2 Algebra and Al Surds TERMINOLOGY
Binomial: A mathematical expression consisting of two terms such as x + 3 or 3 x - 1 Binomial product: The product of two binomial expressions such as ( x + 3) (2 (2 x - 4) Expression: A mathematical statement involving numbers, pronumerals and symbols e.g. 2 x - 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. 2 y - 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. 2 x , - 3 Trinomial: An expression with three terms such as 2 3 x - 2x + 1
Algeb ebra ra and Su Surds rds Chapter 2 Alg
45
INTRODUCTION THIS CHAPTER REVIEWS ALGEBRA skills, including simplifying expressions,
removing grouping symbols, factorising, completing the square and simplifying algebraic fractions. fractions. Operations with surds, surds, including rationalising the denominator denominator,, are also studied in this chapter. 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 �rst word in this title.
Simplifying Expressions Addition and subtraction
EXAMPLES Simplify 1. 7x - x
Solution
Here x is called a pronumeral.
7x - x = 7x - 1x = 6x 2. 4x 2
-
3x 2 + 6x 2
Solution 4x 2
-
3x 2 + 6x 2 = x 2 + 6x 2 2 = 7x
CONTINUED
Algeb ebra ra and Su Surds rds Chapter 2 Alg
45
INTRODUCTION THIS CHAPTER REVIEWS ALGEBRA skills, including simplifying expressions,
removing grouping symbols, factorising, completing the square and simplifying algebraic fractions. fractions. Operations with surds, surds, including rationalising the denominator denominator,, are also studied in this chapter. 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 �rst word in this title.
Simplifying Expressions Addition and subtraction
EXAMPLES Simplify 1. 7x - x
Solution
Here x is called a pronumeral.
7x - x = 7x - 1x = 6x 2. 4x 2
-
3x 2 + 6x 2
Solution 4x 2
-
3x 2 + 6x 2 = x 2 + 6x 2 2 = 7x
CONTINUED
46
Maths In Focus Mathematics Extension 1 Preliminary Course
3. x 3 - 3x - 5x + 4 Only add or subtract ‘like’ terms. These have the same pronumeral pronumeral (for example, 3 x and 5 x ).
Solution x 3 - 3x - 5x + 4
=
x3
-
8x + 4
4. 3a - 4b - 5a - b
Solution 3a - 4b - 5a - b = 3a - 5a - 4b - b = - 2a - 5b
2.1 Exercises 2.1 Simplify 1.
2x + 5x
16. 7b + b - 3b
2.
9a - 6 a
17. 3b - 5b + 4b + 9b
3.
5z - 4z
18.
4.
5a + a
19. 6x - 5y
5.
4b - b
20. 8a + b - 4b - 7a
6.
2r
21. xy + 2y + 3xy
7.
-
4 y
8.
-
2x - 3x
9.
2a - 2a
10.
-
-
5r +
-
5x + 3x - x - 7x -
y
22. 2ab 2 - 5ab 2 - 3ab 2
3y
23. m 2
-
5m - m + 12
24. p 2 - 7p + 5p - 6
4k + 7 k
25. 3x + 7y + 5x - 4y
11. 3t + 4t + 2t
26. ab + 2b - 3ab + 8b
12. 8w - w
27. ab + bc
+
3w
-
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 xy 2 + 2y 3
15. 8h - h - 7h
30. 3x 3 - 4x 2 - 3x + 5x 2 - 4x - 6
Algeb ebra ra and Su Surds rds Chapter 2 Alg
47
Multiplication EXAMPLES Simplify 1.
-
5x # 3y
#
2x
Solution -
5x # 3y
2.
-
#
2x
= -
30xyx
= -
30x 2 y
3x 3 y 2 # - 4xy 5
Solution -
3x 3 y 2
# -
Use index laws to simplify this question.
4xy 5 = 12x 4 y 7
2.2 Exercises Simplify
^ h
1.
5 # 2b
11. 2x 2
5
2.
2x # 4y
12. 2ab 3
#
3.
5 p # 2p
13. 5a 2 b # - 2ab
4.
-
3z
5.
-
5a # - 3 b
6.
x # 2y
7.
8ab # 6c
17. k 3 p # p 2
8.
4d
18.
9.
3a # 4 a # a
19. 7m 6 # - 2m 5
10.
^ 3 y h
20.
-
#
2w
#
#
7z
3d
3
14. 7 pq 2
#
3a
3p 2 q 2
15. 5ab # a 2 b 2 16. 4h 3
# -
2h 7
^ 3t h -
-
3 4
2x 2
#
3x 3 y
# -
4xy 2
48
Maths In Focus Mathematics Extension 1 Preliminary Course
Division Use cancelling or index laws to simplify divisions.
EXAMPLES Simplify 1. 6v 2 y ' 2vy
Solution By cancelling, 2
6v y '
6v 2 y 2vy = 2vy 6 3 # v # v 1 # y 1 =
2 1 # v # y 1
= 3v Using index laws, 6v 2 y ' 2vy = 3v 2 1y 1 1 1 0 = 3v y = 3v -
-
5a 3 b 2. 15ab 2
Solution 5a 3 b 15ab 2
=
=
=
1 a3 3
-
1 a2 b 3 a2 3b
1
b1
-
2
1
-
2.3 Exercises Simplify 1.
30x ' 5
2.
2 y ' y
3. 4. 5.
8a 2 2 8a 2 a 8a 2 2a
6.
xy 2x
7.
12 p 3 ' 4p 2
8.
3a 2 b 2 6ab
9.
20x 15xy
10.
9x 7 3x 4
-
Chapter 2 Algebra and Surds
11.
15ab ' - 5b
-
16.
2ab 12. 6a 2 b 3
7 pq 3
17. 5a 9 b 4 c 2 a -
14. 14cd 2 ' 21c 3 d 3
2
'
2
4x y z
-
9
5 2
4
2
-
20a 5 b 3 c 1 -
-
1
19.
-
5x 4 y 7 z ' 15xy 8 z
20.
-
3 9 ^ a 4 b 1 h ' -18a 1 b 3
2xy 2 z 3 3
-
^ hb 18. 4a ^ b h
8 p 13. 4 pqs -
15.
42 p 5 q 4
-
-
2
-
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
49
50
Maths In Focus Mathematics Extension 1 Preliminary Course
2.
-
] 2x 5 g -
Solution -
(2x - 5)
1 (2x - 5)
= -
1 # 2x - 1 # - 5 = - 2x + 5 = -
]
g
3. 5a 2 4 + 3ab - c
Solution 5a 2 (4 + 3ab - c) = 5a 2 # 4 + 5a 2 # 3ab - 5a 2 # c 2 3 2 = 20a + 15a b - 5a c
^
4. 5 - 2 y + 3
h
Solution 5 - 2 ( y
+
3) = 5 - 2 # y - 2 # 3 = 5 - 2 y - 6 = - 2 y - 1
]
5. 2 b - 5
g ]b 1g -
+
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.
] g 3 ] 2h 3 g 2 x-4 +
]
g 3h
5 a-2
3.
