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EDEXCEL INTERNATIONAL GCSE (9 –1)
FURTHER PURE MATHEMATICSStudent Book Ali Datoo
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EDEXCEL INTERNATIONAL GCSE (9 –1)
FURTHER PURE MATHEMATICSStudent Book
Ali Datoo
Greg Attwood, Keith Pledger, David Wilkins, Alistair Macpherson, Bronwen Moran, Joseph Petran and Geoff Staley
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CONTENTS ii
PREFACE
CHAPTER 1
CHAPTER 2
CHAPTER 3
CHAPTER 4
CHAPTER 5
CHAPTER 6
CHAPTER 7
CHAPTER 8
CHAPTER 9
CHAPTER 10
CHAPTER 11
GLOSSARY
INDEX
VI
2
32
60
192
312
328
332
334
344
356
367
382
390
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THE QUADRATIC FUNCTION CHAPTER 2 23
The path followed by a bottlenose dolphin jumping out of the water is called a parabola. A parabola is a visual realisation of a quadratic function of the form y 5 2x2 1 k. Using this, scientists can calculate the height of a dolphin’s jump (in the y-axis) and the distance travelled (in the x-axis).
There is no scientific agreement about why dolphins jump. Some scientists believe it is because they are trying to conserve energy, some believe it is to help them find food, and others believe they do it just for fun.
The parabola is a beautiful and elegant shape, commonly seen in nature. It is also seen in many man-made structures such as bridges and other buildings.
2 THE QUADRATIC FUNCTION
LEARNING OBJECTIVES
• Factorise quadratic expressions where the coefficient of x2 is greater than 1
• Complete the square and use this to solve quadratic equations
• Solve quadratic equations using the quadratic formula
• Understand and use the discriminant to identify whether the roots are (i) equal and real, (ii) unequal and real or (iii) not real
• Understand the roots α and β and how to use them
BASIC PRINCIPLES
1 ▶ Factorise
a 6 x 2 1 9x b 2 b 2 1 8b c 9q m 2 2 27m
d 9x y 2 1 36 x 2 y e 24x 2 64 x 2
2 ▶ Factorise
a x 2 1 9x 1 18 b x 2 2 7x 1 12 c x 2 2 2x 2 3
d x 2 1 15x 1 36 e x 2 1 12x 1 27
3 ▶ Factorise
a x 2 2 9 b x 2 2 25
c 9 x 2 2 16 d 25 x 2 2 16
HINT These questions are all examples of the difference of two squares.
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24 CHAPTER 2 THE QUADRATIC FUNCTION
2.1 FACTORISE QUADRATIC EXPRESSIONS WHERE THE COEFFICIENT OF x 2 IS GREATER THAN 1
SKILLS: DECISION MAKING, CRITICAL THINKING
Factorise 3 x 2 1 5x 2 2
We need to find two numbers:
They need to add together to make 15 (the coefficient of x), and they need to multiply together to give 26 (the coefficient of x2 3 the constant term, in this case 3 3 22)
6 3 21 5 26 and 6 1 (21) 5 5
These numbers are then used to split the 5x into two terms, 6x and 21x
3 x 2 1 5x 2 2 5 3 x 2 1 6x 2 x 2 2
5 3x (x 1 2) 2 (x 1 2)
5 (x 1 2) (3x 2 1)
SKILLS: CRITICAL THINKING, DECISION MAKING
Factorise 6 x 2 2 5x 2 4
Find two numbers that add together to make 25 and multiply together to give 224
