# Maths in Focus - Margaret Grove - ch2

August 12, 2017 | Author: Sam Scheding | Category: Fraction (Mathematics), Numbers, Rational Number, Factorization, Algebra

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Descripción: Mathematics Preliminary Course - 2nd Edition...

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2 Algebra and Surds TERMINOLOGY Binomial: A mathematical expression consisting of two terms such as x + 3 or 3x - 1 Binomial product: The product of two binomial expressions such as (x + 3) (2x - 4) Expression: A mathematical statement involving numbers, pronumerals and symbols e.g. 2x - 3 Factorise: The process of writing an expression as a product of its factors. It is the reverse operation of expanding brackets i.e. take out the highest common factor in an expression and place the rest in brackets e.g. 2y - 8 = 2 (y - 4) Pronumeral: A letter or symbol that stands for a number

Rationalising the denominator: A process for replacing a surd in the denominator by a rational number without altering its value Surd: From ‘absurd’. The root of a number that has an irrational value e.g. 3 . It cannot be expressed as a rational number Term: An element of an expression containing pronumerals and/or numbers separated by an operation such as + , - , # or ' e.g. 2x, - 3 Trinomial: An expression with three terms such as 3x 2 - 2x + 1

Chapter 2 Algebra and Surds

45

INTRODUCTION THIS CHAPTER REVIEWS ALGEBRA skills, including simplifying expressions, removing grouping symbols, factorising, completing the square and simplifying algebraic fractions. Operations with surds, including rationalising the denominator, are also studied in this chapter.

DID YOU KNOW? One of the earliest mathematicians to use algebra was Diophantus of Alexandria. It is not known when he lived, but it is thought this may have been around 250 AD. In Baghdad around 700–800 AD a mathematician named Mohammed Un-Musa Al-Khowarezmi wrote books on algebra and Hindu numerals. One of his books was named Al-Jabr wa’l Migabaloh, and the word algebra comes from the first word in this title.

Simplifying Expressions Addition and subtraction

EXAMPLES DID YOU KNOW? Simplify

7x Box 1. text...

-x

Solution

Here x is called a pronumeral.

7x - x = 7x - 1 x = 6x 2. 4x 2 - 3x 2 + 6x 2

Solution 4x 2 - 3x 2 + 6x 2 = x 2 + 6 x 2 = 7x 2

CONTINUED

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Maths In Focus Mathematics Preliminary Course

3. x 3 - 3x - 5x + 4 Only add or subtract ‘like’ terms. These have the same pronumeral (for example, 3x and 5x).

Solution x 3 - 3 x - 5x + 4 = x 3 - 8 x + 4 4. 3a - 4b - 5a - b

Solution 3a - 4b - 5a - b = 3a - 5a - 4b - b = - 2a - 5b

2.1 Exercises Simplify 1.

2x + 5x

16. 7b + b - 3b

2.

9a - 6a

17. 3b - 5b + 4b + 9b

3.

5z - 4z

18. - 5x + 3x - x - 7x

4.

5a + a

19. 6x - 5y - y

5.

4b - b

20. 8a + b - 4b - 7a

6.

2r - 5r

21. xy + 2y + 3xy

7.

- 4y + 3y

22. 2ab 2 - 5ab 2 - 3ab 2

8.

- 2x - 3x

23. m 2 - 5m - m + 12

9.

2a - 2a

24. p 2 - 7p + 5p - 6

10. - 4k + 7k

25. 3x + 7y + 5x - 4y

11. 3t + 4t + 2t

26. ab + 2b - 3ab + 8b

12. 8w - w + 3w

27. ab + bc - ab - ac + bc

13. 4m - 3m - 2m

28. a 5 - 7x 3 + a 5 - 2x 3 + 1

14. x + 3x - 5x

29. x 3 - 3xy 2 + 4x 2 y - x 2 y + xy 2 + 2y 3

15. 8h - h - 7h

30. 3x 3 - 4x 2 - 3x + 5x 2 - 4x - 6

Chapter 2 Algebra and Surds

47

Multiplication EXAMPLES Simplify 1. - 5x # 3y # 2x

Solution - 5x # 3y # 2x = - 30xyx = - 30x 2 y 2. - 3x 3 y 2 # - 4xy 5

Solution

Use index laws to simplify this question.

- 3x 3 y 2 # - 4xy 5 = 12x 4 y 7

2.2 Exercises Simplify 1.

5 # 2b

5 11. ^ 2x 2h

2.

2x # 4y

12. 2ab 3 # 3a

3.

5p # 2p

13. 5a 2 b # - 2ab

4.

- 3z # 2w

14. 7pq 2 # 3p 2 q 2

5.

- 5a # - 3b

15. 5ab # a 2 b 2

6.

x # 2y # 7z

16. 4h 3 # - 2h 7

7.

8ab # 6c

17. k 3 p # p 2

8.

4d # 3d

4 18. ^ - 3t 3 h

9.

3a # 4a # a

19. 7m 6 # - 2m 5

10. ^ - 3y h3

20. - 2x 2 # 3x 3 y # - 4xy 2

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Maths In Focus Mathematics Preliminary Course

Division Use cancelling or index laws to simplify divisions.

EXAMPLES Simplify 1. 6v 2 y ' 2vy

Solution By cancelling, 6v 2 y ' 2vy = =

6v 2 y 2vy 63 # v # v1 # y1 21 # v # y1

= 3v Using index laws, 6v 2 y ' 2vy = 3v 2 - 1 y 1 - 1 = 3v 1 y 0 = 3v 2.

5a 3 b 15ab 2

Solution 5a 3 b = 1 a3 -1 b1- 2 3 15ab 2 = 1 a 2 b -1 3 a2 = 3b

2.3 Exercises Simplify 1.

30x ' 5

2.

2y ' y

3. 4. 5.

8a 2

6.

xy 2x

7.

12p 3 ' 4p 2

8.

3a 2 b 2 6ab

9.

20x 15xy

10.

- 9x 7 3x 4

2

8a 2 a 8a 2 2a

Chapter 2 Algebra and Surds

11. -15ab ' - 5b 12.

2ab 6a 2 b 3

13.

- 8p 4pqs

16.

7pq 3

17. 5a 9 b 4 c - 2 ' 20a 5 b -3 c -1 2 ^ a -5 h b 4 2

18.

14. 14cd 2 ' 21c 3 d 3 15.

42p 5 q 4

-1

4a - 9 ^ b 2 h

19. - 5x 4 y 7 z ' 15xy 8 z - 2

2xy 2 z 3

20. - 9 ^ a 4 b -1 h ' -18a -1 b 3 3

4x 3 y 2 z

Removing grouping symbols The distributive law of numbers is given by

a ] b + c g = ab + ac

EXAMPLE 7 # (9 + 11) = 7 # 20 = 140 Using the distributive law, 7 # (9 + 11) = 7 # 9 + 7 # 11 = 63 + 77 = 140

This rule is used in algebra to help remove grouping symbols.

EXAMPLES Expand and simplify. 1. 2 ] a + 3 g

Solution 2 (a + 3) = 2 # a + 2 # 3 = 2a + 6

CONTINUED

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Maths In Focus Mathematics Preliminary Course

2. - ] 2x - 5 g

Solution -(2x - 5) = -1 (2x - 5) = -1 # 2x - 1 # - 5 = - 2x + 5 3. 5a 2]4 + 3ab - c g

Solution 5a 2 (4 + 3ab - c) = 5a 2 # 4 + 5a 2 # 3ab - 5a 2 # c = 20a 2 + 15a 3 b - 5a 2 c 4. 5 - 2 ^ y + 3 h

Solution 5 - 2 (y + 3 ) = 5 - 2 # y - 2 # 3 = 5 - 2y - 6 = - 2y - 1 5. 2 ] b - 5 g - ] b + 1 g

Solution 2 (b - 5) - (b + 1) = 2 # b + 2 # - 5 - 1 # b -1 # 1 = 2b - 10 - b - 1 = b - 11

2.4 Exercises Expand and simplify 1.

2]x - 4 g

7.

ab ] 2a + b g

2.

3 ] 2h + 3 g

8.

5n ] n - 4 g

3.

-5 ] a - 2 g

9.

3x 2 y _ xy + 2y 2 i

4.

x ^ 2y + 3 h

10. 3 + 4 ] k + 1 g

5.

x]x - 2 g

11. 2 ] t - 7 g - 3

6.

