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IGCSE Maths CIE
2. Algebra & Graphs
CONTENTS
2.1 Algebra Toolkit
2.1.1 Algebraic Notation & Vocabulary
2.1.2 Algebra Basics
2.2 Algebraic Roots & Indices
2.2.1 Algebraic Roots & Indices
2.3 Expanding & Factorising Brackets
2.3.1 Expanding Brackets
2.3.2 Factorising
2.3.3 Factorising Quadratics
2.4 Linear Equations & Inequalities
2.4.1 Solving Linear Equations
2.4.2 Solving Linear Inequalities
2.5 Quadratic Equations
2.5.1 Completing the Square
2.5.2 Solving Quadratic Equations
2.5.3 Quadratic Equation Methods
2.6 Rearranging Formula
2.6.1 Rearranging Formulae
2.7 Simultaneous Equations
2.7.1 Simultaneous Equations
2.8 Algebraic Fractions
2.8.1 Algebraic Fractions
2.8.2 Solving Algebraic Fractions
2.9 Forming & Solving Equations
2.9.1 Forming Equations
2.9.2 Equations & Problem Solving
2.10 Functions
2.10.1 Functions Toolkit
2.10.2 Composite & Inverse Functions
2.11 Sequences
2.11.1 Sequences
2.11.2 nth Term
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2.12 Coordinate Geometry
2.12.1 Coordinate Geometry
2.13 Linear Graphs
2.13.1 Straight Line Graphs (y = mx + c)
2.13.2 Parallel & Perpendicular Lines
2.14 Quadratic Graphs
2.14.1 Quadratic Graphs
2.15 Further Graphs & Tangents
2.15.1 Types of Graphs
2.15.2 Using Graphs
2.16 Solving & Graphing Inequalities
2.16.1 Graphical Inequalities
2.17 Real-Life Graphs
2.17.1 Conversion Graphs
2.17.2 Distance-Time & Speed-Time Graphs
2.17.3 Rates of Change of Graphs
2.18 Differentiation
2.18.1 Differentiation
2.18.2 Applications of Differentiation
2.18.3 Problem Solving with Differentiation
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2.1 Algebra Toolkit
2.1.1 Algebraic Notation & Vocabulary
Algebraic Notation
What is algebraic notation?
When writing expressions in algebra (as opposed to sums in numbers) there are
conventions and symbols that are used that take on a particular meaning
This is what we mean by algebraic notation
In number work, for adding and subtracting, we use + and –
In algebra, we still do!
Examples:
a+b
c+d– e
However for multiplication, no symbol is used, and for division, fractions are used
Examples:
ab (means a × b)
a
(means a ÷ b)
b
3ab (means 3 × a × b)
We can have combinations of these of course
Examples:
c
ab + (means a×b + c÷3)
3
The order of operations still apply in algebra so a×b and c÷d would happen before the
addition
Powers (indices) and roots are used the same way as with numbers
Examples:
a2 means a × a
4a2 means 4 × a2
With the order of operations, a2 will happen before multiplying by 4
Brackets also work in the same way as with numbers
Examples
3( a + b ) means 3 × ( a + b ) but with brackets taking priority, when known, a + b
would be worked out first, then we would multiply by 3
5x (2x + 3) means 5 × x × (2 × x + 3) , again with the brackets worked out first
How will I need to use algebraic notation?
Algebraic expressions are used in many parts of the course
You need to be able to understand their meaning and work with them
e.g. for rearranging a formula
You may be given a situation in words that you then have to write in algebra
Which could lead to an equation you may then have to solve
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Worked Example
At the start of a competition, Raheem has p conkers and Howard has 2q conkers.
At the end of the competition Raheem still has the same number of conkers he
started with but Howard won 6 and lost none.
(a)
Write down an expression for the number of conkers that Howard has at the end of
the competition.
Howard has 2q + 6
At the end of the competition, Raheem and Howard have a total of 40 conkers.
(b)
Write an equation in terms of p and q that shows the number of conkers Raheem
and Howard have in total at the end of the competition.
p + 2q + 6 = 40
This can be simplified to p + 2q = 34
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Algebraic Vocabulary
You need to know the meanings of the word term and factor, as they are the basic building
blocks in algebra.
You need to know the differences between an expression, equation, formula and inequality in
order to fully understand algebra and proofs. You may be asked to identify which is which.
What is a term?
A term is either…
…a letter (variable) on its own, e.g. x
..a number on its own, e.g. 20
…a number multiplied by a letter, e.g. 5x
The number in front of a letter is called a coefficient
The coefficient of x in the term 6x is 6
The coefficient of y in the term -5y is -5
Terms that are just numbers (with no letters) are called constants
Terms can include powers and more than one letter
E.g. 6xy, 4x2, ab3c, …
What is a factor?
A factor is any number or letter that divides a term exactly (with no remainder)
E.g. all the factors of 4xy are 1, 2, 4, x, 2x, 4x, y, 2y, 4y, xy, 2xy and 4xy
A term can be separated into factors that multiply together to give that term
E.g. two factors of 5x are 5 and x
To factorise means to write something as a multiplication of factors
A common factor is one that divides both terms
The highest (or greatest) common factor is 2x
E.g. the common factors of 6xy and 4x are 2, x and 2x
What is an expression?
An expression is an algebraic statement that does not have an equals sign
There is nothing to solve
An expression is made by adding, subtracting, multiplying or dividing terms
6y
E.g. 2x + 5y, b2 – 2cd,
,…
5t
A single term can be an expression
Expressions can be simplified (made easier)
E.g. x + x + x simplifies to 3x
What is an equation?
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An equation is an algebraic statement with an equals sign between a left-hand side and a
right-hand side
Both sides are equal in value
E.g. if 2x has the same value as 10, then 2x = 10
An equation can be solved by finding the missing values of the letters that make the lefthand side equal to the right-hand side
x = 5 is called the solution
E.g. the equation 2x = 10 is solved by x = 5
What is a formula?
A formula is a worded rule, definition or relationship between different quantities, written in
shorthand using letters
The formula is w = mg
E.g. weight, w, is mass, m, multiplied by gravitational acceleration, g
It is common to substitute numbers into a formula, but a formula on its own cannot be
solved
To turn a formula into an equation, more information is needed
E.g. In the formula w = mg, if w = 50 and m = 5 then the equation 50 = 5g can be formed
What is an inequality?
An inequality compares a left-hand side to a right-hand side and states which one is
bigger
x > y means x is greater than y
x ≥ y means x is greater than, or equal to, y
x < y means x is less than y
x ≤ y means x is less than, or equal to, y
E.g. x ≥ 8 means x can take any value that is greater than, or equal to, 8
This is the same as saying “8 or more”, or "at least 8"
The solutions of inequalities are usually, themselves, inequalities
x + 10 < 15 solves to give x < 5, so x is any number less than 5
ExamTo Tip
fully understand the wording of an exam question you need to know the
difference between an expression, equation, formula and inequality.
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Worked Example
(a)
From the list below, write down the
(i)
expression,
(ii)
term,
(iii)
equation.
2x + 5 = 4
7x – 9 x = vt – w 3x 4x – 1 ≥ 0
(i)
An expression does not have an equals or inequality sign
7x – 9 is the expression
(ii)
A term is a number, letter or combination of both
3x is the term
(iii)
An equation has an equals sign and can be solved
2x + 5 = 4 is the equation
(b)
Write down correct names for the two remaining items on the list.
x = vt – w is a group of different quantities forming a relationship
x = vt – w is a formula
4x – 1 ≥ 0 compares the size of two quantities using an inequality sign
4x – 1 ≥ 0 is an inequality
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2.1.2 Algebra Basics
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What is substitution?
Substitution
Substitution is where we replace letters in a formula with their values
This allows you to find one other value that is in the formula
How do we substitute numbers into a formula?
Write down the formula, if not clearly stated in question
Substitute the numbers given, using brackets around negative numbers, (-3), (-5) etc
Simplify any calculations if you can
Rearrange the formula if necessary (it is usually easier to substitute first)
Work out the calculation (use a calculator if allowed)
Are there any common formulae to be aware of?
The formula for the equation of a straight line is often used
y = mx + c
Formulae for accelerating objects are often used
v = u + at
v 2 = u 2 + 2as
s = ut +
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1 2
at
2
The letters mean the following:
t stands for the amount of time something accelerates for (in seconds)
u stands for its initial speed (in m/s) - the speed at the beginning
v stands for its final speed (in m/s) - the speed after t seconds
a stands for its acceleration (in m/s2) during in that time
s stands for the distance covered in t seconds
You do not need to memorise these formulae, but you should know how to substitute
numbers into them
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Worked Example
(a)
Find the value of the expression 2x (x + 3y ) when x = 2 and y = − 4 .
Substitute the numbers given.
Use brackets () around negative numbers that you have substituted so that you
don't forget about them.
It is a good idea to show every step of working to make sure that you are following
the order of operations correctly.
2 × 2 × (2 + 3 × (−4) )
= 2 × 2 × (2 − 12)
= 2 × 2 × (−10)
= 4 × − 10
− 40
(b)
The formula P = 2l + 2w is used to find the perimeter, P, of a rectangle of length l
and width w .
Given that the rectangle has a perimeter of 20 cm and a width of 4 cm, find its
length.
Substitute the values you are given into the formula.
20 = 2 × l + 2 × 4
Simplify.
20 = 2l + 8
Subtract 8 from both sides.
12 = 2l
Divide both sides by 2.
l = 6 cm
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Collecting Like Terms
How do we collect like terms?
TERMS are separated by + or –
The sign belongs to the coefficient of the term after the symbol
If there is no symbol in front of the first term then this is a positive term
2x - 3y means +2 x's and -3 y's
“LIKE” terms must have exactly the same LETTERS AND POWERS (the COEFFICIENT can be
different)
Examples of like terms:
2x and 3x
2x2 and 3x2
2xy and 3xy
4(x + y) and 5(x + y)
Examples that are NOT like terms
2x and 3y (different letters)
2x2 and 3x4 (different powers)
2xy and 3xyz (different letters)
4(x + y) and 5(x + y)2 (different powers)
Remember multiplication can be done in any order
xy and yx are like terms
Add the COEFFICIENTS of like terms
If the answer is a positive answer then put "+" in front if there are other terms before it
x - 2y + 5y = x + 3y
If the answer is a negative number then put "-" in front
x - 5y + 2y = x - 3y
Tip
ExamA “Coefficient”
answers the question “how many?”
For example:
the coefficient of x in 2x2 – 5x + 2 is -5
and:
the coefficient of x in ax2 + bx + c is b
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Worked Example
Simplify x − 3xy + 2x + 4x − 2xy − x + 7.
2
2
Reorder the terms so that the x 2 's are together, as are the xy 's, x 's and the
constants.
Make sure that you keep the same sign in front of them when you reorder them.
x 2 + 2x 2 − 3xy − 2xy + 4x − x + 7
Add the coefficients of the 'like' terms.
3 x 2 − 5 xy + 3 x + 7
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2.2 Algebraic Roots & Indices
2.2.1 Algebraic Roots & Indices
Algebraic Roots & Indices
Can I use the laws of indices with algebra?
Laws of indices work with numerical and algebraic terms
These can be used to simplify expressions where terms are multiplied or divided
Deal with the number and algebraic parts separately
(3x 7) × (6x 4) = (3 × 6) × (x 7 × x 4) = 18x11
3x 7 3 x 7 1 3
= ×
= x
6x 4 6 x 4 2
(3x 7) 2 = (3) 2 × (x 7) 2 = 9x14
The index laws you need to know and use are summarised here:
How can I solve equations when the unknown is in the index?
If two powers (bigger than 1) are equal and the base numbers are the same then the indices
must be the same
If ax = ay then x = y
If the unknown is part of the index then write both sides with the same base number
Then you can ignore the base number and make the indices equal and solve that
equation
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52x = 125
52x = 53
2x = 3
3
x=
2
In more complicated questions you might have to use negative and fractional indices
You may also have to rewrite both sides with the same base number
1
4
1
(23) x = 2
2
23x = 2−2
3x = − 2
2
x=−
3
8x =
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Worked Example
(a)
Simplify
(3x 2) (2x 3y 2)
.
(6x 2y )
Multiply out the brackets in the numerator.
Rearrange the numerator so that you are multiplying the numbers together, the x
terms together and the y terms together.
3 × 2 × x2 × x3 × y2
6x 2y
Simplify the numerator.
Multiply the constants together and add the powers of the x terms together.
6x 5y 2
6x 2y
Divide the constants.
Subtract the power of the x term in the denominator from the x term in the
numerator: x 5 −2 = x 3 .
Subtract the power of the y term in the denominator from the y term in the
numerator: y 2 −1 = y 1 .
x3 y
(b)
1
⎛ 54x 7 ⎞ −
Simplify ⎜⎜ 4 ⎟⎟ 3 .
⎝ 2x ⎠
Simplify the expression inside the brackets.
Cancel down the constants.
Subtract the power of the x term in the denominator from the x term in the
numerator: x 7 −4 = x 3 .
( 27x 3)
−
1
3
Apply the negative index outside the brackets by 'flipping' the fraction inside the
brackets.
1
⎛⎜ 1 ⎞⎟ 3
⎜
⎟
⎝ 27x 3 ⎠
Apply the fractional index outside the brackets to everything inside the brackets.
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1
13
1
1
3×
3
27 x 3
Simplify.
1
3x
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2.3 Expanding & Factorising Brackets
2.3.1 Expanding Brackets
Expanding One Bracket
How do I expand a bracket?
The expression 3x(x + 2) means 3x multiplied by the bracket (x + 2)
3x is the term outside the bracket (sometimes called a "factor") and x + 2 are the terms
inside the bracket
Expanding the brackets means multiplying the term on the outside by each term on the
inside
This will remove / "get rid of" the brackets
3x(x + 2) expands to 3x × x + 3x × 2 which simplifies to 3x 2 + 6x
Beware of minus signs
Remember the basic rules of multiplication with signs
−×−=+
−×+=−
It helps to put brackets around negative terms
Example
Worked
(a)
Expand x (2x − 3) .
Multiply the 4x term outside the brackets by both terms inside the brackets, watch
out for negatives!
4x × 2x + 4x × (−3)
Simplify.
8 x 2 − 12 x
(b)
Expand −7x (4 − 5x ) .
Multiply the −7x outside the brackets by both terms inside the brackets, watch out
for negatives!
(−7x ) × 4 + (−7x ) × (−5x )
Simplify.
− 28 x + 35 x 2
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Expand & Simplify
How do I simplify an expression when there is more than one term in
brackets?
Look out for two or more terms that contain brackets in an expression that are being
added/subtracted
E.g. 4(x + 7) + 5x (3 − x )
Notice that the two sets of brackets are connected by a + sign, so you are not
multiplying the brackets together
STEP 1: Expand each set of brackets separately by multiplying the term on the outside of
the brackets by each of the terms on the inside, be careful with negative terms
E.g. the first set of brackets expands to 4 × x + 4 × 7 , and simplifies to 4x + 28, the
second set of brackets expands to 5x × 3 + 5x × (−x ) and simplifies to 15x − 5x 2
So, 4(x + 7) + 5x (3 − x ) = 4x + 28 + 15x − 5x 2
STEP 2: Collect together like terms
E.g. 4(x + 7) + 5x (3 − x ) = 19x + 28 − 5x 2
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Worked Example
(a)
Expand and simplify 2(x + 5) + 3x (x − 8) .
Expand each set of brackets separately by multiplying the term outside the
brackets by each of the terms inside the brackets.
Keep negative terms inside brackets so that you don't miss them!
2 × x + 2 × 5 + 3x × x + 3x × (−8)
Simplify.
2x + 10 + 3x 2 − 24x
Collect 'like' terms.
− 22 x + 10 + 3 x 2
(b)
Expand and simplify 3x (x + 2) − 7(x − 6) .
Expand each set of brackets separately by multiplying the term outside the
brackets by each of the terms inside the brackets.
Keep negative terms inside brackets so that you don't miss them!
3x × x + 3x × 2 + (−7) × x + (−7) × (−6)
Simplify.
3x 2 + 6x − 7x + 42
Collect 'like' terms.
3 x 2 − x + 42
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Expanding Two Brackets
How do I expand two brackets using FOIL?
Every term in the first bracket must be multiplied by every term in the second bracket
To expand (x + 1)(x + 3) will need 4 multiplications in total
A good way to remember all the multiplications is FOIL
F = First: multiply together the first terms in each bracket
O = Outside: multiply the first term in the first bracket by the second term in the second
bracket (visually, these are the "outer" terms)
I = Inside: multiply the second term in the first bracket by the first term in the second
bracket (visually, these are the "inner" terms)
L = Last: multiply together the last terms in each bracket
It helps to put negative terms in brackets when multiplying
Simplify the final answer by collecting like terms (if there are any)
How do I expand two brackets using a grid?
To expand (x + 1)(x + 3), write out the brackets as headings in a grid (in either direction)
x
+1
x
+3
For each cell in the middle, multiply the term in the row heading by the term in the column
heading
x
+1
x
x2
3x
x
+3
3
Add together all the terms inside the grid to get the answer
x2 + x + 3x + 3
collect like terms
x2 + 4x + 3
How do I expand a bracket squared?
Write (x + 3)2 as (x + 3)(x + 3) and use one of the methods above
for example, with FOIL: (x + 3)(x + 3) = x2 + 3x + 3x + 9
collect like terms to get the final answer
x2 + 6x + 9
Do not make the common mistake of saying (x + 3)2 is x2 + 32
This cannot be true, as substituting x = 1, for example, gives (1 + 3)2 = 42 = 16 on the left,
but 12 + 32 = 1 + 9 = 10 on the right
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Worked Example
(a)
Expand (2x − 3) (x + 4) .
Using FOIL, multiply together: the first terms, the outside terms, the inside terms
and the last terms.
F
O
I
L
2x × x + 2x × 4 + (−3) × x + (−3) × 4
Simplify.
2x 2 + 8x − 3x − 12
Collect 'like' terms.
2 x 2 + 5 x − 12
(b)
Expand (x − 3) (3x − 5) .
Using FOIL, multiply together: the first terms, the outside terms, the inside terms
and the last terms.
F
O
I
L
x × 3x + x × (−5) + (−3) × 3x +
(−3) × (−5)
Simplify.
3x 2 − 5x − 9x − 15
Collect 'like' terms.
3 x 2 − 14 x − 15
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Worked Example
Expand (2x + 3) .
2
Rewrite the expression as two separate brackets multiplied together.