-
4.
x 2y +
5.
x x-2
6.
2a 3 a - 8 b
^
]
]
g
]
7.
ab 2a + b
8.
5n n - 4
]
g
_ i 10. 3 4 ] k 1 g 11. 2 ] t 7 g 3 9.
3x 2 y xy + 2y 2 +
+
-
g
g
^
12. y 4y
+
-
3
h
+
8y
Chapter 2 Algebra and Surds
]
g 14. 3 ] 2x 5 g 15. 5] 3 2m g 7 ] m 2 g 16. 2 ] h 4 g 3 ] 2h 9 g
]
13. 9 - 5 b + 3 -
20. 2ab 3 - a
-
+
+
-
+
]
] g ] 5d 3 g 3a 4 h 18. a ] 2a 1 g ^ a 17. 3 2d - 3
-
+
2
]
19. x 3x - 4
+
-
-
+
3
]
+
-
g ] t 1 g -
25. 4 + 3 a + 5
-
g 5 ]x 1 g -
g
-
24. 2 3t - 4
-
-
-
^ 1 h y 23. ] a b g ] a b g 22. 8 - 4 2 y
-
+
]
g b ] 4a 1 g
21. 5x - x - 2
-
51
+
+
3
g ]a 7 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 x - 2 . Each term in the �rst bracket is multiplied by each term in the second bracket.
]
g]
g
]a bg^x y h +
+
=
ax + ay + bx + by
Proof
] a b g ] c d g a ] c d g b ] c d g +
+
=
=
+
ac
+
+
+
ad + bc + bd
EXAMPLES Expand and simplify
^
h^
1. p + 3 q - 4
h
Solution
^ p 3 h ^ q 4 h +
-
]
2. a + 5
=
pq - 4p + 3q - 12
g
2
Solution
]a 5 g +
2
(a + 5)(a + 5) 2 = a + 5a + 5a + 25 2 = a + 10a + 25
=
Can you see a quick way of doing this?
52
Maths In Focus Mathematics Extension 1 Preliminary Course
The rule below is not a binomial product (one expression is a trinomial), but it works the same way.
]a bg^x +
+
y+z
h
=
ax
+
ay + az
+
bx + by
+
bz
EXAMPLE
]
Expand and simplify x + 4
g ^ 2x
-
h
3y - 1 .
Solution (x + 4) (2x - 3 y
-
1) = 2x 2 - 3xy - x + 8x - 12y - 4 2 = 2x - 3xy + 7x - 12y 4
2.5 Exercises Expand and simplify 1. 2. 3. 4. 5. 6. 7. 8. 9.
]a 5g]a 2g ]x 3 g]x 1 g +
]
g] g 18. ^ 3x 4y h^ 3x 4y h 19. ]x 3g]x 3g 20. ^ y 6h^ y 6h 21. ] 3a 1 g ] 3a 1 g 17. a + 2b a - 2b
+
+
-
-
^2 y 3h^y 5h -
+
]m 4g]m 2g ]x 4g]x 3g -
-
+
+
+
-
-
]x 5g]x 5g +
-
] g ] 3a 1 g 11. ^2 y 3h^ 4y 3h 12. ]x 4g^ y 7h 10. 5a - 4
-
+
-
-
+
^ h] g 14. ]n 2g]n 2g 15. ]2x 3g]2x 3g 13. x 2 + 3 x - 2 +
-
+
^
-
h^
-
-
+
] 23. ]x 24. ] b 25. ]x 26. ]a 27. ]a 28. ]k 29. ]x
-
g]
22. 2z - 7 2z + 7
-
-
+
+
^ y 2h^y 5h ]2x 3g]x 2g ]h 7g]h 3g +
+
h
16. 4 - 7 y 4 + 7y
+ + -
g^ h 3 g ] 2a 2 b 1 g 2g^x 2x 4h 3g^a 3a 9h 9 x - 2y + 2 +
2
2
+
9
-
4
+
g
g
2
g 2g
^
30. y - 7
2
2
h
2
] g 32. ]2t 1g 31. 2x + 3 -
2
2
-
-
+
+
+
Chapter 2 Algebra and Surds
]
33. 3a + 4b
^
] g 38. ] a b g
g
2
37. a + b
h bg
34. x - 5y 2
]
35. 2a +
-
2
2
39. ] a + b g ^ a 2 - ab + b 2 h
2
40. ] a - b g ^ a 2 + ab + b 2 h
36. ] a - b g ] a + b g
Some binomial products have special results and can be simpli�ed 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 - b 2 Proof (a + b) (a - b) = a 2 - ab + ab 2 2 = a - b
-
b2
Perfect squares
] a + b g2 = a 2 + 2ab + b 2 Proof
]a bg +
2
(a + b)(a + b) 2 2 = a + ab + ab + b 2 2 = a + 2ab + b =
]a bg -
Proof
]a bg -
2
(a - b)(a - b) 2 2 = a - ab - ab + b 2 2 = a - 2ab + b =
2
=
a 2 - 2ab + b 2
53
54
Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLES Expand and simplify
]
1. 2x - 3
g
2
Solution
] 2 x 3 g ] 2x g 2
-
2
=
=
^
4x 2
-
-
2 (2x) 3 + 3 2
12x + 9
h^
2. 3 y - 4 3y + 4
h
Solution
^ h
(3 y - 4) (3y + 4) = 3y 2 - 4 2 y 2 - 16 = 9
2.6 Exercises Expand and simplify 1. 2. 3. 4. 5.
]t 4g ]z 6 g +
-
2
16. ^ p + 1 h ^ p - 1 h
2
17. r + 6 r - 6
] g] g
] x - 1 g2
18. ] x - 10 g ] x + 10 g
^ y 8h ^ q 3h
19. 2a + 3 2a - 3
]
g] 20. ^ x 5y h^x
2
+
2
+
-
+
g 5y h
6.
]k 7 g
2
21. ] 4a + 1 g ] 4a - 1 g
7.
] n + 1 g2
22. 7 - 3 x 7 + 3x
8. 9.
-
] g] g 2h^ x 2h 23. ^ x 24. ^ x 5h 25. ]3ab 4cg]3ab 4c g
]2b 5g ]3 x g -
2
2
+
2
2
10. ^ 3 y - 1 h2
^
2
+
-
2
+
-
h
11. x + y 2 12. ] 3a - b g2
+
2 26. x + x
b
l
b
lb
2
l
] g 14. ]t 4g]t 4g 2
1 1 27. a - a a + a
-
28. _ x + 6 y - 2 @ i _ x - 6 y - 2 @ i
15. ] x - 3 g ] x + 3 g
29. 6 a + b
13. 4d + 5e +
] g
+
2 c @
55
Chapter 2 Algebra and Surds
2 30. 7 ] x + 1 g - y A
36. ] x - 4 g3
31. ] a + 3 g2 - ] a - 3 g2
37. b x
] g] g ]3 x 1g 4
^
34. x + y
]
+
2
]
39. 2a + 5
-
h x ^2 yh 2
-
35. 4n - 3
+
-
-
4) (x
-
4) 2 .