2a ] 3a - 8 b g

12. y ^ 4y + 3 h + 8y

Chapter 2 Algebra and Surds

13. 9 - 5 ] b + 3 g

20. 2ab ] 3 - a g - b ] 4a - 1 g

14. 3 - ] 2x - 5 g

21. 5x - ] x - 2 g - 3

15. 5] 3 - 2m g + 7 ] m - 2 g

22. 8 - 4 ^ 2y + 1 h + y

16. 2 ] h + 4 g + 3 ] 2h - 9 g

23. ] a + b g - ] a - b g

17. 3 ] 2d - 3 g - ] 5d - 3 g

24. 2 ] 3t - 4 g - ] t + 1 g + 3

18. a ] 2a + 1 g - ^ a 2 + 3a - 4 h

25. 4 + 3 ] a + 5 g - ] a - 7 g

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19. x ] 3x - 4 g - 5 ] x + 1 g

Binomial Products A binomial expression consists of two numbers, for example x + 3. A set of two binomial expressions multiplied together is called a binomial product. Example: ] x + 3 g ] x - 2 g. Each term in the ﬁrst bracket is multiplied by each term in the second bracket.

] a + b g ^ x + y h = ax + ay + bx + by

Proof ]a + bg]c + d g = a ]c + d g + b ]c + d g = ac + ad + bc + bd

EXAMPLES Expand and simplify 1. ^ p + 3h^ q - 4h

Solution ^ p + 3 h ^ q - 4 h = pq - 4p + 3q - 12 2. ]a + 5g2

Solution ] a + 5 g2 = (a + 5)(a + 5) = a 2 + 5a + 5a + 25 = a 2 + 10a + 25

Can you see a quick way of doing this?

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Maths In Focus Mathematics Preliminary Course

The rule below is not a binomial product (one expression is a trinomial), but it works the same way.

] a + b g ^ x + y + z h = ax + ay + az + bx + by + bz

EXAMPLE Expand and simplify ] x + 4 g ^ 2x - 3y - 1 h .

Solution (x + 4) (2x - 3y - 1) = 2x 2 - 3xy - x + 8x - 12y - 4 = 2x 2 - 3xy + 7x - 12y - 4

2.5 Exercises Expand and simplify 1.

]a + 5g]a + 2g

17. ]a + 2bg]a - 2bg

2.

]x + 3g]x - 1g

18. ^ 3x - 4y h^ 3x + 4y h

3.

^ 2y - 3h^ y + 5h

19. ]x + 3g]x - 3g

4.

]m - 4g]m - 2g

20. ^ y - 6h^ y + 6h

5.

]x + 4g]x + 3g

21. ] 3a + 1 g ] 3a - 1 g

6.

^ y + 2h^ y - 5h

22. ]2z - 7g]2z + 7g

7.

]2x - 3g]x + 2g

23. ]x + 9g^ x - 2y + 2h

8.

]h - 7g]h - 3g

24. ] b - 3 g ] 2a + 2b - 1 g

9.

]x + 5g]x - 5g

25. ]x + 2g^ x 2 - 2x + 4h

10. ] 5a - 4 g ] 3a - 1 g

26. ]a - 3g^ a 2 + 3a + 9h

11. ^ 2y + 3h^ 4y - 3h

27. ]a + 9g2

12. ]x - 4g^ y + 7h

28. ]k - 4g2

13. ^ x 2 + 3h]x - 2g

29. ]x + 2g2

14. ]n + 2g]n - 2g

30. ^ y - 7h2

15. ]2x + 3g]2x - 3g

31. ]2x + 3g2

16. ^ 4 - 7y h^ 4 + 7y h

32. ]2t - 1g2

Chapter 2 Algebra and Surds

33. ]3a + 4bg2

37. ] a + b g2

34. ^ x - 5y h2

38. ] a - b g2

35. ]2a + bg2

39. ] a + b g ^ a 2 - ab + b 2 h

36. ] a - b g ] a + b g

40. ] a - b g ^ a 2 + ab + b 2 h

Some binomial products have special results and can be simplified quickly using their special properties. Binomial products involving perfect squares and the difference of two squares occur in many topics in mathematics. Their expansions are given below.

Difference of 2 squares ] a + b g ] a - b g = a2 - b2

Proof (a + b) (a - b) = a 2 - ab + ab - b 2 = a2 - b2

Perfect squares

] a + b g2 = a 2 + 2ab + b 2

Proof ] a + b g2 = (a + b) (a + b) = a 2 + ab + ab + b 2 = a 2 + 2ab + b 2

]a - bg2 = a 2 - 2ab + b 2

Proof ] a - b g2 = (a - b) (a - b) = a 2 - ab - ab + b 2 = a 2 - 2ab + b 2

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Maths In Focus Mathematics Preliminary Course

EXAMPLES Expand and simplify 1. ]2x - 3g2

Solution ] 2x - 3 g2 = ] 2x g2 - 2 (2x) 3 + 3 2 = 4x 2 - 12x + 9 2. ^ 3y - 4h^ 3y + 4h

Solution (3y - 4) (3y + 4) = ^ 3y h2 - 4 2 = 9y 2 - 16

2.6 Exercises Expand and simplify 1.

]t + 4g2

16. ^ p + 1 h ^ p - 1 h

2.

]z - 6g2

17. ]r + 6g]r - 6g

3.

] x - 1 g2

18. ] x - 10 g ] x + 10 g

4.

^ y + 8h2

19. ]2a + 3g]2a - 3g

5.

^ q + 3h2

20. ^ x - 5y h^ x + 5y h

6.

]k - 7g2

21. ] 4a + 1 g ] 4a - 1 g

7.

] n + 1 g2

22. ]7 - 3xg]7 + 3xg

8.

]2b + 5g2

23. ^ x 2 + 2h^ x 2 - 2h

9.

]3 - xg2

2 24. ^ x 2 + 5h

10. ^ 3y - 1 h2

25. ]3ab - 4cg]3ab + 4c g

11. ^ x + y h2

2 2 26. b x + x l

12. ] 3a - b g2 13. ]4d + 5eg2

1 1 27. b a - a lb a + a l

14. ]t + 4g]t - 4g

28. _ x + 6 y - 2 @ i _ x - 6 y - 2 @ i

15. ] x - 3 g ] x + 3 g

29. 6]a + bg + c @2

Chapter 2 Algebra and Surds

30. 7 ] x + 1 g - y A

36. ] x - 4 g3

2

55

Expand (x - 4) (x - 4) 2 .

31. ] a + 3 g2 - ] a - 3 g2

1 2 1 2 37. b x - x l - b x l + 2

32. 16 - ]z - 4g]z + 4g

38. _ x 2 + y 2 i - 4x 2 y 2

33. 2x + ]3x + 1g2 - 4

39. ]2a + 5g3

34. ^ x + y h2 - x ^ 2 - y h

40. ] 2x - 1 g ] 2x + 1 g ] x + 2 g2

2

35. ] 4n - 3 g ] 4n + 3 g - 2n 2 + 5

PROBLEM Find values of all pronumerals that make this true. a b d f e i i i h i i c c

c e b g b

#

Try c = 9.

Factorisation Simple factors Factors are numbers that exactly divide or go into an equal or larger number, without leaving a remainder.

EXAMPLES The numbers 1, 2, 3, 4, 6, 8, 12 and 24 are all the factors of 24. Factors of 5x are 1, 5, x and 5x.

To factorise an expression, we use the distributive law.

ax + bx = x ] a + b g

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Maths In Focus Mathematics Preliminary Course

EXAMPLES Factorise 1. 3x + 12

Solution Divide each term by 3 to find the terms inside the brackets.

The highest common factor is 3. 3x + 12 = 3 ] x + 4 g 2. y 2 - 2y

Solution Check answers by expanding brackets.

The highest common factor is y. y 2 - 2y = y ^ y - 2 h 3. x 3 - 2x 2

Solution x and x2 are both common factors. We take out the highest common factor which is x2. x 3 - 2x 2 = x 2 ] x - 2 g 4. 5] x + 3 g + 2y ] x + 3 g

Solution The highest common factor is x + 3. 5 ] x + 3 g + 2y ] x + 3 g = ] x + 3 g ^ 5 + 2 y h 5. 8a 3 b 2 - 2ab 3

Solution There are several common factors here. The highest common factor is 2ab2. 8a 3 b 2 - 2ab 3 = 2ab 2 ^ 4a 2 - bh

Chapter 2 Algebra and Surds

2.7 Exercises Factorise 1.

2y + 6

19. x ] m + 5 g + 7 ] m + 5 g

2.

5x - 10

20. 2 ^ y - 1 h - y ^ y - 1 h

3.

3m - 9

21. 4^ 7 + y h - 3x ^ 7 + y h

4.

8x + 2

22. 6x ]a - 2g + 5]a - 2g

5.

24 - 18y

23. x ] 2t + 1 g - y ] 2t + 1 g

6.

x 2 + 2x

7.

m 2 - 3m

24. a ] 3x - 2 g + 2b ] 3x - 2 g - 3c ] 3x - 2 g

8.

2y 2 + 4y

9.