(2x + 3) (2x + 3)
Using FOIL, multiply together: the first terms, the outside terms, the inside terms
and the last terms.
F
O
I
L
2x × 2x + 2x × 3 + 3 × 2x + 3 × 3
Simplify.
4x 2 + 6x + 6x + 9
Collect 'like' terms.
4 x 2 + 12 x + 9
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Expanding Three Brackets
How do I expand three brackets?
Multiply out any two brackets using a standard method and simplify this answer (collect
any like terms)
Replace the two brackets above with one long bracket containing the expanded result
Expand this long bracket with the third (unused) bracket
This step often looks like (x + a)(x2 + bx + c)
Every term in the first bracket must be multiplied with every term in the second bracket
This leads to six terms
A grid can often help to keep track of all six terms, for example (x + 2)(x2 + 3x + 1)
x2
+3x
+1
x
x3
2x2
3x2
x
+2
6x
2
add all the terms inside the grid (diagonals show like terms) to get x3 + 2x2 + 3x2 +
6x + x + 2
collect like terms to get the final answer of x3 + 5x2 + 7x + 2
Simplify the final answer by collecting like terms (if there are any)
It helps to put negative terms in brackets when multiplying
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Worked Example
(a)
Expand (2x − 3) (x + 4) (3x − 1) .
Start by expanding the first two sets of brackets using the FOIL method and
simplify by collecting 'like' terms.
(2x − 3) (x + 4)
= 2x × x + 2x × 4 + (−3) × x + (−3) × 4
= 2x 2 + 8x − 3x − 12
= 2x 2 + 5x − 12
Rewrite the original expression with the first two brackets expanded.
(2x 2 + 5x − 12) (3x − 1)
Multiply all of the terms in the first set of brackets by all of the terms in the second
set of brackets.
2x 2 × 3x + 5x × 3x + (−12) × 3x + 2x 2 × (−1) + 5x × (−1) + (−12) × (−1)
Simplify.
6x 3 + 15x 2 − 36x − 2x 2 − 5x + 12
Collect 'like' terms.
6 x 3 + 13 x 2 − 41 x + 12
(b)
Expand (x − 3) (x + 2) (2x − 1) .
Start by expanding the first two sets of brackets using the FOIL method and
simplify by collecting 'like' terms.
(x − 3) (x + 2)
= x × x + x × 2 + (−3) × x + (−3) × 2
= x 2 + 2x − 3x − 6
= x2 − x − 6
Rewrite the original expression with the first two brackets expanded.
(x 2 − x − 6) (2x − 1)
Multiply all of the terms in the first set of brackets by all of the terms in the second
set of brackets.
x 2 × 2x + (−x ) × 2x + (−6) × 2x + x 2 × (−1) + (−x ) × (−1) + (−6) × (−1)
Simplify.
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YOUR NOTES
2x 3 − 2x 2 − 12x − x 2 + x + 6
Collect 'like' terms.
2 x 3 − 3 x 2 − 11 x + 6
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2.3.2 Factorising
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What is factorisation?
Basic Factorising
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A factorised expression is one written as the product (multiplication) of two, or more, terms
(factors)
3(x + 2) is factorised, as it is 3 × (x + 2)
3x + 6 is not factorised as it is "something" + "something"
3xy is factorised as it is 3 × x × y
12 can also be factorised: 12 = 2 x 2 x 3
In algebra, factorisation is the opposite of expanding brackets
it's "putting it into" brackets
How do I factorise two terms?
To factorise 12x2 + 18x
The highest common factor of 12 and 18 is 6
The highest common factor of x2 and x is x
this is the largest letter that divides both x2 and x
Multiply both to get the common factor
6x
Rewrite each term in 12x2 + 18x as "common factor × something"
6x × 2x + 6x × 3
"Take out" the common factor by writing it outside brackets
Put the remaining 2x + 3 inside the brackets
Answer: 6x(2x + 3)
Check this expands to give 12x2 + 18x
ExamYouTip
can always check that your factorisation is correct by simply expanding the
brackets in your answer!
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Worked Example
(a)
Factorise 5x + 15
Find the highest common factor of 5 and 15
5
There is no x in the second term, so no highest common factor in x needed
Write each term as 5 × "something"
5 ×x+5 ×3
"Take out" the 5
5(x + 3)
5(x + 3)
(b)
Factorise fully 30x2 - 24x
Find the highest common factor of 30 and 24
6
Find the highest common factor of x2 and x
x
Find the common factor (by multiplying these together)
6x
Write each term as 6x × "something"
6x × 5x - 6x × 4
"Take out" the 6x
6x(5x - 4)
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6x(5x - 4)
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Factorising by Grouping
How do I factorise expressions with common brackets?
To factorise 3x(t + 4) + 2(t + 4), both terms have a common bracket, (t + 4)
the whole bracket, (t + 4), can be "taken out" like a common factor
(t + 4)(3x + 2)
this is like factorising 3xy + 2y to y(3x + 2)
y represents (t + 4) above
How do I factorise by grouping?
Some questions may require you to form the common bracket yourself
for example, factorise xy + px + qy + pq
"group" the first pair of terms, xy + px, and factorise, x(y + p)
"group" the second pair of terms, qy + pq, and factorise, q(y + p),
now factorise x(y + p) + q(y + p) as above
(y + p)(x + q)
This is called factorising by grouping
The groupings are not always the first pair of terms and the second pair of terms, but two
terms with common factors
ExamAs Tip
always, once you have factorised something, expand it by hand to check
your answer is correct.
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Worked Example
Factorise ab + 3b + 2a + 6
Method 1
Notice that ab and 3b have a common factor of b
Notice that 2a and 6 have a common factor of 2
Factorise the first two terms, using b as a common factor
b(a + 3) + 2a + 6
Factorise the second two terms, using 2 as a common factor
b(a + 3) + 2(a + 3)
(a + 3) is a common bracket
We can factorise using (a + 3) as a factor
(a + 3)(b + 2)
Method 2
Notice that ab and 2a have a common factor of a
Notice that 3b and 6 have a common factor of 3
Rewrite the expression grouping these terms together
ab + 2a + 3b + 6
Factorise the first two terms, using a as a common factor
a(b + 2) + 3b + 6
Factorise the second two terms, using 3 as a common factor
a(b + 2) + 3(b + 2)
(b + 2) is a common bracket
We can factorise using (b + 2) as a factor
(b + 2)(a + 3)
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2.3.3 Factorising Quadratics
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Factorising Simple Quadratics
What is a quadratic expression?
A quadratic expression is in the form:
ax2 + bx + c (as long as a ≠ 0)
If there are any higher powers of x (like x3 say) then it is not a quadratic
If a = 1 e.g. x 2 − 2x − 8 , it can be called a “monic” quadratic expression
If a ≠ 1 e.g. 2x 2 − 2x − 8 , it can be called a “non-monic” quadratic expression
Method 1: Factorising a monic (a = 1) quadratic expression "by inspection"
This is shown easiest through an example; factorising x 2 − 2x − 8
We need a pair of numbers that for x 2 + bx + c
multiply to c
which in this case is -8
and add to b
which in this case is -2
-4 and +2 satisfy these conditions
Write these numbers in a pair of brackets like this:
(x + 2) (x − 4)
Method 2: Factorising a monic (a = 1) quadratic expression "by grouping"
This is shown easiest through an example; factorising x 2 − 2x − 8
We need a pair of numbers that for x 2 + bx + c
multiply to c
which in this case is -8
and add to b
which in this case is -2
2 and -4 satisfy these conditions
Rewrite the middle term by using 2x and -4x
x 2 + 2x − 4x − 8
Group and factorise the first two terms, using x as the highest common factor, and
group and factorise the second two terms, using -4 as the factor
x (x + 2) − 4(x + 2)
Note that these now have a common factor of (x + 2) so this whole bracket can be
factorised out
(x + 2) (x − 4)
Method 3: Factorising a monic (a = 1) quadratic expression "by using a grid"
This is shown easiest through an example; factorising x 2 − 2x − 8
We need a pair of numbers that for x 2 + bx + c
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multiply to c
which in this case is -8
and add to b
which in this case is -2
-4 and +2 satisfy these conditions
Write the quadratic equation in a grid (as if you had used a grid to expand the brackets),
splitting the middle term as -4x and 2x
The grid works by multiplying the row and column headings, to give a product in the
boxes in the middle
x2
+2x
-4x
-8
Write a heading for the first row, using x as the highest common factor of x2 and -4x
x
x2
+2x
-4x
-8
You can then use this to find the headings for the columns, e.g. “What does x need to be
multiplied by to give x2?”
x
x
x2
+2x
-4
-4x
-8
We can then fill in the remaining row heading using the same idea, e.g. “What does x need to
be multiplied by to give +2x?”
x
+2
x
x2
+2x
-4
-4x
-8
We can now read-off the factors from the column and row headings
(x + 2) (x − 4)
Which method should I use for factorising simple quadratics?
The first method, by inspection, is by far the quickest so is recommended in an exam for
simple quadratics (where a = 1)
However the other two methods (grouping, or using a grid) can be used for harder (nonmonic) quadratic equations where a ≠ 1 so you should learn at least one of them too
ExamAs Tip
a check, expand your answer and make sure you get the same expression as
the one you were trying to factorise.
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Worked Example
(a) Factorise x − 4x − 21.
2
We will factorise by inspection.
We need two numbers that:
multiply to -21, and sum to -4
-7, and +3 satisfy this
Write down the brackets.
(x + 3)(x - 7)
(b) Factorise x 2 − 5x + 6 .
We will factorise by splitting the middle term and grouping.
We need two numbers that:
multiply to 6, and sum to -5
-3, and -2 satisfy this
Split the middle term.
x2 - 2x - 3x + 6
Factorise x out of the first two terms.
x(x - 2) - 3x +6
Factorise -3 out of the last two terms.
x(x - 2) - 3(x - 2)
These have a common factor of (x - 2) which can be factored out.
(x - 2)(x - 3)
(c) Factorise x 2 − 2x − 24 .
We will factorise by using a grid.
We need two numbers that:
multiply to -24, and sum to -2
+4, and -6 satisfy this
Use these to split the -2x term and write in a grid.
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x2
-6x
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+4x
-24
Write a heading using a common factor for the first row:
x
x2
-6x
+4x
-24
Work out the headings for the rows, e.g. “What does x need to be multiplied by to
make x2?”
x
x
x2
-6x
+4
+4x
-24
Repeat for the heading for the remaining row, e.g. “What does x need to be
multiplied by to make -6x?”
x
-6
x
x2
-6x
+4
+4x
-24
Read-off the factors from the column and row headings.
(x + 4)(x - 6)
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Factorising Harder Quadratics
How do I factorise a non-monic quadratic expression?
Factorising a non-monic (a ≠ 1) quadratic expression "by grouping"
This is shown easiest through an example; factorising 4x 2 − 25x − 21
We need a pair of numbers that for ax 2 + bx + c
multiply to ac
which in this case is 4 × -21 = -84
and add to b
which in this case is -25
-28 and +3 satisfy these conditions
Rewrite the middle term using -28x and +3x
4x 2 − 28x + 3x − 21
Group and factorise the first two terms, using 4x as the highest common factor, and
group and factorise the second two terms, using 3 as the factor
4x (x − 7) + 3(x − 7)
Note that these terms now have a common factor of (x - 7) so this whole bracket can
be factorised out, leaving 4x + 3 in its own bracket
(x − 7) (4x + 3)
Factorising a non-monic (a ≠ 1) quadratic expression "by using a grid"
This is shown easiest through an example; factorising 4x 2 − 25x − 21
We need a pair of numbers that for ax 2 + bx + c
multiply to ac
which in this case is 4 × -21 = -84
and add to b
which in this case is -25
-28 and +3 satisfy these conditions
Write the quadratic equation in a grid (as if you had used a grid to expand the brackets),
splitting the middle term as -28x and +3x
The grid works by multiplying the row and column headings, to give a product in the
boxes in the middle
4x2
+3x
-28x
-21
Write a heading for the first row, using 4x as the highest common factor of 4x2 and -28x
4x
4x2
+3x
-28x
-21
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You can then use this to find the headings for the columns, e.g. “What does 4x need to
be multiplied by to give 4x2?”
4x
x
4x2
+3x
-7
-28x
-21
We can then fill in the remaining row heading using the same idea, e.g. “What does x
need to be multiplied by to give +3x?”
4x
+3
x
4x2
+3x
-7
-28x
-21
We can now read-off the factors from the column and row headings
(x − 7) (4x + 3)
Tip
Exam
As a check, expand your answer and make sure you get the same expression as the
one you were trying to factorise.
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Worked Example
(a) Factorise 6x − 7x − 3 .
2
We will factorise by splitting the middle term and grouping.
We need two numbers that:
multiply to 6 × -3 = -18, and sum to -7
-9, and +2 satisfy this
Split the middle term.
6x2 + 2x - 9x - 3
Factorise 2x out of the first two terms.
2x(3x + 1) - 9x - 3
Factorise -3 of out the last two terms.
2x(3x + 1) - 3(3x + 1)
These have a common factor of (3x + 1) which can be factored out.
(3x + 1)(2x - 3)
(b) Factorise 10x 2 + 9x − 7 .
We will factorise by using a grid.
We need two numbers that:
multiply to 10 × -7 = -70, and sum to +9
-5, and +14 satisfy this
Use these to split the 9x term and write in a grid.
10x2
+14x
-5x
-7
Write a heading using a common factor for the first row:
5x
10x2
+14x
-5x
-7
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Work out the headings for the rows, e.g. “What does 5x need to be multiplied by to
make 10x2?”
5x
2x
10x2
+14x
-1
-5x
-7
Repeat for the heading for the remaining row, e.g. “What does 2x need to be
multiplied by to make +14x?”
5x
+7
2x
10x2
+14x
-1
-5x
-7
Read-off the factors from the column and row headings.
(2x - 1)(5x + 7)
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Difference Of Two Squares
What is the difference of two squares?
When a "squared" quantity is subtracted from another "squared" quantity, you get the
difference of two squares
for example,
a2 - b2
92 - 52
(x + 1)2 - (x - 4)2
4m2 - 25n2, which is (2m)2 - (5n)2
How do I factorise the difference of two squares?
Expand the brackets (a + b)(a - b)
= a2 - ab + ba - b2
ab is the same quantity as ba, so -ab and +ba cancel out
= a2 - b2
From the working above, the difference of two squares, a2 - b2, factorises to
(a + b ) (a − b )
It is fine to write the second bracket first, (a - b)(a + b)
but the a and the b cannot swap positions
a2 - b2 must have the a's first in the brackets and the b's second in the brackets
ExamTheTip
difference of two squares is a very important rule to learn as it often appears
in harder questions involving factorisation, e.g. in algebraic fractions
The word difference in maths means a subtraction, it should remind you that
you are subtracting one squared term from another
You should be able to recognise factorised difference of two squares
expressions
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Worked Example
(a)
Factorise 9x 2 − 16.
Recognise that 9x 2 and 16 are both squared terms and the second term is
subtracted from the first term - you can factorise using the difference of two
squares.
9x 2 − 16 = (3x ) 2 − (4) 2
Rewrite the expression with the square root of each term added together in the first
bracket and subtracted from each other in the second bracket.
(3x + 4) (3x − 4)
(b)
Factorise 4x 2 − 25.
Recognise that 4x 2 and 25 are both squared terms and the second term is
subtracted from the first term - you can factorise using the difference of two
squares.
4x 2 − 25 = (2x ) 2 − (5) 2
Rewrite the expression with the square root of each term added together in the first
bracket and subtracted from each other in the second bracket.
(2 x + 5) (2 x − 5)
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Quadratics Factorising Methods
How do I know if it factorises?
Method 1: Use a calculator to solve the quadratic expression equal to 0
If the solutions are integers or fractions (without square roots), then the quadratic
expression factorises
Method 2: Find the value under the square root in the quadratic formula, b2 – 4ac (called the
discriminant)
If this number is a perfect square number, then the quadratic expression factorises
Which factorisation method should I use for a quadratic expression?
Does it have 2 terms only?
Yes, like x 2 − 7x
Use "basic factorisation" to take out the highest common factor
x (x − 7)
Yes, like x 2 − 9
Use the "difference of two squares" to factorise
(x + 3) (x − 3)
Does it have 3 terms?
Yes, starting with x2 (monic) like x 2 − 3x − 10
Use "factorising simple quadratics" by finding two numbers that add to -3 and
multiply to -10
(x + 2) (x − 5)
Yes, starting with ax2 (non-monic) like 3x 2 + 15x + 18
Check to see if the 3 in front of x2 is a common factor for all three terms (which it is
in this case), then use "basic factorisation" to factorise it out first
3(x 2 + 5x + 6)
The quadratic expression inside the brackets is now x2 +... (monic), which
factorises more easily
3(x + 2) (x + 3)
Yes, starting with ax2 (non-monic) like 3x 2 − 5x − 2
The 3 in front of x2 is not a common factor for all three term
Use "factorising harder quadratics", for example factorising by grouping or
factorising using a grid
(3x + 1) (x − 2)
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Worked Example
Factorise −8x + 100x − 48.
2
Spot the common factor of -4 and put outside a set of brackets, work out the terms
inside the brackets by dividing the terms in the original expression by -4.
−8x 2 + 100x − 48 = − 4(2x 2 − 25x + 12)
Check the discriminant for the expression inside the brackets, (b 2 − 4ac) , to see if it
will factorise.
(−25) 2 − 4 × 2 × 12
= 625 − 96
= 529
529 = 132 , it is a perfect square so the expression will factorise.
Proceed with factorising 2x 2 − 25x + 12 as you would for a harder quadratic, where
a ≠ 1.
"+12" means the signs will be the same.
"-25" means that both signs will be negative.
a × c = 2 × 12 = 24
The only numbers which multiply to give 24 and follow the rules for the signs above
are:
(−1) × (−24) and (−2) × (−12) and (−3) × (−8) and (−4) × (−6)
but only the first pair add to give −25.
Split the −25x term into −24x − x .
2x 2 − 24x − x + 12
Group and factorise the first two terms, using 2x as the highest common factor and
group and factorise the last two terms using 1 as the highest common factor.
2x (x − 12) + 1(x − 12)
These factorised terms now have a common term of (x − 12) , so this can now be
factorised out.
(2x + 1) (x − 12)
Put it all together.
−8x 2 + 100x − 48 = − 4(2x 2 − 25x + 12) = − 4(2x − 1) (x − 12)
− 4 (2 x − 1) ( x − 12)
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2.4 Linear Equations & Inequalities
2.4.1 Solving Linear Equations
Solving Linear Equations
What are linear equations?
