019
© VIDLEARN® 2019
CONSIDER…1. x2 - 8x = -132. 2x2 + 10x + 5 = 03. x2 - 3x
- 11 = 04. 3x2 + 2 = 9xx = 5.73 (2dp) or
x = 2.27 (2dp)
Turning point (4,-3) x = -0.56 (2dp) or
x = -4.44 (2dp)
Turning point (-2.5,-3.75)x = 5.14 (2dp) or
x = -2.14 (2dp)
Turning point (1.5,-13.25)x = 2.76 (2dp) or
x = 0.24 (2dp)
Turning point (1.5-1.58)
44
Session Objectives

SLIDE NUMBER 45
May 2019
© VIDLEARN® 2019
squares.
45
The quadratic formula
Quadratic Equations
SLIDE NUMBER 46
May 2019
© VIDLEARN® 2019
By completing the square on the quadratic equation
ax2 + bx + c = 0
a quadratic formula can be found which will solve most
quadratic equations which do not factorise.
This is an alternative method of solving equations other than by

completing the square.
The quadratic formula for solving quadratic equations is
x = -b ± b2 - 4ac
2a
√
46
Quadratic Equations
SLIDE NUMBER 47
May 2019
© VIDLEARN® 2019
x = -b ± b2 - 4ac
2a
√
x2 - 2x - 6 = 0
ax2 + bx + c = 0
then a = 1, b = -2 and c = -6
x = -(-2) ± (-2)2 – 4(1)(-6)
2(1)
√
x = 2 ± 4 – -24
2
√
x = 2 ± 28
2

√
x = 2 ± 2 7
2
√
x = 1 ± 7
√
x = 3.645751311 or
x = -1.645751311
x = 3.65 (2dp) or
x = -1.65 (2dp)
47
Quadratic Equations
SLIDE NUMBER 48
May 2019
© VIDLEARN® 2019
x = -b ± b2 - 4ac
2a
√
3 x2 + 4x - 2 = 0
ax2 + bx + c = 0
then a = 3, b = 4 and c = -2
x = -(4) ± (4)2 – 4(3)(-2)
2(3)
√
x = -4 ± 16 – -24
6
√
x = -4± 40

6
√
x = -4 ±2 10
6
√
x = 0.3874258867 or
x = -1.72075922
x = -2 ± 10
3
√
x = 0.39 (2dp) or
x = -1.72 (2dp)
48
Use the quadratic formula to solve the following quadratics
SLIDE NUMBER 49
May 2019
© VIDLEARN® 2019
CONSIDER…1. 3x2 + 6x - 7 = 02. x2 + 3x + 1 = 03. 8x2 -
6x + 1 = 04. 4x2 + 7x - 6 = 0x = 0.83 (2dp) or
x = -2.83 (2dp)2. x = -0.38 (2dp) or
x = -2.62 (2dp)3. x = 0.5 or x = 0.25x = 0.63 (2dp) or
x = -2.38 (2dp)
49
Session Objectives

SLIDE NUMBER 50
May 2019
© VIDLEARN® 2019
squares.
50
Solving Quadratic inequalities
Quadratic Equations
SLIDE NUMBER 51
May 2019
© VIDLEARN® 2019
x2 + 2x -15
(x+5)(x-3)
Roots: (-5,0)and (3,0)
If …
x2 + 2x -15 ≥ 0
above x-axis
x ≥ 3 and x ≤ -5

51
Quadratic Equations
SLIDE NUMBER 52
May 2019
© VIDLEARN® 2019
x2 + 2x -15
(x+5)(x-3)
Roots: (-5,0)and (3,0)
If …
x2 + 2x -15 ≤ 0
below x-axis
-5 ≤ x ≤ 3

52
Quadratic Equations
SLIDE NUMBER 53
May 2019
© VIDLEARN® 2019
x2 + 12x + 82 > 22 – 4x
x2 + 16x + 60 > 0
(x + 6)(x + 10)
Roots: (-6,0)and (-10,0)
x2 + 16x + 60 > 0
above x-axis
x > -6 and x < -10
53

Quadratic Equations
SLIDE NUMBER 54
May 2019
© VIDLEARN® 2019
5x2 < 80
x2 < 16
x2 - 16 < 0
(x + 4)(x - 4)
Roots: (-4,0)and (4,0)
x2 - 16 < 0
below x-axis
-4 < x < 4
{x: -4 < x < 4}
54
Use the quadratic formula to solve the following quadratics
SLIDE NUMBER 55

