ch2 algebra and surdskjtfc
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TERMINOLOGY
2Algebra andSurds
Binomial:A mathematical expression consisting oftwo terms such as 3x + or x3 1-
Binomial product:The product of two binomialexpressions such as ( 3) (2 4)x x+ -
Expression:A mathematical statement involving numbers,pronumerals and symbols e.g. x2 3-
Factorise:The process of writing an expression as aproduct of its factors. It is the reverse operation ofexpanding brackets i.e. take out the highest common
factor in an expression and place the rest in bracketse.g. 2 8 2( 4)y y= --
Pronumeral:A letter or symbol that stands for a number
Rationalising the denominator:A process for replacing asurd in the denominator by a rational number withoutaltering its value
Surd:From absurd. The root of a number that has anirrational value e.g. 3 . It cannot be expressed as arational number
Term:An element of an expression containingpronumerals and/or numbers separated by an operationsuch as , , or# '+ - e.g. 2 , 3x -
Trinomial:An expression with three terms such asx x3 2 1
2- +
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INTRODUCTION
THIS CHAPTER REVIEWS ALGEBRAskills, including simplifying expressions,
removing grouping symbols, factorising, completing the square and
simplifying algebraic fractions. Operations with surds, including rationalisingthe 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 700800 AD a mathematician named Mohammed Un-Musa
Al-Khowarezmiwrote books on algebra and Hindu numerals. One of his books was named
Al-Jabr wal Migabaloh, and the word algebracomes from the rst word in this title.
Simplifying Expressions
Addition and subtraction
EXAMPLES
Simplify
1. x x7 -
Solution
7 7 1
6
x x x x
x
- = -
=
2. x x x4 3 62 2 2- +
Solution
4 3 6 6
7
x x x x x
x
2 2 2 2 2
2
- + = +
=
Here x is called a
pronumeral.
CONTINUED
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3. x x x3 5 43 - - +
Solution
x x x x x3 5 4 8 43 3- - + = - +
4. a b a b3 4 5- - -
Solution
3 4 5 3 5 4
2 5
a b a b a a b b
a b
- - - = - - -
= - -
Only add or subtract like
terms. These have the
same pronumeral (for
example, 3x and 5x).
1. 2 5x x+
2. 9 6a a-
3. 5 4z z-
4. 5a a+
5. b b4 -
6. r r2 5-
7. y y4 3- +
8. x x2 3- -
9. 2 2a a-
10. k k4 7- +
11. 3 4 2t t t+ +
12. w w w8 3- +
13. m m m4 3 2- -
14. 3 5x x x+ -
15. 8 7h h h- -
16. b b b7 3+ -
17. 3 5 4 9b b b b- + +
18. x x x x5 3 7- + - -
19. x y y6 5- -
20. a b b a8 4 7+ - -
21. 2 3xy y xy + +
22. 2 5 3ab ab ab2 2 2- -
23. m m m5 122 - - +
24. 7 5 6p p p2 - + -
25. 3 7 5 4x y x y + + -
26. 2 3 8ab b ab b+ - +
27. ab bc ab ac bc + - - +
28. a x a x7 2 15 3 5 3- + - +
29. 3 4 2x xy x y x y xy y 3 2 2 2 2 3- + - + +
30. 3 4 3 5 4 6x x x x x3 2 2- - + - -
2.1 Exercises
Simplify
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Multiplication
EXAMPLES
Simplify
1. x y x5 3 2# #-
Solution
5 3 2 30
30
x y x xyx
x y2# #- = -
= -
2. 3 4x y xy 3 2 5#- -
Solution
x y xy x y 3 4 123 2 5 4 7#- - =
Use index laws
to simplify this
question.