2
-
2
-
+
2
4x 2 y 2
g
3
40. ] 2x - 1 g ] 2x + 1 g ] x + 2 g2
-
g]4n 3 g
2
l b 1x l
38. _ x 2 + y2 i
32. 16 - z - 4 z + 4 33. 2x +
-
1 x
Expand (x
2n 2 + 5
PROBLEM Find values of all pronumerals that make this true. a b c
#
d e f e b i i i h g i i c c b
Try c
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
=
9.
56
Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLES Factorise 1. 3x + 12
Solution Divide each term by 3 to �nd 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
+
]
2 y x + 3
g
Solution The highest common factor is x + 3. 5 x + 3 + 2 y x + 3 = x + 3 5 + 2y
]
g
]
g ]
g^
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 - b
^
h
Chapter 2 Algebra and Surds
2.7 Exercises Factorise 1.
2 y + 6
19. x ] m + 5 g + 7 ] m + 5 g
2.
5x - 10
20. 2 ^ y
3.
3m - 9
21. 4 7 + y
4.
8x + 2
5.
24 - 18 y
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. 9.
2 y
2
+
10. ab
+
h 22. 6x ]a 2g 5]a 2 g
4y
28. 4x 3 - 24x 2
16. 3q
5
17. 5b
3
+
30. 24a 2 b 5 + 16ab 2
2
14. 6ab + 3a - 2a -
29. 35m 3 n 4 - 25m 2 n
9mn
13. 8x z - 2xz
15. 5x
31. 2rr 2 + 2rrh
2
]
32. x - 3
2x + xy 2q
-
27. 15a 4 b 3 + 3ab
2
2
+
26. 3 pq 5 - 6q 3
2
ab
+
^
3x 7 + y
-
25. 6x 3 + 9x 2
11. 4x y - 2xy 12. 3mn
h
-
2
3
1 h - y ^ y - 1 h
^
15a - 3a 2
-
2
+
33. y 2 x + 4
2
15b
]
g
]
5 x-3
g
g 2]x 4g +
+
34. a ] a + 1 g - ] a + 1 g2
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)
57
58
Maths In Focus Mathematics Extension 1 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 ) 3xy = 2 (x - 2) - 3y (- 2 + x) = 2 (x - 2 ) - 3y (x - 2) =
or 2x - 4 + 6y -
=
(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
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
9.
ay + a + y + 1
20. 2a + 6 - ab - 3b
+
10xy - 3xy
-
-
4y
10. x 2 + 5x - x - 5
21. x 3 - 3x 2 + 6x - 18
11. y + 3 + ay + 3a
22. pq - 3p + q 2
-
15y 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 + 6 ab + 18b
25. xy + 7x - 4y 28
29. 5 y
26. x 4 - 4x 3 - 5x + 20
30.
rr
2
15 + 10xy
+
-
59
30x
2rr - 3r - 6
Trinomials A trinomial is an expression with three terms, for example x 2 - 4x + 3. Factorising a trinomial usually gives a binomial product.
]
g
]
x 2 + a + b 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
'
`
m
2
-
5m + 6
] g ] g]
[m + - 2 ] [m + m-3 = m - 2
=
g
] 3 g]
-
3.
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 y - 1 + y - 2 = y + 2 ` y
'
^
h^
h
+
2 and -1.
Guess and check by trying 2 and -1 or - 2 and 1.
60
Maths In Focus Mathematics Extension 1 Preliminary Course
2.9 Exercises Factorise x 2 + 4x + 3
1.
2. y 2
+
14. a 2 - 4a + 4 15. x 2 + 14x - 32
7y + 12
3.
m 2 + 2m + 1
16. y 2 - 5y - 36
4.
t2
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.
v2
20. k 2
8.
t 2 - 6t + 9
21. x 2 + x - 12
9.
x2
22. m 2
+
-
+
8t + 16
8v + 15
9x - 10
-
7k + 10
-
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
25. l 2 - 11l
+
9y - 36
+
18
13. x 2 - 5x - 24
]
g
]
g]
g
The result x 2 + a + b x + ab = x + a x + b only works when the coef�cient of x 2 (the number in front of x 2) is 1. When the coef�cient 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. 5 y 2 - 13y + 6
Solution—guess and check For 5 y 2, one bracket will have 5 y and the other y : 5 y y . Now look at the constant (term without y in it): + 6.
^
h^ h
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 13 y so the numbers cannot be both positive. They must both be negative. 5 y - y To get a product of 6, the numbers must be 2 and 3 or 1 and 6. Guess 2 and 3 and check: 5 y - 2 y - 3 = 5y 2 - 15y - 2y + 6
^
h^ h
^
h^
h
=
5 y 2 - 17y + 6
This is not correct. Notice that we are mainly interested in checking the middle two terms, y and - 2y . -15 Try 2 and 3 the other way around: 5 y - 3 y - 2 . Checking the middle terms: -10 y - 3y = -13y This is correct, so the answer is 5 y - 3 y - 2 . Note: If this did not check out, do the same with 1 and 6.
^
h^
h
^
h^
h
Solution —cross method Factors of 5 y 2 are 5 y and y . Factors of 6 are -1 and - 6 or - 2 and - 3. Possible combinations that give a middle term of -13 y are 2
5 y
-
3
y
-
5 y
-
y
-
3
5 y
-
1
5 y
-
2
y
-
6
y
-
By guessing and checking, we choose the correct combination. -3 5 y # - 2 = -10y 5 y y y `
-
# -
3
3y y -13
= -
2
5 y 2 - 13y + 6
=
^5y 3 h ^ y 2 h -
-
Solution — PSF method P: Product of �rst and last terms S: Sum or middle term F: Factors of P that give S - 3y 30 y 2 -10 y y -13
30 y 2 -13 y - 3 y, - 10y
)
`
5 y 2 - 13y
+
6
5y 2 - 3y - 10y + 6 = y 5y - 3 - 2 5y - 3 y - 3 y - 2 = 5
=
^
^
h ^ h^ h
h CONTINUED
6 1
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Maths In Focus Mathematics Extension 1 Preliminary Course
2.
4 y 2 + 4y - 3
Solution—guess and check For 4 y 2, both brackets will have 2 y or one bracket will have 4 y and the other y . Try 2 y in each bracket: ^ 2 y 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. ^ 2 y - h ^ 2y + h To get a product of 3, the numbers must be 1 and 3. Guess and check: ^ 2 y - 3 h ^ 2y + 1 h Checking the middle terms: 2 y - 6y = - 4y This is almost correct, as the sign is wrong but the coef�cient is right (the number in front of y ). Swap the signs around: ^ 2 y - 1 h ^ 2y + 3 h = 4y 2 + 6y - 2y - 3 2 = 4 y + 4y - 3 This is correct, so the answer is ^ 2 y
-
1 h ^ 2y + 3 h .