A linear equation is an equation that will produce a straight line when plotted on a graph
The greatest power of x in a linear equation is 1
This means there are no terms of x 2 or a higher order
A linear equation is normally in form ax + b = c
where a, b , and c are constants and x is a variable
How do I solve a linear equation?
To solve a linear equation you need to isolate the variable, usually x , by carrying out inverse
operations to both sides of the equation
Inverse operations are just the opposite operations to what has already happened to
the variable
The order in which the inverse operations are carried out is important
Most of the time, this will be BIDMAS in reverse
However it depends on the order in which the operations were applied to the variable
to form the equation
How do I solve a linear equation of the form ax + b = c?
The operations that have been applied to x here are:
STEP 1
Multiply by a
STEP 2
Add b
To solve this, you must carry out the inverse operations in reverse order
STEP 1
Subtract b
STEP 2
Divide by a
For example, to solve the equation 2x + 1 = 9
STEP 1
Subtract 1
2x + 1 = 9
( − 1)
=
( − 1)
2x = 8
STEP 2
Divide by 2
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2x = 8
( ÷ 2)
( ÷ 2)
=
x=4
Be extra careful if any of the terms have negatives
For example, to solve the equation 2 − 3x = 10
STEP 1
Subtract 2
2 − 3x = 10
( − 2)
( − 2)
=
− 3x = 8
Be careful not to drop the negative sign
STEP 2
Divide by -3
−3x = 8
( ÷ − 3)
( ÷ − 3)
=
x =−
8
3
How do I solve a linear equation with the unknown variable, x, on both
sides?
If a linear equation contains the unknown variable, usually x on both sides start by
collecting these terms together on one side of the equation
Moving the x term with the smallest coefficient (number in front of x ) is easiest
For example, to solve the equation 4x − 7 = 11 + x
STEP 1
Move the x term with the smallest coefficient of x
The coefficients are 4 and 1 so move the x -term on the right hand side
4x − 7 = 11 + x
(−x )
(−x )
=
3x − 7 = 11
STEP 2
Solve the linear equation using the method above
3x − 7 = 11
3x = 18
x=6
How do I solve a linear equation that contains brackets?
If a linear equation contains brackets on one, or both, sides start by expanding the brackets
For example, to solve the equation 2(x − 1) = 10 − (3 + x )
STEP 1
Expand the brackets on both sides
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2x − 2 = 10 − 3 − x
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STEP 2
Collect the x terms to one side by subtracting the term with the smaller coefficient of x
2x − 2 = 7 − x
− (−x )
− (−x )
=
3x − 2 = 7
Be extra careful if any of the terms have negatives
STEP 3
Solve the linear equation using the method above
3x − 2 = 7
3x = 9
x=3
How do I solve a linear equation that contains fractions?
If a linear equation contains a fraction on one or both sides, remove the fractions by
multiplying both sides by everything on the denominator
Remember to put brackets around the expression first
For example, to solve the equation
2
3
=
x+3
4 − 2x
of the equation by both (x + 3) and (4 − 2x )
STEP 1
Multiply both sides by (x + 3)
2
x+3
(x + 3) =
you will need to multiply both sides
3
(x + 3)
4 − 2x
3(x + 3)
2=
4 − 2x
STEP 2
Multiply both sides by (4 − 2x )
3(x + 3)
(4 − 2x )
4 − 2x
2(4 − 2x ) = 3(x + 3)
2(4 − 2x ) =
STEP 3
Expand the brackets on both sides
8 − 4x = 3x + 9
STEP 4
Collect the x terms to one side by subtracting the term with the smaller coefficient of x
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8 − 4x = 3x + 9
− (−4x )
− (−4x )
=
8 = 7x + 9
STEP 5
Solve the equation
You can swap the sides if it makes solving the equation easier
7x + 9 = 8
7x = − 1
1
x= −
7
Tip
ExamIf you
have time in the exam, you should substitute your answer back into the
equation to check you got it right
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Worked Example
a)
Solve the equation.
3x − 2
= −1
4−x
Get rid of the fraction by multiplying both sides by the denominator (4 − x ) .
3x − 2 = − (4 − x )
Expand the brackets.
3x − 2 = − 4 + x
Bring the x terms to one side of the equation by subtracting x from both sides.
2x − 2 = − 4
Get the x term by itself by adding 2 to both sides.
2x = − 2
Solve by dividing both sides by 2.
x = −1
b)
Solve the equation.
2+
x+1
=4
3
Isolate the fraction by subtracting 2 from both sides.
x+1
=2
3
Get rid of the fraction by multiplying both sides by the denominator (3).
x+1=6
Get the x term by itself by subtracting 1 from both sides.
x = 5
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2.4.2 Solving Linear Inequalities
Solving Linear Inequalities
What is a linear inequality?
An inequality tells you that one expression is greater than (“>”) or less than (“<”) another
“⩾” means “greater than or equal to”
“⩽” means “less than or equal to”
A Linear Inequality just has an x (and/or a y) etc in it and no x2 terms or terms with higher
powers of x
For example, 3x + 4 ⩾ 7 would be read “3x + 4 is greater than or equal to 7”
How do I solve linear inequalities?
Solving linear inequalities is just like Solving Linear Equations
Follow the same rules, but keep the inequality sign throughout
If you change the inequality sign to an equals sign you are changing the meaning of the
problem
When you multiply or divide both sides by a negative number, you must flip the sign of the
inequality
e.g. 1 < 2 → [times both sides by (–1)] → –1 > –2 (sign flips)
Never multiply or divide by a variable (x) as this could be positive or negative
The safest way to rearrange is simply to add & subtract to move all the terms onto one side
You also need to know how to use Number Lines and deal with “Double” Inequalities
How do I represent linear inequalities on a number line?
Inequalities such as x < a and x > a can be represented on a normal number line using an
open circle and an arrow
For < , the arrow points to the left of a
For > , the arrow points to the right of a
Inequalities such as x ≤ a and x ≥ a can be represented on a normal number line using a
solid circle and an arrow
For ≤ , the arrow points to the left of a
For ≥ , the arrow points to the right of a
Inequalities such as a < x < b and a ≤ x ≤ b can be represented on a normal number
line using two circles at a and b and a line between them
For < or > use an open circle
For ≤ or ≥ , use a solid circle
Disjoint inequalities such as "x < a or x > b " can be represented with two circles at a and b ,
an arrowed line pointing left from a and an arrowed line pointing right from b , and a blank
space between a and b
How do I solve double inequalities?
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Inequalities such as a < 2x < b can be solved by doing the same thing to all three parts
of the inequality
Use the same rules as solving linear inequalities
ExamDoTip
not change the inequality sign to an equals when solving linear inequalities,
you will lose marks in an exam for doing this.
Example
Worked
(a)
Solve the inequality −7 ≤ 3x − 1 < 2 , illustrating your answer on a number line.
This is a double inequality, so any operation carried out to one side must be done to
all three parts.
Use the expression in the middle to choose the inverse operations needed to
isolate x.
Add 1 to all three parts.
Remember not to change the inequality signs.
−6 ≤ 3x < 3
Divide all three parts by 3.
3 is positive so there is no need to flip the signs.
−2 ≤ x < 1
Illustrate the final answer on a number line, using an open circle at 1 and a closed
circle at -2.
(b)
Give your answer to part (a) in set notation
Rewrite your answer using the set notation rules discussed above
{ x : x ≥ − 2} ∩ { x : x < 1}
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2.5 Quadratic Equations
2.5.1 Completing the Square
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Completing the Square
How can I rewrite the first two terms of a quadratic expression as the
difference of two squares?
Look at the quadratic expression x2 + bx + c
The first two terms can be written as the difference of two squares using the following rule
x 2 + bx is the same as (x + p ) 2 − p 2 where p is half of b
Check this is true by expanding the right-hand side
Is x 2 + 2x the same as (x + 1) 2 − 12 ?
Yes: (x + 1)(x + 1) - 12 = x2 + 2x + 1 - 1 = x2 + 2x
This works for negative values of b too
x 2 − 20x can be written as (x − 10) 2 − (−10) 2 which is (x − 10) 2 − 100
A negative b does not change the sign at the end
How do I complete the square?
Completing the square is a way to rewrite a quadratic expression in a form containing a
squared-bracket
To complete the square on x2 + 10x + 9
Use the rule above to replace the first two terms, x2 + 10x, with (x + 5)2 - 52
add 9: (x + 5)2 - 52 + 9
simplify the numbers: (x + 5)2 - 25 + 9
answer: (x + 5)2 - 16
How do I complete the square when there is a coefficient in front of the x2
term?
You first need to take a out as a factor of the x2 and x terms only
⎡⎢
⎤
⎢⎢ 2 b ⎥⎥⎥
2
ax + bx + c = a ⎢⎢ x + x ⎥⎥ + c
a ⎦
⎣
Use square-shaped brackets here to avoid confusion with curly brackets later
b
Then complete the square on the bit inside the square-brackets: x 2 + x
a
⎡
⎤
2
2
⎢
⎥
This gives a ⎣ (x + p ) − p ⎦ + c
b
where p is half of
a
Finally multiply this expression by the a outside the square-brackets and add the c
a (x + p ) 2 − ap 2 + c
This looks far more complicated than it is in practice!
Usually you are asked to give your final answer in the form a (x + p ) 2 + q
For quadratics like −x 2 + bx + c , do the above with a = -1
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How do I find the turning point by completing the square?
Completing the square helps us find the turning point on a quadratic graph
If y = (x + p ) 2 + q then the turning point is at (−p ,q )
Notice the negative sign in the x-coordinate
This links to transformations of graphs (translating y = x 2 by p to the left and q up)
If y = a (x + p ) 2 + q then the turning point is still at (−p ,q )
It's at a minimum point if a > 0
It's at a maximum point if a < 0
It can also help you create the equation of a quadratic when given the turning point
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It can also be used to prove and/or show results using the fact that any "squared term", i.e.
the bracket (x ± p)2, will always be greater than or equal to 0
You cannot square a number and get a negative value
ExamTo Tip
know if you have completed the square correctly, expand your answer to
check.
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Worked Example
(a)
By completing the square, find the coordinates of the turning point on the graph of
y = x 2 + 6x − 11.
Find half of +6 (call this p)
p=
6
=3
2
Write x2 + 6x in the form (x + p)2 - p2
x 2 + 6x is the same as (x + 3) 2 − 32
Put this result into the equation of the curve
y = (x + 3) 2 − 32 − 11
Simplify the numbers
y = (x + 3) 2 − 20
Use that the turning point of y = (x + p ) 2 + q is at (−p ,q )
p = 3 and q = -20
turning point at (-3, -20)
(b)
Write −3x 2 + 12x + 24 in the form a (x + p ) 2 + q
Factorise -2 out of the first two terms only
Use square-shaped brackets
−3 ⎡⎢⎣ x 2 − 4x ⎤⎥⎦ + 24
Complete the square on the x2 - 4x inside the brackets (write in the form (x + p)2 - p2
where p is half of -4)
−3 ⎡⎢⎣ (x − 2) 2 − (−2) 2 ⎤⎥⎦ + 24
Simplify the numbers inside the brackets
(-2)2 is 4
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−3 ⎡⎢⎣ (x − 2) 2 − 4 ⎤⎥⎦ + 24
Multiply -3 by all the terms inside the square-shaped brackets
−3(x − 2) 2 + 12 + 24
Simplify the numbers
−3(x − 2) 2 + 36
This is now in the form a(x + p)2 + q where a = -3, p = -2 and q = 36
− 3 ( x − 2) 2 + 36
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2.5.2 Solving Quadratic Equations
Solving Quadratics by Factorising
How do I solve a quadratic equation using factorisation?
Rearrange it into the form ax2 + bx + c = 0
zero must be on one side
it is easier to use the side where a is positive
Factorise the quadratic and solve each bracket equal to zero
If (x + 4)(x - 1) = 0, then either x + 4 = 0 or x - 1 = 0
Because if A × B = 0, then either A = 0 or B = 0
To solve (x − 3) (x + 7) = 0
…solve “first bracket = 0”:
x–3=0
add 3 to both sides: x = 3
…and solve “second bracket = 0”
x+7=0
subtract 7 from both sides: x = -7
The two solutions are x = 3 or x = -7
The solutions have the opposite signs to the numbers in the brackets
To solve (2x − 3) (3x + 5) = 0
…solve “first bracket = 0”
2x – 3 = 0
add 3 to both sides: 2x = 3
divide both sides by 2: x =
3
2
…solve “second bracket = 0”
3x + 5 = 0
subtract 5 from both sides: 3x = -5
divide both sides by 3: x = −
The two solutions are x =
5
3
3
5
or x = −
2
3
To solve x (x − 4) = 0
it may help to think of x as (x – 0) or (x)
…solve “first bracket = 0”
(x) = 0, so x = 0
…solve “second bracket = 0”
x–4=0
add 4 to both sides: x = 4
The two solutions are x = 0 or x = 4
It is a common mistake to divide both sides by x at the beginning - you will lose a
solution (the x = 0 solution)
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ExamUseTipa calculator to check your final solutions!
Calculators also help you to factorise (if you're struggling with that step)
A calculator gives solutions to 6x 2 + x − 2 = 0 as x = −
2
1
and x =
3
2
"Reverse" the method above to factorise!
6x 2 + x − 2 = (3x + 2) (2x − 1)
Warning: a calculator gives solutions to 12x2 + 2x – 4 = 0 as x = −
1
2
2
and x =
3
But 12x2 + 2x – 4 ≠ (3x + 2) (2x − 1) as these brackets expand to 6x2 +
... not 12x2 + ...
Multiply by 2 to correct this
12x2 + 2x – 4 = 2(3x + 2) (2x − 1)
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Worked Example
(a)
Solve (x − 2) (x + 5) = 0
Set the first bracket equal to zero
x–2=0
Add 2 to both sides
x=2
Set the second bracket equal to zero
x+5 =0
Subtract 5 from both sides
x = -5
Write both solutions together using “or”
x = 2 or x = -5
(b)
Solve (8x + 7) (2x − 3) = 0
Set the first bracket equal to zero
8x + 7 = 0
Subtract 7 from both sides
8x = -7
Divide both sides by 8
7
x=− 8
Set the second bracket equal to zero
2x - 3 = 0
Add 3 to both sides
2x = 3
Divide both sides by 2
x=
3
2
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Write both solutions together using “or”
7
3
x = − 8 or x = 2
(c)
Solve x (5x − 1) = 0
Do not divide both sides by x (this will lose a solution at the end)
Set the first “bracket” equal to zero
(x) = 0
Solve this equation to find x
x=0
Set the second bracket equal to zero
5x - 1 = 0
Add 1 to both sides
5x = 1
Divide both sides by 5
1
x= 5
Write both solutions together using “or”
1
x = 0 or x = 5
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Solving by Completing the Square
How do I solve a quadratic equation by completing the square?
To solve x2 + bx + c = 0
replace the first two terms, x2 + bx, with (x + p)2 - p2 where p is half of b
this is called completing the square
x2 + bx + c = 0 becomes
(x + p)2 - p2 + c = 0 where p is half of b
rearrange this equation to make x the subject (using ±√)
For example, solve x2 + 10x + 9 = 0 by completing the square
x2 + 10x becomes (x + 5)2 - 52
so x2 + 10x + 9 = 0 becomes (x + 5)2 - 52 + 9 = 0
make x the subject (using ±√)
(x + 5)2 - 25 + 9 = 0
(x + 5)2 = 16
x + 5 = ±√16
x = ±4 - 5
x = -1 or x = -9
If the equation is ax2 + bx + c = 0 with a number in front of x2, then divide both sides by a first,
before completing the square
How does completing the square link to the quadratic formula?
The quadratic formula actually comes from completing the square to solve ax2 + bx + c = 0
a, b and c are left as letters, to be as general as possible
You can see hints of this when you solve quadratics
For example, solving x2 + 10x + 9 = 0
by completing the square, (x + 5)2 = 16 so x = ± 4 - 5 (from above)
by the quadratic formula, x =
−10 ±
2
64
= −5 ±
8
= ± 4 - 5 (the same
2
structure)
Tip
ExamWhen
making x the subject to find the solutions at the end, don't expand the
squared brackets back out again!
Remember to use ±√ to get two solutions
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Worked Example
Solve 2x − 8x − 24 = 0 by completing the square
2
Divide both sides by 2 to make the quadratic start with x2
x 2 − 4x − 12 = 0
Halve the middle number, -4, to get -2
Replace the first two terms, x2 - 4x, with (x - 2)2 - (-2)2
(x − 2) 2 − (−2) 2 − 12 = 0
Simplify the numbers
(x − 2) 2 − 4 − 12 = 0
(x − 2) 2 − 16 = 0
Add 16 to both sides
(x − 2) 2 = 16
Square root both sides
Include the ± sign to get two solutions
x − 2 = ± 16 = ± 4
Add 2 to both sides
x = ±4 +2
Work out each solution separately
x = 6 or x = -2
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Quadratic Formula
How do I use the quadratic formula to solve a quadratic equation?
A quadratic equation has the form:
ax2 + bx + c = 0 (as long as a ≠ 0)
you need "= 0" on one side
The quadratic formula is a formula that gives both solutions:
−b ± b 2 − 4 ac
x=
2a
Read off the values of a, b and c from the equation
Substitute these into the formula
write this line of working in the exam
Put brackets around any negative numbers being substituted in
To solve 2x2 - 7x - 3 = 0 using the quadratic formula:
a = 2, b = -7 and c = -3
x=
− (−7) ±
(−7) 2 − 4 × 2 × (−3)
2×2
Type this into a calculator
once with + for ± and once with - for ±
The solutions are x = 3.886 and x = -0.386 (to 3 dp)
Rounding is often asked for in the question
The calculator also gives these solutions in exact form (surd form), if required
x=
7+
73
4
and x =
7−
73
4
What is the discriminant?
The part of the formula under the square root (b2 – 4ac) is called the discriminant
The sign of this value tells you if there are 0, 1 or 2 solutions
If b2 – 4ac > 0 (positive)
then there are 2 different solutions
If b2 – 4ac = 0
then there is only 1 solution
sometimes called "two repeated solutions"
If b2 – 4ac < 0 (negative)
then there are no solutions
If your calculator gives you solutions with i terms in, these are "complex" and not
what we are looking for
Interestingly, if b2 – 4ac is a perfect square number ( 1, 4, 9, 16, …) then the quadratic
expression could have been factorised!
Can I use my calculator to solve quadratic equations?