(the coefficient of x 2 multiplied by the constant term, in this case 6 3 24)
3 3 28 5 224 and 3 1 28 5 25
6 x 2 2 5x 2 4 5 6 x 2 1 3x 2 8x 2 4
5 3x (2x 1 1) 2 4 (2x 1 1)
5 (2x 1 1) (3x 2 4)
SKILLS: CRITICAL THINKING, DECISION MAKING
1 ▶ Factorise
a 3 x 2 2 7x 2 6 b 2 x 2 1 11x 1 5
c 2 x 2 2 7x 2 4 d 5 x 2 2 16x 1 3
e 6 x 2 1 x 2 12 f 9 x 2 2 6x 1 1
g 9 x 2 2 18x 2 7
2 ▶ Solve
a 2 x 2 1 5x 1 2 5 0 b 4 x 2 1 17x 1 4 5 0 c 3 x 2 2 13x 5 2 4
d 6 x 2 2 10x 1 4 5 0 e 2 (2 x 2 1 3) 5 2 15x f 3 ( x 2 2 2) 5 2 17x
g 15 x 2 1 42x 2 9 5 0 h 3x (3x 2 4) 5 2 4
EXAMPLE 1
The 5x term has been split into 6x and 2x
The 3x2 1 6x term and the 2x 2 2 term have both been factorised
EXAMPLE 2
The 25x term has been split into 3x and 28x
The 6x2 1 3x term and the 28x 2 4 term have both been factorised
EXERCISE 1
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THE QUADRATIC FUNCTION CHAPTER 2 25
2.2 COMPLETE THE SQUARE AND USE THIS TO SOLVE QUADRATIC EQUATIONS
A perfect square quadratic is in the form:
x 2 1 2bx 1 b 2 5 (x 1 b) 2 or x 2 2 2bx 1 b 2 5 (x 2 b) 2
To turn a non-perfect square into a perfect square you need to manipulate the expression.
To complete the square of the function x 2 1 2bx you need a further term b 2 .
So the completed square form is:
x 2 1 2bx 5 (x 1 b) 2 2 b 2
Similarly
x 2 2 2bx 5 (x 2 b) 2 2 b 2
Completing the square: x 2 1 bx 5 (x 1 b __ 2 )
2
2 ( b __ 2 )
2
SKILLS: REASONING, EXECUTIVE FUNCTION
Complete the square for
a x 2 1 12x b 2 x 2 2 10x
a x 2 1 12x 2b 5 12 so b 5 6
5 (x 1 6) 2 2 6 2
5 (x 1 6) 2 2 36
b 2 x 2 2 10x Here the coefficient of x 2 is 2
5 2( x 2 2 5x) So take out the coefficient of x 2
5 2 [ (x 2 5 __ 2
) 2
2 ( 5 __ 2 )
2
] Complete the square on ( x 2 2 5x)
5 2 (x 2 5 __ 2 )
2
2 25 ___ 2
By completing the square we can solve any quadratic equation.
Complete the square to solve
a x 2 1 8x 1 10 5 0 b 2 x 2 2 8x 1 7 5 0
a x 2 1 8x 1 10 5 0 Check coefficient of x 2 5 1
x 2 1 8x 5 2 10 Subtract 10 to get LHS in the form a x 2 1 bx
(x 1 4) 2 2 4 2 5 2 10 Complete the square for ( x 2 1 8x)
(x 1 4) 2 5 2 10 1 16 Add 4 2 to both sides
(x 1 4) 2 5 6
(x 1 4) 5 ± √ __
6 Square root both sides
x 5 2 4 ± √ __
6 Subtract 4 from both sides
Then the solutions of x 2 1 8x 1 10 5 0 are either
x 5 2 4 1 √ __
6 or x 5 2 4 2 √ __
6 Leave your answer in surd form
EXAMPLE 3
EXAMPLE 4
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26 CHAPTER 2 THE QUADRATIC FUNCTION
b 2 x 2 2 8x 1 7 5 0 The coefficient of x 2 is 2
x 2 2 4x 1 7 __ 2 5 0 So divide by 2
x 2 2 4x 5 2 7 __ 2 Subtract 7 __
2 from both sides
(x 2 2) 2 2 (2) 2 5 2 7 __ 2 Complete the square for x 2 2 4x
(x 2 2) 2 5 2 7 __ 2 1 4 Add 2 2 to both sides
(x 2 2) 2 5 1 __ 2 Combine the RHS
(x 2 2) 5 ± √ __
1 __ 2 Square root both sides
x 5 2 ± 1 ___ √
__ 2 Add 2 to both sides
So the roots are either
x 5 2 1 1 ___ √
__ 2 or x 5 2 2 1 ___
√ __
2
SKILLS: PROBLEM SOLVING
1 ▶ Complete the square for these expressions.