Solution —cross method Factors of 4 y 2 are 4 y and y or 2 y and 2 y . Factors of 3 are -1 and 3 or - 3 and 1. Trying combinations of these factors gives 3 2 y 2 y # - 1 = - 2y 2 y `
1
2 y # 3
-
2 4 y + 4y - 3
=
=
6y 4 y
^ 2y + 3 h ^ 2y - 1 h
Solution — PSF method 2 of �rst and last terms y -12 4 y S: Sum or middle term F: Factors of P that give S y, - 2y +6 2 + 6 y y -12 y -2 + 4 y 2 2 y + 4y - 3 = 4y + 6y - 2y - 3 ` 4 = 2 y ^ 2y + 3 h - 1 ^ 2y + 3 h
P: Product
)
=
^ 2 y + 3 h ^ 2y - 1 h
Chapter 2 Algebra and Surds
Exercises
2.10 Factorise 1.
2a 2
2.
5 y 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 - 5 b + 3
20. 4x 2
-
4x - 15
6.
7x 2 - 9x + 2
21. 6 y 2
-
13y + 2
7.
3 y 2 + 5y - 2
22. 6 p 2
-
5p - 6
8.
2x 2 + 11x + 12
23. 8x 2
+
31x + 21
9.
5 p 2 + 13p - 6
24. 12b 2 - 43b + 36
+
11a + 5
16. 4n 2
-
11n + 6
10. 6x 2 + 13x + 5
25. 6x 2 - 53x - 9
11. 2 y 2 - 11y - 6
26. 9x 2 + 30x + 25
12. 10x 2
3x - 1
27. 16 y 2 + 24y + 9
13. 8t 2 - 14t + 3
28. 25k 2 - 20k + 4
14. 6x 2 - x - 12
29. 36a 2
15. 6 y 2 + 47y - 8
30. 49m 2
+
-
12a + 1
+
84m + 36
Perfect squares You have looked at some special binomial products, including a + b 2 = a 2 + 2ab + b 2 and a - b 2 = a 2 - 2ab + b 2 . When factorising, use these results the other way around.
] g
] g
] ]
a 2 + 2ab + b 2 = a + b a 2 - 2ab + b 2 = a - b
g g
2 2
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Maths In Focus Mathematics Extension 1 Preliminary Course
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 2 = x - 4
=
]
g
2. 4a 2 + 20a + 25
Solution 4a 2 + 20a + 25
= =
2.11
] 2a g 2 (2a) (5) ] 2a 5 g 2
+
+
+
52
2
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.
x2
16. t 2 + t +
6.
4x 2 + 12x + 9
7.
16b 2 - 8b + 1
8.
9a 2
9.
25x 2 - 40x + 16
-
12x + 36
+
10. 49 y 2
12a + 4
+
14y + 1
11. 9 y 2 - 30y + 25
1 4
4x 4 + 3 9 6 y 1 18. 9 y 2 + + 5 25 17. x 2 -
19. x 2 + 2 + 20. 25k 2
-
1 x2
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 - b2. a2 - b 2 = ] 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
] g
9b 2 - 1 = 3b 2 - 1 2 = (3b + 1) (3b - 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) =
2.12
(a + b + 2)(a - b + 4)
Exercises
Factorise 1.
a2 - 4
7.
1 - 4z 2
2.
x2 - 9
8.
25t 2 - 1
9.
9t 2 - 4
3. y 2
-
1
4.
x 2 - 25
10. 9 - 16x 2
5.
4x 2 - 49
11. x 2 - 4y 2
6.
16 y 2 - 9
12. 36x 2 - y 2
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Maths In Focus Mathematics Extension 1 Preliminary Course
13. 4a 2 - 9b 2 14. x
2
15. 4a
2
y 2 20. 9
2
100y
-
81b
16. x + 2
2
-
2
19. x 2
-
-
-
1
]
21. x + 2
2
2
-
-
+
2
g ^2y 1 h 2
-
22. x 4 - 1
g y ]b 2 g 17. ] a 1 g ]1 w g 18. z
]
2
-
23. 9x 6 - 4y 2
2
24. x 4 - 16y 4 25. a 8 - 1
1 4
Sums and differences of 2 cubes
]
a 3 + b3 = a + b
g^a
2
-
ab + b2
h
Proof (a + b) (a 2 - ab + b 2) = a3 - a 2 b + ab2 + a2 b - ab2 + b3 3 3 = a + b
]
a3 - b3 = a - b
g^a
2
+
ab + b2
Proof (a - b) (a 2 + ab + b 2) = a3 + a 2 b + ab2 - a2 b - ab2 - b3 3 3 = a - b
EXAMPLES Factorise 1. 8x 3 + 1
Solution 8x 3 + 1
=
] 2x g
3
+
13
] g
(2x + 1) [ 2x 2 - (2x)(1) + 1 2] 2 = (2x + 1) (4x - 2x + 1) =
h
+
2
Chapter 2 Algebra and Surds
2. 27a 3 - 64b 3
Solution
] g ] g (3a 4b)[] 3a g
27a 3 - 64b3 = 3 a 3 - 4b
3
-
2
] g
(3a)(4 b) + 4 b 2] 2 2 = (3a - 4 b)(9 a + 12 ab + 16 b ) =
+
Exercises
2.13 Factorise 1.
b3 - 8
2.
x 3 + 27 3
1
12.
x3 8
13.
1000 a3
+
]
g
-
27 1 b3
3.
t +
4.
a 3 - 64
14. x + 1
5.
1 - x3
15. 125x 3 y 3 + 216z 3
6.
8 + 27 y 3
16. a - 2
]
7. y 3 + 8z 3 8. 9.
17. 1 -
x 3 - 125y 3 8x 3
3
10. a b
+ 3
3
27y
-
3
11. 1000 + 8t
y 3
-
g ]a 1 g 3
-
]
18. y 3 + 3 + x
]
]
+
3
x3 27
19. x + 1
1
3
g
3
g ^ y 2 h 3
20. 8 a + 3
+
g
3
-
-
3
b3
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 Extension 1 Preliminary Course
Exercises
2.14 Factorise 1.
2x 2 - 18
16. x 3 - 3x 2
-
10x
2.
3 p 2 - 3p - 36
17. x 3 - 3x 2
-
9x + 27
3.
5 y 3
18. 4x 2 y 3 - y
4.
4a 3 b + 8a 2 b2 - 4ab2 - 2a2 b
19. 24 - 3b 3
5.
5a 2
20. 18x 2 + 33x - 30
6.
-
7.
3z 3 + 27z 2
8.
9ab - 4a 3 b3
23. z 3 + 6z 2 + 9z
9.
x3
24. 4x 4 - 13x 2 + 9
5
-
10a + 5
-
2x 2 + 11x - 12
-
+
21. 3x 2 - 6x + 3 22. x 3 + 2x 2 - 25x - 50
60 z
x
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
]
g ]x 4 g 13. y ^ y 5 h 16 ^ y 5 h 12. x - 3 2
2
-
+
14. x 4
-
x3
15. x 6
-
1
+
-
+
2
+
8x - 8
27. 40x - 5x 4 28. a 4 - 13a 2 + 36 29. 4k 3 + 40k 2 + 100 k 30. 3x 3
+
9x 2 - 3x - 9
DID YOU KNOW? Long division can be used to �nd factors of an expression. For example, x - 1 is a factor of x 3 + 4x - 5. We can �nd the other factor by dividing x 3 + 4x - 5 by x - 1. x 2 + x + 5 x - 1 x 3 + 4 x - 5
g
x 3
-
x 2
x 2 x You will study this in Chapter 12.
