aqa level 2 further mathematics - schudio · 2020. 3. 23. · aqa level 2 further mathematics...

AQA Level 2 Further Mathematics

© MEI, 09/08/12 1/1

Further algebra

Section 1: Equations

Crucial points

1. Check your answer when solving equations When you solve an equation, always check your solution by substituting it into the original equation to make sure that it fits.

2. When solving problems, make sure your answer makes sense Always look at your answers to problems in the light of the original question. If there are two solutions, do they both make sense, or should one be discarded? Think carefully about the meaning of a negative solution, since this may or may not be a valid solution.


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Further algebra


Notes and Examples These notes contain subsections on

Linear equations

Solving quadratic equations by factorisation

Solving quadratic equations that cannot be factorised

Problem solving

Linear equations A linear equation involves only terms in x (or whatever variable is being used) and numbers. So it has no terms involving x², x³ etc. Equations like these are called linear because the graph of an expression involving only terms in x and numbers (e.g. y = 2x + 1) is always a straight line. Solving a linear equation may involve simple algebraic techniques such as gathering like terms and multiplying out brackets. Example 1 shows a variety of techniques that you might need to use. Example 1

Solve these equations.

(i) 5 2 3 8x x

(ii) 3(2 1) 4 2( 3)y y

(iii) 2 1

2 33

aa

Solution

(i) 5 2 3 8

5 3 8 2

5 3 10

5 3 10

2 10

5

x x

x x

x x

x x

x

x

Add 2 to each side

Subtract 3x from each side

Divide each side by 2

AQA FM Further algebra Section 1 Notes & Examples

© MEI, 09/08/12 2/7

(ii)

138

3(2 1) 4 2( 3)

6 3 4 2 6

6 3 10 2

6 13 2

8 13

y y

y y

y y

y y

y

y

(iii) 2 1

2 33

2 1 3(2 3)

2 1 6 9

2 6 10

4 10

2.5

aa

a a

a a

a a

a

a

You can look at a demonstration of solving simple linear equations using the Flash resource Solving linear equations. You can also look at the Linear equations video which demonstrates the solution of a wide range of linear equations. In Example 2, the problem is given in words and you need to express this algebraically before solving the equation. Example 2

Sarah has a choice of two tariffs for text messages on her mobile phone.

Tariff A: 10p for the first 5 messages each day, 2p for all others

Tariff B: 4p per message

How many messages would Sarah need to send each day for the two tariffs to cost the

same? (She always sends at least 5!)

Solution Let the number of messages Sarah sends per day be n.

Under Tariff A, she has to pay 10p for each of 5 messages and 2p for each of n - 5

messages.

Cost = 50 2( 5)n

Under Tariff B, she has to pay 4p for each of n messages.

Cost = 4n

For the cost to be the same

Multiply out the brackets

Add 3 to each side

Add 2y to each side

Divide each side by 8

Multiply both sides by 3

Multiply out the brackets

Add 1 to each side

Subtract 6a from each side

Divide both sides by -4


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50 2( 5) 4

50 2 10 4

40 2 4

40 2

20

n n

n n

n n

n

n

She needs to send 20 messages per day for the two tariffs to cost the same.

For practice in examples like this one, try the interactive questions Forming and solving linear equations.

Solving quadratic equations by factorisation Solving quadratic equations is important not just from the algebraic point of view, but because it gives you information about the graph of a quadratic function. The solutions of the equation ax² + bx + c = 0 tells you where the graph of the function y = ax² + bx + c crosses the x-axis, since these are the points where y = 0. Some quadratic equations can be solved by factorising. Example 3

Solve these quadratic equations by factorising.

(a) x² + 2x – 8 = 0 (b) 2x² + 11x + 12 = 0

Solution

(a) x² + 2x – 8 = 0

(x + 4)(x – 2) = 0

x + 4 = 0 or x – 2 = 0

x = –4 or 2

(b) 2x² + 11x + 12 = 0

(2x + 3)(x + 4) = 0

2x + 3 = 0 or x + 4 = 0

x = 23 or –4

You can see further examples using the Flash resources Quadratic equations 1 (in which the coefficient of x² is always 1) and Quadratic equations 2 (in which the coefficient of x² is greater than 1). For practice in examples like the ones above, try the interactive questions Solving quadratics by factorisation.

For this expression to be zero, one or other of the factors must be zero.