a x 2 1 4x b x 2 2 16x c 3 x 2 2 24x d x 2 2 x 2 12
e x 2 1 x 2 1 f 3 x 2 2 6x 1 1 g 2 x 2 1 3x 2 1 h 4 x 2 1 6x 2 1
2 ▶ Solve these quadratic equations by completing the square.
Leave your answer in surd form where appropriate.
a 6 x 2 2 11x 2 10 5 0 b x 2 1 2x 2 2 5 0 c 2 x 2 2 6x 1 1 5 0
d 2 x 2 1 3x 2 6 5 0 e 4 x 2 2 59x 2 15 5 0 f 4 x 2 1 8x 2 9 5 0
g 15 2 6x 2 2 x 2 5 0 h 4 x 2 2 x 2 8 5 0
3 ▶ Show by completing the square that the solutions to a x 2 1 bx 1 c 5 0 are x 5 2 b ± √ _______
b 2 2 4ac _______________________ 2a
2.3 SOLVE QUADRATIC EQUATIONS USING THE QUADRATIC FORMULAThe quadratic formula can be used to solve any quadratic equation of the form a x 2 1 bx 1 c 5 0
x 5 2 b ± √ _________
b 2 2 4ac _______________ 2a
SKILLS: PROBLEM SOLVING
Use the quadratic formula to solve 4 x 2 2 3x 2 2 5 0
x 5 2 ( 2 3) ± √
_____________ ( 2 3) 2 2 4(4)( 2 2) _________________________
2 3 4 Use x 5 2 b ± √
_______ b 2 2 4ac ______________
2a , where a 5 4, b 5 2 3, c 5 2 2
x 5 3 ± √
______ (9 1 32) ____________
8 2 4 3 4 3 2 5 2 32
x 5 3 ± √ ___
41 _______ 8
Then x 5 3 1 √ ___
41 _______ 8
or x 5 3 2 √ ___
41 _______ 8 Leave your answer in surd form
EXERCISE 2
EXAMPLE 5
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THE QUADRATIC FUNCTION CHAPTER 2 27
SKILLS: EXECUTIVE FUNCTION
1 ▶ Solve these equations using the quadratic formula.
Leave your answer in surd form where appropriate.
a x 2 5 6x b 2 x 2 1 50x 5 0 c x 2 2 x 2 6 5 0
d p 2 2 4p 1 2 5 0 e m 2 1 2m 2 2 5 0 f x 2 1 7x 1 10 5 0
g t 2 2 5t 2 6 5 0 h x 2 1 2x 2 35 5 0 i n 2 2 4n 1 4 5 0
j x 2 1 6x 1 6 5 0
2 ▶ Solve these equations, leaving your answer in surd form:
a 9 x 2 2 6x 1 25 5 0 b 3 x 2 2 6x 1 2 5 0 c 6 x 2 2 5x 2 6 5 0
d 2 x 2 1 3x 2 1 5 0 e 2 x 2 1 7x 1 3 5 0 f 3 x 2 1 7x 1 1 5 0
g 4 x 2 2 16x 1 15 5 0 h 7 x 2 1 5x 2 3 5 0 i 10 x 2 2 15x 2 8 5 0
j 6 x 2 1 3x 2 0 5 0 k (x 2 7) 2 5 36 l 4 x 2 1 15x 1 13 5 0
3 ▶ The diagram shows the floor plan of a bedroom.
The total area is 35.5m Find the value of m.