Yes to check your final answers, but a method must still be shown as above
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Tip
ExamMake
sure the quadratic equation has "= 0" on the right-hand side, otherwise it
needs rearranging first
Always look for how the question wants you to leave your final answers
for example, correct to 2 decimal places
Example
Worked
Use the quadratic formula to find the solutions of the equation 3x2 - 2x - 4 = 0,
giving your answers correct to 3 significant figures.
Write down the values of a, b and c
a = 3, b = -2, c = -4
Substitute these values into the quadratic formula, x =
−b ±
b 2 − 4ac
2a
Put brackets around any negative numbers
x=
− (−2) ±
(−2) 2 − 4 × 3 × (−4)
2×3
Input this into a calculator
Use + for ± to get the first solution
x = 1.53518...
Input this into a calculator a second time
Use - for ± to get the second solution
x = 0.86851...
Present both answers together (using the word "or" between them)
Round the answers correct to 3 significant figures (note how this affects the
number of decimal places)
x = 1.54 or x = 0.869
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2.5.3 Quadratic Equation Methods
Quadratic Equation Methods
If you have to solve a quadratic equation but are not told which method to use, here is a guide as
to what to do
When should I solve by factorisation?
When the question asks to solve by factorisation
For example, part (a) Factorise 6x2 + 7x – 3, part (b) Solve 6x2 + 7x – 3 = 0
When solving two-term quadratic equations
For example, solve x2 – 4x = 0
…by taking out a common factor of x to get x(x – 4) = 0
...giving x = 0 and x = 4
For example, solve x2 – 9 = 0
…using the difference of two squares to factorise it as (x + 3)(x – 3) = 0
...giving x = -3 and x = 3
(Or by rearranging to x2 = 9 and using ±√ to get x = = ±3)
When should I use the quadratic formula?
When the question says to leave solutions correct to a given accuracy (2 decimal places, 3
significant figures etc)
When the quadratic formula may be faster than factorising
It's quicker to solve 36x2 + 33x – 20 = 0 using the quadratic formula then by
factorisation
If in doubt, use the quadratic formula - it always works
When should I solve by completing the square?
When part (a) of a question says to complete the square and part (b) says to use part (a) to
solve the equation
When making x the subject of harder formulae containing x2 and x terms
For example, make x the subject of the formula x2 + 6x = y
Complete the square: (x + 3)2 – 9 = y
Add 9 to both sides: (x + 3)2 = y + 9
Take square roots and use ±: x + 3 = ± y + 9
Subtract 3: x = − 3 ± y + 9
Tip
ExamCalculators
can solve quadratic equations so use them to check your solutions
If the solutions on your calculator are whole numbers or fractions (with no
square roots), this means the quadratic equation does factorise
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Worked Example
(a)
Solve x 2 − 7x + 2 = 0 , giving your answers correct to 2 decimal places
“Correct to 2 decimal places” suggests using the quadratic formula
Substitute a = 1, b = -7 and c = 2 into the formula, putting brackets around any
negative numbers
x=
− (−7) ±
(−7) 2 − 4 × 1 × 2
2×1
Use a calculator to find each solution
x = 6.70156… or 0.2984...
Round your final answers to 2 decimal places
x = 6.70 or x = 0.30
(b)
Solve 16x 2 − 82x + 45 = 0
Method 1
If you cannot spot the factorisation, use the quadratic formula
Substitute a = 16, b = -82 and c = 45 into the formula, putting brackets around any
negative numbers
x=
− (−82) ±
(−82) 2 − 4 × 16 × 45
2 × 16
Use a calculator to find each solution
9
5
x = 2 or x = 8
Method 2
If you do spot the factorisation, (2x – 9)(8x – 5), then use that method instead
(2x − 9) (8x − 5) = 0
Set the first bracket equal to zero
2x − 9 = 0
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Add 9 to both sides then divide by 2
2x = 9
9
x=
2
Set the second bracket equal to zero
8x − 5 = 0
Add 5 to both sides then divide by 8
8x = 5
5
x=
8
9
5
x = 2 or x = 8
(c)
By writing x 2 + 6x + 5 in the form (x + p ) 2 + q , solve x 2 + 6x + 5 = 0
This question wants you to complete the square first
Find p (by halving the middle number)
p=
6
=3
2
Write x2 + 6x as (x + p)2 - p2
x 2 + 6x = (x + 3) 2 − 32
= (x + 3) 2 − 9
Replace x2 + 6x with (x + 3)2 – 9 in the equation
(x + 3) 2 − 9 + 5 = 0
(x + 3) 2 − 4 = 0
Make x the subject of the equation (start by adding 4 to both sides)
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(x + 3) 2 = 4
Take square roots of both sides (include a ± sign to get both solutions)
x +3 = ± 4 = ±2
Subtract 3 from both sides
x = ±2 −3
Find each solution separately using + first, then - second
x = - 5, x = - 1
Even though the quadratic factorises to (x + 5)(x + 1), this is not the method asked
for in the question
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2.6 Rearranging Formula
2.6.1 Rearranging Formulae
Simple Rearranging
What are formulae?
A formula (plural, formulae) is a mathematical relationship consisting of variables,
constants and an equals sign
You will come across many formulae in your IGCSE course, including
the formulae for areas and volumes of shapes
equations of lines and curves
the relationship between speed, distance and time
Some examples of formulae you should be familiar with are
The equation of a straight line
y = mx + c
The area of a trapezium
(a + b ) h
Area =
2
Pythagoras' theorem
a2 + b 2 = c2
You will also be expected to rearrange formulae that you are not familiar with
How do I rearrange formulae where the subject appears only once?
Rearranging formulae can also be called changing the subject
The subject is the variable that you want to find out, or get on its own on one side of the
formula
The method for changing the subject is the same as the method used for solving linear
equations
STEP 1
Remove any fractions or brackets
Remove fractions by multiplying both sides by anything on the denominator
Expand any brackets only if it helps to release the variable, if not it may be easier to
leave the bracket there
STEP 2
Carry out inverse operations to isolate the variable you are trying to make the subject
This works in the same way as with linear equations, however you will create
expressions rather than carry out calculations
(a + b ) h
For example, to rearrange A =
so that h is the subject
2
Multiply by 2
2A = ( a + b ) h
Expanding the bracket will not help here as we would end up with the subject
appearing twice, so instead divide by the whole expression ( a + b )
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2A
=h
a+b
You can now rewrite this with the subject (h ) on the left hand side
h=
2A
a+b
How do I rearrange formulae that include powers or roots?
If the formula contains a power of n, use the nth root to reverse this operation
For example to make x the subject of y = ax 5
Divide both sides by a first
y
= x5
a
Then take the 5th root of both sides
5
y
=x
a
If n is even then there will be two answers: a positive and a negative
For example if y = x 2 then x = ± y
If the formula contains an nth root, reverse this operation by raising both sides to the power
of n
For example to make a the subject of m = 3 2ab
Raise both sides to the power of 3 first
m3 = 2ab
Divide both sides by 2b
a=
m3
2b
Are there any common formulae to be aware of?
The formula for the equation of a straight line is often used
y = mx + c
Formulae for accelerating objects are often used
v = u + at
v 2 = u 2 + 2as
s = ut +
1 2
at
2
The letters mean the following:
t stands for the amount of time something accelerates for (in seconds)
u stands for its initial speed (in m/s) - the speed at the beginning
v stands for its final speed (in m/s) - the speed after t seconds
a stands for its acceleration (in m/s2) during in that time
s stands for the distance covered in t seconds
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You do not need to memorise these formulae, but you should know how to substitute
numbers into them
Tip
ExamIf you
are unsure about the order in which you would carry out the inverse
operations, try substituting numbers in and reverse the order that you would
carry out the substitution
Worked Example
Make x the subject of y =
ax 2 − b .
Use inverse operations to isolate x .
Square both sides.
y 2 = ( ax 2 − b ) 2
y 2 = ax 2 − b
Add b to both sides.
y 2 = ax 2 − b
(+b )
=
2
y + b = ax 2
(+b )
Divide both sides by a .
y 2 + b = ax 2
( ÷ a)
( ÷ a)
=
y2 + b
a
= x2
Square root both sides.
y2 + b
=
a
x2
y2 + b
=x
a
The equation is fully correct as it is and will gain full marks, however the two sides
can be swapped if preferred. Remember that when you square root you get two
answers (a positive and a negative).
x=±
y2 + b
a
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Subject Appears Twice
How do I rearrange formulae where the subject appears twice?
YOUR NOTES
If the subject appears twice, you will need to factorise at some point
Factorising means putting an expression into brackets, with the subject on the outside
of the brackets
If the subject appears inside a set of brackets, you will need to expand these brackets
before you can begin rearranging
If the subject appears on two sides of a formula, you will need to bring those terms to the
same side before you can factorise
Worked Example
Rearrange the formula p =
2 − ax
to make x the subject.
x−b
Get rid of the fraction by multiplying both sides by the expression on the
denominator.
p (x − b ) = 2 − ax
Expand the brackets on the left hand side to 'release' the x .
px − pb = 2 − ax
Bring the terms containing x to one side of the equals sign and any other terms to
the other side.
px − pb = 2 − ax
(+ax )
=
px − pb + ax = 2
(+pb )
=
px + ax = 2 + pb
(+ax )
(+pb )
Factorise the left-hand side to bring x outside of the brackets, so that it appears
only once.
x (p + a) = 2 + pb
Isolate x by dividing by the whole expression (p + a) .
x =
2 + pb
p +a
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2.7 Simultaneous Equations
2.7.1 Simultaneous Equations
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Linear Simultaneous Equations
What are linear simultaneous equations?
When there are two unknowns (say x and y) in a problem, we need two equations to be able
to find them both: these are called simultaneous equations
you solve two equations to find two unknowns, x and y
for example, 3x + 2y = 11 and 2x - y = 5
the solutions are x = 3 and y = 1
If they just have x and y in them (no x2 or y2 or xy etc) then they are linear simultaneous
equations
How do I solve linear simultaneous equations by elimination?
"Elimination" completely removes one of the variables, x or y
To eliminate the x's from 3x + 2y = 11 and 2x - y = 5
Multiply every term in the first equation by 2
6x + 4y = 22
Multiply every term in the second equation by 3
6x - 3y = 15
Subtract the second result from the first to eliminate the 6x's, leaving 4y - (-3y) = 22 15, i.e. 7y = 7
Solve to find y (y = 1) then substitute y = 1 back into either original equation to find x (x =
3)
Alternatively, to eliminate the y's from 3x + 2y = 11 and 2x - y = 5
Multiply every term in the second equation by 2
4x - 2y = 10
Add this result to the first equation to eliminate the 2y's (as 2y + (-2y) = 0)
The process then continues as above
Check your final solutions satisfy both equations
How do I solve linear simultaneous equations by substitution?
"Substitution" means substituting one equation into the other
Solve 3x + 2y = 11 and 2x - y = 5 by substitution
Rearrange one of the equation into y = ... (or x = ...)
For example, the second equation becomes y = 2x - 5
Substitute this into the first equation (replace all y's with 2x - 5 in brackets)
3x + 2(2x - 5) = 11
Solve this equation to find x (x = 3), then substitute x = 3 into y = 2x - 5 to find y (y = 1)
Check your final solutions satisfy both equations
How do you use graphs to solve linear simultaneous equations?
Plot both equations on the same set of axes
to do this, you can use a table of values or rearrange into y = mx + c if that helps
Find where the lines intersect (cross over)
The x and y solutions to the simultaneous equations are the x and y coordinates of the
point of intersection
e.g. to solve 2x - y = 3 and 3x + y = 4 simultaneously, first plot them both (see graph)
find the point of intersection, (2, 1)
the solution is x = 2 and y = 1
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How do I solve linear simultaneous equations from worded contexts?
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Tip
ExamAlways
check that your final solutions satisfy the original simultaneous
equations - you will know immediately if you've got the right solutions or not.
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Worked Example
Solve the simultaneous equations
5x + 2y = 11
4x - 3y = 18
Number the equations.
5x + 2y = 11
4x − 3y = 18
(1)
(2)
Make the y terms equal by multiplying all parts of equation (1) by 3 and all parts of
equation (2) by 2.
This will give two 6y terms with different signs. The question could also be done by
making the x terms equal by multiplying all parts of equation (1) by 4 and all parts of
equation (2) by 5, and subtracting the equations.
15x + 6y = 33
8x − 6y = 36
(3)
(4)
The 6y terms have different signs, so they can be eliminated by adding equation (4)
to equation (3).
15x + 6y = 33
+(
8x − 6y = 36)
23x
= 69
Solve the equation to find x by dividing both sides by 23.
x=
69
=3
23
Substitute x = 3 into either of the two original equations.
(1)
5(3) + 2y = 11
Solve this equation to find y.
15 + 2y = 11
2y = 11 − 15
2y = − 4
−4
y=
= −2
2
Substitute x = 3 and y = - 2 into the other equation to check that they are correct
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(2) 4x − 3y = 18
4(3) − 3(−2) = 18
12 − (−6) = 18
18 = 18
x = 3, y = − 2
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Quadratic Simultaneous Equations
What are quadratic simultaneous equations?
When there are two unknowns (say x and y) in a problem, we need two equations to be able
to find them both: these are called simultaneous equations
If there is an x2 or y2 or xy in one of the equations then they are quadratic (or non-linear)
simultaneous equations
How do I solve quadratic simultaneous equations?
Use the method of substitution
Substitute the linear equation, y = ... (or x = ...), into the quadratic equation
Do not try to substitute the quadratic equation into the linear equation
2
Solve x + y2 = 25 and y - 2x = 5
Rearrange the linear equation into y = 2x + 5
Substitute this into the quadratic equation, replacing all y's with (2x + 5) in brackets
x2 + (2x + 5)2 = 25
Expand and solve this quadratic equation (x = 0 and x = -4)
Substitute each value of x into the linear equation, y = 2x + 5, to get their value of y
Present your solutions in a way that makes it obvious which x belongs to which y
x = 0, y = 5 or x = -4, y = -3
Check your final solutions satisfy both equations
How do you use graphs to solve quadratic simultaneous equations?
Plot both equations on the same set of axes
to do this, you can use a table of values (or, for straight lines, rearrange into y = mx + c if
it helps)
Find where the lines intersect (cross over)
The x and y solutions to the simultaneous equations are the x and y coordinates of the
point of intersection
e.g. to solve y = x2 + 3x + 1 and y = 2x + 1 simultaneously, first plot them both (see graph)
find the points of intersection, (-1, -1) and (0, 1)
the solutions are x = -1 and y = -1 or x = 0 and y = 1
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ExamIf theTipresulting quadratic has a repeated root then the line is a tangent to the
curve. If the resulting quadratic has no roots then the line does not intersect
with the curve – or you have made a mistake!
When giving your final answer, make sure you indicate which x and y values go
together. If you don’t make this clear you can lose marks for an otherwise
correct answer.
Don't make the common mistake of thinking each squared term in x2 + y2 = 25
can be square-rooted to give x + y = 5 (they can't, the most you can do is
x 2 + y 2 = ± 5 , but you shouldn't be making x or y the subject of this anyway!)
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Worked Example
Solve the equations
x2 + y2 = 36
x = 2y + 6
Number the equations.
x 2 + y 2 = 36
x = 2y + 6
(1)
(2)
There is one quadratic equation and one linear equation so this must be done by
substitution.
Equation (2) is equal to x so this can be eliminated by substituting it into the x part
for equation (1).
Substitute x = 2y + 6 into equation (1).
(2y + 6) 2 + y 2 = 36
[1]
Expand the brackets, remember that a bracket squared should be treated the same
as double brackets.
(2y + 6) (2y + 6) + y 2 = 36
4y 2 + 6(2y ) + 6(2y ) + 62 + y 2 = 36
[1]
Simplify.
4y 2 + 12y + 12y + 36 + y 2 = 36
5y 2 + 24y + 36 = 36
[1]
Rearrange to form a quadratic equation that is equal to zero.
5y 2 + 24y + 36 − 36 = 0
5y 2 + 24y = 0
The question does not give a specified degree of accuracy, so this can be
factorised.
Take out the common factor of y .
y (5y + 24) = 0
Solve to find the values of y .
Let each factor be equal to 0 and solve.
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y1 = 0
5y 2 + 24 = 0 ⇒ y 2 = −
24
= − 4.8
5
[1]
Substitute the values of y into one of the equations (the linear equation is easier) to
find the values of x .
x 1 = 2(0) + 6 = 6
⎛ 24 ⎞
x 2 = 2 ⎜⎜− ⎟⎟ + 6 = − 9 . 6 + 6
⎝ 5 ⎠
x1 = 6, y1 = 0
x 2 = − 3 . 6 , y 2 = − 4 . 8 [1]
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2.8 Algebraic Fractions
2.8.1 Algebraic Fractions
Simplifying Algebraic Fractions
What is an algebraic fraction?
An algebraic fraction is a fraction with an algebraic expression on the top (numerator)
and/or the bottom (denominator)
How do you simplify an algebraic fraction?
Factorise fully top and bottom
Cancel common factors (including common brackets)
Tip
ExamIf you
are asked to simplify an algebraic fraction and have to factorise the top or
bottom, it is very likely that one of the factors will be the same on the top and the
bottom – you can use this to help you factorise difficult quadratics!
Worked Example
4x + 6
Simplify
2x 2 − 7x − 15
Factorise the top, by using 2 as a common factor
2(2x + 3)
2x 2 − 7x − 15
Factorise the bottom using your preferred method
Using the fact that the top factorised to (2x + 3) may help!
2(2x + 3)
(2x + 3) (x − 5)
The common factors on the top and bottom reduce to 1 (cancel out)
2(2x + 3)
(2x + 3) (x − 5)
=
2
(x − 5)
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Adding & Subtracting Algebraic Fractions
How do I add (or subtract) two algebraic fractions?