2
+
4x
-
x
5 x - 5 5 x - 5 0 So the other factor of x 3 + 4x - 5 is x 2 + x + 5 3 2 ` x + 4x - 5 = (x - 1) (x + 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
]
g
2
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: a2
] ] ]
2ab + b 2 = a + b x 2 + 2x 3 + 3 2 = x + 3 x 2 + 6x + 9 = x + 3 +
]g
g g g
2 2 2
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 + b2 = a - b 2 n 2 - 2n 5 + 5 2 = n - 5 2 n 2 - 10n + 25 = n - 5 2
]g
] ] ]
g g g
To complete the square on a 2 + pa, divide p by 2 and square it. a
2
+
pa +
2
dn d p 2
=
p a+ 2
2
n
70
Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLES 1. Complete the square on x 2 + 12x.
Solution Divide 12 by 2 and square it: x
2
+
12x +
2
c m 12 2
=
x 2 + 12x + 6 2
x 2 + 12x + 36 2 = x + 6 =
]
g
2. Complete the square on y 2 - 2y .
Solution Divide 2 by 2 and square it: 2 2 2 2 y 2 - 2y + = y - 2y + 1 2 2 = y - 2y + 1 2 = y - 1
c m
^
2.15
h
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 5.
m 2 - 14m
6.
q 2 + 18q 2
2x
15. x 2 + 9x 16. y 2
-
5 y 2
17. k 2
-
11k 2
7.
x
8.
t 2 - 16t
18. x 2 + 6xy
9.
x 2 - 20x
19. a 2
+
10. w 2 + 44w 11. x 2 - 32x
-
4ab
20. p 2 - 8pq
Chapter 2 Algebra and Surds
71
Algebraic Fractions Simplifying fractions EXAMPLES Simplify 1.
4x + 2 2
Solution 4x + 2 2
2.
]
2 2x + 1 2 = 2x + 1 =
g
Factorise �rst, then cancel.
2x 2 - 3x - 2 x3 - 8
Solution 2x 2 - 3x - 2 x3 - 8
=
] 2x 1 g ] x 2 g ] x 2 g ^ x 2x 4 h +
-
=
-
2
+
+
2x + 1 x 2 + 2x + 4
Exercises
2.16 Simplify 1.
5a + 10 5
9.
2.
6t - 3 3
2 p 2 + 7p - 15 10. 6 p - 9
3.
8 y + 2 6
4.
8 4d - 2
5.
6.
11.
12.
x2 5x
2
-
2x 13.
y - 4 y 2 - 8y + 16 2
7.
2ab - 4a a 2 - 3a
8.
s2 + s - 2 s 2 + 5s + 6
14.
15.
b3 - 1 b2 - 1
a2 - 1 a 2 + 2a - 3
]
3 x-2
g y ]x 2 g +
x3
-
-
8
x 3 + 3x 2 - 9x - 27 x 2 + 6x + 9 2 p 2 - 3p - 2 8 p 3
+
1
ay - ax + by - bx 2ay - by - 2ax + bx
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Maths In Focus Mathematics Extension 1 Preliminary Course
Operations with algebraic fractions
EXAMPLES Simplify 1.
x-1 5
-
x+3 4
Solution Do algebraic fractions the same way as ordinary fractions.
x -1 5
-
x+3 4
=
=
=
2.
]
4 x -1
g 5 ]x 3 g -
+
20 4x - 4 - 5x - 15 20 - x - 19 20
2a 2 b + 10ab a 2 - 25 ' 4b + 12 b 3 + 27
Solution 2a 2 b + 10ab a 2 - 25 ' 4b + 12 b 3 + 27
=
=
=
3.
2 x-5
+
2a 2 b + 10ab 4b + 12 # b 3 + 27 a 2 - 25 2ab a + 5
]
g
] b 3 g ^ b 3b 9 h ] 8ab ] a 5 g ^ b 3b 9 h +
2
-
+
-
2
-
+
1 x+2
Solution 2 x-5
+
1 x+2
=
=
=
]
#
g ] g ] g] g ] g] g ] g] g 2 x+2 + 1 x-5 x-5 x+2 2x + 4 + x - 5 x-5 x+2 3x - 1 x-5 x+2
]
g
4 b+3 a+5 a-5
g]
g
Chapter 2 Algebra and Surds
2.17
Exercises
1. Simplify x 3x (a) + 2 4 y + 1 2y (b) + 5 3 a+2 a (c) 3 4 p - 3 p + 2 (d) + 6 2 x-5 x-1 (e) 2 3 2. Simplify 3 (a) b+2 (b) (c) (d)
(f) (g) (h) (i) (j)
q 2 + 2q +
(c)
ax - ay + bx - by
x 3 + y 3
x 2 - y 2
ab 2 + a 2 b
(d) x -
a
2
4
+
1 2a + 1
y + 2
y+3
-
-
16
1
+
1
-
2 x
x+2
-
2
2
3
a+1 +
5 y - 1
7 x - x - 12 2
#
#
y 2 - 9 6x - 24
#
x 2 - 2x - 8 y 3 + 27
3
2x + 8 x2 - 9
+
x-3
x 2 + 3x 4x - 16
#
(d)
5b b2 ' 2b + 6 b2 + b - 6
(e)
x 2 - 8x + 15 x 2 - 9 x 2 + 5x + 6 ' # 2x - 10 5x 2 + 10x 10x 2
-
b b+1
5. Simplify
3. Simplify 2 3 (a) x + x
(c) 1 +
x
-
2 3a - 15 y - y - 2 a 2 - 5a (b) 2 ' # 5ay y - 4y + 4 y 2 - 4
q3 + 1 # 1 p + 2
3ab 2 12ab - 6a ' 5xy x 2 y + 2xy 2
x-1
2 2
x-3
3x 2 (a) 4 y - 12
b 2 + 2b # 6a - 3
p 2 - 4
1
x+1
1
+
4. Simplify
x 2 - 6x + 9 x 2 - 5x + 6 (e) ' x 2 - 25 x 2 + 4x - 5
(b)
1
(a) -
2 x
(b)
3
(c)
a+b x2
(d)
x+2
(e) p - q +
1 p + q
x
2
x
2
-
1 7x + 10
5 -
4
-
2 p
2
+
pq
a a+b
-
-
3 x-2
+
x
2
-
2 2x - 15
2
-
x+2
3 pq - q 2 b
a-b
x + y x (e) x - y + y - x
+
-
1 a
2
-
b2
y y
2
-
x2
Substitution Algebra is used in writing general formulae or rules. For example, the formula A = lb is used to �nd the area of a rectangle with length l and breadth b. We can substitute any values for l and b to �nd the area of different rectangles.