15a - 3a 2

25. 6x 3 + 9x 2 26. 3pq 5 - 6q 3 27. 15a 4 b 3 + 3ab

10. ab 2 + ab

28. 4x 3 - 24x 2

11. 4x 2 y - 2xy

29. 35m 3 n 4 - 25m 2 n

12. 3mn 3 + 9mn

30. 24a 2 b 5 + 16ab 2

13. 8x 2 z - 2xz 2 14. 6ab + 3a - 2a

31. 2rr 2 + 2rrh

2

32. ]x - 3g2 + 5]x - 3g

15. 5x 2 - 2x + xy

33. y 2 ]x + 4g + 2]x + 4g

16. 3q 5 - 2q 2

34. a ] a + 1 g - ] a + 1 g2

17. 5b 3 + 15b 2

35. 4ab ^ a 2 + 1 h - 3 ^ a 2 + 1 h

18. 6a 2 b 3 - 3a 3 b 2

Grouping in pairs If an expression has 4 terms, it may be factorised in pairs.

ax + bx + ay + by = x(a + b) + y (a + b) = ( a + b) ( x + y)

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Maths In Focus Mathematics Preliminary Course

EXAMPLES Factorise 1. x 2 - 2x + 3x - 6

Solution x 2 - 2x + 3x - 6 = x (x - 2) + 3 (x - 2) = (x - 2) (x + 3) 2. 2x - 4 + 6y - 3xy

Solution 2x - 4 + 6y - 3xy = 2 (x - 2) + 3y (2 - x) = 2 ( x - 2) - 3y ( x - 2 ) = (x - 2) (2 - 3y) or 2x - 4 + 6y - 3xy = 2 (x - 2) - 3y (- 2 + x) = 2 ( x - 2) - 3y ( x - 2 ) = (x - 2) (2 - 3y)

2.8 Exercises Factorise 1.

2x + 8 + bx + 4b

12. m - 2 + 4y - 2my

2.

ay - 3a + by - 3b

13. 2x 2 + 10xy - 3xy - 15y 2

3.

x 2 + 5x + 2x + 10

14. a 2 b + ab 3 - 4a - 4b 2

4.

m 2 - 2m + 3m - 6

15. 5x - x 2 - 3x + 15

5.

ad - ac + bd - bc

16. x 4 + 7x 3 - 4x - 28

6.

x 3 + x 2 + 3x + 3

17. 7x - 21 - xy + 3y

7.

5ab - 3b + 10a - 6

18. 4d + 12 - de - 3e

8.

2xy - x 2 + 2y 2 - xy

19. 3x - 12 + xy - 4y

9.

ay + a + y + 1

20. 2a + 6 - ab - 3b

10. x 2 + 5x - x - 5

21. x 3 - 3x 2 + 6x - 18

11. y + 3 + ay + 3a

22. pq - 3p + q 2 - 3q

Chapter 2 Algebra and Surds

23. 3x 3 - 6x 2 - 5x + 10

27. 4x 3 - 6x 2 + 8x - 12

24. 4a - 12b + ac - 3bc

28. 3a 2 + 9a + 6ab + 18b

25. xy + 7x - 4y - 28

29. 5y - 15 + 10xy - 30x

26. x 4 - 4x 3 - 5x + 20

30. rr 2 + 2rr - 3r - 6

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Trinomials A trinomial is an expression with three terms, for example x 2 - 4x + 3. Factorising a trinomial usually gives a binomial product. x 2 + ] a + b g x + ab = ] x + a g ] x + b g

Proof x 2 + (a + b) x + ab = x 2 + ax + bx + ab = x(x + a) + b(x + a) = (x + a) (x + b)

EXAMPLES Factorise 1. m 2 - 5m + 6

Solution a + b = - 5 and ab = + 6 -2 +6 ' -3 -5 Numbers with sum - 5 and product + 6 are - 2 and - 3. ` m 2 - 5m + 6 = [m + ] - 2 g] [m + ] - 3 g] = ]m - 2g]m - 3g

Guess and check by trying - 2 and - 3 or -1 and - 6.

2. y 2 + y - 2

Solution a + b = + 1 and ab = - 2 +2 -2 ' -1 +1 Two numbers with sum + 1 and product - 2 are + 2 and -1. ` y2 + y - 2 = ^ y + 2 h ^ y - 1 h

Guess and check by trying 2 and -1 or - 2 and 1.

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Maths In Focus Mathematics Preliminary Course

2.9 Exercises Factorise 1.

x 2 + 4x + 3

14. a 2 - 4a + 4

2.

y 2 + 7y + 12

15. x 2 + 14x - 32

3.

m 2 + 2m + 1

16. y 2 - 5y - 36

4.

t 2 + 8t + 16

17. n 2 - 10n + 24

5.

z2 + z - 6

18. x 2 - 10x + 25

6.

x 2 - 5x - 6

19. p 2 + 8p - 9

7.

v 2 - 8v + 15

20. k 2 - 7k + 10

8.

t 2 - 6t + 9

21. x 2 + x - 12

9.

x 2 + 9x - 10

22. m 2 - 6m - 7

10. y 2 - 10y + 21

23. q 2 + 12q + 20

11. m 2 - 9m + 18

24. d 2 - 4d - 5

12. y 2 + 9y - 36

25. l 2 - 11l + 18

13. x 2 - 5x - 24

The result x 2 + ] a + b g x + ab = ] x + a g ] x + b g only works when the coefficient of x 2 (the number in front of x 2) is 1. When the coefficient of x 2 is not 1, for example in the expression 5x 2 - 2x + 4, we need to use a different method to factorise the trinomial. There are different ways of factorising these trinomials. One method is the cross method. Another is called the PSF method. Or you can simply guess and check.

EXAMPLES Factorise 1. 5y 2 - 13y + 6

Solution—guess and check For 5y2, one bracket will have 5y and the other y: ^ 5y h ^ y h . Now look at the constant (term without y in it): + 6.

Chapter 2 Algebra and Surds

The two numbers inside the brackets must multiply to give + 6. To get a positive answer, they must both have the same signs. But there is a negative sign in front of 13y so the numbers cannot be both positive. They must both be negative. ^ 5y - h ^ y - h To get a product of 6, the numbers must be 2 and 3 or 1 and 6. Guess 2 and 3 and check: ^ 5y - 2 h ^ y - 3 h = 5y 2 - 15y - 2y + 6 = 5y 2 - 17y + 6 This is not correct. Notice that we are mainly interested in checking the middle two terms, -15y and - 2y. Try 2 and 3 the other way around: ^ 5y - 3 h ^ y - 2 h . Checking the middle terms: -10y - 3y = -13y This is correct, so the answer is ^ 5y - 3 h ^ y - 2 h . Note: If this did not check out, do the same with 1 and 6.

Solution—cross method Factors of 5y 2 are 5y and y. Factors of 6 are -1 and - 6 or - 2 and - 3. Possible combinations that give a middle term of -13y are 5y

-2

5y

-3

5y

-1

5y

-6

y

-3

y

-2

y

-6

y

-1

By guessing and checking, we choose the correct combination. 5y

-3

y

-2

5y # - 2 = -10y y # - 3 = - 3y -13y

` 5y 2 - 13y + 6 = ^ 5y - 3 h ^ y - 2 h

Solution—PSF method P: Product of first and last terms S: Sum or middle term F: Factors of P that give S - 3y 30y 2 ) -10y -13y

30y 2 -13y - 3y, -10y

` 5y 2 - 13y + 6 = 5y 2 - 3y - 10y + 6 = y ^ 5y - 3 h - 2 ^ 5 y - 3 h = ^ 5y - 3 h ^ y - 2 h

CONTINUED

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2. 4y 2 + 4y - 3

Solution—guess and check For 4y2, both brackets will have 2y or one bracket will have 4y and the other y. Try 2y in each bracket: ^ 2y h ^ 2y h . Now look at the constant: - 3. The two numbers inside the brackets must multiply to give - 3. To get a negative answer, they must have different signs. ^ 2y - h ^ 2y + h To get a product of 3, the numbers must be 1 and 3. Guess and check: ^ 2y - 3 h ^ 2 y + 1 h Checking the middle terms: 2y - 6y = - 4y This is almost correct, as the sign is wrong but the coefﬁcient is right (the number in front of y). Swap the signs around: ^ 2y - 1 h ^ 2 y + 3 h = 4y 2 + 6 y - 2 y - 3 = 4y 2 + 4y - 3 This is correct, so the answer is ^ 2y - 1 h ^ 2y + 3 h .

Solution—cross method Factors of 4y 2 are 4y and y or 2y and 2y. Factors of 3 are -1 and 3 or - 3 and 1. Trying combinations of these factors gives 3 2y 2y #- 1 = - 2y 2y # 3 = 6y 4y ` 4y 2 + 4y - 3 = ^ 2 y + 3 h ^ 2 y - 1 h 2y

-1

Solution—PSF method P: Product of ﬁrst and last terms -12y 2 S: Sum or middle term 4y F: Factors of P that give S + 6y, - 2y 2 + 6y -12y ) -2y + 4y ` 4y 2 + 4y - 3 = 4 y 2 + 6 y - 2 y - 3 = 2y ^ 2y + 3 h - 1 ^ 2 y + 3 h = ^ 2y + 3 h ^ 2y - 1 h

Chapter 2 Algebra and Surds

2.10

Exercises

Factorise 1.