The rules are the same as fractions with numbers:
1. Find the lowest common denominator (LCD)
The LCD of x - 2 and x + 5 is found by multiplying them together: LCD = (x - 2)(x + 5)
this is the same as with numbers, where the LCD of 2 and 9 is 2 × 9 = 18
The LCD of x and 2x is not found by multiplying them together, as 2x already includes an
x , so the LCD is just 2x
this is the same as with numbers, where the LCD of 2 and 4 is just 4, not 2 × 4 = 8
The LCD of x + 2 and (x + 2)(x - 1) is just (x + 2)(x - 1), as this already includes an (x + 2)
The LCD of x + 1 and (x + 1)2 is just (x + 1)2, as this already includes an (x + 1)
The LCD of (x + 3)(x - 1) and (x + 4)(x - 1) is three brackets: (x + 3)(x - 1)(x + 4), without
repeating the (x - 1)
2. Write each fraction over this lowest common denominator
3. Multiply the numerators of each fraction by the same amount as the denominators
4. Write as a single fraction over the lowest common denominator (by adding or subtracting
the numerators, taking care to use brackets when subtracting)
5. Check at the end to see if the top factorises and cancels
Tip
ExamLeaving
the top and bottom of the fraction in factorised form will help you see if
anything cancels at the end
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Worked Example
(a) Express
x
x +4
−
3
x −1
as a single fraction
The lowest common denominator is (x + 4) (x − 1)
Write each fraction over this common denominator, remember to multiply the top
of the fractions too
x (x − 1)
3(x + 4)
−
( x + 4) ( x − 1)
(x − 1) (x + 4)
Simplify the numerators
x2 − x
3x + 12
−
( x + 4) ( x − 1)
(x − 1) (x + 4)
Combine the fractions, as they have the same denominator
x 2 − x − (3x + 12) x 2 − x − 3x − 12
x 2 − 4x − 12
=
=
(x + 4) (x − 1)
(x + 4) (x − 1)
(x + 4) (x − 1)
Factorise the top
(x + 2) (x − 6)
(x + 4) (x − 1)
There are no terms which would cancel here, so this is the final answer
( x + 2) ( x − 6)
( x + 4) ( x − 1)
(b) Express
x −4
x −1
as a single fraction
−
2(x − 3)
2x
The lowest common denominator is 2x (x − 3) (You could also use 4x (x − 3) but
this wouldn't be the lowest common denominator)
Write each fraction over this common denominator, remember to multiply the top
of the fractions too
x (x − 4)
(x − 1) (x − 3)
−
2x (x − 3)
2x (x − 3)
Simplify the numerators
x 2 − 4x
x 2 − 4x + 3
−
2x (x − 3)
2x (x − 3)
Combine the fractions, as they have the same denominator
x 2 − 4x − (x 2 − 4x + 3) x 2 − 4x − x 2 + 4x − 3
−3
=
=
2x (x − 3)
2x (x − 3)
2x (x − 3)
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There is nothing else that can be factorised on the numerator, so this is the final
answer
−3
2 x ( x − 3)
Multiplying & Dividing Algebraic Fractions
How do I multiply algebraic fractions?
1. Simplify both fractions first by fully factorising, then cancelling any common brackets on
top or bottom (from either fraction)
2. Multiply the tops together
3. Multiply the bottoms together
4. Check for any further factorising and cancelling
How do I divide algebraic fractions?
Flip ("reciprocate") the second fraction and replace ÷ with ×
a
b
So ÷ becomes ×
b
a
Then follow the same rules for multiplying two fractions
Example
Worked
x +3
2x + 6
Divide
x −4
by
x 2 − 16
, giving your answer as a simplified fraction
Division is the same as multiplying by the reciprocal (the fraction flipped)
x +3
2x + 6
x + 3 x 2 − 16
÷ 2
=
×
x − 4 x − 16 x − 4
2x + 6
It can often help to factorise first, as there may be factors that cancel out
x + 3 x 2 − 16 x + 3 (x − 4) (x + 4)
×
=
×
x −4
2x + 6
x −4
2(x + 3)
Multiply the numerators and denominators, and cancel any terms that are the same
on the top and bottom
=
(x + 3) (x − 4) (x + 4)
(x + 3) (x − 4) (x + 4)
=
2(x − 4) (x + 3)
2(x − 4) (x + 3)
=
x +4
2
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2.8.2 Solving Algebraic Fractions
Solving Algebraic Fractions
How do I solve an equation that contains algebraic fractions?
There are two methods for solving equations that contain algebraic fractions
One method is to deal with the algebraic fractions by adding or subtracting them first and
then solving the equation
Follow the rules for solving a linear equation containing a fraction on one or both sides
Remove the fractions first by multiplying both sides by everything on the
denominator
Remember to put brackets around any expression that you multiply by
The second method is to begin by multiplying everything in the fraction by each of the
expressions on the denominator
This will remove the denominators of the fractions, leaving you with either a linear or a
quadratic equation to solve
Multiplying everything in the fraction by the common denominator is a way of carrying
out this process in one go
For example, to solve the equation
4
+
5
x−3
x+1
term in the equation by both (x − 3) and (x + 1)
STEP 1
Multiply every term by (x − 3)
4
5
(x − 3) = 5(x − 3)
x+1
4
5(x − 3)
(x − 3) +
= 5(x − 3)
(x − 3)
x+1
5(x − 3)
4+
= 5(x − 3)
x+1
x−3
(x − 3) +
= 5 you will need to multiply every
STEP 2
Multiply every term by (x + 1)
5(x − 3)
(x + 1) = 5(x − 3) (x + 1)
x+1
5(x − 3)
4(x + 1) +
(x + 1) = 5(x − 3) (x + 1)
(x + 1)
4(x + 1) + 5(x − 3) = 5(x − 3) (x + 1)
4(x + 1) +
STEP 3
Expand the brackets on both sides and simplify
4x + 4 + 5x − 15 = 5(x 2 + x − 3x − 3)
9x − 11 = 5(x 2 − 2x − 3)
9x − 11 = 5x 2 − 10x − 15
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STEP 4
Rearrange the equation so that it is in a form that can be solved
9x − 11 = 5x 2 − 10x − 15
− (9x − 11)
=
− (9x − 11)
0 = 5x 2 − 10x − 9x − 15 + 11
0 = 5x 2 − 19x − 4
STEP 5
Solve the equation
You can swap the sides if it makes solving the equation easier
5x 2 − 19x − 4 = 0
(5x + 1) (x − 4) = 0
x= −
1
or x = 4
5
Tip
ExamMultiplying
by both denominators at once can speed up the process, but be
careful with the algebra if choosing this technique
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Worked Example
2
5
− = 6p
p +3 p
Show that this equation can be written as 6p 3 + 18p 2 + 3p + 15 = 0 .
To clear the fractions, we multiply both sides by the denominators.
We can do this one denominator at a time. We can start by multiplying by (p + 3) .
2−
5(p + 3)
= 6p (p + 3)
p
Now multiply by p
2(p ) − 5(p + 3) = 6p (p + 3) (p )
Now expand brackets
2p − 5(p + 3) = 6p 2 (p + 3)
2p − 5p − 15 = 6p 3 + 18p 2
Collect like terms
−3p − 15 = 6p 3 + 18p 2
Add the terms on the left hand side to the right hand side, to complete the question
0 = 6p 3 + 18p 2 + 3p + 15
6 p 3 + 18 p + 3 p + 15 = 0
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2.9 Forming & Solving Equations
2.9.1 Forming Equations
Forming Equations
Before solving an equation you may need to form it from the information given in the question.
How do I form an expression?
An expression is an algebraic statement without an equals sign e.g. 3x + 7 or 2(x 2 − 14)
Sometimes we need to form expressions to help us express unknown values
If a value is unknown you can represent it by a letter such as x
You can turn common phrases into expressions
Here you can represent the "something" by any letter
2 less than "something"
Double the amount of
"something"
5 lots of "something"
3 more than "something"
Half the amount of
"something"
x −2
2x
5x
x +3
1
x
x or
2
2
You might need to use brackets to show the correct order
"something" add 1 then multiplied by 3
(x + 1) × 3 which simplifies to 3(x + 1)
"something" multiplied by 3 then add 1
(x × 3) + 1 which simplifies to 3x + 1
To make the expression as easy as possible choose the smallest value to be represented
by a letter
If Adam is 10 years younger than Barry, then Barry is 10 years older than Adam
Represent Adam's age as x then Barry's age is x + 10
This makes the algebra easier, rather than calling Barry's age x and Adam's age
x − 10
If Adam's age is half of Barry's age then...
Barry's age is double Adam's age
So if Adam's age is x then Barry's age is 2x
This makes the algebra easier, rather than using x for Barry's age and
1
x for
2
Adams's age
How do I form an equation?
An equation is simply an expression with an equals sign that can then be solved
You will first need to form an expression and make it equal to a value or another expression
It is useful to know alternative words for basic operations:
For addition: sum, total, more than, increase, etc
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For subtraction: difference, less than, decrease, etc
For multiplication: product, lots of, times as many, double, triple etc
For division: shared, split, grouped, halved, quartered etc
Using the first example above
If Adam is 10 years younger than Barry and the sum of their ages is 25, you can find out
how old each one is
Represent Adam's age as x then Barry's age is x + 10
We can solve the equation x + x + 10 = 25 or 2x + 10 = 25
Sometimes you might have two unrelated unknown values (x and y) and have to use the
given information to form two simultaneous equations
Example
Worked
At a theatre the price of a child's ticket is £ x and the price of an adult's ticket is £ y .
Write equations to represent the following statements:
a)
An adult's ticket is double the price of a child's ticket.
b)
A child's ticket is £7 cheaper than an adult's ticket.
c)
The total cost of 3 children's tickets and 2 adults' tickets is £45.
a)
Adult = 2 × Child
y = 2x
1
equivalently you could put x = y
2
b)
Rewrite as:
Adult = Child + £7
y =x +7
equivalently you could put x = y − 7 or y − x = 7
c)
Total means add
3 × Child + 2 × Adult = £45
3 x + 2 y = 45
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Forming Equations from Shapes
YOUR NOTES
Many questions involve having to form and solve equations from information given about things
relating to shapes, like lengths or angles.
How do I form an equation involving the area or perimeter of a 2D shape?
Read the question carefully to decide if it involves area, perimeter or angles
If no diagram is given it is almost always a good idea to quickly sketch one
Add any information given in the question to the diagram
This information will normally involve expressions in terms of one or two variables
If the question involves perimeter, figure out which sides are equal length and add these
together
Consider the properties of the given shape to decide which sides will have equal
lengths
In a square or rhombus, all four sides are equal
In a rectangle or parallelogram, opposite sides are equal
If a triangle is given, are any of the sides equal length?
If the question involves area, write down the necessary formula for the area of that shape
If it is an uncommon shape you may need to split it up into two or more common
shapes that you can work out areas for
In the case you will have to split the length and width up accordingly
Remember that a regular polygon means all the sides are equal length
For example, a regular pentagon with side length 2x – 1 has 5 equal sides so its
perimeter is 5(2x – 1)
If one of the shapes is a circle or part of a circle, use π throughout rather than multiplying by
it and ending up with long decimals
How do I form an equation involving angles in a 2D shape?
If no diagram is given it is almost always a good idea to quickly sketch one
Add any information given in the question to the diagram
This information will normally involve expressions in terms of one or two variables
Consider the properties of angles within the given shape to decide which sides will have
equal lengths
If a triangle is given, how many of the angles are equal?
An isosceles triangle has two equal angles
An equilateral triangle has three equal angles
Consider angles in parallel lines (alternative, corresponding, co-interior)
In a parallelogram or rhombus, opposite angles are equal and all four sum to 360°
A kite has one equal pair of opposite angles
If the question involves angles, use the formula for the sum of the interior angles of a
polygon
For a polygon of n sides, the sum of the angles will be 180°×(n - 2)
Remember that a regular polygon means all the angles are equal
If a question involves an irregular polygon, assume all the angles are different unless told
otherwise
Look out for key information that can give more information about the angles
For example, a trapezium "with a line of symmetry" will have two pairs of equal angles
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How do I form an equation involving the surface area or volume of a 3D
shape?
Read the question carefully to decide if it involves surface area or volume
Mixing these up is a common mistake made in GCSE exams
If no diagram is given it is almost always a good idea to quickly sketch one
Add any information given in the question to the diagram
This information will normally involve expressions in terms of one or two variables
Consider the properties of the given shape to decide which sides will have equal lengths
In a cube all sides are equal
All prisms have the same shape (cross section) at the front and back
Pyramids normally have 1/3 in the formula
If the question involves volume, write down the necessary formula for the area of that
shape
If it is an uncommon shape the exam question will give you the formula that you need
Substitute the expressions for the side lengths into the formula
Remember to include brackets around any expression that you substitute in
It the question involves surface area,
STEP 1
Write down the number of faces the shape has and if any are the same
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STEP 2
Identify the 2D shape of each face and write down the formula for the area of each one
STEP 3
Substitute the given expressions into the formula for each one, being careful to identify
the correct expression for the dimension
You may need to add or subtract some expressions
STEP 4
Add the expressions together, double checking that you have one for each of the
faces
Remember to consider any faces that may be hidden in the diagram
ExamUseTippencil to annotate the diagrams carefully
You may find that most of your working for a question is on the diagram itself
Read the question carefully - don't find the area if it wants the perimeter, don't
find the volume if it wants the surface area, etc!
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Worked Example
A rectangle has a length of 3x + 1 cm and a width of 2x − 5 cm.
Its perimeter is equal to 22 cm.
a)
Use the above information to form an equation in terms of x.
The perimeter of a rectangle is 2(length) × 2(width).
P = 2(3x + 1) + 2(2x – 5)
Expand the brackets.
2(3x + 1) + 2(2x – 5) = 6x + 2 + 4x - 10
Simplify.
6x + 2 + 4x – 10 = 10x – 8
Set equal to the value given for the perimeter.
10x – 8 = 22
This equation can be simplified.
5x – 4 = 11
b)
Solve the equation from part (a) to find the value of x.
Add 4 to both sides.
5x – 4 = 11
5x = 15
Divide both sides by 5.
x=3
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2.9.2 Equations & Problem Solving
Equations & Problem Solving
What is problem solving?
Problem solving in mathematics involves using several stages, across a variety of topics, to
answer a question
In this set of notes all the problems will involve equations
These could be linear equations, quadratic equations or simultaneous equations and
other, relatively straightforward equations
You may notice there are not many subheadings in these notes
That is deliberate so the examples are not labelled or dealt with in an order
This is the nature of problem solving questions!
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You never know exactly what’s coming ... !
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In an ordinary mathematics question you would be given an equation to solve
In a problem solving question you would have to generate the equation ...
... using information from the question
... using your knowledge of standard mathematical results
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A key feature of problem solving questions is to interpret the answer in context
An answer on a calculator may be 2
If the question was about money then your final answer should be £1.20
A quadratic equation can have two solutions
Only one may be valid if only positive values are relevant (eg distance)
In problem solving questions you are typically given less information about the type of
maths involved
It is impossible to list every type of problem solving question you could see
There are endless contexts questions can be set in
There is no one-fits-all step-by-step method to solving problems
Practice, experience and familiarity are the keys to solving problems successfully
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Do not necessarily expect whole number (integer) or “nice” solutions
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Especially where a calculator is allowed
Rounding appropriately may be one of the skills being tested
eg Rounding a value in cm only needs to be to one decimal place;
so it indicates mm
ExamDoTip
not start by focusing on what the question has asked you to find, but on
what maths you can do.If your attempt turns out to be unhelpful, that’s fine, you
may still pick up some marks.If your attempt is relevant it could nudge you
towards the full solution – and full marks!Add information to a diagram as you
work through a problem. If there is no diagram, try sketching one.
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Worked Example
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2.10 Functions
2.10.1 Functions Toolkit
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What is a function?
Introduction to Functions
YOUR NOTES
A function is a combination of one or more mathematical operations that takes a set of
numbers and changes them into another set of numbers
It may be thought of as a mathematical “machine”
Putting 3 in to the function would give 2 × 3 + 1 = 7
Putting -4 in would give 2 × (-4) + 1 = -7
Putting x in would give 2x + 1 For example, if the function (rule) is “double the number
and add 1”, the two mathematical operations are "multiply by 2 (×2)" and "add 1 (+1)"
The number being put into the function is often called the input
The number coming out of the function is often called the output
What does a function look like?
A function f can be written as f(x) = … or f : x ↦ …
These two different types of notation mean exactly the same thing
Other letters can be used. g, h and j are common but any letter can technically be used
Normally, a new letter will be used to define a new function in a question
For example, the function with the rule “triple the number and subtract 4” would be written
f (x ) = 3x – 4 or f: x ↦ 3x – 4
In such cases, x would be the input and f (x ) would be the output
Sometimes functions don’t have names like f and are just written as y = …
eg. y = 3x – 4
How does a function work?
A function has an input (x ) and output (f (x ) or y )
Whatever goes in the bracket (instead of x )with f, replaces the x on the other side
This is the input
If the input is known, the output can be calculated
For example, given the function f (x ) = 2x + 1
f (3) = 2 × 3 + 1 = 7
f ( − 4) = 2 × ( − 4) + 1 = − 7
f ( a) = 2a + 1
If the output is known, an equation can be formed and solved to find the input
For example, given the function f (x ) = 2x + 1
If f (x ) = 15, the equation 2x + 1 = 15 can be formed
Solving this equation gives an input of 7
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Worked Example
A function is defined as f (x) = 3x − 2x + 1.
2
(a)
Find f (7) .
The input is x = 7 , so substitute 7 into the expression everywhere you see an x .
f (7) = 3(7) 2 − 2(7) + 1
Calculate.
f (7) = 3( 49) − 14 + 1
= 147 − 14 + 1
f (7) = 134
(b)
Find f (x + 3) .
The input is x = x + 3 so substitute x + 3 into the expression everywhere you
see an x .
f (x + 3) = 3(x + 3) 2 − 2(x + 3) + 1
Expand the brackets and simplify.
f (x + 3) = 3(x 2 + 6x + 9) − 2(x + 3) + 1
= 3x 2 + 18x + 27 − 2x − 6 + 1
= 3x 2 + 16x + 22
f ( x + 3) = 3 x 2 + 16 x + 22
A second function is defined g : x ↦ 3x – 4 .
(c)
Find the value of x for which g : x ↦ − 16.
Form an equation by setting the function equal to -16.
3x − 4 = − 16
Solve the equation by first adding 4 to both sides, then dividing by 3.
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3x − 4 = − 16
3x = − 12
12
x= −
3
x = −4
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2.10.2 Composite & Inverse Functions
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Composite Functions
What is a composite function?
A composite function is one function applied to the output of another function
Composite functions may also referred to as compound functions
What do composite functions look like?
The notation you will see for a composite function is fg(x)
This can be written as f(g(x)) and means “f applied to the output of g(x)”
i.e. g(x) happens first
Always apply the function on the outside to the output of the function on the inside
gf(x) means g(f(x)) and means “g applied to the output of f(x)”
i.e. f(x) happens first
How does a composite function work?