+
4 x +x- 6 2
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Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLES 1. P = 2l + 2b is the formula for �nding the perimeter of a rectangle with length l and breadth b. Find P when l = 1.3 and b = 3.2.
Solution P = 2l + 2b
] g ] g
2 1.3 + 2 3.2 = 2.6 + 6.4 =
=
9
2. V = rr 2 h is the formula for �nding 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 2 (8.7) = r 2.1 = 120.5 correct to 1 decimal place
] g
] g
9C + 32 is the formula for changing degrees Celsius °C into 5 degrees Fahrenheit °F �nd F when C = 25. 3. If F =
] g
Solution 9C + 32 5 9 25 = + 32 5 225 = + 32 5 225 + 160 = 5 385 = 5 = 77 This means that 25°C is the same as 77°F.
F =
] g
Chapter 2 Algebra and Surds
Exercises
2.18 1.
Given a = 3.1 and b = - 2.3 �nd, correct to 1 decimal place. (a) ab (b) 3b (c) 5a 2 (d) ab 3 (e) a + b 2
g
]
(f) (g) 2.
a-b -
b2
]
g
T = a + n - 1 d is the formula for �nding 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, �nd y if m = 3, x = - 2 and b = - 1.
4.
If h = 100t - 5t 2 is the height of a particle at time t , �nd h when t = 5.
5.
Given vertical velocity v = - gt , �nd v when g = 9.8 and t = 20.
6.
If y = 2 x + 3 is the equation of a function, �nd y when x = 1.3, correct to 1 decimal place.
7.
S = 2r r r + h is the formula for the surface area of a cylinder. Find S when r = 5 and h = 7, correct to the nearest whole number.
]
g
2
8. A = rr is the area of a circle with radius r . Find A when r = 9.5, correct to 3 signi�cant �gures. n-1
9. Given u n = ar is the nth term of a geometric series, �nd u n if a = 5, r = - 2 and n = 4.
10. Given V = 1 lbh is the volume 3 formula for a rectangular pyramid, �nd V if l = 4.7, b = 5.1 and h = 6.5. 11. The gradient of a straight line is y 2 - y 1 given by m = x - x . Find m 2 1 if x 1 = 3, x 2 y 2 = 5.
]
1, y 1 = - 2 and
= -
g
12. If A = 1 h a + b gives the area 2 of a trapezium, �nd 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 3 and t = 5 . , u= 1 4 a= 5 6 a , �nd S if a = 5 15. Given S = 1 - r and r = 2 . S is the sum to in�nity 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, �nd the exact value of y when x = 2.
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Maths In Focus Mathematics Extension 1 Preliminary Course
18. Find the value of E in the energy equation E = mc 2 if m = 8.3 and c = 1.7.
c
19. A = P 1 +
20. If S =
^
h
a r n - 1 is the sum of r - 1
a geometric series, �nd S if a = 3, r = 2 and n = 5.
m
r n is the formula 100
21. Find the value of
for �nding compound interest. Find A when P = 200, r = 12 and n = 5, correct to 2 decimal places.
a3 b2 if c 2
cm
2 3 1 4 , b = 2 and c = . a= 3 4 3 2
c m
c m
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
=
a' b
=
^ xh
2
=
ab a b
x2
a b
=
=
x
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#
=
3 5
5
2. Simplify 3 40 .
Solution 3 40
=
Find a factor of 40 that is a perfect square.
3 4 # 10
3 # 4 # 10 10 = 3 # 2 # = 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
78
Maths In Focus Mathematics Extension 1 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. Evaluate x if
(h) 15 8
(a)
(i) 7 40
=
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.
x
x x x
=
=
=
=
x
7 3 =
x
6 23 =
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
2. Simplify
=
7 2 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
79
Exercises
2.20 Simplify 1.
5
2.
3 2
3.
3
4.
7 3
5.
5
6.
4 6
7.
2
-
8 2
8.
5
+
4 5
+
3 5
9.
2
-
2 2
-
3 2
10.
5
+
11.
8
12.
3
13.
12
+
2 5
-
+
5 3
-
-
2 2
50
-
32
15.
28
+
63
16. 2 8
-
18
17. 3 54
4 3
90
18.
4 5
+
-
19. 4 48
6
-
14.
2 24
5 40
+
2 10
-
3 147
+
20. 3 2
+
8
21.
63
-
28
-
50
22.
12
-
45
-
48
45
23.
150
-
2
24.
32
+
48
25.
80
-
5 12
12
-
-
45
+
24
-
243
-
50
-
3 245
+
+
5
+
147
2 50
27
Multiplication and division
a# b
=
a b #c d
=
a# a
=
a b
=
To get a b # c d = ac bd , multiply surds with surds and rationals with rationals.
ab ac bd a2
=
a
a b
EXAMPLES Simplify 1. 2 2 # -5 7
Solution 2 2 # -5 7
10 14
= -
CONTINUED
80
Maths In Focus Mathematics Extension 1 Preliminary Course
2. 4 2 #5 18
Solution 4 2 #5 18
3.
20 36 = 20 #6 = 120 =
2 14 4 2
Solution 2 14 4 2
=
2 2
4 2 7 2
=
4.
7
#
3 10 15 2
Solution 3 10 15 2
=
=
5.
d
3# 5 # 2 15 2 5 5
10 3
2
n
Solution
d
10 3
n ^^ 10hh 3 2
=
10 3 1 = 3 3 =
2
2
Chapter 2 Algebra and Surds
Exercises
2.21 Simplify 1.
7
3
2.
3# 5
3.
2 #3 3
4.
5 7 #2 2
#
3 3 #2 2
5.
-
6.
5 3 #2 3
7.
-
8.
2 7# 7
9.
2 3 # 5 12
23.
24.
25.
26.
4 5 # 3 11 27.
28. 10.
6# 2
11.
8 #2 6
29.
5 8 10 2 16 2 2 12 10 30 5 10 2 2 6 20 4 2 8 10 3 3 15 2 8
12. 3 2 # 5 14 13.
10 # 2 2
14. 2 6 # -7 6
^ 2h 16. ^ 2 7 h 15.
2
2
17.
31.
32.
3# 5# 2
18. 2 3 # 7 # - 5 19.
30.
2 # 6 #3 3
20. 2 5 # -3 2 # -5 5
33.
34.
3 15 6 10 5 12 5 8 15 18 10 10 15 2 6 2
d n 2 3
2
21.
22.
4 12 2 2
5n 35. d 7
12 18 3 6
Expanding brackets The same rules for expanding brackets and binomial products that you use in algebra also apply to surds.