2a 2 + 11a + 5

16. 4n 2 - 11n + 6

2.

5y 2 + 7y + 2

17. 8t 2 + 18t - 5

3.

3x 2 + 10x + 7

18. 12q 2 + 23q + 10

4.

3x 2 + 8x + 4

19. 8r 2 + 22r - 6

5.

2b 2 - 5b + 3

20. 4x 2 - 4x - 15

6.

7x 2 - 9x + 2

21. 6y 2 - 13y + 2

7.

3y 2 + 5y - 2

22. 6p 2 - 5p - 6

8.

2x 2 + 11x + 12

23. 8x 2 + 31x + 21

9.

5p 2 + 13p - 6

24. 12b 2 - 43b + 36

10. 6x 2 + 13x + 5

25. 6x 2 - 53x - 9

11. 2y 2 - 11y - 6

26. 9x 2 + 30x + 25

12. 10x 2 + 3x - 1

27. 16y 2 + 24y + 9

13. 8t 2 - 14t + 3

28. 25k 2 - 20k + 4

14. 6x 2 - x - 12

29. 36a 2 - 12a + 1

15. 6y 2 + 47y - 8

30. 49m 2 + 84m + 36

Perfect squares You have looked at some special binomial products, including ]a + bg2 = a 2 + 2ab + b 2 and ]a - bg2 = a 2 - 2ab + b 2 . When factorising, use these results the other way around.

a 2 + 2ab + b 2 = ] a + b g2 a 2 - 2ab + b 2 = ] a - b g2

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EXAMPLES In a perfect square, the constant term is always a square number.

Factorise 1. x 2 - 8x + 16

Solution x 2 - 8x + 16 = x 2 - 2 (4) x + 4 2 = ] x - 4 g2 2. 4a 2 + 20a + 25

Solution 4a 2 + 20a + 25 = ] 2a g2 + 2 (2a) (5) + 5 2 = ] 2a + 5 g2

2.11

Exercises

Factorise 1.

y 2 - 2y + 1

12. 16k 2 - 24k + 9

2.

x 2 + 6x + 9

13. 25x 2 + 10x + 1

3.

m 2 + 10m + 25

14. 81a 2 - 36a + 4

4.

t 2 - 4t + 4

15. 49m 2 + 84m + 36

5.

x 2 - 12x + 36

16. t 2 + t +

6.

4x 2 + 12x + 9

7.

16b 2 - 8b + 1

8.

9a 2 + 12a + 4

4x 4 + 3 9 6y 1 18. 9y 2 + + 5 25

9.

25x 2 - 40x + 16

19. x 2 + 2 +

10. 49y 2 + 14y + 1 11. 9y 2 - 30y + 25

1 4

17. x 2 -

1 x2

20. 25k 2 - 20 +

4 k2

Chapter 2 Algebra and Surds

Difference of 2 squares A special case of binomial products is ] a + b g ] a - b g = a 2 - b 2. a2 - b2 = ] a + b g ] a - b g

EXAMPLES Factorise 1. d 2 - 36

Solution d 2 - 36 = d 2 - 6 2 = ]d + 6 g]d - 6 g 2. 9b 2 - 1

Solution 9b 2 - 1 = ] 3b g2 - 1 2 = ( 3 b + 1) ( 3 b - 1 ) 3. (a + 3) 2 - (b - 1) 2

Solution ] a + 3 g2 - ] b - 1 g2 = [(a + 3) + (b - 1)] [(a + 3) - (b - 1)] = (a + 3 + b - 1) ( a + 3 - b + 1)

= ( a + b + 2 ) (a - b + 4 )

2.12

Exercises

Factorise 1.

a2 - 4

7.

1 - 4z 2

2.

x2 - 9

8.

25t 2 - 1

3.

y2 - 1

9.

9t 2 - 4

4.

x 2 - 25

10. 9 - 16x 2

5.

4x 2 - 49

11. x 2 - 4y 2

6.

16y 2 - 9

12. 36x 2 - y 2

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13. 4a 2 - 9b 2

20.

14. x 2 - 100y 2 15. 4a - 81b 2

21. ] x + 2 g2 - ^ 2y + 1 h2

2

22. x 4 - 1

16. ]x + 2g2 - y 2 17. ] a - 1 g - ] b - 2 g 2

2

18. z - ] 1 + w g 2

19. x 2 -

y2 -1 9

2

1 4

23. 9x 6 - 4y 2 24. x 4 - 16y 4 25. a 8 - 1

Sums and differences of 2 cubes

a 3 + b 3 = ] a + b g ^ a 2 - ab + b 2 h

Proof (a + b) (a 2 - ab + b 2) = a 3 - a 2 b + ab 2 + a 2 b - ab 2 + b 3 = a3 + b3 a 3 - b 3 = ] a - b g ^ a 2 + ab + b 2 h

Proof (a - b) (a 2 + ab + b 2) = a 3 + a 2 b + ab 2 - a 2 b - ab 2 - b 3 = a3 - b3

EXAMPLES Factorise 1. 8x 3 + 1

Solution 8x 3 + 1 = ] 2x g3 + 1 3 = (2x + 1) [] 2x g2 - (2x) (1) + 1 2] = (2x + 1 ) (4 x 2 - 2 x + 1 )

Chapter 2 Algebra and Surds

2. 27a 3 - 64b 3

Solution 27a 3 - 64b 3 = ] 3a g3 - ] 4b g3 = (3a - 4b) [] 3a g2 + (3a) (4b) + ] 4b g2] = (3a - 4b) (9a 2 + 12ab + 16b 2)

2.13

Exercises

Factorise 1.

b3 - 8

2.

x 3 + 27

3.

12.

x3 - 27 8

t3 + 1

13.

1000 1 + 3 3 a b

4.

a 3 - 64

14. ] x + 1 g3 - y 3

5.

1 - x3

15. 125x 3 y 3 + 216z 3

6.

8 + 27y 3

16. ]a - 2g3 - ]a + 1g3

7.

y 3 + 8z 3

8.

x 3 - 125y 3

9.

8x 3 + 27y 3

10. a 3 b 3 - 1 11. 1000 + 8t 3

17. 1 -

x3 27

18. y 3 + ]3 + xg3 19. ] x + 1 g3 + ^ y - 2 h3 20. 8]a + 3g3 - b 3

Mixed factors Sometimes more than one method of factorising is needed to completely factorise an expression.

EXAMPLE Factorise 5x 2 - 45.

Solution 5x 2 - 45 = 5 (x 2 - 9) = 5 (x + 3) (x - 3)

(using simple factors) (the difference of two squares)

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Maths In Focus Mathematics Preliminary Course

2.14

Exercises

Factorise 1.

2x 2 - 18

16. x 3 - 3x 2 - 10x

2.

3p 2 - 3p - 36

17. x 3 - 3x 2 - 9x + 27

3.

5y 3 - 5

18. 4x 2 y 3 - y

4.

4a 3 b + 8a 2 b 2 - 4ab 2 - 2a 2 b

19. 24 - 3b 3

5.

5a 2 - 10a + 5

20. 18x 2 + 33x - 30

6.

- 2x 2 + 11x - 12

21. 3x 2 - 6x + 3

7.

3z 3 + 27z 2 + 60z

22. x 3 + 2x 2 - 25x - 50

8.

9ab - 4a 3 b 3

23. z 3 + 6z 2 + 9z

9.

x3 - x

24. 4x 4 - 13x 2 + 9

10. 6x 2 + 8x - 8

25. 2x 5 + 2x 2 y 3 - 8x 3 - 8y 3

11. 3m - 15 - 5n + mn

26. 4a 3 - 36a

12. ] x - 3 g2 - ] x + 4 g2

27. 40x - 5x 4

13. y 2 ^ y + 5 h - 16 ^ y + 5 h

28. a 4 - 13a 2 + 36

14. x 4 - x 3 + 8x - 8

29. 4k 3 + 40k 2 + 100k

15. x 6 - 1

30. 3x 3 + 9x 2 - 3x - 9

DID YOU KNOW? Long division can be used to find factors of an expression. For example, x - 1 is a factor of x 3 + 4x - 5. We can find the other factor by dividing x 3 + 4x - 5 by x - 1. x2 + x + 5 x - 1 x3 + 4x - 5

g

x3

-

x2 x 2 + 4x x2

-

x 5x - 5 5x - 5

0 So the other factor of x 3 + 4x - 5 is x 2 + x + 5 ` x 3 + 4x - 5 = (x - 1) (x 2 + x + 5)

Chapter 2 Algebra and Surds

69

Completing the Square Factorising a perfect square uses the results a 2 ! 2ab + b 2 = ] a ! b g2

EXAMPLES 1. Complete the square on x 2 + 6x.