If you are putting a number into fg(x)
STEP 1
Put the number into g(x)
STEP 2
Put the output of g(x) into f(x)
For example, if f (x ) = 2x + 1 and g(x ) =
1
x
1
1
fg(2) = f ( ) = 2 ×
+1=2
2
2
gf(2) = g(2 × 2 + 1) = g(5) =
1
5
If you are using algebra, to find an expression for a composite function
STEP 1
For fg(x) put g(x) wherever you see x in f(x)
STEP 2
Simplify if necessary
For example, if f (x ) = 2x + 1 and g(x ) =
1
x
1
1
2
)=2×
+1=
+1
x
x
x
1
gf(x ) = g(2x + 1) =
2x + 1
fg(x ) = f (
Tip
ExamMake
sure you are applying the functions in the correct order
The letter nearest the bracket is the function applied first
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Worked Example
In this question, f (x) = 2x − 1 and g(x) = (x + 2) .
2
(a)
Find fg(4) .
g is on the inside of the composite function so apply g first.
fg(4) = f (g(4) ) = f ( (4 + 2) 2) = f (62) = f ( 36)
Apply f to the output of g.
f ( 36) = 2( 36) − 1
= 72 − 1
fg (4) = 71
(b)
Find g f (x ) .
f is on the inside of the composite function so apply f first by substituting the
function f(x) into g(x).
g f (x ) = g ( f (x ) ) = g (2x − 1) = ( (2x − 1) + 2) 2
Simplify.
g f (x ) = (2x − 1 + 2) 2
gf ( x ) = (2 x + 1) 2
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Inverse Functions
What is an inverse function?
An inverse function does the exact opposite of the function it came from
For example, if the function “doubles the number and adds 1” then its inverse is
“subtract 1 and halve the result”
It is the inverse operations in the reverse order
How do I write inverse functions?
An inverse function f-1 can be written as f −1 (x ) = … or f −1 : x ↦ …
For example, if f (x ) = 2x + 1 its inverse can be written as
(x – 1)
(x – 1)
f −1 (x ) =
or f −1: x ↦
2
2
How do I find an inverse function?
The easiest way to find an inverse function is to 'cheat' and swap the x and y variables
Note that this is a useful method but you MUST remember not to do this in any other
circumstances in maths
STEP 1
Write the function in the form y = …
STEP 2
Swap the x s and y s to get x = …
STEP 3
Rearrange the expression to make y the subject again
STEP 4
Write as f-1(x) = … (or f-1 : x ↦ …)
y should not exist in the final answer
For example, if f (x ) = 2x + 1 its inverse can be found as follows
STEP 1
Write the function in the form y = 2x + 1
STEP 2
Swap the x and y to get x = 2y + 1
STEP 3
Rearrange the expression to make y the subject again
x − 1 = 2y
x−1
=y
2
→ y =
x−1
2
STEP 4
Rewrite using the correct notation for an inverse function
x−1
f −1 (x ) =
2
How does a function relate to its inverse?
If f (3) = 10 then the input of 3 gives an output of 10
The inverse function undoes f(x)
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An input of 10 into the inverse function gives an output of 3
If f (3) = 10 then f −1 ( 10) = 3
ff−1 (x ) = f −1 f (x ) = x
If you apply a function to x, then immediately apply its inverse function, you get x
Whatever happened to x gets undone
f and f-1 cancel each other out when applied together
If f (x ) = 2x and you want to solve f −1 (x ) = 5
Finding the inverse function f −1 (x ) in this case is tricky (impossible if you haven't
studied logarithms)
instead, take f of both sides and use that ff−1 cancel each other out:
ff−1 (x ) = f (5)
x = f (5)
x = 25 = 32
Example
Worked
Find the inverse of the function f (x ) = 5 − 3x .
Write the function in the form y = 5 − 3x and then swap the x and y .
y = 5 − 3x
x = 5 − 3y
Rearrange the expression to make y the subject again.
x = 5 − 3y
x + 3y = 5
3y = 5 − x
5−x
y=
3
Rewrite using the correct notation for an inverse function
f −1 ( x ) =
5 − x
3
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2.11 Sequences
2.11.1 Sequences
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Introduction to Sequences
YOUR NOTES
What are sequences?
A sequence is an ordered set of (usually) numbers
Each number in a sequence is called a term
The location of a term within a sequence is called its position
The letter n is often used for (an unknown) position
Subscript notation is used to talk about a particular term
a1 would be the first term in a sequence
a7 would be the seventh term
an would be the nth term
What is a position-to-term rule?
A position-to-term rule gives the nth term of a sequence in terms of n
This is a very powerful piece of mathematics
With a position-to-term rule the 100th term of a sequence can be found without having
to know or work out the first 99 terms!
What is a term-to-term rule?
A term-to-term rule gives the (n+1)th term in terms of the nth term
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ie an+1 is given in terms of an
If a term is known, the next one can be worked out
How do I use the position-to-term and term-to-term rules?
These can be used to generate a sequence
From a given sequence the rules can be deduced
Recognising and being aware of the types of sequences helps
Linear and quadratic sequences
Geometric sequences
Fibonacci sequences
Other sequences
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Tip
ExamWrite
the position numbers above (or below) each term in a sequence.
This will make it much easier to recognise and spot common types of
sequence.
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Worked Example
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Types of Sequences
What other sequences are there?
Linear and quadratic sequences are particular types of sequence covered their own notes
Other sequences include geometric and Fibonacci sequences, which are looked at in
more detail below
Other sequences include cube numbers and triangular numbers
Another common type of sequence in exam questions, is fractions with combinations of
the above
Look for anything that makes the position-to-term and/or the term-to-term rule easy
to spot
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What is a geometric sequence?
A geometric sequence can also be referred to as a geometric progression and sometimes
as an exponential sequence
In a geometric sequence, the term-to-term rule would be to multiply by a constant, r
an+1 = r.an
r is called the common ratio and can be found by dividing any two consecutive terms, or
r = an+1 / an
In the sequence 4, 8, 16, 32, 64, ... the common ratio, r, would be 2 (8 ÷ 4 or 16 ÷ 8 or 32 ÷ 16
and so on)
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What is a Fibonacci sequence?
THE Fibonacci sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...
The sequence starts with the first two terms as 1
Each subsequent term is the sum of the previous two
ie The term-to-term rule is an+2 = an+1 + an
Notice that two terms are needed to start a Fibonacci sequence
Any sequence that has the term-to-term rule of adding the previous two terms is called a
Fibonacci sequence but the first two terms will not both be 1
Fibonacci sequences occur a lot in nature such as the number of petals of flowers
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Problem solving and sequences
When the type of sequence is known it is possible to find unknown terms within the
sequence
This can lead to problems involving setting up and solving equations
Possibly simultaneous equations
Other problems may involve sequences that are related to common number sequences
such as square numbers, cube numbers and triangular numbers
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Worked Example
The 3rd and 6th terms in a Fibonacci sequence are 7 and 31 respectively.
Find the 1st and 2nd terms of the sequence.
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How do I identify a sequence?
Is it obvious?
Does it tell you in the question?
Is there is a number that you multiply to get from one term to the next?
If so then it is a geometric sequence
Next, look at the differences between the terms
If 1st differences are constant – it is a linear sequence
If 2nd differences are constant – it is a quadratic sequence
Special cases to be aware of:
If the differences repeat the original sequence
It is a geometric sequence with common ratio 2
Fibonacci sequences also have differences that repeat the original sequence
However questions usually indicate if a Fibonacci sequence is involved
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Worked Example
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2.11.2 nth Term
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What is a linear sequence?
Linear Sequences
YOUR NOTES
A linear sequence is one where the terms go up (or down) by the same amount each time
eg 1, 4, 7, 10, 13, … (add 3 to get the next term)
15, 10, 5, 0, -5, … (subtract 5 to get the next term)
A linear sequence is often referred to as an arithmetic sequence
If we look at the differences between the terms, we see that they are constant
What should we be able do with linear sequences?
You should be able to recognise and continue a linear sequence
You should also be able to find a formula for the nth term of a linear sequence in terms of n
This formula will be in the form:
nth term = dn + b, where;
d is the common difference
b is a constant that makes the first term “work”
How do I find the nth term of a linear (arithmetic) sequence?
Find the common difference between the terms- this is d
Put the first term and n = 1 into the formula, then solve to find b
ExamIf aTip
sequence is going up by d each time, then its nth term contains dn
e.g. 5, 7, 9, 11, is going up by 2 each term so the nth term contains 2n
(the complete nth term for this example is 2n + 3)
If a sequence is going down by d each time, then its nth term contains −dn
e.g. 5, 3, 1, -1, ... is going down by 2 each term then the nth term contains −2n
(the complete nth term for this example is −2n + 7)
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Worked Example
Given the sequence 5, 7, 9, 11, 13, ...
(a)
Find the next three terms.
Looking at the difference between the terms, we see that they are all 2. So this is a
linear sequence with common difference 2
So the next three terms are
13 + 2 = 15
15 + 2 = 17
17 + 2 = 19
15, 17, 19
(b)
Find a formula for the nth term.
In part (a) we established that the common difference is 2. So d = 2
nth term = 2n + b
The first term is 5. Substitute this and n = 1 into the formula, and solve for b
Now we can write the nth term
5 = 2×1 + b
5 =2+b
b=3
2n + 3
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Quadratic Sequences
What is a quadratic sequence?
Unlike in a linear sequence, in a quadratic sequence the differences between the terms (the
first differences) are not constant
However, the differences between the differences (the second differences) are constant
Another way to think about this is that in a quadratic sequence, the sequence of first
differences is a linear sequence
eg Sequence: 2, 3, 6, 11, 18, …
1st Differences: 1 3 5 7 (a Linear Sequence)
2nd Differences: 2 2 2 (Constant)
If the second differences there are constant, we know that the example is a quadratic
sequence
What should we be able to do with quadratic sequences?
You should be able to recognise and continue a quadratic sequence
You should also be able to find a formula for the nth term of a quadratic sequence in terms of
n
This formula will be in the form:
nth term = an2 + bn + c
(The process for finding a, b, and c is given below)
How do I find the nth term of a quadratic sequence?
1. Work out the sequences of first and second differences
Note: check that the first differences are not constant and the second differences are
constant, to make sure you have a quadratic sequence!
e.g.
sequence: 1, 10, 23, 40, 61
first difference: 9, 13, 17, 21, ...
second differences: 4, 4, 4, ...
2. a = [the second difference] ÷ 2
e.g. a = 4 ÷ 2 = 2
3. Write out the first three or four terms of an2 with the first three or four terms of the given
sequence underneath.
Work out the difference between each term of an2 and the corresponding term of the given
sequence.
e.g. an2 = 2n2 = 2, 8, 18, 32, ...
sequence = 1, 10, 23, 40, ...
difference = -1, 2, 5, 8, ...
4. Work out the linear nth term of these differences. This is bn + c.
e.g. bn + c = 3n − 4
5. Add this linear nth term to an2. Now you have an2 + bn + c.
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e.g. an2 + bn + c = 2n2 + 3n − 4
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Exam Tip
Before doing the very formal process to find the nth term, try comparing the
sequence to the square numbers 1, 4, 9, 16, 25, … and see if you can spot the
formula
For example:
Sequence 4, 7, 12, 19, 28, …
Square Numbers 1 4 9 16 25
We can see that each term of the sequence is 3 more than the equivalent
square number so the formula is
nth term = n2 + 3
This could save you a lot of time!
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Worked Example
For the sequence 5, 7, 11, 17, 25, ....
(a)
Find a formula for the nth term.
Start by finding the first and second differences
Hence
Sequence: 5, 7, 11, 17, 25
First differences: 2, 4, 6, 8, ...
Second difference: 2, 2, 2, ...
a=2÷2=1
Now write down an2 ( just n2 in this case as a = 1) with the sequence underneath, and
on the next line write the difference between an2 and the sequence
an2. : 1, 4, 9, 16, ...
sequence: 5, 7, 11, 17, ...
difference: 4, 3, 2, 1, ...
Work out the nth term of these differences to give you bn + c
bn + c = −n + 5
Add an2 and bn + c together to give you the nth term of the sequence
nth term = n2 − n + 5
(b)
Hence find the 20th term of the sequence.
Substitute n = 20 into n2 − n + 5
(20)2 − 20 + 5 = 400 − 15
20th term = 385
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2.12 Coordinate Geometry
2.12.1 Coordinate Geometry
What is coordinate geometry?
Coordinate geometry is the study of geometric figures like lines and shapes, using
coordinates.
Given two points, at GCSE, you are expected to know how to find;
1. Midpoint of a Line
2. Gradient of a Line
3. Length of a Line
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Midpoint of a Line
How do I find the midpoint of a line in two dimensions (2D)?
The midpoint of a line will be the same distance from both endpoints
You can think of a midpoint as being the average (mean) of two coordinates
The midpoint of (x 1, y 1) and (x 2, y 2) is
⎛⎜ x 1 + x 2 y 1 + y 2 ⎞⎟
⎜⎜
⎟⎟
,
2 ⎠
⎝ 2
How can I extend the idea of the midpoint in two dimensions (2D)?
The midpoint of AB splits AB in the ratio 1 : 1.
If you are asked to find the point that divides AB in the ratio m : n, then you need to find the
point that lies
m
of the way from A to B.
m+n
E.g. dividing AB in the ratio 2 : 3 means finding the point that is
2
of the way from A to B.
5
Normally an exam question will ask you to find the point that divides AB in the ratio 1 : n, so;
find the difference between the x coordinates,
divide this difference by [1 + n], and add the result to the x coordinate of A,
repeat for the y coordinates.
How do I find the midpoint of a line in three dimensions (3D)?
Finding the midpoint of a line in 3D involves a simple extension of the formula used for 2D
midpoints
Similar to before, you can think of a midpoint as being the average (mean) of three x and
three y coordinates
The midpoint of (x 1, y 1, z1) and (x 2, y 2, z2) is
⎛⎜ x 1 + x 2 y 1 + y 2 z1 + z2 ⎞⎟
⎜⎜
⎟⎟
,
,
2
2 ⎠
⎝ 2
Tip
ExamIf working
in 2D (most questions!) making a quick sketch of the two points will
help you know roughly where the midpoint should be, which can be helpful to
check your answer
If working in 3D (some questions!), just check that your midpoint's x coordinate
lies between the two given x coordinates, and so on for the y and z coordinates
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Worked Example
The coordinates of A are (−4, 3) and the coordinates of B are (8, −12).
(a)
Find M, the midpoint of AB.
⎛⎜ x 1 + x 2 y 1 + y 2 ⎞⎟
⎟⎟
,
2
2
⎝
⎠
The midpoint can be found using M = ⎜⎜
Fill in the values of x and y from each coordinate
Simplify
⎛⎜ −4 + 8 3 + − 12 ⎞⎟ ⎛⎜ 4 −9 ⎞⎟
⎜
⎟= ⎜ ,
⎟
,
2
⎝ 2
⎠ ⎝2 2 ⎠
M = (2, −4.5)
(b)
A point N divides AB in the ratio 1 : 2.
Find the coordinates of N.
Calculate the difference between the x coordinate of A and the x coordinate of B
8 − (−4) = 12
Divide this difference by 3 (as there are 1 + 2 = 3 parts in the ratio 1 : 2)
12 ÷ 3 = 4
Add 4 to the x coordinate of A
x coordinate of N = − 4 + 4 = 0
Repeat for y
−12 − 3 = − 15
(−15) ÷ 3 = − 5
y coordinate of N = 3 + (−5) = − 2
Alternatively, you may find it helpful to sketch the coordinates A and B and find N
intuitively. Note that in the sketch, the coordinates do not need to be placed in the
correct orientation to one another, simply along a straight line:
Write the final answer as a coordinate point
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N = (0, −2)
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Gradient of a Line
What is the gradient of a line?
The gradient is a measure of how steep a 2D line is
A large value for the gradient means the line is steeper than for a small value of the
gradient
A gradient of 3 is steeper than a gradient of 2
A gradient of −5 is steeper than a gradient of −4
A positive gradient means the line goes upwards from left to right
A negative gradient means the line goes downwards from left to right
In the equation for a straight line, y = mx + c , the gradient is represented by m
The gradient of y = − 3x + 2 is −3
How do I find the gradient of a line?
The gradient can be calculated using
gradient =
change in y
change in x
rise
instead
run
For two coordinates (x 1 , y 1) and (x 2 , y 2) the gradient of the line joining them is
You may see this written as
y2 − y1
x2 − x1
or
y1 − y2
x1 − x2
The order of the coordinates must be consistent on the top and bottom
i.e. (Point 1 – Point 2) or (Point 2 – Point 1) for both the top and bottom
How do I draw a line with a given gradient?
A line with a gradient of 4 could instead be written as
4
.
1
change in y
, this would mean for every 1 unit to the right (x direction),
change in x
the line moves upwards (y direction) by 4 units.
−4
Notice that 4 also equals
, so for every 1 unit to the left, the line moves downwards
−1
As gradient =
by 4 units
If the gradient was −4, then
rise −4
+4
=
or
. This means the line would move
run +1
−1
downwards by 4 units for every 1 unit to the right.
2
, we can think of this as either
3
2
For every 1 unit to the right, the line moves upwards by , or
3
If the gradient is a fraction, for example
For every 3 units to the right, the line moves upwards by 2.
(Or for every 3 units to the left, the line moves downwards by 2.)
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If the gradient was −
2
this would mean the line would move downwards by 2 units for
3
every 3 units to the right
Once you know this, you can select a point (usually given, for example the y -intercept) and
then count across and upwards or downwards to find another point on the line, and then
join them with a straight line
ExamBeTip
very careful with negative numbers when calculating the gradient; write
down your working rather than trying to do it in your head to avoid mistakes
For example,
(−3) − (−9)
(−18) − (7)
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Worked Example
(a)
Find the gradient of the line joining (-1, 4) and (7, 28)
Using gradient =
change in y
:
change in x
28 − 4
7− −1
Simplify:
28 − 4
24
=
=3
7− −1
8
Gradient = 3
(b)
On the grid below, draw a line with gradient −2 that passes through (0, 1).
Mark the point (0, 1) and then count 2 units down for every 1 unit across
(c)
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On the grid below, draw a line with gradient
2
that passes through (0,-1)
3
Mark the point (0,-1) and then count 2 units up for every 3 units across
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Length of a Line
How do I calculate the length of a line?