1. b5 2#
2. x y2 4#
3. p p5 2#
4. z w3 2#-
5. a b5 3#- -
6. x y z2 7# #
7. ab c8 6#
8. d d4 3#
9. a a a3 4# #
10. y3 3-^ h
11. 2x2 5
^ h 12. ab a2 33 #
13. a b ab5 22 # -
14. pq p q7 32 2 2#
15. ab a b5 2 2#
16. h h4 23 7# -
17. k p p3 2#
18. t3 34
-^ h 19. m m7 26 5# -
20. x x y xy 2 3 42 3 2# #- -
2.2 Exercises
Simplify
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Division
Use cancelling or index laws to simplify divisions.
EXAMPLES
Simplify
1. v y vy 6 22 '
Solution
By cancelling,
v y vy vy
v y
v y
v v y
v
6 22
6
2
6
3
2
2
1 1
3 1 1
'
# #
# # #
=
=
=
Using index laws,
v y vy v y
v y
v
6 2 3
3
3
2 2 1 1 1
1 0
' =
=
=
- -
2.15
5
ab
a b2
3
Solution
3
3
aba b a b
a b
b
a
155
3
2
3
3 1 1 2
2 1
2
=
=
=
- -
-
1
1
1. x30 5'
2. y y2 '
3.2
8a2
4.8aa2
5.
a
a
2
8 2
6.x
xy
2
7. p p12 43 2'
8.6
3
ab
a b2 2
9.1520
xy
x
10. x
x
3
94
7-
2.3 Exercises
Simplify
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11. ab b15 5'- -
12.6
2
a b
ab2 3
13.pqs
p
4
8-
14. cd c d 14 212 3 3'
15.4
2
x y z
xy z3 2
2 3
16.pq
p q
7
423
5 4
17. a b c a b c 5 209 4 2 5 3 1'- - -
18. a b
a b
4
29 2 1
5 2 4
- -
-
^^
hh
19. x y z xy z5 154 7 8 2'- -
20. a b a b9 184 13 1 3
'- -- -^ h
Removing grouping symbols
The distributive law of numbers is given by
a b c ab ac + = +] g
EXAMPLE
( )7 9 11 7 20
140
# #+ =
=
Using the distributive law,
( )7 9 11 7 9 7 1163 77
140
# # #+ = +
= +
=
EXAMPLES
Expand and simplify.1. a2 3+] gSolution
2( 3) 2 2 3
2 6
a a
a
# #+ = +
= +
This rule is used in algebra to help remove grouping symbols.
CONTINUED
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2. x2 5- -] gSolution
( ) ( )x x
xx
2 5 1 2 5
1 2 1 52 5
# #
- - = - -
= - - -
= - +
3. a ab c 5 4 32 + -] gSolution
( )a ab c a a ab a c
a a b a c
5 4 3 5 4 5 3 5
20 15 5
2 2 2 2
2 3 2
# # #+ - = + -
= + -
4. y5 2 3- +
^ h
Solution
( )y y
y
y
5 2 3 5 2 2 3
5 2 6
2 1
# #- + = - -
= - -
= - -
5. b b2 5 1- - +] ]g gSolution
( ) ( )b b b bb b
b
2 5 1 2 2 5 1 1 12 10 1
11
# # # #- - + = + - - -
= - - -
= -
1. x2 4-] g2. h3 2 3+] g3. a5 2- -] g4. x y2 3+^ h5. x x 2-] g6. a a b2 3 8-] g
7. ab a b2 +] g8. n n5 4-] g9. x y xy y 3 22 2+_ i10. k3 4 1+ +] g11. t2 7 3- -] g 12. y y y4 3 8+ +^ h
2.4 Exercises
Expand and simplify
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13. b9 5 3- +] g14. x3 2 5- -] g15. m m5 3 2 7 2- + -] ]g g
16. h h2 4 3 2 9+ + -
] ]g g17. d d3 2 3 5 3- - -] ]g g18. a a a a2 1 3 42+ - + -] ^g h19. x x x3 4 5 1- - +] ]g g
20. ab a b a2 3 4 1- - -] ]g g21. x x5 2 3- - -] g 22. y y8 4 2 1- + +^ h
23. a b a b+ --
] ]g g24. t t2 3 4 1 3- - + +] ]g g 25. a a4 3 5 7+ + --] ]g g
Binomial Products
A binomialexpression consists of two numbers, for example 3.x +
A set of two binomial expressions multiplied together is called a binomialproduct.