2.4 UNDERSTAND AND USE THE DISCRIMINANT TO IDENTIFY WHETHER THE ROOTS ARE (i) EQUAL AND REAL, (ii) UNEQUAL AND REAL OR (iii) NOT REAL
The equation a x 2 1 bx 1 c 5 0 has two solutions:
x 5 2 b 1 √ _________
b 2 2 4ac _______________ 2a
and x 5 2 b 2 √ _________
b 2 2 4ac _______________ 2a
These two solutions may be the same or different, real numbers or not real numbers.
The nature of the roots (solutions) of the equation will clearly depend on the expression b 2 2 4ac under the square root sign. This expression b 2 2 4ac is called the discriminant, as it allows us to identify (discriminate) whether the roots of a particular equation are equal and real, unequal and real or not real at all.
If:
• b 2 2 4ac . 0 the roots of the equation are unequal and real (two roots)
• b 2 2 4ac 5 0 the roots of the equation are equal and real (one root)
• b 2 2 4ac , 0 there are no real roots of the equation (no roots)
SKILLS: REASONING
Say what can you deduce from the values of the discriminants of these equations, and find the roots where possible.
a 2 x 2 2 3x 1 5 5 0 b 3 x 2 2 x 2 1 5 0 c 4 x 2 2 12x 1 9 5 0
EXERCISE 3
2m
3m
m 1 1
1.5
EXAMPLE 6
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28 CHAPTER 2 THE QUADRATIC FUNCTION
a 2 x 2 2 3x 1 5 5 0
a 5 2, b 5 2 3, c 5 5
b 2 2 4ac
(23) 2 2 4 3 2 3 5 5 2 31
Therefore there are no real roots.
b 3 x 2 2 x 2 1 5 0
a 5 3, b 5 2 1, c 5 2 1
b 2 2 4ac 5 (21) 2 2 4 3 3 3 (21) 5 13
Therefore there are two unequal roots.
x 5 2 (21) 6 √
___ 13 _____________
2 3 3 so Use the quadratic formula to find solutions.
x 5 0.768, 20.434
c 4 x 2 2 12x 1 9 5 0
a 5 4, b 5 2 12, c 5 9
(212) 2 2 4 3 4 3 9 5 0 Calculate the discriminant.
Therefore the roots are real and equal.
x 5 2 (212)
_______ 2 3 4
and x 5 3 __ 2 When the discriminant is 0 you can always factorise:
In this case: 4 x 2 2 12x 1 9 5 (2x 2 3) 2 5 0
The equation k x 2 2 2x 2 8 5 0 has two real roots.
What can you deduce about the value of the constant k?
Since the equation has two real roots, we know that the discriminant b2 2 4ac must be greater than zero.
We substitute a 5 k, b 5 (2 2) and c 5 (2 8) into the inequality b2 2 4ac . 0, giving
(22) 2 2 4 3 k 3 2 8 . 0
4 1 32k . 0
k . 2 4 ___ 32
k . 2 1 __ 8
SKILLS: REASONING
1 ▶ Use the discriminant to determine whether these equations have one root, two roots or no roots.
a x 2 2 2x 1 1 5 0 b x 2 2 3x 2 2 5 0 c 2 x 2 2 3x 2 4 5 0
d 2 x 2 2 4x 1 5 5 0 e 2 x 2 2 4x 1 2 5 0 f 2 x 2 2 7x 1 3 5 0
g 3 x 2 2 6x 1 5 5 0 h 7 x 2 2 144x 1 57 5 0 i 16 x 2 2 2x 1 3 5 0
j x 2 1 22x 1 121 5 0 k 5 x 2 2 4x 1 81 5 0
EXAMPLE 7
EXERCISE 4
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THE QUADRATIC FUNCTION CHAPTER 2 29
2 ▶ The equation p x 2 2 2x 2 7 5 0 has two real roots. What can you deduce about the value of p?
3 ▶ The equation 3 x 2 1 2x 1 m 5 0 has equal roots. Find the value of m.