The distance between two points with coordinates (x 1 , y 1) and (x 2 , y 2) can be found
using the formula
d=
2
2
(x 1 − x 2) + (y 1 − y 2)
This formula is really just Pythagoras’ Theorem a2 = b 2 + c2 , applied to the difference in the
x -coordinates and the difference in the y -coordinates;
You may be asked to find the length of a diagonal in 3D space. This can be answered using
3D Pythagoras
ExamAs Tip
we are squaring the difference in and in the formula, it does not matter if they
are positive or negative
32 is the same as (-3)2, this may help to speed up your working
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Worked Example
Point A has coordinates (3, -4) and point B has coordinates (-5, 2).
Calculate the distance of the line segment AB.
Using the formula for the distance between two points,
d = (x 1 − x 2) 2 + (y 1 − y 2) 2
Substituting in the two given coordinates:
d = (3 − − 5) 2 + (−4 − 2) 2
Simplify:
d=
(8) 2 + (−6) 2 =
64 + 36 = 100 = 10
Answer = 10 units
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2.13 Linear Graphs
2.13.1 Straight Line Graphs (y = mx + c)
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Finding Equations of Straight Lines
Why do we want to know about straight lines and their equations?
Straight Line Graphs (Linear Graphs) have lots of uses in mathematics – one use is in
navigation
We may want to know the equation of a straight line so we can program it into a computer
that will plot the line on a screen, along with several others, to make shapes and graphics
How do we find the equation of a straight line?
The general equation of a straight line is y = mx + c where;
m is the gradient,
c is the y -axis intercept (or simply, the y -intercept)
To find the equation of a straight line you need TWO things:
1. the gradient, m , which you can put straight into y = mx + c
get this from the question directly, or from two points using
rise
or the gradient
run
formula
2. any point on the line- substitute this point into y = mx + c (as you already know m ) and
solve to find c
if given two points which you used to find the gradient, just choose either one of
them for the point to find c
You may be asked to give the equation in the form ax + by + c = 0
(especially if m is a fraction)
If in doubt, SKETCH IT!
What if the line is not in the form y=mx+c?
A line could be given in the form ax + by + c = 0
It is harder to identify the gradient and intercept in this form
We can rearrange the equation into y = mx + c , so it is easier to identify the gradient and
intercept
ax + by + c = 0
Subtract ax from both sides
by + c = − ax
Subtract c from both sides
by = − ax − c
Divide both sides by b
ax c
y=−
−
c
b
a
c
y=− x−
c
b
a
c
In this case, the gradient is − and the y -intercept is −
c
b
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Worked Example
(a)
Find the equation of the straight line with gradient 3 that passes through (5, 4).
We know that the gradient is 3 so the line takes the form
y = 3x + c
To find the value of c, substitute (5, 4) into the equation
4 = 3(5) + c
4 = 15 + c
c = − 11
Replace c with −11 to complete the equation of the line
y = 3x − 11
(b)
Find the equation of the straight line that passes through (-2, 6) and (8, 1).
You may find it helpful to sketch the information given
First find m, the gradient
6−1
−2 − 8
5
=
−10
1
=−
2
m=
We know that the line takes the form
y=−
1
x +c
2
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To find the value of c, substitute either of the given points into this equation. Here
we will pick (8, 1) as it is doesn't contain negative numbers so is easier to work with
1
(8) + c
2
1= −4+ c
c =5
1= −
Replace c with −11 to complete the equation of the line
y=−
1
x +5
2
We can check against our sketch that this equation looks correct- it has a negative
gradient and it crosses the y-axis between 1 and 6
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Drawing Linear Graphs
How do we draw the graph of a straight line from an equation?
Before you start trying to draw a straight line, make sure you understand how to find the
equation of a straight line – that will help you understand this
How we draw a straight line depends on what form the equation is given in
There are two main forms you might see:
y = mx + c and ax + by = c
Different ways of drawing the graph of a straight line:
1. From the form y = mx + c
(you might be able to rearrange to this form easily)
plot c on the y-axis
go 1 across, m up (and repeat until you can draw the line)
2. From ax + by = c
put x = 0 to find y-axis intercept
put y = 0 to find x-axis intercept
(You may prefer to rearrange to y = mx + c and use above method)
Tip
ExamIt might
be easier just to plot ANY two points on the line (a third one as a check is
not a bad idea either)
or use the TABLE function on your calculator.
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Worked Example
On the axes below, draw the graphs of y = 3x − 1 and 3x + 5y = 15.
For y = 3x − 1 , first plot c, which is (0, −1)
Then, as m = 3,
rise 3
= . So plot a point 3 up and 1 right. Repeat at least once more
run 1
and then join the points with a straight line. Extend the line to the edges of the grid.
The steps for 3x + 5y = 15 are the same, but first we need to rearrange into the form
y = mx + c
5y = − 3x + 15
3
y = − x +3
5
Now we can plot c, which is (0, 3)
For the gradient,
rise −3
. So plot a point 3 down and 5 right. There isn't space
=
run
5
on the grid to repeat this so join the points with a straight line. Extend the line to the
edges of the grid. The answer is shown above
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2.13.2 Parallel & Perpendicular Lines
What are parallel lines?
Parallel Lines
Parallel lines are lines that have the same gradient, but are not the same line
Parallel lines do not intersect with each other
You can easily spot that two lines are parallel when they are written in the form y = mx + c ,
as they will have the same value of m (gradient)
y = 3x + 7 and y = 3x − 4 are parallel
y = 2x + 3 and y = 3x + 3 are not parallel
y = 4x + 9 and y = 4x + 9 are not parallel; they are the exact same line
How do I find the equation of a line parallel to another line?
As parallel lines have the same gradient, a line of the form y = mx + c will be parallel to a line
in the form y = mx + d , where m is the same for both lines
If c = d then they would be the same line and therefore not parallel
If you are asked to find the equation of a line parallel to y = mx + c , you will also be given
some information about a point that the parallel line, y = mx + d passes through; (x 1 , y 1)
You can then substitute this point into y = mx + d and solve to find d
Example
Worked
Find the equation of the line that is parallel to y = 3x + 7 and passes through (2,1)
As the gradient is the same, the line that is parallel will be in the form:
y = 3x + d
Substitute in the coordinate that the line passes through:
1 = 3(2) + d
Simplify:
1=6+d
Subtract 6 from both sides:
−5 = d
Final answer:
y = 3x − 5
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Perpendicular Lines
What are perpendicular lines?
You should already know that parallel lines have equal gradients
Perpendicular lines meet each other at right angles – ie they meet at 90°
What’s the deal with perpendicular gradients (and lines)?
Before you start trying to work with perpendicular gradients and lines, make sure you
understand how to find the equation of a straight line – that will help you do the sorts of
questions you will meet
Gradients m1 and m2 are perpendicular if m1 × m2 = −1
For example
1 and −1
1
and 3
3
2
3
− and
3
2
We can use m2 = −1 ÷ m1 to find a perpendicular gradient. This is called the negative
reciprocal.
If in doubt, SKETCH IT!
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Worked Example
The line L has equation y = 2x − 2 .
Find an equation of the line perpendicular to L which passes through the point
(2, − 3) .
Leave your answer in the form ax + by + c = 0 where a , b and c are integers.
L is in the form y = mx + c so we can see that its gradient is 2
m1 = 2
Therefore the gradient of the line perpendicular to L will be the negative reciprocal
of 2
m2 = −
1
2
Now we need to find c for the line we're after. Do this by substituting the point
(2, − 3) into the equation y = −
1
x + c and solving for c
2
1
×2+ c
2
−3= −1+ c
c = −2
−3 = −
Now we know the line we want is
y=−
1
x −2
2
But this is not in the form asked for in the question. So rearrange into the form
ax + by + c = 0 where a , b and c are integers
1
x +2=0
2
2y + x + 4 = 0
y+
Write the final answer
x + 2y + 4 = 0
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2.14 Quadratic Graphs
2.14.1 Quadratic Graphs
Quadratic Graphs
A quadratic is a function of the form y = ax 2 + bx + c where a is not zero
They are a very common type of function in mathematics, so it is important to know their key
features
What does a quadratic graph look like?
The shape made by a quadratic graph is known as a parabola
The parabola shape of a quadratic graph can either look like a “u-shape” or an “n-shape”
A quadratic with a positive coefficient of x 2 will be a u-shape
A quadratic with a negative coefficient of x 2 will be an n-shape
A quadratic will always cross the y -axis
A quadratic may cross the x -axis twice, once, or not at all
The points where the graph crosses the x -axis are called the roots
If the quadratic is a u-shape, it has a minimum point (the bottom of the u)
If the quadratic is an n-shape, it has a maximum point (the top of the n)
Minimum and maximum points are both examples of turning points
How do I sketch a quadratic graph?
We could create a table of values for the function and then plot it accurately, however we
often only require a sketch to be drawn, showing just the key features
The most important features of a quadratic are
Its overall shape; a u-shape or an n-shape
Its y -intercept
Its x -intercept(s), these are also known as the roots
Its minimum or maximum point (turning point)
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If it is a positive quadratic ( a in ax 2 + bx + c is positive) it will be a u-shape
If it is a negative quadratic ( a in ax 2 + bx + c is negative) it will be an n-shape
The y -intercept of y = ax 2 + bx + c will be (0, c)
The roots, or the x -intercepts will be the solutions to y = 0 ; ax 2 + bx + c = 0
You can solve a quadratic by factorising, completing the square, or using the quadratic
formula
There may be 2, 1, or 0 solutions and therefore 2, 1, or 0 roots
The minimum or maximum point of a quadratic can be found by;
Completing the square
Once the quadratic has been written in the form y = p (x − q ) 2 + r , the minimum or
maximum point is given by (q , r)
Be careful with the sign of the x-coordinate. E.g. if the equation is y = (x − 3) 2 + 2
then the minimum point is (3, 2) but if the equation is y = (x + 3) 2 + 2 then the
minimum point is (−3, 2)
Using differentiation
dy
= 0 will find the x -coordinate of the minimum or maximum point
Solving
dx
You can then substitute this into the equation of the quadratic to find the y coordinate
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Worked Example
a)
Sketch the graph of y = x 2 − 5x + 6 showing the x and y intercepts
It is a positive quadratic, so will be a u-shape
The +c at the end is the y -intercept, so this graph crosses the y -axis at (0,6)
Factorise
y = (x − 2) (x − 3)
Solve y = 0
(x − 2) (x − 3) = 0
x = 2 or x = 3
So the roots of the graph are
(2,0) and (3,0)
b)
Sketch the graph of y = x 2 − 6x + 13 showing the y -intercept and the turning point
It is a positive quadratic, so will be a u-shape
The +c at the end is the y -intercept, so this graph crosses the y-axis at
(0,13)
We can find the minimum point (it will be a minimum as it is a positive quadratic) by
completing the square:
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x 2 − 6x + 13 = (x − 3) 2 − 9 + 13 = (x − 3) 2 + 4
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This shows that the minimum point will be
(3,4)
As the minimum point is above the x -axis, this means the graph will not cross the x
-axis i.e. it has no roots
We could also show that there are no roots by trying to solve x 2 − 6x + 13 = 0
If we use the quadratic formula, we will find that x is the square root of a negative
number, which is not a real number, which means there are no real solutions, and
hence no roots
c)
Sketch the graph of y = − x 2 − 4x − 4 showing the root(s), y -intercept, and turning
point
It is a negative quadratic, so will be an n-shape
The +c at the end is the y -intercept, so this graph crosses the y -axis at (0, -4)
We can find the maximum point (it will be a maximum as it is a negative quadratic) by
completing the square:
−x 2 − 4x − 4 = − 1(x 2 + 4x + 4) = − 1( (x + 2) 2 − 4 + 4) = − (x + 2) 2
This shows that the maximum point will be
(-2, 0)
As the maximum is on the x -axis, there is only one root
We could also show that there is only one root by solving −x 2 − 4x − 4 = 0
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If you use the quadratic formula, you will find that the two solutions for x are the
same number; in this case -2
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2.15 Further Graphs & Tangents
2.15.1 Types of Graphs
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Types of Graphs
Why do we need to know what graphs look like?
Graphs are used in various aspects of mathematics – but in the real world they can take on
specific meanings
For example a linear (straight line) graph could be the path a ship needs to sail along to get
from one port to another
An exponential graph (y = k x ) can be used to model population growth – for instance to
monitor wildlife conservation projects
What are the shapes of graphs that we need to know?
Recalling facts alone won’t do much for boosting your IGCSE Mathematics grade!
But being familiar with the general shapes of graphs will help you quickly recognise the sort
of maths you are dealing with and features of the graph a question may refer to
Below the basic form of the five types of function (other than trig graphs) you need to
recognise;
linear (y = ± x )
quadratic (y = ± x 2 )
cubic (y = ± x 3 )
1
)
x
exponential (y = k ±x )
reciprocal (y = ±
In addition, you need to recognise the three basic trigonometric graphs- but these are
dealt with in another section
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Worked Example
Match the graphs to the equations.
Graphs:
A
B
C
D
E
Equations:
(1) y = 0 . 6x + 2 , (2) y = 3x , (3) y = − 0 . 7x 3 , (4) y =
4
, (5) y = − x 2 + 3x + 2
x
Starting with the equations,
(1) is a linear equation (y = mx + c) so matches the only straight line, graph (D)
(2) is an exponential equation with a positive coefficient so matches graph (A)
(3) is a cubic equation with a negative coefficient so matches graph (E)
(4) is a reciprocal equation (notice that it takes the same form as inverse
proportion) with a positive coefficient so matches graph (B)
(5) is a quadratic equation with a negative coefficient so matches graph (C)
Graph (A) → Equation (2)
Graph (B) → Equation (4)
Graph (C) → Equation (5)
Graph (D) → Equation (1)
Graph (E) → Equation (3)
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Drawing Graphs Using a Table
How do we draw a graph using a table of values in a non-calculator
exam?
Before you start, think what the graph might look like- see the previous notes on being
familiar with shapes of graphs
Using the rules of BIDMAS/ order of operations, substitute each x - value into the given
function
PLOT POINTS and join with a SMOOTH CURVE
If there are any points that don't seem to fit with the shape of the rest of the curve, check
your calculations for them again!
How do we draw a graph using a table of values in a calculator exam?
Before you start, think what the graph might look like – see the previous notes on being
familiar with shapes of graphs
Find the TABLE function on your CALCULATOR
Enter the FUNCTION – f(x)
(use ALPHA button and x or X, depending on make/model)
(Press = when finished)
(If you are asked for another function, g(x), just press enter again)
Enter Start, End and Step (gap between x values)
Press = and scroll up and down to see y values
PLOT POINTS and join with a SMOOTH CURVE
To avoid errors always put negative numbers in brackets and use the (-) key rather than the
subtraction key
If your calculator does not have a TABLE function, then you will have to work out each y
value separately using the normal mode on your calculator
Tip
ExamWhen
using the TABLE function of your calculator, double-check that your
calculator's y-values are the same as any that are given in the question
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Worked Example
Calculator Allowed
(a)
Complete the table of values for the function y = x 3 − 5x + 2 .
x
y
−3
−2
−1
0
1
2
4
3
14
Use the TABLE function on your calculator for f (x ) = x 2 − x − 6 , starting at -3,
ending at 3 and with steps of 1
If your calculator does not have a TABLE function then substitute the values of x
into the function one by one for the missing values, being careful to put negative
numbers in brackets, e.g.
x = − 3, y = (−3) 3 − 5(−3) + 2 = − 10
x
y
−3
−2
−1
− 10
4
6
0
2
1
−2
2
0
3
14
(b)
On the grid provided, draw the graph of y = x 3 − 5x + 2 for values of x from −3 to 3
.
Carefully plot the points from your table of values in (a) on the grid, noting the
different scales on the and axes
For example, the first column represents the point (−3, − 10)
After plotting the points, join them with a smooth curve- do not use a ruler!
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It is best practice to label the curve with its equation
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2.15.2 Using Graphs
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Solving Equations Using Graphs
How do we use graphs to solve equations?
Solutions are always read off the x-axis
Solutions of f(x) = 0 are where the graph of y = f(x) crosses the x-axis
If asked to use the graph of y = f(x) to solve a different equation (the question will say
something like “by drawing a suitable straight line”) then:
Rearrange the equation to be solved into f(x) = mx + c and draw the line y = mx + c
Solutions are the x-coordinates of where the line (y = mx + c) crosses the curve (y = f(x))
E.g. if given the curve for y = x3 + 2x2 + 1 and asked to solve x3 + 2x2 − x − 1 = 0, then;
1. rearrange x3 + 2x2 − x − 1 = 0 to x3 + 2x2 + 1 = x + 2
2. draw the line y = x + 2 on the curve y = x3 + 2x2 + 1
3. read the x-values of where the line and the curve cross (in this case there would be
3 solutions, approximately x = -2.2, x = -0.6 and x = 0.8);
Note that solutions may also be called roots
How do we use graphs to solve linear simultaneous equations?
Plot both equations on the same set of axes using straight line graphs y = mx + c
Find where the lines intersect (cross)
The x and y solutions to the simultaneous equations are the x and y coordinates of the
point of intersection
e.g. to solve 2x - y = 3 and 3x + y = 4 simultaneously, first plot them both (see graph)
find the point of intersection, (2, 1)
the solution is x = 2 and y = 1
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How do we use graphs to solve simultaneous equations where one is
quadratic?
e.g. to solve y = x2 + 4x − 12 and y = 1 simultaneously, first plot them both (see graph)
find the two points of intersection (by reading off your scale), (-6.1 , 1) and (-2.1, 1) to 1
decimal place
the solutions from the graph are approximately x = -6.1 and y = 1 and x = 2.1 and y = 1
note their are two pairs of x, y solutions
to find exact solutions, use algebra
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Tip
ExamIf solving
an equation, give the x values only as your final answer
If solving a pair of linear simultaneous equations give an x and a y value as your
final answer
If solving a pair of simultaneous equations where one is linear and one is
quadratic, give two pairs of x and y values as your final answer
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Worked Example
The graph of y = x + x − 3x − 1 is shown below.
3
2
Use the graph to estimate the solutions of the equation x 3 + x 2 − 4x = 0 . Give your
answers to 1 decimal place.
We are given a different equation to the one plotted so we must rearrange it to
f (x ) = mx + c (where f (x ) is the plotted graph)
x 3 + x 2 − 4x = 0
+x −1
= +x −1
3
2
x + x − 3x + 1 = x − 1
Now plot y = x − 1 on the graph- this is the solid red line on the graph below
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The solutions are the x coordinates of where the curve and the straight line cross so
x = − 2 . 6, x = 0, x = 1 . 6
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Finding Gradients of Tangents
What is the gradient of a graph?