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You can also see examples of solving quadratic equations using the formula in the Flash resource Quadratic equations 3. The Solving quadratics video looks at solving quadratics by all the methods covered. Try the Quadratics puzzle, either on your own or with one or two others. Cut out all the pieces and match up each equation with its solution. The pieces will form a large hexagon.

Example 5

For each of the following quadratic equations, solve the equation, where possible, by

a suitable method.

y = x² + x – 3 2 4 13b ac

Two real roots

y = x² + 2x + 1 2 4 0b ac

One real root

y = x² + 2x + 2 2 4 4b ac

No real roots


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(i) 015²2 xx (ii) 01011²6 xx

Solution (i) a = 2, b = -5, c = 1

² 4

2

5 17

2 2

5 17

4

b b acx

a

(ii) a = 3, b = -2, c = 4

2 4 ( 2)² 4 3 4 4 48 44b ac

There are no real solutions.

Problem solving Some problems, when translated into algebra, involve quadratic equations. Example 6

A rectangular box has width 2 cm greater than its length, and height 3 cm less than its

length. The total surface area of the box is 548 cm².

What are the dimensions of the box?

Solution

Let the length of the box be x cm.

The width of the box is x + 2 cm, and the height is x – 3 cm.

The surface are of the box is given by 2x(x + 2) + 2x(x – 3) + 2(x + 2)(x – 3)

2x(x + 2) + 2x(x – 3) + 2(x + 2)(x – 3) = 548

x(x + 2) + x(x – 3) + (x + 2)(x – 3) = 274

x² + 2x + x² – 3x + x² – x – 6 = 274

3x² – 2x – 280 = 0

(3x + 28)(x – 10) = 0

x = 10

The length of the box is 10 cm, the width is 12 cm and the height is 7 cm.

Notice that in Example 10, you could discard one of the possible solutions as a negative solution did not make sense in the context. This is not always the case. In some situations, a negative solution can have a practical meaning.

Divide through by 2

The discriminant is 3364, which is 58²,

so this must factorise

3x + 28 = 0 gives a negative value of x, which does not make

sense in this context. So the solution must be x – 10 = 0.


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For example if the height of a stone thrown from the edge of a cliff is negative, this simply means that the stone is below the level of the cliff at that point. However, if the stone was thrown from level ground, then a negative height does not make sense. Some problems leading to quadratic equations do have two possible solutions. Always consider whether your solution(s) make sense in the context. For practice in examples like the ones above, try the interactive questions Forming and solving quadratics.


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Further algebra


Section overview

Background Solving equations, including quadratic equations, are essential skills within mathematics. You need to ensure that you are confident with the work in this section.

Specification content Solving linear equations. Solving quadratic equations by factorisation, by completing the square or by the formula.

Key points

The quadratic formula: 2 4

2

b b acx

a

Additional resources Flash resources Linear equations shows examples of solving linear equations. Quadratic equations 1 shows examples of solving quadratic equations by factorising, in cases where the coefficient of x² is 1. Quadratic equations 2 shows examples of solving quadratic equations by factorising, in cases where the coefficient of x² is greater than 1. Quadratic equations 3 shows solving quadratic equations by using the formula. Quadratic equations 4 shows solving quadratic equations by completing the square. Interactive questions Forming and solving linear equations tests you on solving a problem given in words by first expressing the problem as an equation. Solving quadratics by factorisation tests you on solving a quadratic equation in cases where the equation can be factorised. Forming and solving quadratic equations tests you on solving quadratic equations by any appropriate method. Active learning The Quadratics puzzle is a hexagonal jigsaw which deals with solving quadratic equations by any of the possible methods. Match up the equations with the solutions.

AQA FM Further algebra Section 1 Overview

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Mathscentre videos The Linear equations video looks at examples of linear equations, including some involving brackets or fractions. The Solving quadratics video looks at solving quadratics by factorising, completing the square or using the quadratic formula.


© MEI, 15/10/12 1/2

Further algebra


Exercise

1. Solve the following equations:

(i) 2x – 3 = 8 (ii) 3y + 2 = y – 5

(iii) 3 – 2a = 3a – 1 (iv) 3(p – 3) = 2(2p + 1)

(v) 2(1 – z) + 3(z + 3) = 4z + 1 (vi) 2 1 3

5 4

b b

2. The largest angle of a triangle is three times as big as the smallest angle.

The third angle is 20° greater than the smallest angle.