81
82
Maths In Focus Mathematics Extension 1 Preliminary Course
Simplifying surds by removing grouping symbols uses these general rules.
a
^
b
h
c
+
=
ab
ac
+
Proof a
^
b+ c
h
=
a# b
=
ab
a# c
+
ac
+
Binomial product:
^
a
+
b
h^
c
d
+
h
ac
=
ad
+
bc +
+
bd
Proof
^
a
+
b
h^
c
+
d
h
=
a# c
=
ac
+
a# d
+
ad
bc
+
b# c
+
bd
+
Perfect squares:
^
a
+
b
h
2
=
a + 2 ab + b
Proof
^
a
+
b
h ^ 2
h^
h
a+ b a+ b 2 a + ab + ab + b 2 = = a + 2 ab + b
=
^
a
-
b
h
2
=
a - 2 ab + b
Proof
^
a
-
b
h ^ 2
h^
h
a- b a- b 2 a - ab - ab + b 2 = = a - 2 ab + b
=
Difference of two squares:
^
a
+
b
h^
a- b
h
=
Proof
^
a
+
b
h^
a- b
h
=
a2
=
a-b
-
ab
+
ab - b2
a
-
b
+
b# d
Chapter 2 Algebra and Surds
83
EXAMPLES Expand and simplify
^
2
1.
5
h
2
+
Solution 2( 5
+
2) = 2 # 5 10 10
= =
^
2. 3 7 2 3
-
4
+ +
3 2
2# 2
+
2
h
Solution 3 7 (2 3
3.
^
2
+
-
3 2 ) = 3 7 #2 3
3 5
h^
=
6 21
3
-
2
3 7 #3 2
-
9 14
-
h
Solution ( 2
4.
+
^
3 5)( 3
5
+
2 3
2)
-
h^
5
-
=
2# 3
=
6
2 3
-
2# 2
-
2 + 3 15
-
+
3 5# 3
-
3 5# 2
3 10
h
Solution ( 5
+
2 3 )( 5
-
2 3) = 5 # 5
-
5 - 2 15 = 5 - 12 =
+
5 #2 3 2 15
-
+
2 3# 5
-
2 3 #2 3
4 #3
7
= -
Another way to do this question is by using the difference of two squares. ( 5
+
2 3)( 5
-
2 3) = =
^ 5 h ^2 3 h 2
-
5 - 4 #3 7
= -
2
Notice that using the difference of two squares gives a rational answer.
84
Maths In Focus Mathematics Extension 1 Preliminary Course
Exercises
2.22 1.
^ 5 2 h^2 (n) ^ 5 2h (o) ^ 2 2 3h (p) ^ 3 2 7h (q) ^ 2 3 3 5 h (r) ^ 7 2 5 h (s) ^ 2 8 3 5 h (t) ^ 3 5 2 2 h
Expand and simplify
^
(a)
2
5
(b)
3 2 2
^
h
3
+
5
-
(m) 2 11
^ 3 2 5h (d) 7 ^ 5 2 2 3 h (e) 3^ 2 4 6h (f) 3 ^ 5 11 3 7 h (g) 3 2 ^ 2 4 3 h (h) 5 ^ 5 5 3 h (i) 3 ^ 12 10 h (j) 2 3 ^ 18 3h (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 2h (o) 2 3 ^ 2 12 h (c) 4 3
-
3.
+
-
-
-
+
-
2.
^ 2 3h^ 5 3 3 h (b) ^ 5 2 h^ 2 7h (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 3h (g) ^ 7 3 h^ 7 3h (h) ^ 2 3 h^ 2 3h (i) ^ 6 3 2 h^ 6 3 2 h (j) ^ 3 5 2 h^3 5 2h (k) ^ 8 5 h^ 8 5h (l) ^ 2 9 3 h^ 2 9 3 h +
+
-
+
-
-
+
-
+
-
-
+
+
-
+
+
-
+
-
g
^
h
^
(b) 2 2 - 5 = a + b 10 5.
h^
^
h^
6. Evaluate k if 2 7- 3 2 7
^
h^
_
7. Simplify 2 x 8.
2
-
^
If 2 3 - 5 a and b.
h
2
+ =
i
+
y
h
3 5
Expand and simplify (a) a+3-2 a+3 2 (b) p - 1 - p
_
-
-
g]
4. Evaluate a and b if 2 (a) 2 5 + 1 = a + b
-
+
g
]
Expand and simplify (a)
If a = 3 2 , simplify (a) a2 (b) 2a3 (c) (2a)3 (d) a + 1 2 (e) a + 3 a – 3
]
+
-
-
2
+
+
-
2
-
+
-
2
-
+
-
2
+
-
h
2
+
-
5 2
2
-
+
-
2
+
h
11
+
3
h
i_
+
=
x
2
h
k.
-
i
3 y .
a - b , evaluate
9. Evaluate a and b if 2 7 2 - 3 =a+b 2.
^
h
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
^ xh
2
=
x. Multiplying by
b a # b b
=
a b b
b
b is the same as multiplying by 1.
Proof b a # b b
=
=
a b b2 a b b
EXAMPLES 1. Rationalise the denominator of
3 . 5
Solution 5 3 # 5 5
=
3 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 Extension 1 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
+
c
+
b
, multiply by d
Proof a+ b c # c + d c
d = d
-
=
=
^ a b h^ c ^ c d h ^ c ^ a b h^ c ^ c h ^ d h ^ a b h^ c +
-
+
-
+
h d h dh dh d
-
2
-
+
2
-
c - d
EXAMPLES 1. Write with a rational denominator 5 2 Multiply by the conjugate surd 2 + 3.
-
3
.
Solution 5 2
-
3
#
2
+
3
2
+
3
=
5
^
^ 2h
2 2
+
3
h
32 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 3
+
+
5
4 2
#
3
-
4 2
3
-
4 2
=
=
^2
h^ 3 4 2 h ^ 3 h ^4 2 h
3
+
5
2
-
-
2 # 3 - 8 6 + 15 3 - 16 #2
2
-
4 10
c
-
d
c
-
d
Chapter 2 Algebra and Surds
=
=
6-8 6
15 - 4 10 - 29 15 + 4 10 -6 + 8 6 29
3 3
3. Evaluate a and b if
a + b.
=
3- 2
+
Solution 3 3 3
2
-
#
3
+
2
3
+
2
=
^
3 3
3+ 2
^ 3 2 h^ 3 3 9 3 6 ^ 3h ^ 2h -
=
h 2
+
h
+
2
2
-
3#3+3 6 3-2 9+3 6 = 1 = 9 + 3 6 9# 6 = 9 + 54 = 9 + =
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
^
2 3
h
-
2
3
+
-
+
2
4+
^ 3h
h^
2
5
^
3
+
h
2
h
3 -2 15 + 2 5
22 2 3 - 4 + 15 + 2 5 = 3-4 2 3 - 4 + 15 + 2 5 = -1 15 - 2 5 = -2 3 + 4 -
87
88
Maths In Focus Mathematics Extension 1 Preliminary Course
2.23 1.
Express with rational denominator (a) (b) (c) (d) (e) (f) (g) (h) (i) (j)
2.