Solution Using a 2 + 2ab + b 2: a=x 2ab = 6x Substituting a = x: 2xb = 6x b=3

Notice that 3 is half of 6.

To complete the square: a 2 + 2ab + b 2 = ] a + b g2 2 x + 2x ] 3 g + 3 2 = ] x + 3 g2 x 2 + 6x + 9 = ] x + 3 g2 2. Complete the square on n 2 - 10n.

Solution Using a 2 - 2ab + b 2: a=n 2ab = 10x Substituting a = n: 2nb = 10n b=5

Notice that 5 is half of 10.

To complete the square: a 2 - 2ab + b 2 = ] a - b g2 n 2 - 2n ] 5 g + 5 2 = ] n - 5 g2 n 2 - 10n + 25 = ] n - 5 g2

To complete the square on a 2 + pa, divide p by 2 and square it. p 2 p 2 a 2 + pa + d n = d a + n 2 2

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Maths In Focus Mathematics Preliminary Course

EXAMPLES 1. Complete the square on x 2 + 12x.

Solution Divide 12 by 2 and square it: x 2 + 12x + c

12 2 m = x 2 + 12x + 6 2 2 = x 2 + 12x + 36 = ]x + 6g2

2. Complete the square on y 2 - 2y.

Solution Divide 2 by 2 and square it: 2 2 y 2 - 2y + c m = y 2 - 2 y + 1 2 2 = y 2 - 2y + 1 = ^ y - 1 h2

2.15

Exercises

Complete the square on 1.

x 2 + 4x

12. y 2 + 3y

2.

b 2 - 6b

13. x 2 - 7x

3.

x 2 - 10x

14. a 2 + a

4.

y 2 + 8y

15. x 2 + 9x

5.

m 2 - 14m

16. y 2 -

6.

q 2 + 18q

5y 2

7.

x 2 + 2x

17. k 2 -

11k 2

8.

t 2 - 16t

18. x 2 + 6xy

9.

x 2 - 20x

19. a 2 - 4ab

10. w 2 + 44w 11. x 2 - 32x

20. p 2 - 8pq

Chapter 2 Algebra and Surds

71

Algebraic Fractions Simplifying fractions EXAMPLES Simplify 4x + 2 2

1.

Solution 2 ] 2x + 1 g 4x + 2 = 2 2 = 2x + 1

Factorise first, then cancel.

2x 2 - 3x - 2 x3 - 8

2.

Solution ] 2x + 1 g ] x - 2 g 2x 2 - 3x - 2 = 3 ] x - 2 g ^ x 2 + 2x + 4 h x -8 2x + 1 = 2 x + 2x + 4

2.16

Exercises

Simplify 1.

5a + 10 5

2. 3. 4.

9.

b3 - 1 b2 - 1

6t - 3 3

10.

8y + 2 6

2p 2 + 7p - 15 6p - 9

11.

a2 - 1 a + 2a - 3

8 4d - 2 2

5.

6.

x 5x 2 - 2x y-4

12.

13.

y - 8y + 16

2

3 ]x - 2g + y ]x - 2g x3 - 8 x 3 + 3x 2 - 9x - 27 x 2 + 6x + 9

2

7.

2ab - 4a 2 a 2 - 3a

8.

s2 + s - 2 s 2 + 5s + 6

14.

15.

2p 2 - 3p - 2 8p 3 + 1 ay - ax + by - bx 2ay - by - 2ax + bx

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Maths In Focus Mathematics Preliminary Course

Operations with algebraic fractions

EXAMPLES Simplify 1.

x+3 x-1 5 4

Solution Do algebraic fractions the same way as ordinary fractions.

4 ]x - 1 g - 5 ]x + 3 g x -1 x +3 = 5 4 20 4x - 4 - 5x - 15 = 20 - x - 19 = 20

2.

2a 2 b + 10ab a 2 - 25 ' 3 4b + 12 b + 27

Solution 2a 2 b + 10ab a 2 - 25 2a 2 b + 10ab 4b + 12 ' = # 2 4b + 12 b 3 + 27 b 3 + 27 a - 25 2ab ] a + 5 g 4 ]b + 3 g = # 2 ] a + 5 g]a - 5 g ] b + 3 g ^ b - 3b + 9 h 8ab = ] a - 5 g ^ b 2 - 3b + 9 h

3.

2 1 + x-5 x+2

Solution 2 ]x + 2g + 1 ]x - 5g 2 1 + = x-5 x+2 ]x - 5g]x + 2g 2x + 4 + x - 5 = ]x - 5g]x + 2g 3x - 1 = ]x - 5g]x + 2g

Chapter 2 Algebra and Surds

2.17 1.

2.

Exercises

Simplify x 3x (a) + 4 2 y + 1 2y (b) + 5 3 a+2 a (c) 4 3 p-3 p+2 (d) + 6 2 x-5 x-1 (e) 2 3 4.

Simplify 3 b 2 + 2b # (a) b + 2 6a - 3

1 1 + x+1 x-3

(g)

3 2 x 2 + x -4

(h)

1 1 + a 2 + 2a + 1 a + 1

(i)

5 2 1 + y+2 y+3 y-1

(j)

2 7 x 2 - 16 x 2 - x - 12

2

Simplify (a)

y2 - 9 3x 2 x 2 - 2x - 8 # # 4y - 12 6x - 24 y 3 + 27

q3 + 1 (b) 2 # q + 2q + 1 p + 2

(b)

2 a 2 - 5a 3a - 15 y - y - 2 ' # 5ay y 2 - 4y + 4 y2 - 4

3ab 2 12ab - 6a (c) ' 2 5xy x y + 2xy 2

(c)

3 x 2 + 3x 2x + 8 + 2 # x-3 4x - 16 x -9

(d)

5b b2 b ' 2 2b + 6 b 1 + b +b-6

(e)

x 2 - 8x + 15 x 2 - 9 x 2 + 5x + 6 ' # 2 2x - 10 5x + 10x 10x 2

p2 - 4

(d)

ax - ay + bx - by x2 - y2

#

x3 + y3 ab 2 + a 2 b

x 2 - 6x + 9 x 2 - 5x + 6 (e) ' x 2 - 25 x 2 + 4x - 5 3.

(f)

5.

Simplify 2 3 (a) x + x

Simplify (a)

1 2 4 + x 2 - 7x + 10 x 2 - 2x - 15 x 2 + x - 6

1 2 x-1 x

(b)

3 5 2 + 2 2 x x x -4

(c) 1 +

3 a+b

(c)

3 2 + p 2 + pq pq - q 2

(d) x -

x2 x+2

(d)

a b 1 + a + b a - b a2 - b2

(b)

(e) p - q +

1 p+q

2

x+y y x (e) x - y + y - x - 2 y - x2

Substitution Algebra is used in writing general formulae or rules. For example, the formula A = lb is used to find the area of a rectangle with length l and breadth b. We can substitute any values for l and b to find the area of different rectangles.

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EXAMPLES 1. P = 2l + 2b is the formula for finding the perimeter of a rectangle with length l and breadth b. Find P when l = 1.3 and b = 3.2.

Solution P = 2 l + 2b = 2 ] 1 . 3 g + 2 ] 3 .2 g = 2 .6 + 6 . 4 =9 2. V = rr 2 h is the formula for finding the volume of a cylinder with radius r and height h. Find V (correct to 1 decimal place) when r = 2.1 and h = 8.7.

Solution V = rr 2 h = r ] 2.1 g2 (8.7) = 120.5 correct to 1 decimal place

9C + 32 is the formula for changing degrees Celsius ] °C g into 5 degrees Fahrenheit ] °F g find F when C = 25. 3. If F =

Solution 9C + 32 5 9 ] 25 g = + 32 5 225 = + 32 5 225 + 160 = 5 385 = 5 = 77 This means that 25°C is the same as 77°F. F=

Chapter 2 Algebra and Surds

2.18 1.

Exercises

Given a = 3.1 and b = - 2.3 find, correct to 1 decimal place. (a) ab (b) 3b (c) 5a 2 (d) ab 3 (e) ]a + bg2 (f)

a-b

(g) - b 2 2.

T = a + ] n - 1 g d is the formula for finding the term of an arithmetic series. Find T when a = - 4, n = 18 and d = 3.

3.

Given y = mx + b, the equation of a straight line, find y if m = 3, x = - 2 and b = - 1.

4.

If h = 100t - 5t 2 is the height of a particle at time t, find h when t = 5.

5.

Given vertical velocity v = - gt, find v when g = 9.8 and t = 20.

6.

If y = 2 x + 3 is the equation of a function, find y when x = 1.3, correct to 1 decimal place.

7.

S = 2r r ] r + h g is the formula for the surface area of a cylinder. Find S when r = 5 and h = 7, correct to the nearest whole number.

8.

A = rr 2 is the area of a circle with radius r. Find A when r = 9.5, correct to 3 significant figures.

9.

n-1

Given u n = ar is the nth term of a geometric series, find u n if a = 5, r = - 2 and n = 4.