May 2019
© VIDLEARN® 2019
CONSIDER…1. x2 - 2x - 48 ≥ 02. x2 - 49 ≤ 03. 24 < 10x -
x2 4. x2 + 7x - 10 > 4xx ≤ -6 and x ≥ 82. -7 ≤ x ≤ 73. 4 < x <
6x > 2 and x < -5
55
Session Objectives
SLIDE NUMBER 56
May 2019
© VIDLEARN® 2019
squares.
56
Solving using an iterative method

Quadratic Equations
SLIDE NUMBER 57
May 2019
© VIDLEARN® 2019
Use the iteration formula xn+1 = 2 + 1
xn
to find solutions to x2 - 2x = 1 to 3dp. Use the starting value x0
= 3
Press 3 then =
2 + 1
ANS
x1 = 2.3333
x2 = 2.4286
x3 = 2.4118 x4 = 2.4146
x5 = 2.4141
x6 = 2.4142
x25 = 2.4142
x = 2.414 to 3dp
57
SLIDE NUMBER 58
May 2019
© VIDLEARN® 2019
CONSIDER…
Use the iteration formula xn+1 = 1 - 2

xn
to find solutions to x2 + 2x = 1 to 3dp. Use the starting value x0
= 1
Use the iteration formula xn+1 = 3xn + 1
2
to find solutions to -2x2 + 3x + 1 = 0 to 1dp. Use the starting
value x0 = 1
Use the iteration formula xn+1 = 2x2n + 11
to find solutions to 2x2 = 11 to 3dp. Use the starting value x0 =
3
√
4xn
x = -2.414
x = 1.8
x = 2.345
58
Session Objectives
SLIDE NUMBER 59
May 2019
© VIDLEARN® 2019

squares.
59
SLIDE NUMBER 60
May 2019
© VIDLEARN® 2019
CONSIDER…
End of Presentation
End of presentation
Rebecca Wigfull
SLIDE NUMBER 61
May 2019
© VIDLEARN® 2019
GCSE Mathematics – Quadratic Equations

61
GCSE Mathematics – Algebra – Linear Equations
SLIDE NUMBER 1
April 2019
© VIDLEARN® 2019
Elliott Wade
1
Session Objectives
Differentiate between Expressions, Equations and Formula
Solve simple Linear Equations
Solve Linear Equations involving multi-step skills
Solve Linear Inequalities
SLIDE NUMBER 2
April 2019
© VIDLEARN® 2019

Solve
Linear
Equation
Expression
Equation
Inequality
2
CONSIDER…
and ask yourself:
learning
SLIDE NUMBER 3
April 2019
© VIDLEARN® 2019
3
Session Objectives
Solve simple Linear equations
SLIDE NUMBER 4

April 2019
© VIDLEARN® 2019
Solve
Linear
Equation
Expression
Equation
Inequality
4
Definitions
Put into practice
Exam style questions
Linear Equations
SLIDE NUMBER 5
April 2019
© VIDLEARN® 2019
Oracy
Linear
Equation
Expression
Equation
Inequality
5

Definitions
Equation – States that two things are equal.
For example: 6a = 12
Identity – An equation that is true for any given value
substituted in.
For example: xy = yx
Formula – A mathematical rule usually containing an equals
sign.
For example: Area of a circle
Expression – a group of numbers, letters and operations.
For example: 5a + 9
Linear Equations
SLIDE NUMBER 6
April 2019
© VIDLEARN® 2019
B4L is key. Time to embed into practice. Students need to be
comfortable with using all four key words under this topic. And
so should teachers!
Formula
Expression
Identity
Equation
Equal
6

Linear Equations
SLIDE NUMBER 7
April 2019
© VIDLEARN® 2019
Put into practice
Equation
Formula
Expression
x(x-4) = 12
x + x + x + x
V = u +at
C = d
2x +14
15t3 + 3v = 146
E=MC2
B4L is key. Time to embed into practice.
Resilience
Independence
Algebra
Equation
Deep understanding
7
Linear Equations
SLIDE NUMBER 8
April 2019
© VIDLEARN® 2019

A group of numbers, letters and operations
Dan: x
Harry : x + 5
Regan 2x
Mean = Total sum ÷ number of sellers
x + x + 5 + 2x
Expression = (4x + 5) ÷ 3
Resilience
Independence
Algebra
Equation
Deep understanding
8
CONSIDER…
Consider the following question:
SLIDE NUMBER 9
April 2019
© VIDLEARN® 2019
9
Linear Equations