Find all three angles of the triangle.

3. In a restaurant, there are 24 tables, some of which seat four people, and the rest

seat 6 people. The restaurant can hold 114 people altogether.

How many tables seat four people?

4. Michelle is doing a Statistics project on the heights of students in her class.

She has written:

Mean height of boys = 165 cm

Mean height of girls = 159 cm

Mean height of whole class = 162.2 cm

There are 30 students in Michelle’s class.

How many boys and how many girls are there?

5. Solve these quadratic equations by factorising.

(i) x² + 4x + 3 = 0 (ii) x² + 5x – 6 = 0

(iii) x² – 6x + 8 = 0 (iv) x² – 7x – 18 = 0

(v) 2x² + 5x + 3 = 0 (vi) 2x² + x – 6 = 0

(vii) 4x² – 3x – 10 = 0 (viii) 6x² – 19x + 10 = 0

6. Solve the following quadratic equations, where possible. Give answers in exact

form.

(i) x² + 2x – 2 = 0 (ii) x² – 3x + 5 = 0

(iii) 2x² + x – 4 = 0 (iv) 2x² – 5x – 12 = 0

(v) x² – 5x – 3 = 0 (vi) 3x² + x + 1 = 0

(vii) 4x² + 12x + 9 = 0 (viii) 4x² + 10x + 5 = 0

7. The length of a rectangle is 3 cm greater than its width. The area of the rectangle is 40 cm². Find the length and width of the rectangle.

AQA FM Further algebra Section 1 Exercise

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8. ABCD is a rectangular field with width 20 m and length 50 m. Alistair walks

from D to B by walking a distance of x m along DC to E then walking to B in a

straight line.

The total distance which Alistair walks is 60 m. Find the value of x.

9. The area between the x-axis, the y-axis, the line 5 2y x and the line x a is

shaded in the diagram below.

The shaded area is 3 square units.

a) Show that 0 2.5a

b) Find the exact value of a.


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Further algebra


Solutions to Exercise

1. (i) 2 3 8

2 11

5.5

x

x

x

(ii) 3 2 5

2 2 5

2 7

3.5

y y

y

y

y

(iii) 3 2 3 1

3 5 1

4 5

0.8

a a

a

a

a

(iv) 3( 3) 2(2 1)

3 9 4 2

9 2

11

11

p p

p p

p

p

p

(v)

103

2(1 ) 3( 3) 4 1

2 2 3 9 4 1

11 4 1

11 3 1

10 3

z z z

z z z

z z

z

z

z

(vi)

1113

2 1 3

5 4

4(2 1) 5(3 )

8 4 15 5

13 4 15

13 11

b b

b b

b b

b

b

b

AQA FM Further algebra Section 1 Exercise solutions

© MEI, 15/10/12 2/5

2. Let the smallest angle be x°.

The largest angle is 3x°.

The third angle is (x + 20)°.

The three angles add up to 180°.

3 ( 20) 180

5 20 180

5 160

32

x x x

x

x

x

The angles are 32°, 96° and 52°.

(Check: 32 + 96 + 52 = 180).

3. Let the number of tables which seat 4 people be x.

The number of tables which seat 6 people is 24 – x.

Total number of seats 4 6(24 )x x

4 6(24 ) 114

4 144 6 114

30 2

15

x x

x x

x

x

There are 15 tables which seat 4 people.

(Check: 15 4 9 6 60 54 114)

4. Let x be the number of boys in the class

So number of girls is 30 – x.

Total of boys’ heights 165 x

Total of girls’ heights 159(30 )x

Total of heights for whole class 162.2 30 4866

165 159(30 ) 4866

165 4770 159 4866

6 96

16

x x

x x

x

x

There are 16 boys and 14 girls in the class.

5. (i)

2 4 3 0

( 3)( 1) 0

3 or 1

x x

x x

x x


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(ii)

2 5 6 0

( 6)( 1) 0

6 or 1

x x

x x

x x

(iii)

2 6 8 0

( 2)( 4) 0

2 or 4

x x

x x

x x

(iv)

2 7 18 0

( 9)( 2) 0

9 or 2

x x

x x

x x

(v)

2

32

2 5 3 0

(2 3)( 1) 0

or 1

x x

x x

x x

(vi)

2

32

2 6 0

(2 3)( 2) 0

or 2

x x

x x

x x

(vii)

2

54

4 3 10 0

(4 5)( 2) 0

or 2

x x

x x

x x

(viii)

2

2 53 2

6 19 10 0

(3 2)(2 5) 0

or

x x

x x

x x

6. (i)

2 2 2 0

1, 2, 2

x x

a b c

2 4 2 12 2 2 3

1 32 2 2

b b acx

a

(ii)

2 3 5 0

1, 3, 5

x x

a b c

2 24 ( 3) 4 1 5 9 20 11b ac

Negative so there are no real roots.