Example: x x3 2+ -] ]g g.Each term in the rst bracket is multiplied by each term in the second
bracket.
a b x y ax ay bx by + + = + + +] ^g h
Proof
a b c d a c d b c d
ac ad bc bd
+ + = + + +
= + + +] ] ] ]g g g g
EXAMPLES
Expand and simplify
1. 3 4p q+ -^ ^h hSolution
p q pq p q3 4 4 3 12+ - = - + -^ ^h h
2. 5a 2+] g Solution
( 5)( 5)
5 5 25
10 25
a a a
a a a
a a
5 2
2
2
+ = + +
= + + +
= + +
] g Can you see a quick
way of doing this?
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The rule below is not a binomial product (one expression is a trinomial), but itworks the same way.
a b x y z ax ay az bx by bz+ + + = + + + + +] ^g h
EXAMPLE
Expand and simplify .x x y4 2 3 1+ - -] ^g h Solution
( ) ( )x x y x xy x x y
x xy x y
4 2 3 1 2 3 8 12 4
2 3 7 12 4
2
2
+ - - = - - + - -
= - + - -
1. 5 2a a+ +] ]g g2. x x3 1+ -] ]g g3. 2 3 5y y- +^ ^h h4. 4 2m m- -] ]g g5. 4 3x x+ +] ]g g6. 2 5y y+ -^ ^h h7. 2 3 2x x- +] ]g g8. 7 3h h- -] ]g g9. 5 5x x+ -] ]g g10. a a5 4 3 1- -
] ]g g
11. 2 3 4 3y y+ -^ ^h h12. 4 7x y- +] g h13. 3 2x x2 + -^ ]h g14. 2 2n n+ -] ]g g15. 2 3 2 3x x+ -] ]g g16. 4 7 4 7y y- +^ ^h h
17. 2 2a b a b+ -] ]g g18. 3 4 3 4x y x y - +^ ^h h19. 3 3x x+ -] ]g g20. 6 6y y- +^ ^h h21. a a3 1 3 1+ -] ]g g22. 2 7 2 7z z- +] ]g g23. 9 2 2x x y+ - +] g h24. b a b3 2 2 1- + -] ]g g25. 2 2 4x x x2+ - +] g h26. 3 3 9a a a2- + +
] ^g h
27. 9a 2+] g 28. 4k 2-] g 29. 2x 2+] g 30. 7y 2-^ h 31. 2 3x 2+] g 32. 2 1t 2-] g
2.5 Exercises
Expand and simplify
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33. 3 4a b 2+] g 34. 5x y2-^ h 35. 2a b 2+] g
36. a b a b- +
] ]g g
37. a b 2+] g 38. a b 2-] g 39. a b a ab b2 2+ - +] ^g h
40. a b a ab b2 2
- + +
] ^g h
Some binomial products have special results and can be simplied 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 a b a b2 2+ - = -] ]g g
Proof
( ) ( )a b a b a ab ab b
a b
2 2
2 2
+ - = - + -
= -
a b a ab b22 2 2+ = + +] g
Perfect squares
Proof
( )( )
2
a b a b a b
a ab ab b
a ab b
2
2 2
2 2
+ = + +
= + + +
= + +
] g
2a b a ab b2 2 2- = - +] g
Proof
( )( )
2
a b a b a b
a ab ab b
a ab b
2
2 2
2 2
- = - -
= - - +
= - +
] g