Exercises
1 7
(b)
2 2 2 3
(c)
5 6 7
(d)
5 2 1+
2 3 -
(b) (c) (d) (e) (f)
1 2
+
2
1 5
+
2
2
-
7
2
+
3
5
(f) z 2
+
2 2
(g)
5 3 2
-
4
(h)
8+3 2
(i)
4 5 4 3
-
2 2
(j)
7 5
(l)
4 3
2
+
3 2
-
+
7
(a)
2 6
3
-
4
3
+
4
2 3
+
2
2 5
+ +
(b) (c)
5
-
3 3
3
+
3 3 2
+
6
-
2
+
3
3 +
3
5 6
+
2
2
+
7
4+
4 3
2
3
2 2 3
-
-
2
3
-
2
2
3 2
(d) (e)
3 4 2
6
-
3
+
3 2 5 3 2 4-
a b
=
a 6 b
+
1
2 7 7
-
4
2
+
3
2
-
1
1
2
=
2 5
-
1 3
+
-
2 5
2
+
3
2
+
+
5
3
5
-
3
2+ 3
3 +
1
4. Find a and b if
2 3 5
(k)
2
1 where z = 1 + z2
3 2
2
3
+
#
1
-
1 where t = t
-
2 7
2 -
3
-
2 5
1
1
+
2
(e) t +
Express with rational denominator (a)
Express as a single fraction with rational denominator (a)
3
6
3.
=
a+b 5
=
a+b 7
=
a+
b
2 2 -
2 6
-
1
Chapter 2 Algebra and Surds
5.
Show that
2
-
1
2
+
1
+
4 is 2
7.
5
rational. 6.
If x =
3
+
1 (a) x + x (b) x 2 +
b
+
5
+
2
+
1 5
-
2
-
1
as a single fraction with 3 rational denominator.
2, simplify 8.
1 x2
1 (c) x + x
2
Write
Show that
2 3+2 2
+
8 is 2
rational.
l
2
9.
1 If 2 + x = 3 , where x ! 0, �nd x as a surd with rational denominator.
10. Rationalise the denominator of b +2 b!4 b -2
]
g
89
90
Maths In Focus Mathematics Extension 1 Preliminary Course
Test Yourself 2 1. Simplify (a) 5 y - 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 2. Factorise (a) x 2 - 36 (b) a 2 + 2a - 3 (c) 4ab 2 - 8ab (d) 5 y - 15 + xy (e) 4n - 2p + 6 (f) 8 - x 3 3.
] ^
^
3x
]
^
]
g g] g g ] g
g h^
h^
g
Simplify
8.
If a = 4, b = - 3 and c = - 2, �nd the value of (a) ab 2 (b) a - bc (c) a (d) bc 3 (e) c 2a + 3b
(b)
h^
h
3 x-2
+
1 x+3
-
2 . x +x-6 2
g
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.
h
h
7.
^
9. Simplify 3 12 (a) 6 15
11. Rationalise the denominator of 2 (a) 5 3
h
Simplify 4a - 12 10b (a) # 3 3 5b a - 27 (b)
5.
]
(a) Expand and simplify 2 5+ 3 2 5- 3 . (b) Rationalise the denominator of 3 3 . 2 5+ 3
] g ]
Expand and simplify (a) b + 3 b - 2 (b) 2x - 1 x + 3 (c) 5 m + 3 - m - 2 (d) 4x - 3 2 (e) p - 5 p + 5 (f) 7 - 2 a + 4 - 5a (g) 3 2 2 - 5 (h) 3 + 7 3-2
]
4.
-
6.
(b)
5m + 10 m2 - 4 ' m 2 - m - 2 3m + 3 3
The volume of a cube is V = s . Evaluate V when s = 5.4.
1+ 3 2
12. Expand and simplify (a) 3 2 - 4 3 - 2 2 (b) 7+2
^ ^
h
h^
13. Factorise fully (a) 3x 2 - 27 (b) 6x 2 - 12x - 18 (c) 5 y 3 + 40
h
Chapter 2 Algebra and Surds
14. Simplify 3x 4 y (a) 9xy 5 5 (b) 15x - 5
22. Expand and simplify (a) 2 2 3 + 2 (b) 5 7 - 3 5 2 2 - 3 (c) 3 + 2 3 - 2 (d) 4 3 - 5 4 3 + 5 2 (e) 3 7 - 2
^ ^ ^ ^
15. Simplify
^ h ^ h
(a) 3 11 3 (b) 2 3
g] g g
(b)
17. Factorise (a) a 2 - 2ab + b 2 (b) a 3 - b 3
(d)
19. Simplify 4 3 (a) a + b -
20. Simplify
x-2 5 3
5
+
2
-
2 2 2
-
1
, writing
your answer with a rational denominator. 21. Simplify (a) 3 8 (b) - 2 2 # 4 3 (c) 108 - 48 (d)
h
2
(e)
2 2 3 2
+
3
5
+
2
4 5
-
3 3
24. Simplify 3x x-2 (a) 5 2 2a - 3 a+2 (b) + 7 3 1 2 (c) 2 x +1 x -1 4 1 (d) 2 + k + 2k - 3 k + 3 (e)
3
(b)
112
(d) 4 147
(g) 3x - 2y - x - y
7
+
2 18 2m n 6m 2 n 5
3- 2
25. Evaluate n if (a) 108 - 12 (c) 2 8
3
5
-
2+ 5
8 6
(e) 5a # - 3b # - 2a (f)
h^ h
h
5 3 2 (c) 5 -1
1 18. If x = 3 + 1, simplify x + x and give your answer with a rational denominator.
x-3 2
h^
h
23. Rationalise the denominator of 3 (a) 7
g
(b)
h^
h
2
16. Expand and simplify (a) a + b a - b (b) a + b 2 (c) a - b 2
] ] ]
^
+
=
=
200
=
3 75 180 (e) 2 245 + 2 +
n n n = =
n n
91
92
Maths In Focus Mathematics Extension 1 Preliminary Course
26. Evaluate x 2 +
1+2 3 1 if x = x2 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 5 - x + 7 20 x+7 20 x + 17 20 - x + 17 20
28. Simplify (a) (b) (c) (d)
] ]
-
g
x +1 . 4
2
] (b) ] (c) ] (d) ]
g] g] g] g]
(a)
(b) (c)
^ h] g ^ h] g ] g ] g] g]
30. Simplify 3 2 (a) 5 2 (b) 5 10 (c) 17 2 (d) 10 2
+
2 98 .
g
1
-
x+2
.
g g g g
32. Simplify 5ab - 2 a2 - 7ab - 3 a2 . (a) 2ab + a 2 (b) - 2ab - 5a 2 (c) - 13a 3 b (d) - 2ab + 5a 2
(a)
29. Factorise x 3 - 4x 2 - x + 4 (there may be more than one answer). (a) x 2 - 1 x - 4 (b) x 2 + 1 x - 4 (c) x 2 x - 4 (d) x - 4 x + 1 x - 1
2
+
x-2 x -4 x+5 x+2 x-2 x+1 x+2 x-2 x+9 x+2 x-2 x-3 x+2 x-2
33. Simplify
g
3
31. Simplify
(d)
80 . 27
4 5 3 3 4 5 9 3 8 5 9 3 8 5 3 3
^
h
34. Expand and simplify 3x - 2y 2 . (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 2 (b) a 2 - 16a + 64 = a - 8 2 (c) a 2 - 16a + 8 = a - 4 2 (d) a 2 - 16a + 4 = a - 2 2
^ ^
^ ^
h h
h h