The gradient of a graph at any point is equal to the gradient of the tangent to the curve at
that point
Remember that a tangent is a line that just touches a curve (and doesn’t cross it)
How do I estimate the gradient under a graph?
To find an estimate for the gradient:
Draw a tangent to the curve
Find the gradient of the tangent using Gradient = RISE ÷ RUN
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In the example above, the gradient at x = 4 would be
−2 . 5
4
= − 0 . 625
It is an estimate because the tangent has been drawn by eye and is not exact
(To find the exact gradient we would need to use differentiation)
What does the gradient represent?
In a y-x graph, the gradient represents the rate of change of y against x
This has many practical applications, for example;
in a distance-time graph, the gradient (rate of change of distance against time) is the
speed
in a speed-time graph, the gradient (rate of change of speed against time) is the
acceleration
ExamThisTipis particularly useful when working with Speed-Time and Distance-Time
graphs if they are curves and not straight lines
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Worked Example
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2.16 Solving & Graphing Inequalities
2.16.1 Graphical Inequalities
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Finding Regions using Inequalities
How do we draw inequalities on a graph?
First, see Straight Line Graphs (y = mx + c)
To graph an inequality;
1. DRAW the line (as if using “=”) for each inequality
Use a solid line for ≤ or ≥ (to indicate the line is included)
Use dotted line for < or > (to indicate the line is not included)
2. DECIDE which side of line is wanted.
Below line if "y ≤ ..." or "y < ..."
Above line if "y ≥ ..." or "y > ..."
Use a point that's not on the line as a test if unsure; substitute its x and y value into the
inequality to examine whether the inequality holds true on that side of the line
3. Shade UNWANTED side of each line (unless the question says otherwise)
This is because it is easier, with pen/ pencil/ paper at least, to see which region has not
been shaded than it is to look for a region that has been shaded 2-3 times or more
(Graphing software often shades the area that is required but this is easily overcome
by reversing the inequality sign)
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Worked Example
On the axes given below, show the region that satisfies the three inequalities;
3x + 2y ≥ 12 y < 2x x < 3
Label the region R.
First draw the three straight lines, 3x + 2y = 12, y = 2x and x = 3 , using your
knowledge of Straight Line Graphs (y = mx + c). You may wish to rearrange
3x + 2y = 12 to the form y = mx + c first:
2y = − 3x + 12
3
y = − x +6
2
The line 3x + 2y ≥ 12 takes a solid line because of the "≥" while the lines y < 2x and
x < 3 take dotted lines because of the "<"
Now we need to shade the unwanted regions
3
2
For 3x + 2y ≥ 12 (or y ≥ − x + 6 ), the unwanted region is below the line. We can
check this with the point (0, 0);
"3(0) + 2(0) ≥ 12" is false therefore (0, 0) does lie in the unwanted region for
3x + 2y ≥ 12
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For y < 2x , the unwanted region is above the line. If unsure, check with another
point, for example (1, 0)
"0 < 2(1) " is true, so (1, 0) lies in the wanted (i.e. unshaded) region for y < 2x
For x < 3 , shade the unwanted region to the right of x = 3 . If unsure, check with a
point
Finally, don't forget to label the region R
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Interpreting Graphical Inequalities
How do we interpret inequalities on a graph?
First, see Straight Line Graphs (y = mx + c)
To interpret inequalities/ to find a region defined by inequalities;
1. Write down the EQUATION of each line on the graph
2. REMEMBER that lines are drawn with:
A solid line for ≤ or ≥ (to indicate line included in region)
A dotted line for < or > (to indicate line not included)
3. REPLACE = sign with:
≤ or < if shading below line
≥ or > if shading above line
(Use a point to test if not sure)
If the question asks you to find the maximum in a region, find the coordinate furthest to the
top-right (largest values of x and y)
If the question asks you to find the minimum in a region, find the coordinate furthest to the
bottom-left (smallest values of x and y)
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Worked Example
Write down the three inequalities which define the shaded region on the axes
below.
First, using your knowledge of Straight Line Graphs (y = mx + c), define the three
lines as equations, ignoring inequality signs;
Now decide which inequality signs to use
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For y = x , the shaded region is above the line, and the line is dotted, so the
inequality is
y>x
Check by substituting a point within the shaded region into this inequality. For
example, using (2, 4) as marked on the graph above;
"4 > 2 " is true, so the inequality y > x is correct
For y = − x + 7 , the shaded region is below the line, and the line is solid, so the
inequality is
y ≤ −x +7
or y + x ≤ 7
Again, check by substituting (2, 4) into the inequality;
"4 ≤ − 2 + 7 " is true, so the inequality y ≤ − x + 7 is correct
For x = 1 , the shaded region is to the right of the solid line so the inequality is
x≥2
(Vertical and horizontal inequality lines probably do not need checking with a point,
though do so if you are unsure)
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2.17 Real-Life Graphs
2.17.1 Conversion Graphs
Conversion Graphs
What is a conversion graph?
A conversion graph is a linear graph which shows the relationship between two quantities
They allow you to convert between the two units or quantities by reading values off the
graph at different points on the line
They may be used for
Converting between units of temperature e.g. degrees Celsius (°C) and degrees
Fahrenheit (°F)
Converting between currencies e.g. Dollars ($) and Yen (¥)
Converting between units of volume e.g. Litres and Gallons
Calculating a charge based on usage e.g. a taxi driver charging per kilometre driven
How do I use a conversion graph?
Using the below graph, which converts temperatures between Celsius (°C) and Fahrenheit
(°F), as an example
Suppose we want to convert 100 °F into °C
Identify the axis you are converting from
In this example, we are converting from °F, which is on the y-axis
Using a ruler, draw a line from the value you are interested in, towards the graphed line
In this example, we have drawn a horizontal line from 100 °F
Using a ruler, draw a line from the point of intersection with the graphed line, to the other
axis, and read off the result
In this example, we have shown that 100 °F is equal to around 37.8 °C
Tip
ExamBefore
reading a value from an axis, determine the scale that is being used
1 square on the graph could represent 1, 2, 5, or any other number of units
The scales on the x and y axes may not be the same
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Worked Example
The graph below shows the price charged by a plumber for the time spent on a
particular job. The y-axis shows the price in dollars, and the x-axis shows the
number of hours spent working on the job.
a)
Find the price charged for a job which took 3 hours.
Draw a vertical line from the x-axis at 3 hours, and read off the graph on the y-axis.
Approximately $225
b)
A particular job cost $320. Find the length of time the job took.
Draw a horizontal line from the y-axis at $320, and read off the graph on the x-axis
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Approximately 4.5 hours
c)
The plumber charges a callout fee; a fee charged for travelling to the customer and
inspecting the job before starting any work. Find the price of the callout fee.
This would be the price charged for 0 hours; the y-intercept of the graph.
Read this value off the graph at the point where the line graph meets the y axis.
Approximately $45.
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2.17.2 Distance-Time & Speed-Time Graphs
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Distance-Time Graphs
How does a distance-time graph work?
Distance-time graphs show distance from a fixed point at different times
Distance is on the vertical axis, and time is on the horizontal axis.
The gradient of the graph is the speed
speed =
distance
rise
=
time
run
A positive gradient represents the object (or person) moving away from the starting point
If the graph is a horizontal line the object is stationary (not moving)
A negative gradient represents the object (or person) moving towards the starting point
If the graph is a straight line the speed is constant
If the graph is a curve you can draw the tangent at a point on the graph and find its gradient
This will be an estimate of the speed at that point
ExamIt isTip
easy to get confused between different types of graph.
Look at the label on the vertical axis to make sure you are looking at a
DISTANCE-time graph (not speed-time)
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Worked Example
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Speed-Time Graphs
What is a speed-time graph?
Speed-time graphs show speed at different times
Speed is on the vertical axis, and time is on the horizontal axis
The gradient of the graph is the acceleration
Acceleration =
speed
rise
=
time
run
If the graph is a curve you can draw the tangent at a point on the graph and find its gradient
This will be an estimate of the acceleration at that point
A positive gradient shows positive acceleration (speeding up)
A horizontal line on a speed-time graph shows constant speed (no acceleration)
A negative gradient shows negative acceleration, or deceleration (slowing down)
The distance covered can be found by finding the area under the graph
ExamIt isTip
easy to get confused between different types of graph.
Look at the label on the vertical axis to make sure you are looking at a
SPEED-time graph (not distance-time)
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Worked Example
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2.17.3 Rates of Change of Graphs
Rates of Change of Graphs
‘Real life graphs’ won’t just relate to distance, speed, and time. They can show the relationship
between any two variables, although they most commonly show how something changes with
time.
What is a rate of change?
A rate of change describes how a variable changes with time (or another variable)
All of the following are examples of rates of change
Speed (change in distance divided by time)
Acceleration (change in speed divided by time)
The depth of water in a container as it is filled with water (change in depth divided by
time)
The volume of air inside an inflating balloon as the radius of the balloon increases
(change in volume divided by change in radius)
The number of ice-creams sold as the weather gets warmer (change in ice-creams
sold divided by change in temperature)
How can I use a graph to find rates of change?
We can use the same methods that were used with distance-time and speed-time graphs
To find the rate of change of a unit on the y-axis per unit change in the x-axis (often time) we
can find the gradient of the graph
The units of the rate of change will be the units of the y-axis, divided by the units of the xaxis
If the graph showed volume in cm3 on the y-axis and time in seconds on the x-axis, the
rate of change would be measured in cm3/s or cm3s-1
If the graph is a straight line the rate of change is constant
If the graph is horizontal, the rate of change is zero (y is not changing as x changes)
If the graph is a curve, you can draw a tangent at a point on the graph and find its gradient
This will be an estimate of the rate of change at that point
The rate of change is larger when the graph is steeper (a higher gradient)
In the below image, tangents drawn at points A and B show that the graph is steeper at
B
Therefore the rate of change at B is greater
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Tip
ExamWhen
considering the rate of change of a curved graph, sketch tangents on the
graph at different points to help you
The units of the gradient can help you understand what is happening in the
context of the question
For example, if the y-axis is in dollars and the x-axis is in hours, the gradient
represents dollars per hour
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Worked Example
(a)
Each of the graphs below show the depth of water, d cm, in a container that is being
filled with water at a constant rate.
Match each of the graphs 1, 2, 3, 4 with the containers A, B, C, D
Considering graph 1, the gradient is constant, so the rate of change is
constant. So the depth increases at the same rate throughout.
This matches container D which has vertical sides, so the rate of filling does not
change.
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Graph 1 is Container D
Considering graph 2, the gradient starts shallow and becomes steeper,
meaning that the depth increases faster and faster.
This matches container A, gets narrower towards the top, meaning the rate of
filling increases.
Graph 2 is Container A
Considering graph 3, the gradient starts steep and becomes shallower,
meaning that the depth increases at a slower and slower rate.
This matches container B, which gets wider towards the top, meaning the rate
of filling decreases.
Graph 3 is Container B
Considering graph 4, the gradient starts steep, then becomes shallow, then
becomes steep again. This means that the depth increases quickly, then
slowly, then quickly again.
This matches container C, which is narrow at the bottom (fast filling), gets
wider in the middle (slow filling) and narrow again at the top (faster filling)
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Graph 4 is Container C
(b)
The graph below shows a model of the volume, v litres, of diesel in the tank of
George’s truck after it has travelled a distance of d kilometres.
(i)
Find the gradient of the graph, stating its units.
Gradient =
change in y
change in x
Gradient =
−54 litres
600 kilometres
-0.09 litres per kilometre
(ii)
Interpret what the gradient of the graph represents.
Consider the units of the gradient; litres per kilometre
The gradient represents the amount of diesel used to travel each kilometre.
Travelling 1km requires 0.09 litres of diesel.
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(iii)
Give one reason why this model may not be realistic.
The consumption of fuel may not be linear (a straight line); it is more likely to be
curved.
A reason for this could be that as fuel is used up the truck becomes lighter, so
becomes more fuel efficient.
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2.18 Differentiation
2.18.1 Differentiation
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What is differentiation?
Differentiation
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Differentiation is part of the branch of mathematics called Calculus
It is concerned with the rate at which changes takes place – so has lots of real‑world uses:
The rate at which a car is moving (its speed)
The rate at which a virus spreads amongst a population
To begin to understand differentiation you’ll need to understand gradients
How are gradients related to rates of change?
Gradient generally means steepness.
For example, the gradient of a road up the side of a hill is important to lorry drivers
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On a graph the gradient refers to how steep a line or a curve is
It is really a way of measuring how fast y changes as x changes
This may be referred to as the rate at which y
So gradient describes the rate at which change happens
How do I find the gradient of a curve using its graph?
For a straight line the gradient is always the same (constant)
Recall y = mx + c, where m is the gradient
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For a curve the gradient changes as the value of x changes
At any point on the curve, the gradient of the curve is equal to the gradient of the tangent
at that point
A tangent is a straight line that touches the curve at one point
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How do I find the gradient of a curve using algebra?
This is really where the fun begins!
Drawing tangents each time you want the gradient of a curve is too much effort
It would be great if you could do it using algebra instead
The equation of a curve can be given in the form y = f (x )
Inputting x-coordinates gives outputs of y-coordinates
It is possible to create an algebraic function that take inputs of x-coordinates and gives
outputs of gradients
All of this is done without needing to sketch any graphs
This type of function has a few commonly used names:
The gradient function
The derivative
The derived function
dy
The way to write this function is
dx
This is pronounced "dy by dx"
In function notation, it can be written f'(x )
pronounced f-dashed-of-x
dy
= f'(x ) you need to do an operation called differentiation
To get from y = f (x ) to
dx
Differentiation turns curve equations into gradient functions
The main rule for differentiation is shown
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This looks worse than it is!
For powers of x
STEP 1 Multiply the number in front by the power
STEP 2 Take one off the power (reduce the power by 1)
2x6 differentiates to 12x5
Note the following:
kx differentiates to k
so 10x differentiates to 10
any number on its own differentiates to zero
so 8 differentiates to 0
How do I use the gradient function to find gradients of curves?
Find the x-coordinate of the point on the curve you're interested in
dy
Use differentiation to find the gradient (derived) function,
dx
Substitute the x-coordinate into the gradient (derived) function to find the gradient
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Tip
ExamWhen
differentiating long awkward expressions, write each step out fully and
simplify the numbers after
Don't forget to write the left-hand sides of y = .... and
dy
= ... to avoid mixing up
dx
the curve equation with the gradient function
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Worked Example
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2.18.2 Applications of Differentiation
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Finding Stationary Points & Turning Points
What is a turning point?
The easiest way to think of a turning point is that it is a point at which a curve changes from
moving upwards to moving downwards, or vice versa
Turning points are also called stationary points
stationary means the gradient is zero (flat) at these points
At a turning point the gradient of the curve is zero.
If a tangent is drawn at a turning point it will be a horizontal line
Horizontal lines have a gradient of zero
This means substituting the x-coordinate of a turning point into the gradient function (aka
derived function or derivative) will give an output of zero
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How do I find the coordinates of a turning point?
STEP 1: Solve the equation of the gradient function (derivative / derived function) equal to
zero
dy
ie. solve
=0
dx
This will find the x-coordinate of the turning point
STEP 2: To find the y-coordinate of the turning point, substitute the x-coordinate into the
equation of the graph, y = ...
not into the gradient function
Tip
ExamRemember
to read the questions carefully (sometimes only the x-coordinate of
a turning point is required)
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Worked Example
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Classifying Stationary Points
What are the different types of stationary points?
You can see from the shape of a curve the different types of stationary points
You need to know two different types of stationary points (turning points):
Maximum points (this is where the graph reaches a “peak”)
Minimum points (this is where the graph reaches a “trough”)
These are sometimes called local maximum/minimum points as other parts of the graph
may still reach higher/lower values
How do I use graphs to classify which is a maximum point and which is a
minimum point?
You can see and justify which is a maximum point and which is a minimum point from the
shape of a curve...
... either from a sketch given in the question
... or a sketch drawn by yourself
(You may even be asked to do this as part of a question)
... or from the equation of the curve
For parabolas (quadratics) it should be obvious ...
... a positive parabola (positive x2 term) has a minimum point
... a negative parabola (negative x2 term) has a maximum point
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Cubic graphs are also easily recognisable ...
... a positive cubic has a maximum point on the left, minimum on the right
... a negative cubic has a minimum on the left, maximum on the right
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How do I use the second derivative to classify which is a maximum point
and which is a minimum point?
d2y
, is the derivative-of-the-derivative
dx 2
dy
d2y
differentiate the expression for
to get the expression for 2
dx
dx
The second derivative,
this is the same as differentiating the original equation for y twice
A quick algebraic test to find out the turning point (that does not require sketching) is as
follows
d2y
If the stationary point is at x = a , substitute x = a into the expression for 2 to get a
dx
numerical value...
d2y
...if this value is negative, 2 < 0 , the stationary point is a maximum point
dx
d2y
> 0 , the stationary point is a minimum point
...if this value is positive,
dx 2
d2y
If the value is zero,
= 0 , then unfortunately the test has failed
dx 2
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a zero means it could be any out of a max, min, or other types (stationary
points-of-inflection)
go back to sketching the graph to classify the stationary point(s)
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Worked Example
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2.18.3 Problem Solving with Differentiation
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Problem Solving with Differentiation
What problems could involve differentiation?
Differentiation allows analysis of how one quantity changes as another does
The derived function (gradient function / derivative) gives a measure of the rate of
change
Problems involving a variable quantity can involve differentiation
How the area of a rectangle changes as its length varies
How the volume of a cylinder changes as its radius varies
How the position of a car changes over time (i.e. its speed)
Problems based on the graph of a curve may also arise
The distance between two turning points
The area of a shape formed by points on the curve such as turning points and axes
intercepts
How do I solve problems involving differentiation?
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Problems generally fall into two categories:
1. Graph-based problems
These problems are based around the graph of a curve and its turning points
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2. Maximum/Minimum problems
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The maximum or minimum values have a meaning in the question
e.g. the maximum volume of a box made from a flat sheet of material
e.g. the minimum height of water in a reservoir
These are sometimes called optimisation problems
The maximum or minimum value gives the optimal (ideal/best) solution to the problem
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Tip
ExamDiagrams
can help – if you are not given one, sketch one and add to it as you go
along
Make sure you know how to find the areas and volumes of basic shapes, eg.
area of squares, rectangles, triangles, circles, volume of cubes, cuboids and
cylinders.
Early parts of questions often ask you to “show that” a result is true – even if you
can’t do this part of the question, you can use the answer shown to continue
with the rest of the question
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Worked Example
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