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EXAMPLES
Expand and simplify
1. 2 3x 2-] g Solution
( )x x x
x x
2 3 2 2 2 3 3
4 12 9
2 2 2
2
- = - +
= - +
] ]g g
2. 3 4 3 4y y- +^ ^h hSolution
(3 4)(3 4) 4
9 16
y y y
y
3 2 2
2
- + = -
= -
^ h
1. 4t 2+] g 2. 6z 2-] g 3. x 1 2-] g
4. 8y 2+^ h 5. 3q 2+^ h 6. 7k 2-] g 7. n 1 2+] g
8. 2 5b 2+] g 9. 3 x 2-
] g
10. y3 1 2-^ h
11. x y2+^ h 12. a b3 2-] g
13. 4 5d e 2+] g 14. 4 4t t+ -] ]g g15. x x3 3- +] ]g g
16. p p1 1+ -^ ^h h
17. 6 6r r+ -] ]g g18. x x10 10- +] ]g g
19. 2 3 2 3a a+ -] ]g g20. 5 5x y x y - +^ ^h h21. a a4 1 4 1+ -] ]g g
22. 7 3 7 3x x- +] ]g g23. 2 2x x2 2+ -^ ^h h24. 5x2
2+
^ h
25. 3 4 3 4ab c ab c - +] ]g g26.
2x x
2
+b l
27.1 1
a a a a- +b bl l
28. x y x y 2 2+ - - -_ _i i6 6@ @
29. a b c2
+ +] g6 @
2.6 Exercises
Expand and simplify
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30. x y12
+ -] g7 A
31. a a3 32 2+ - -] ]g g
32. 16 4 4z z- - +] ]g g33. 2 3 1 4x x
2
+ + -] g 34. 2x y x y 2+ - -^ ^h h35. n n n4 3 4 3 2 52- + - +] ]g g
36. x 4 3-] g
37. x x x1 1
22 2
- - +b bl l 38. x y x y 42 2
2 2 2+ -_ i
39. 2 5a3
+
] g 40. x x x2 1 2 1 2 2- + +] ] ]g g g
Expand (x 4) (x 4) .- - 2
PROBLEM
Find values of all pronumerals that make this true.
i i c c b
a b c
d e
f e b
i i i h g
#
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 5xare 1, 5, xand 5x.
To factorise an expression, we use the distributive law.
a bax bx x ++ = ] g
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EXAMPLES
Factorise
1. 3 12x +
Solution
The highest common factor is 3.
x x3 12 3 4+ = +] g
2. 2y y2 -
Solution
The highest common factor isy.
y y y y2 22 - = -^ h
3. 2x x3 2-
Solution
x and x2are both common factors. We take out the highest common
factor which is x2.
x x x x2 23 2 2- = -] g
4. x xy5 3 32+ ++] ]g gSolution
The highest common factor is 3x + .
x x x y y5 3 3 3 5 22+ + + ++ =] ] ] ^g g g h
5. 8 2a b ab3 2 3-
Solution
There are several common factors here. The highest common
factor is 2ab2.
8 2 2 4a b ab ab a b3 2 3 2 2
- = -^ h
Check answers by
expanding brackets.
Divide each term by 3 to
nd the terms inside the
brackets.