2.5 UNDERSTAND THE ROOTS α AND β AND HOW TO USE THEM
If α and β are the roots of the equation a x 2 1 bx 1 c 5 0 then we deduce that
(x 2 α) (x 2 β) 5 0
We can rewrite this as x 2 2 x (α 1 β) 1 αβ 5 0
Comparing this with a x 2 1 bx 1 c 5 0 we can see that
α 1 β 5 2 b ___ a and αβ 5 c __ a
For the equation a x 2 1 bx 1 c 5 0
• The sum of roots, α 1 β 5 2 b ___ a
• The product of the roots, αβ 5 c __ a
KEY POINTS
SKILLS: REASONING, CRITICAL THINKING
1 ▶ The roots of the equation 3 x 2 1 x 2 6 5 0 are α and β
a Find an expression for α 1 β and an expression for αβ
b Hence find an expression for α2 1 β2 and an expression for α2β2
c Find a quadratic equation with roots α2 and β2
a 3 x 2 1 x 2 6 5 0
x 2 1 1 __ 3 x 2 2 5 0
Divide the equation by 3 to obtain an equation where the coefficient of x2 is 1.
Therefore sum of the roots α 1 β 5 2 1 __ 3 The sum of roots : α 1 β 5 2 b ___ a
Note: Sometimes you will need to manipulate the expressions to help you solve the questions.
Product of the roots αβ 5 22 The product of the roots: αβ 5 c __ a
b (α 1 β) 2 5 α 2 1 2αβ 1 β 2
Therefore α 2 1 β 2 5 (α 1 β) 2 2 2αβ
Substituting the results from part a, we find
α 2 1 β 2 5 (2 1 __ 3 )
2
2 2 (2 2)
5 4 1 __ 9
or 37 ___ 9
α 2 β 2 5 (αβ) 2 5 (2 2) 2 5 4
EXAMPLE 8
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30 CHAPTER 2 THE QUADRATIC FUNCTION
c Let the equation be x 2 1 px 1 q 5 0
p 5 ( α 2 1 β 2 ) 5 2 37 ___ 9 Use p 5 2sum of the roots
q 5 α 2 β 2 5 4 q 5 product of the roots
So the equation is: x 2 2 37 ___ 9 x 1 4 5 0 Simplify the equations in order for the
all the coefficients to be integers.
OR 9 x 2 2 37x 1 36 5 0
Note: This question can be answered without finding α and β.Sometimes α and β are not even real numbers.
The roots of the equation x 2 2 3x 2 2 5 0 are α and β.
Without finding the value of α and β, find the equations with the roots
a 3α, 3β b 1 __ α , 1 __ β c α 2 , β 2
If α and β are the roots of x 2 2 3x 2 2 5 0 then α 1 β and αβ 5 22
a If the roots are 3α and 3 β then
Sum of the roots 5 3 (α 1 β) 5 3 3 3 5 9 Using α 1 β 5 3 and αβ 5 2 2
Product of the roots 5 3α 3 3β 5 9αβ 5 2 18
Equation is: x 2 2 9x 2 18 5 0
b If the roots are 1 __ α , 1 __ β
Su m of roots 1 __ α 1 1 __ β 5
β 1 α ______
αβ
5 3 ___ 2 2
5 21.5
Product of roots: 1 __ α 3 1 __ β 5 1 ___
αβ
5 1 ___ 2 2
5 0.5
Equation is: x 2 1 3 __ 2 x 2 1 __
2 5 0
i.e. 2 x 2 1 3x 2 1 5 0
c If the roots are α 2 , β 2 then
Sum of roots 5 α 2 1 β 2 5 (α 1 β) 2 2 2αβ
5 3 2 2 2 (2 2) 5 13
Product of roots 5 α 2 β 2 5 αβ 2 5 (2 2) 2 5 4
Therefore equation is x 2 2 13x 1 4 5 0
EXAMPLE 9
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THE QUADRATIC FUNCTION CHAPTER 2 31
SKILLS: REASONING, CRITICAL THINING
1 ▶ The roots of the equation x 2 1 5x 1 2 5 0 are α and β. Find an equation whose roots are
a 2α 1 1 and 2β 1 1
b αβ and α2β2
2 ▶ The roots of the equation x 2 1 6x 1 1 5 0 are α and β. Find an equation whose roots are
a α 1 3 and β 1 3
b α __ β and
β __ α
3 ▶ The roots of the equation x 2 2 x 2 1 5 0 are α and β. Find an equation whose roots are