SLIDE NUMBER 10
April 2019
© VIDLEARN® 2019
Consider task - Answer
T = 5x + 20y
Resilience
Independence
Algebra
Equation
Deep understanding
10
Session Objectives
Solve linear equations involving multi-step skills
Solve linear inequalities
SLIDE NUMBER 11
April 2019
© VIDLEARN® 2019
11

Methods
One step equations
Linear Equations
SLIDE NUMBER 12
April 2019
© VIDLEARN® 2019
12
Methods
Linear Equations
SLIDE NUMBER 13
April 2019
© VIDLEARN® 2019
Balance method
a + 4 = 12

a = 8
Elimination method
a + 4 = 12
To eliminate +4 from the left we move it over the equals sign.
a + 4 = 12 – 4
Whenever moving across the equals sign, invert the operation
a = 8
-4
-4
Weave solving equations into several topics.
13
One step equations
Linear Equations
SLIDE NUMBER 14
April 2019
© VIDLEARN® 2019
÷10

b = 10
x = 7
÷10
10b = 100
x3 = 343
÷2
c = 16.5
÷2
33 = 2c
-12
v = -15
-12
12 + v = - 3
www.mathsgenie.co.uk
14
Exam Style questions
Linear Equations

SLIDE NUMBER 15
April 2019
© VIDLEARN® 2019
-5
-5
y = 7
7
+4
+4
t = 11
11
www.mathsgenie.co.uk
15
Linear Equations
SLIDE NUMBER 16
April 2019

© VIDLEARN® 2019
÷2
÷2
y = 4
4
x3
x3
y = 18
18
16
Task:
Complete the following one step equations:
4y = 10
10g = 37
a – 7 = -3
V + 2 = -6
= 2.7
5y = 24
SLIDE NUMBER 17
April 2019
© VIDLEARN® 2019

CONSIDER…
17
Session Objectives
Consider Task Review
SLIDE NUMBER 18
April 2019
© VIDLEARN® 2019
4y = 10
10g = 37
a – 7 = -3
V + 2 = -6
= 2.7
5y = 24
÷4
÷4
4y = 10
y = 2.5
÷10
÷10
10g = 37
g = 3.7

+7
+7
a – 7 = -3
a = 4
-2
-2
V + 2 = -6
V = - 8
x4
x4
= 2.7
w = 10.8
÷5
÷5
5y = 24
a = 4.8
18

Session Objectives
SLIDE NUMBER 19
April 2019
© VIDLEARN® 2019
19
Multi-step examples
Exam Style Questions
Common misconception
Linear Equations
SLIDE NUMBER 20
April 2019
© VIDLEARN® 2019
20

Multi-Step Examples – Two Step equations
Linear Equations
SLIDE NUMBER 21
April 2019
© VIDLEARN® 2019
4x = 12
x = 3
-3
-3
÷4
÷4
+3
+3
÷4
÷4
4x = 18
x = 4.5
https://variationtheory.com/category/algebra/equations/
Variation
Solve

Multi Step
Equal
Balance
21
Multi-Step Examples – Variables on both sides
Linear Equations
SLIDE NUMBER 22
April 2019
© VIDLEARN® 2019
x = 2
x + 3 = 5
-x
-x
-3
-3
-x
-x
-3
-3
x = 1
2x + 3 = 5

÷2
÷2
2x = 2
Variation
Solve
Multi Step
Equal
Balance
22
Multi-Step Examples - Brackets
Linear Equations
SLIDE NUMBER 23
April 2019
© VIDLEARN® 2019
-2
-2
-x

-x
Expand
2x + 2 = x +5
x + 2 = 5
x = 3
Variation
Solve
Multi Step
Equal
Balance
23
Linear Equations
SLIDE NUMBER 24
April 2019
© VIDLEARN® 2019
Perimeter = 2x + 2x + 10
Perimeter = 4x + 10
34 = 4x + 10
24 = 4x

6 = x
-10
-10
÷4
÷4
https://www.mathsgenie.co.uk/resources/64_forming-and-
solving-equations.pdf
Students can sometimes discover the question themselves and
attempt to answer before knowing the specific question.
Comfort with answer at the end being an uncommon format.
24
Common Misconception
Linear Equations
SLIDE NUMBER 25
April 2019
© VIDLEARN® 2019
It is not uncommon for equations to be given in a format that
could leave the variable on the right instead of the left.
Consider the previous question:
34 = 4x + 10
24 = 4x
6 = x
-10
-10

