Maths in Focus - Margaret Grove - ch2
Short Description
Descripción: Mathematics Preliminary Course - 2nd Edition...
Description
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
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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
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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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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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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
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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 .
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