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1. 2 6y+
2. x5 10-
3. 3 9m -
4. 8 2x +
5. y24 18-
6. 2x x2 +
7. 3m m2 -
8. 2 4y y2
+
9. 15 3a a2-
10. ab ab2 +
11. 4 2x y xy 2 -
12. 3 9mn mn3 +
13. 8 2x z xz2 2-
14. 6 3 2ab a a2+ -
15. 5 2x x xy 2 - +
16. 3 2q q5 2-
17. 5 15b b3 2+
18. 6 3a b a b2 3 3 2-
19. x m m5 7 5+ + +] ]g g
20. y y y2 1 1- - -^ ^h h
21. 4 7 3 7y x y+ - +^ ^h h22. 6 2 5 2x a a- + -] ]g g23. x t y t2 1 2 1+ - +] ]g g
24. a x b x3 2 2 3 2- + -] ]g gc x3 3 2- -] g
25. 6 9x x3 2+
26. 3 6pq q5 3-
27. 15 3a b ab4 3 +
28. 4 24x x3 2-
29. 35 25m n m n3 4 2-
30. 24 16a b ab2 5 2+
31. r rh2 22r r+
32. 3 5 3x x2
- + -] ]g g33. 4 2 4y x x2 + + +] ]g g34. a a a1 1 2+ - +] ]g g
35. ab a a4 1 3 12 2+ - +^ ^h h
2.7 Exercises
Factorise
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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EXAMPLES
Factorise
1. 2 3 6x x x2 - + -
Solution
2 3 6 ( 2) 3( 2)
( 2)( 3)
x x x x x x
x x
2- + - = - + -
= - +
2. 2 4 6 3x y xy - + -
Solution
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
x y xy x y x
x y x
x y
x y xy x y x
x y x
x y
2 4 6 3 2 2 3 2
2 2 3 2
2 2 3
2 4 6 3 2 2 3 2
2 2 3 2
2 2 3
or
- + - = - + -
= - - -
= - -
- + - = - - - +
= - - -
= - -
1. 2 8 4x bx b+ + +
2. 3 3ay a by b- + -
3. x x x5 2 102 + + +
4. 2 3 6m m m2 - + -
5. ad ac bd bc - + -
6. 3 3x x x3 2+ + +
7. ab b a5 3 10 6- + -
8. 2 2xy x y xy 2 2- + -
9. ay a y 1+ + +
10. 5 5x x x2 + - -
11. 3 3y ay a+ + +
12. 2 4 2m y my - + -
13. x xy xy y 2 10 3 152 2+ - -
14. 4 4a b ab a b2 3 2+ - -
15. x x x5 3 152- - +
16. 7 4 28x x x4 3+ - -
17. 7 21 3x xy y - - +
18. 4 12 3d de e+ - -
19. x xy y 3 12 4+- -
20. a ab b2 6 3+ - -
21. x x x3 6 183 2 +- -
22. pq p q q3 32+- -
2.8 Exercises
Factorise
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23. x x x3 6 5 103 2- - +
24. 4 12 3a b ac bc - + -
25. 7 4 28xy x y + - -
26. x x x4 5 204 3
- - +
27. x x x4 6 8 123 2- + -
28. 3 9 6 18a a ab b2 + + +
29. y xy x5 15 10 30+- -
30. r r r2 3 62
r r+ - -
Trinomials
A trinomialis an expression with three terms, for example 4 3.x x2 - +
Factorising a trinomial usually gives a binomial product.
x a b x a x bx ab2 + + ++ + =] ] ]g g g
Proof
( )
( ) ( )
( ) ( )
x a b x ab x ax bx ab
x x a b x a
x a x b
2 2+ + + = + + +
= + + +
= + +
EXAMPLES
Factorise
1. 5 6m m2 - +
Solution
a b 5+ = - and 6ab = +
62
3
5
+-
-
-
'
Numbers with sum 5- and product 6+ are 2- and 3.-
[ ] [ ]m m m m
m m
5 6 2 3
2 3
2` - + = + - + -
= - -
] ]] ]
g gg g
2. 2y y2 + -
Solution
1a b+ = + and 2ab = -
2211
-+
-
+
'
Two numbers with sum 1+ and product 2- are 2+ and 1- .
y y y y2 2 12` + - = + -^ ^h h
Guess and check by
trying 2- and 3-
or 1- and .6-
Guess and check by
trying 2 and 1- or
2- and 1.
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The result x a b x a x bx ab2
+ + + ++ =] ] ]g g gonly works when the coefcientof x2(the number in front of x2) is 1. When the coefcient of x2is not 1, forexample in the expression 5 2 4,x x2 - + 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.