10. Given V = 1 lbh is the volume 3 formula for a rectangular pyramid, find V if l = 4.7, b = 5.1 and h = 6.5. 11. The gradient of a straight line is y2 - y1 given by m = x - x . Find m 2 1 if x 1 = 3, x 2 = -1, y 1 = - 2 and y 2 = 5. 12. If A = 1 h ] a + b g gives the area 2 of a trapezium, find A when h = 7, a = 2.5 and b = 3.9. 13. Find V if V = 4 rr 3 is the volume 3 formula for a sphere with radius r and r = 7.6, to 1 decimal place.

14. The velocity of an object at a certain time t is given by the formula v = u + at. Find v when u = 1 , a = 3 and t = 5 . 4 5 6 a 15. Given S = , find S if a = 5 1-r and r = 2 . S is the sum to infinity 3 of a geometric series. 16. c = a 2 + b 2 , according to Pythagoras’ theorem. Find the value of c if a = 6 and b = 8. 17. Given y = 16 - x 2 is the equation of a semicircle, find the exact value of y when x = 2.

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18. Find the value of E in the energy equation E = mc 2 if m = 8.3 and c = 1.7. 19. A = P c 1 +

20. If S =

a geometric series, find S if a = 3, r = 2 and n = 5.

r n m is the formula 100

for finding compound interest. Find A when P = 200, r = 12 and n = 5, correct to 2 decimal places.

a ^rn - 1h is the sum of r -1

21. Find the value of

a3 b2 if c2

2 3 1 4 a = c 3 m , b = c 2 m and c = c m . 4 3 2

Surds An irrational number is a number that cannot be written as a ratio or fraction (rational). Surds are special types of irrational numbers, such as 2, 3 and 5 . Some surds give rational values: for example, 9 = 3. Others, like 2 , do not have an exact decimal value. If a question involving surds asks for an exact answer, then leave it as a surd rather than giving a decimal approximation.

Simplifying surds

Class Investigations 1. Is there an exact decimal equivalent for 2 ? 2. Can you draw a line of length exactly 2 ? 3. Do these calculations give the same results? (a) 9 # 4 and 9 # 4 (b)

4

and

4 9

(c)

9 9 + 4 and

9 +

4

(d)

9 - 4 and

9 -

4

Here are some basic properties of surds.

a# b =

ab

a' b =

a

^ x h2 =

b

=

x2 = x

a b

Chapter 2 Algebra and Surds

77

EXAMPLES 1. Express in simplest surd form

45 .

45 also equals 3 # 15 but this will not simplify. We look for a number that is a perfect square.

Solution 45 = 9 # 5 = 9 # 5 =3# 5 =3 5 2. Simplify 3 40 .

Solution

Find a factor of 40 that is a perfect square.

3 40 = 3 4 # 10 = 3 # 4 # 10 = 3 # 2 # 10 = 6 10 3. Write 5 2 as a single surd.

Solution 5 2 = =

2.19 1.

25 # 2 50

Exercises

Express these surds in simplest surd form.

(k)

112

(l)

300

(a)

12

(b)

63

(c)

24

(d)

50

(e)

72

(f)

200

(g)

48

(h)

75

(i)

32

(a) 2 27

(j)

54

(b) 5 80

(m) 128

2.

(n)

243

(o)

245

(p)

108

(q)

99

(r)

125

Simplify

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Maths In Focus Mathematics Preliminary Course

(c) 4 98

(g) 3 13

(d) 2 28

(h) 7 2

(e) 8 20

(i) 11 3

(f) 4 56

(j) 12 7

(g) 8 405

4.

(h) 15 8

(a)

(i) 7 40

x =3 5

(b) 2 3 =

x

(c) 3 7 =

x

Write as a single surd.

(d) 5 2 =

x

(a) 3 2

(e) 2 11 =

(b) 2 5

(f)

(c) 4 11

(g) 4 19 =

(d) 8 2

(h)

(e) 5 3

(i) 5 31 =

(f) 4 10

(j)

(j) 8 45 3.

Evaluate x if

x

x =7 3 x

x = 6 23 x

x = 8 15

Addition and subtraction Calculations with surds are similar to calculations in algebra. We can only add or subtract ‘like terms’ with algebraic expressions. This is the same with surds.

EXAMPLES 1. Simplify 3 2 + 4 2 .

Solution 3 2+4 2 =7 2 2. Simplify

3 - 12 .

Solution First, change into ‘like’ surds. 3 - 12 = 3 - 4 # 3 = 3 -2 3 =- 3 3. Simplify 2 2 - 2 + 3 .

Solution 2 2- 2+ 3=

2+ 3

Chapter 2 Algebra and Surds

2.20

79

Exercises

Simplify 1.

5 +2 5

14.

50 -

32

2.

3 2 -2 2

15.

28 +

63

3.

3 +5 3

16. 2 8 -

18

4.

7 3 -4 3

5.

5 -4 5 4 6 -

6.

17. 3 54 + 2 24 18.

90 - 5 40 - 2 10

19. 4 48 + 3 147 + 5 12

6

7.

2 -8 2

20. 3 2 + 8 - 12

8.

5 +4 5 +3 5

21.

63 - 28 - 50

9.

2 -2 2 -3 2

22.

12 - 45 - 48 - 5

10.

5 +

45

23.

150 + 45 + 24

11.

8 -

2

24.

32 - 243 - 50 + 147

12.

3 +

48

25.

80 - 3 245 + 2 50

13.

12 -

27

Multiplication and division To get a b # c d = ac bd , multiply surds with surds and rationals with rationals.

a # b = ab a b # c d = ac bd a# a =

a b

=

a2 = a

a b

EXAMPLES Simplify 1. 2 2 #- 5 7

Solution 2 2 #- 5 7 = -10 14

CONTINUED

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Maths In Focus Mathematics Preliminary Course

2. 4 2 # 5 18

Solution 4 2 # 5 18 = 20 36 = 20 # 6 = 120

3.

2 14 4 2

Solution 2 14 4 2

=

2 2 # 7 2

=

4.

7

4 2

3 10 15 2

Solution 3 10 15 2

=

3# 5 # 2 15 2

5 = 5

5. d

2

10 n 3

Solution 2 ^ 10 h 10 n = 3 ^ 3 h2 10 = 3 =31 3

2

d

Chapter 2 Algebra and Surds

2.21

Exercises

Simplify 1.

7 #

2.

3# 5

3.

2 #3 3

4.

5 7 #2 2

3

5.

-3 3 #2 2

6.

5 3 #2 3

7.

- 4 5 # 3 11

8.

2 7# 7

9.

2 3 # 5 12

10.

6# 2

11.

8 #2 6

23.

24.

25.

26.

27.

28.

5 8 10 2 16 2 2 12 10 30 5 10 2 2 6 20 4 2 8 10 3 3 15 2

29.

8

12. 3 2 # 5 14 13.

10 # 2 2

14. 2 6 #-7 6 15. ^ 2 h

2

2 16. ^ 2 7 h

17.

31.

32.

3 15 6 10 5 12 5 8 15 18 10 10

3# 5# 2

18. 2 3 # 7 #- 5 19.

30.

2 # 6 #3 3

33.

15 2 6 2n 3

35. d

5n 7

20. 2 5 # - 3 2 # - 5 5 21.

22.

4 12 2 2

2

34. d

2

12 18 3 6

Expanding brackets The same rules for expanding brackets and binomial products that you use in algebra also apply to surds.

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Simplifying surds by removing grouping symbols uses these general rules.

a^ b + ch=

ab + ac

Proof a^ b + ch = =

a# b + ab + ac

a# c

Binomial product:

^ a + b h^ c + d h =

ac +

ad +

bc +

bd

Proof ^ a + b h^ c + d h = a # c + a # d + b # c + b # d = ac + ad + bc + bd Perfect squares:

^ a + b h2 = a + 2 ab + b

Proof ^ a + b h2 = ^ a + b h ^ a + b h = a 2 + ab + ab + b 2 = a + 2 ab + b ^ a - b h2 = a - 2 ab + b

Proof ^ a - b h2 = ^ a - b h ^ a - b h = a 2 - ab - ab + b 2 = a - 2 ab + b Difference of two squares:

^ a + b h^ a - b h = a - b

Proof ^ a + b h ^ a - b h = a 2 - ab + ab - b 2 =a-b

Chapter 2 Algebra and Surds

83

EXAMPLES Expand and simplify 1. 2 ^ 5 + 2 h

Solution 2( 5 +

2) = = =

2# 5 + 10 + 4 10 + 2

2# 2

2. 3 7 ^ 2 3 - 3 2 h

Solution 3 7 (2 3 - 3 2 ) = 3 7 # 2 3 - 3 7 # 3 2 = 6 21 - 9 14 3. ^ 2 + 3 5 h ^ 3 -

2h

Solution ( 2 + 3 5)( 3 -

2) = =

2# 3 - 2# 2 +3 5# 3 -3 5# 2 6 - 2 + 3 15 - 3 10

4. ^ 5 + 2 3 h ^ 5 - 2 3 h

Solution ( 5 + 2 3 ) ( 5 - 2 3 ) = 5 # 5 - 5 #2 3 + 2 3 # 5 - 2 3 #2 3 = 5 - 2 15 + 2 15 - 4#3 = 5 - 12 = -7 Another way to do this question is by using the difference of two squares. 2 2 ( 5 + 2 3)( 5 - 2 3) = ^ 5 h - ^2 3 h = 5 - 4#3 = -7

Notice that using the difference of two squares gives a rational answer.