÷4
÷4
Having the variable on the right does not alter the outcome of
the answer.
Consider:
2 + 3 = 5 and 5 = 2 + 3
https://www.mathsgenie.co.uk/resources/64_forming-and-
solving-equations.pdf
Students can sometimes discover the question themselves and
attempt to answer before knowing the specific question.
Comfort with answer at the end being an uncommon format.
25
Consider this angle problem from the ‘Increasingly Difficult
Questions’ website.
SLIDE NUMBER 26
April 2019
© VIDLEARN® 2019
CONSIDER…
http://taylorda01.weebly.com/uploads/4/2/3/8/42387051/angles_
in_triangles_and_quadrilaterals_01_v2.pdf
26

Consider Slide - Answer
Two base angles must be equal (Isosceles Triangle)
Angles in a triangle sum to 180 degrees.
An equation could be formed to show this problem…
Linear Equations
SLIDE NUMBER 27
April 2019
© VIDLEARN® 2019
3n – 10 + 3n – 10 + ? = 180
6n -20 + ? = 180
6n + ? = 200
? = 200 - 6n
Pattern
Misconception
Comfort zone
Oracy
27
Session Objectives
SLIDE NUMBER 28
April 2019

© VIDLEARN® 2019
28
Notation
Examples and approaches
Linear Equations
SLIDE NUMBER 29
April 2019
© VIDLEARN® 2019
29
Notation
Confidence in understanding notation used to show inequalities
is key!
Linear Equations
SLIDE NUMBER 30
April 2019

© VIDLEARN® 2019
Essential information!
Less than
Greater than
Less than or equal to
Greater than or equal to
30
Notation continued
Linear Equations
SLIDE NUMBER 31
April 2019
© VIDLEARN® 2019
Notation

Number line
Inequaltiy
Representation
31
Examples and Approaches
Linear Equations
SLIDE NUMBER 32
April 2019
© VIDLEARN® 2019
The direction of the inequality sign is essential! So do we ever
change it?... When manipulating inequalities most rules are
similar to standard linear equations
Okay to do:
Simplify a side
Add/subtract from both sides
Multiply/divide both sides by a positive number.
Actions that flip the sign:
Multiply/divide both sides by a negative number.
Swapping left and right sides.
32
Tip!
More ink in the symbol means more on the diagram
Examples and Approaches – Solving Example 1

Linear Equations
SLIDE NUMBER 33
April 2019
© VIDLEARN® 2019
+10
+10
÷2
÷2
2t < 20
t < 10
Number line
Negative
Solve
Inequality
Multi Step
33
Examples and Approaches – Solving Example 2
Linear Equations
SLIDE NUMBER 34
April 2019
© VIDLEARN® 2019

-5
-5
÷-5
÷-5
- 5x ≤ - 15
x ≥ 3
Don’t forget to flip the sign here!
Number line
Negative
Solve
Inequality
Multi Step
34
Examples and Approaches – Example 3 - Number line
representation
Linear Equations
SLIDE NUMBER 35
April 2019
© VIDLEARN® 2019
Can you show the following inequality on a number line?

Number line
Negative
Solve
Inequality
Multi Step
35
Task: Complete the following 5 questions from
www.CorbettMaths.com
SLIDE NUMBER 36
April 2019
© VIDLEARN® 2019
CONSIDER…
1.
2.
3.
5. Show the following inequality on a number line.
4. Write the inequality shown on the number line
https://corbettmaths.com/wp-

content/uploads/2018/12/Inequalities-pdf.pdf
36
Session Objectives
Consider task - Answers
SLIDE NUMBER 37
April 2019
© VIDLEARN® 2019
1.
2.
3.
5. Show the following inequality on a number line.
4. Write the inequality shown on the number line
>
X ≤ -10
X is less than or equal to -4
X is less than -3
https://donsteward.blogspot.com/2012/09/linear-
inequalities.html
Promotion of discussion: Is this about inequalities?
What question could you ask relating to inequalities?
What key words in the question help you model the problem?