a 1 __ α and 1 __ β
b α ____ α 1 β
and β ____
α 1 β
EXERCISE 5
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32 CHAPTER 2 THE QUADRATIC FUNCTION32 CHAPTER 2 EXAM PRACTICE
EXAM PRACTICE: CHAPTER 2
1 The equation x 2 1 (p 2 3) x 1 (3 2 2p) 5 0 , where p is a constant, has two distinct real roots.
a Show that p satisfies p 2 1 2p 2 3 . 0 [1]
b Find the possible values of p. [2]
2 a Show that x 2 1 6x 1 11 can be written as (x 1 a) 2 1 b [2]
b Find the value of the discriminant. [2]
3 Factorise completely
a 5 x 2 1 16x 1 3 [3]
b 3 x 2 2 7x 1 4 [3]
4 Solve these equations by completing the square:
a p 2 1 3p 1 2 5 0 [3]
b 3 x 2 1 13x 2 10 5 0 [3]
5 Solve these equations by using the quadratic formula:
a 5 x 2 1 3x 2 1 5 0 [2]
b (2x 2 5) 2 5 7 [3]
6 4x 2 5 2 x 2 5 b 2 (x 1 a) 2 where a and b are integers.
a Find the value of a and b. [2]
b Calculate the discriminant. [2]
7 Solve 4 ______ 2x 1 1
2 3 5 2 1 ________ 4 x 2 2 1
[3]
8 Given that the roots of the equation 2 x 2 2 7x 1 3 5 0 are α and β, find the exact value of
a α 2 1 β 2 [2]
b α 2 β [3]
c α 3 2 β 3 [3]
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CHAPTER SUMMARY CHAPTER 2 33
CHAPTER SUMARY: CHAPTER 21 x 2 2 y 2 5 (x 2 y)(x 1 y ) is known as the difference of two squares.
It is important to spot this in exams.
2 Quadratic equations can be solved by
a factorisation.
b completing the square: x 2 1 bx 5 (x 1 b __ 2 )
2
2 ( b __ 2 )
2
c using the quadratic formula: x 5 2 b ± √ _______
b 2 2 4ac ______________ 2a
3 The discriminant of a quadratic expression is b 2 2 4ac
4 If α and β are the roots of the equation a x 2 1 bx 1 c 5 0 then
i α 1 β 5 2 b __ a
ii αβ 5 c __ a
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EDEXCEL INTERNATIONAL GCSE (9 –1)
FURTHER PURE MATHEMATICSStudent Book Ali Datoo
Pearson Edexcel International GCSE (9–1) Further Pure Mathematics prepares students for the new 2017 International GCSE (9–1) Further Pure Mathematics specification. This book provides comprehensive coverage of the specification and is designed to supply students with the best preparation possible for the examination:
• Written by a team of experienced International GCSE Maths teachers, authors and academics
• Chapters are mapped closely to the specification to provide comprehensive coverage and are enhanced by targeted reading and writing skills sections
• Learning is embedded with differentiated exercises and exam practice throughout• Signposted transferable skills• Track progress with the Pearson Progression Scale• Reviewed by a language specialist to ensure the book is written in a clear and
accessible style for students whose first language may not be English• Glossary of key Maths terminology • eBook included• Teacher support materials available online
For Pearson Edexcel International GCSE Further Pure Mathematics specification (4PM1) for first teaching 2017.
www.pearsonglobalschools.com
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