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Maths In Focus Mathematics Preliminary Course

2.22 1.

Exercises (m)^ 2 11 + 5 2 h^ 2 11 - 5 2 h

Expand and simplify (a)

2^ 5 + 3h

(b)

3 ^2 2 - 5 h

(n) ^ 5 + 2 h

2

2 (o) ^ 2 2 - 3 h

(c) 4 3 ^ 3 + 2 5 h (d)

2 (p) ^ 3 2 + 7 h

7 ^5 2 - 2 3 h

2 (q) ^ 2 3 + 3 5 h

(e) - 3 ^ 2 - 4 6 h (f)

2 (r) ^ 7 - 2 5 h

3 ^ 5 11 + 3 7 h

2 (s) ^ 2 8 - 3 5 h

(g) - 3 2 ^ 2 + 4 3 h (h)

5^ 5 - 5 3h

(i)

3 ^ 12 + 10 h

2 (t) ^ 3 5 + 2 2 h

3.

If a = 3 2 , simplify (a) a2 (b) 2a3 (c) (2a)3 (d) ]a + 1g2 (e) ] a + 3 g ] a – 3 g

4.

Evaluate a and b if 2 (a) ^ 2 5 + 1h = a + b

(j) 2 3 ^ 18 + 3 h (k) - 4 2 ^ 2 - 3 6 h (l) - 7 5 ^ - 3 20 + 2 3 h (m) 10 3 ^ 2 - 2 12 h (n) - 2 ^ 5 + 2 h (o) 2 3 ^ 2 - 12 h 2.

(b) ^ 2 2 - 5 h ^ 2 - 3 5 h = a + b 10

Expand and simplify (a) ^ 2 + 3h^ 5 + 3 3 h

5.

Expand and simplify (a) ^ a + 3 - 2 h ^ a + 3 + 2 h 2 (b) _ p - 1 - p i

6.

Evaluate k if ^ 2 7 - 3 h ^ 2 7 + 3 h = k.

(g) ^ 7 + 3 h^ 7 - 3 h

7.

Simplify _ 2 x + y i _ x - 3 y i .

(h) ^ 2 - 3 h^ 2 + 3 h

8.

If ^ 2 3 - 5 h = a - b , evaluate a and b.

9.

Evaluate a and b if ^ 7 2 - 3 h2 = a + b 2 .

(b) ^ 5 - 2 h^ 2 - 7 h (c) ^ 2 + 5 3 h^ 2 5 - 3 2 h (d) ^ 3 10 - 2 5 h^ 4 2 + 6 6 h (e) ^ 2 5 - 7 2 h^ 5 - 3 2 h (f) ^ 5 + 6 2 h^ 3 5 - 3 h

(i) ^ 6 + 3 2 h^ 6 - 3 2 h (j) ^ 3 5 + 2 h^ 3 5 - 2 h (k) ^ 8 - 5 h^ 8 + 5 h (l) ^ 2 + 9 3 h^ 2 - 9 3 h

2

10. A rectangle has sides 5 + 1 and 2 5 - 1. Find its exact area.

Rationalising the denominator Rationalising the denominator of a fractional surd means writing it with a rational number (not a surd) in the denominator. For example, after 3 5 3 rationalising the denominator, becomes . 5 5

Chapter 2 Algebra and Surds

85

DID YOU KNOW? A major reason for rationalising the denominator used to be to make it easier to evaluate the fraction (before calculators were available). It is easier to divide by a rational number than an irrational one; for example, 3 = 3 ' 2.236 5 3

5 5

This is hard to do without a calculator.

This is easier to calculate.

= 3 # 2.236 ' 5

Squaring a surd in the denominator will rationalise it since ^ x h = x. 2

Multiplying by

b a b a # = b b b

b

b is the same as multiplying by 1.

Proof b a b a # = b b b2 a b = b

EXAMPLES 1. Rationalise the denominator of

Solution

3 . 5

5 3 5 3 # = 5 5 5 2. Rationalise the denominator of

Solution

2 5 3

. Don’t multiply by 5

2 5 3

#

3 3

=

2 3

5 9 2 3 = 5# 3 2 3 = 15

3

as it takes 5 3 longer to simplify.

86

Maths In Focus Mathematics Preliminary Course

When there is a binomial denominator, we use the difference of two squares to rationalise it, as the result is always a rational number.

To rationalise the denominator of

a+ b c+ d

, multiply by

Proof a+ b c+ d

^ a + b h^ c - d h c- d ^ c + d h^ c - d h ^ a + b h^ c - d h = ^ c h2 - ^ d h2 ^ a + b h^ c - d h = c-d c- d

#

=

EXAMPLES 1. Write with a rational denominator 5 2 -3 Multiply by the conjugate surd 2 + 3.

.

Solution 5 2 -3

2 +3

#

2 +3

=

5 ^ 2 + 3h

^ 2 h2 - 3 2 10 + 3 5 = 2-9 10 + 3 5 = -7 10 + 3 5 =7

2. Write with a rational denominator 2 3+ 5 3+4 2

.

Solution 2 3 +

5

3 +4 2

#

3 -4 2 3 -4 2

=

^2 3 + 5 h^ 3 - 4 2 h

^ 3 h2 - ^ 4 2 h2 2 # 3 - 8 6 + 15 - 4 10 = 3 - 16 # 2

c- d c- d

Chapter 2 Algebra and Surds

6 - 8 6 + 15 - 4 10 - 29 - 6 + 8 6 - 15 + 4 10 = 29 =

3. Evaluate a and b if

3 3 3- 2

= a + b.

Solution 3 3 3- 2

#

3+ 2 3+ 2

=

3 3^ 3 + 2h

^ 3 - 2 h^ 3 + 2 h 3 9+3 6 = ^ 3 h2 - ^ 2 h2 3#3+3 6 3-2 9+3 6 = 1 =9+3 6 =

=9+ 9# 6 = 9 + 54 So a = 9 and b = 54. 4. Evaluate as a fraction with rational denominator 2 + 3+2

5 3-2

.

Solution 2 + 3+2

5 3 -2

=

2^ 3 - 2h + 5 ^ 3 + 2h

^ 3 + 2h ^ 3 - 2h 2 3 - 4 + 15 + 2 5 = ^ 3 h2 - 2 2 2 3 - 4 + 15 + 2 5 3-4 2 3 - 4 + 15 + 2 5 = -1 = - 2 3 + 4 - 15 - 2 5 =

87

88

Maths In Focus Mathematics Preliminary Course

2.23 1.

Express with rational denominator (a) (b) (c) (d) (e)

2.

Exercises 3.

1 7

(a)

3

(b)

2 2 2 3

(c)

5 6 7

(d)

5 2 1+

2 3

6 -5

(g)

5 +2 2

8+3 2

(j)

4 3 -2 2

(f)

1 5 +

2

2 -

7

2 +

3

2 +3

4 5 (j)

7 5 (k)

4 3 +

(l) 2

3

4.

2 -7 5 +2 6 3 -4 3 +4 3 3 3 +

(b) (c)

2 +5 2 2

2 5 +3 2

3 2 +

+

3 3 2 -

#

3

6 -

3

2 3 2 +3 5 6 +2 2 +7 4+

2 3 +

3 -2

3

6 +

1 3

+

2

-

2

2

3 -

2

(d) (e)

2 5 3 4 2

2

2 -1

+

+

5 -

3

5

2

3

3 5 3 2 4-

3

2+

3

3 +1

Find a and b if (a)

2 3

-

1 where z = 1 + z2

(h) (i)

1 2 -1

1 where t = t

3 2 +4

2 7

Express with rational denominator

(e)

3

(g)

5

(i)

(d)

2 -

2

3 2 -4

(c)

2

(f) z 2 -

(h)

(b)

1 + 2 +1

(e) t +

(f)

(a)

Express as a single fraction with rational denominator

=

a b

=

a 6 b

2 =a+b 5 5 +1 2 7 7 -4 2 +3 2 -1

=a+b 7 =a+

b

2 -

2 6 -1

Chapter 2 Algebra and Surds

5.

2 -1

Show that

2 +1

+

4 is 2

7.

If x =

(b) x 2 +

8.

1 x2

2

+

1 5 -

2

-

as a single fraction with 3 rational denominator.