37
Session Objectives
From Fluency and Reasoning to Problem Solving.
SLIDE NUMBER 38
April 2019
© VIDLEARN® 2019
For what values of x is the perimeter of the square greater than
the perimeter of the rectangle?
<
P of S = 12x - 8
P of R = 6x + 22
https://donsteward.blogspot.com/2012/09/linear-
inequalities.html
Promotion of discussion: Is this about inequalities?
What question could you ask relating to inequalities?
What key words in the question help you model the problem?
38
Session Objectives
From Fluency and Reasoning to Problem Solving.
SLIDE NUMBER 39
April 2019

© VIDLEARN® 2019
For what values of x is the perimeter of the square greater than
the perimeter of the rectangle?
<
P of S =
P of R =
12x - 8
6x + 22
<
<
12x - 8
6x + 22
39
Session Objectives
SLIDE NUMBER 40
April 2019
© VIDLEARN® 2019
<
<
<
12x - 8
6x + 22
-6x

-6x
+8
+8
<
6x - 8
22
÷6
÷6
6x
<
<
30
5
x
40
Session Objectives
SLIDE NUMBER 41
April 2019
© VIDLEARN® 2019
The purpose of the session was to:

41
SLIDE NUMBER 42
April 2019
© VIDLEARN® 2019
CONSIDER…
End of Presentation
End of presentation
Elliott Wade
SLIDE NUMBER 43
April 2019
© VIDLEARN® 2019
GCSE Mathematics – Algebra – Linear Equations
43

GCSE Mathematics – Simultaneous Equations
SLIDE NUMBER 1
May 2019
© VIDLEARN® 2019
Rebecca Wigfull
1
Session Objectives
Solve simple simultaneous equations.
Solve complex simultaneous equations.
Solve simultaneous equations in context.
Solve linear and non linear simultaneous equations.
Solve simultaneous equations by finding approximate solutions
using a graph.
SLIDE NUMBER 2
May 2019
© VIDLEARN® 2019
2
CONSIDER…

and ask yourself:
learning
SLIDE NUMBER 3
May 2019
© VIDLEARN® 2019
3
Session Objectives
Solve simple simultaneous equations.
Solve complex simultaneous equations.
Solve simultaneous equations in context.
Solve linear and non linear simultaneous equations.
Solve simultaneous equations by finding approximate solutions
using a graph.
SLIDE NUMBER 4
May 2019
© VIDLEARN® 2019
4

Simple simultaneous equations
Simultaneous Equations
SLIDE NUMBER 5
May 2019
© VIDLEARN® 2019
2x - y = 1
2x + 2y = 10
1
2
2x - y = 1
2x + 2y = 10
-3y = -9
y = 3
-
Sub y = 3 into
1
2x - 1x3 = 1
2x - 3 = 1
2x = 4
x = 2
Sub y = 3 and x = 2 into
2
2x2 + 2x3 = 10
4 + 6 = 10
10 = 10
x = 2 and y = 3
(2,3)
Check
Same Sign Subtract (SSS)

5
Simple simultaneous equations
Simultaneous Equations
SLIDE NUMBER 6
May 2019
© VIDLEARN® 2019
2x - 5y = 1
3x + 5y = 14
1
2
2x - 5y = 1
3x + 5y = 14
5x = 15
x = 3
+
Sub x = 3 into
1
2x3 - 5y = 1
6 - 5y = 1
- 5y = -5
y = 1
Sub x = 3 and y = 1 into
2
3x3 + 5x1 = 14
9 + 5 = 14
14 = 14

x = 3 and y = 1
(3,1)
Check
6
Solve the following simultaneous equations
SLIDE NUMBER 7
May 2019
© VIDLEARN® 2019
CONSIDER…
a)
3x - y = 18
3x + 6y = -3
b)
3x - 2y = 57
5x - 2y = 111
c)
2x - y = 7
4x + y = 23
x = 5 and y = -3
(5,-3)
x = 27 and y = 12
(27,12)

x = 5 and y = 3
(5,3)
7
Solution
s
SLIDE NUMBER 8
May 2019
© VIDLEARN® 2019
CONSIDER…
a)
3x – y = 18
3x + 6y = -3
1
2
3x - y = 18
3x + 6y = -3

-7y = 21
y = -3
-
Sub y = -3 into
1
3x - 1x-3 = 18
3x + 3 = 18
3x = 15
x = 5
Sub x = 5 and y = -3 into
2
3x5 + 6x-3 = -3
15 - 18 = -3
-3 = -3
x = 5 and y = -3
(5,-3)
Check
Same Sign Subtract (SSS)

GCSE Mathematics – Basics of numberSLIDE NUMBER 1 May 2019

GCSE Mathematics – Basics of numberSLIDE NUMBER 1 May 2019

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GCSE Mathematics – Basics of numberSLIDE NUMBER 1 May 2019