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(iii)

22 4 0

2, 1, 4

x x

a b c

2 4 1 33 1 33

2 2 2 4

b b acx

a

(iv) 22 5 12 0x x

32

(2 3)( 4) 0

or 4

x x

x x

(v)

2 5 3 0

1, 5 , 3

x x

a b c

2 5 374

2 2

b b acx

a

(vi)

23 1 0

3, 1, 1

x x

a b c

2 24 1 4 3 1 1 12 11b ac Negative so there are no real roots.

(vii) 24 12 9 0x x

2

32

(2 3) 0x

x

(viii)

24 10 5 0

4, 10, 5

x x

a b c

2 4 10 20 10 2 5 5 5

2 2 4 8 4

b b acx

a

7. Let x be the width of the rectangle, so the length is x + 3.

Area ( 3)x x

2

2

( 3) 40

3 40

3 40 0

( 8)( 5 ) 0

8 or 5

x x

x x

x x

x x

x

Dimensions must be positive, so width is 5 cm and length is 8 cm.


© MEI, 15/10/12 5/5

8.

2 2 2

2 2

(60 ) (50 ) 20

3600 120 2500 100 400

700 20

35

x x

x x x x

x

x

9. (a) If a = 0, the vertical line is on the x-axis and there is no area so a > 0.

If a = 2.5, the whole triangle below is shaded.

The area of the triangle is 6.25, which is too big so a < 2.5.

So 0 2.5a

(b) The shaded area is 21(5 5 2 ) (10 2 ) 5

2 2

aa a a a a

2

2

5 3

5 3 0

5 25 12

2

5 13

2

a a

a a

a

a

0 2.5a so 5 13

2


© MEI, 09/08/12 1/2

Further algebra


Multiple Choice Test

In Questions 1 to 4, solve the equations.

1) 3 2 5x x

(a) x = -7 (b) x = -3.5

(c) x = -3 (d) x = -1.5

(e) I don’t know

2) 1 5 3x x

(a) 12

x (b) 23

x

(c) x = 1 (d) 12

x

(e) I don’t know

3) 1 2( 3) 5(2 )y y

(a) y = 1 (b) y = 157

(c) y = 173

(d) y = 5

(e) I don’t know

4) 2 1

3 2

a a

(a) a = 1.4 (b) a = -7

(c) a = -1 (d) a = 0.2

(e) I don’t know

5) A football club has played 15 matches so far this season. They score 3 points for a

win, 1 point for a draw and 0 points for a loss. They have lost 4 matches, and their

points score is 27. How many matches have they won?

(a) 3 (b) 4

(c) 8 (d) 11

(e) I don’t know

AQA FM Further algebra Section 1 MC test

© MEI, 09/08/12 2/2

6) The solutions of the equation x² + 2x – 5 = 0 are

(a) no real solutions (b) 1 6

(c) 1 12 (d) 1 24

(e) I don’t know

7) The solutions of the equation 2x² – 11x + 15 = 0 are

(a) no real solutions (b) 1.5 and 5

(c) 11 241

4

(d) 2.5 and 3

(e) I don’t know

8) The solutions of the equation 3x² – 2x + 4 = 0 are

(a) no real solutions (b) 2 and 32

(c) 1 13

3

(d)

1 11

3

(e) I don’t know

9) The solutions of the equation 2x² + 5x – 4 = 0 are

(a) 5 57

4

(b)

5 7

4

(c) 5 57

4

(d)

5 7

4

(e) I don’t know

10) A rectangle has length (3x – 1) cm and width (x + 3) cm. The area of the rectangle

is 65 cm². What is the value of x?

(a) 3.61 or -6.28 (b) 3.61

(c) 6.28 (d) -3.61 or 6.28

(e) I don’t know

aqa level 2 further mathematics - schudio · 2020. 3. 23. · aqa level 2 further mathematics...

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