3 + 2, simplify

1 (a) x + x

2 5 +

5 +1

rational. 6.

Write

Show that

8 2 + is 3+2 2 2

rational. 2

1 (c) b x + x l

9.

1 If 2 + x = 3 , where x ! 0, find x as a surd with rational denominator.

10. Rationalise the denominator of b +2 ]b ! 4 g b -2

89

90

Maths In Focus Mathematics Preliminary Course

Test Yourself 2 1.

2.

3.

4.

Simplify (a) 5y - 7y 3a + 12 (b) 3 (c) - 2k 3 # 3k 2 y x (d) + 5 3 (e) 4a - 3b - a - 5b (f) 8 + 32 (g) 3 5 - 20 + 45 Factorise (a) x 2 - 36 (b) a 2 + 2a - 3 (c) 4ab 2 - 8ab (d) 5y - 15 + xy - 3x (e) 4n - 2p + 6 (f) 8 - x 3 Expand and simplify (a) b + 3 ] b - 2 g (b) ] 2x - 1 g ] x + 3 g (c) 5 ] m + 3 g - ] m - 2 g (d) ]4x - 3g2 (e) ^ p - 5h^ p + 5h (f) 7 - 2 ] a + 4 g - 5a (g) 3 ^ 2 2 - 5 h (h) ^ 3 + 7 h^ 3 - 2h Simplify 4a - 12 10b (a) # 3 5b 3 a - 27 (b)

5.

5m + 10 m2 - 4 ' 2 m - m - 2 3m + 3

The volume of a cube is V = s 3. Evaluate V when s = 5.4.

6.

(a) Expand and simplify ^ 2 5 + 3 h ^ 2 5 - 3 h. (b) Rationalise the denominator of 3 3 . 2 5+ 3

7.

Simplify

8.

If a = 4, b = - 3 and c = - 2, ﬁnd the value of (a) ab 2 (b) a - bc (c) a (d) ]bcg3 (e) c ] 2a + 3b g

9.

Simplify 3 12 (a) 6 15 (b)

3 1 2 + - 2 . x-2 x+3 x +x-6

4 32 2 2

10. The formula for the distance an object falls is given by d = 5t 2 . Find d when t = 1.5. 11. Rationalise the denominator of 2 (a) 5 3 (b)

1+ 3 2

12. Expand and simplify (a) ^ 3 2 - 4h^ 3 - 2 h 2 (b) ^ 7 + 2h 13. Factorise fully (a) 3x 2 - 27 (b) 6x 2 - 12x - 18 (c) 5y 3 + 40

Chapter 2 Algebra and Surds

14. Simplify 3x 4 y (a) 9xy 5 (b)

5 15x - 5

15. Simplify 2 (a) ^ 3 11 h 3 (b) ^ 2 3 h 16. Expand and simplify (a) ] a + b g ] a - b g (b) ] a + b g 2 (c) ] a - b g 2 17. Factorise (a) a 2 - 2ab + b 2 (b) a 3 - b 3 1 18. If x = 3 + 1, simplify x + x and give your answer with a rational denominator. 19. Simplify 4 3 (a) a + b (b)

x-3 x-2 5 2

20. Simplify

2 3 , writing 5+2 2 2-1

your answer with a rational denominator. 21. Simplify (a) 3 8 (b) - 2 2 # 4 3 (c) 108 - 48 (d)

23. Rationalise the denominator of 3 (a) 7 (b)

2

5 3 2 (c) 5 -1 (d) (e)

2 2 3 2+ 3 5+ 2 4 5-3 3

24. Simplify 3x x-2 (a) 5 2 a+2 2a - 3 (b) + 7 3 1 2 (c) 2 1 x + x -1 4 1 (d) 2 + k + 2k - 3 k + 3 (e)

3 2+ 5

-

5 3- 2

25. Evaluate n if (a) 108 - 12 = (b)

112 + 7 =

n n

8 6

(c) 2 8 + 200 =

2 18

(d) 4 147 + 3 75 = n 180 (e) 2 245 + = n 2

(e) 5a # - 3b # - 2a (f)

22. Expand and simplify (a) 2 2 ^ 3 + 2 h (b) ^ 5 7 - 3 5 h^ 2 2 - 3 h (c) ^ 3 + 2 h^ 3 - 2 h (d) ^ 4 3 - 5 h^ 4 3 + 5 h 2 (e) ^ 3 7 - 2 h

2m 3 n 6m 2 n 5

(g) 3x - 2y - x - y

n

91

92

Maths In Focus Mathematics Preliminary Course

26. Evaluate x 2 +

1+2 3 1 if x = 2 x 1-2 3

27. Rationalise the denominator of

3

2 7 (there may be more than one answer). 21 (a) 28 2 21 (b) 28 21 (c) 14 21 (d) 7 x-3 x +1 . 5 4 -]x + 7 g 20 x+7 20 x + 17 20 - ] x + 17 g 20

28. Simplify (a) (b) (c) (d)

(a) (b) (c) (d)

32. Simplify 5ab - 2a 2 - 7ab - 3a 2 . (a) 2ab + a 2 (b) - 2ab - 5a 2 (c) - 13a 3 b (d) - 2ab + 5a 2 33. Simplify (a) (b) (c)

29. Factorise x 3 - 4x 2 - x + 4 (there may be more than one answer). (a) ^ x 2 - 1 h ] x - 4 g (b) ^ x 2 + 1 h ] x - 4 g (c) x 2 ] x - 4 g (d) ] x - 4 g ] x + 1 g ] x - 1 g 30. Simplify 3 2 + 2 98 . (a) 5 2 (b) 5 10 (c) 17 2 (d) 10 2

3 2 1 + . x-2 x+2 x2 - 4 x+5 ]x + 2g]x - 2g x+1 ]x + 2g]x - 2g x+9 ]x + 2g]x - 2g x-3 ]x + 2g]x - 2g

31. Simplify

(d)

80 . 27

4 5 3 3 4 5 9 3 8 5 9 3 8 5 3 3

34. Expand and simplify ^ 3x - 2y h2 . (a) 3x 2 - 12xy - 2y 2 (b) 9x 2 - 12xy - 4y 2 (c) 3x 2 - 6xy + 2y 2 (d) 9x 2 - 12xy + 4y 2 35. Complete the square on a 2 - 16a. (a) a 2 - 16a + 16 = ^ a - 4 h2 (b) a 2 - 16a + 64 = ^ a - 8 h2 (c) a 2 - 16a + 8 = ^ a - 4 h2 (d) a 2 - 16a + 4 = ^ a - 2 h2

Chapter 2 Algebra and Surds

Challenge Exercise 2 1.

2.

Expand and simplify (a) 4ab ] a - 2b g - 2a 2 ] b - 3a g (b) _ y 2 - 2 i_ y 2 + 2 i (c) ] 2x - 5 g3 Find the value of x + y with rational denominator if x = 3 + 1 and 1 y= . 2 5-3 2 3

2x + y x-y 3x + 2y . + - 2 x-3 x+3 x +x-6

12. (a) Expand ^ 2x - 1 h3. 6x 2 + 5x - 4 (b) Simplify . 8x 3 - 12x 2 + 6x - 1 13. Expand and simplify ] x - 1 g ^ x - 3 h2. 14. Simplify and express with rational 2 +

5

-

5 3

3.

Simplify

4.

b Complete the square on x 2 + a x.

15. Complete the square on x 2 + 2 x. 3

Factorise (a) (x + 4)2 + 5 (x + 4) (b) x 4 - x 2 y - 6y 2 (c) 125x 3 + 343 (d) a 2 b - 2a 2 - 4b + 8

16. If x =

5.

6. 7.

8.

9.

7 6 - 54

.

11. Simplify

denominator

Simplify

d=

4x 2 - 16x + 12

| ax 1 + by 1 + c |

.

Simplify

10. Factorise

^a + 1h a3 + 1

.

a2 4 - 2. 2 x b

.

lx 1 + kx 2

17. Find the exact value with rational 1 denominator of 2x 2 - 3x + x if x = 2 5 . 18. Find the exact value of 1+2 3 1 (a) x 2 + 2 if x = x 1-2 3 (b) a and b if

is the formula for

a2 + b2 the perpendicular distance from a point to a line. Find the exact value of d with a rational denominator if a = 2, b = -1, c = 3, x 1 = - 4 and y 1 = 5. 3

2 -1

, find the value of x when k+l k = 3, l = - 2, x 1 = 5 and x 2 = 4.

Complete the square on 4x 2 + 12x. 2xy + 2x - 6 - 6y

3 +4

3 -4 2+3 3

=a+b 3

19. A = 1 r 2 i is the area of a sector of a 2 circle. Find the value of i when A = 12 and r = 4. 20. If V = rr 2 h is the volume of a cylinder, find the exact value of r when V = 9 and h = 16. 21. If s = u + 1 at 2, find the exact value of s 2 when u = 2, a = 3 and t = 2 3 .

93

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