1. 4 3x x2 + +
2. y y7 122 + +
3. m m2 12 + +
4. t t8 162 + +
5. 6z z2 + -
6. 5 6x x2 - -
7. v v8 152 - +
8. 6 9t t
2- +
9. x x9 102 + -
10. 10 21y y2 - +
11. m m9 182 - +
12. y y9 362 + -
13. 5 24x x2 - -
14. 4 4a a2 - +
15. x x14 322 + -
16. 5 36y y2 - -
17. n n10 242 +-
18. x x10 252 +-
19. p p8 92 + -
20. k k7 102 +-
21. x x 12
2+ -
22. m m6 72 - -
23. 12 20q q2 + +
24. d d4 52 - -
25. l l11 182 +-
2.9 Exercises
Factorise
EXAMPLES
Factorise
1. 5 13 6y y2 - +
Solutionguess and check
For 5y2, one bracket will have 5yand the othery:
.y y5^ ^h h Now look at the constant (term withoutyin it): .6+
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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 13yso the numbers cannot be both
positive. They must both be negative.
y y5 - -
^ ^h hTo get a product of 6, the numbers must be 2 and 3 or 1 and 6.Guess 2 and 3 and check:
3 5 15 2 6
5 17 6
y y y y y
y y
5 2 2
2
- = - - +
= - +
-^ ^h h
This is not correct.
Notice that we are mainly interested in checking the middle two terms,
.y y15 2and- -
Try 2 and 3 the other way around:
.y y5 3 2- -^ ^h h Checking the middle terms: y y y10 3 13- - = -
This is correct, so the answer is .y y5 3 2- -^ ^h h Note: If this did not check out, do the same with 1 and 6.Solutioncross method
Factors of 5y2are 5yandy.
Factors of 6 are 1- and 6- or 2- and .3-
Possible combinations that give a middle term of y13- are
By guessing and checking, we choose the correct combination.
y13-
y y
y y
5 2 10
3 3
#
#
- = -
- = -
y y y y5 13 6 5 3 22` - + = - -^ ^h hSolutionPSF method
P:Product of rst and last terms 30y2
S:Sum or middle term y13-
F:Factors of Pthat give S ,y y3 10- -
y yy
y
30 310
13
2 -
-
-
)
y y y y y
y y y
y y
5 13 6 5 3 10 6
5 3 2 5 3
5 3 2
2 2` - + = - - +
= - - -
= - -
^ ^^ ^
h hh h
5y
y 3-
2- 5y
y 2-
3- 5y
y 6-
1- 5y
y 1-
6-
5y
y 2-
3-
CONTINUED
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2. 4 4 3y y2 + -
Solutionguess and check
For 4y2, both brackets will have 2yor one bracket will have 4yand the
othery.Try 2yin each bracket:
.y y2 2^ ^h 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.
y y2 2 +-^ ^h h
To get a product of 3, the numbers must be 1 and 3.
Guess and check:
y y2 3 2 1+-^ ^h h
Checking the middle terms: y y y2 6 4- = -
This is almost correct, as the sign is wrong but the coefcient is right
(the number in front ofy).
Swap the signs around:4 6 2 3
4 4 3
y y y y y
y y
2 1 2 3 2
2
+ = +
= +
- - -
-
^ ^h h
This is correct, so the answer is .y y2 1 2 3- +^ ^h h
Solutioncross method
Factors of 4y2are 4yandyor 2yand 2y.
Factors of 3 are 1- and 3 or 3- and 1.
Trying combinations of these factors givesy y
y y
y
2 1 2
2 3 6
4
#
#
- = -
=
y y y y4 4 3 2 3 2 12` + - = + -^ ^h h
SolutionPSF method
P:Product of rst and last terms y12 2-
S:Sum or middle term 4y
F:Factors of Pthat give S ,y y6 2+ -
y yy
y
12 62
4
2-+
-
+
) y y y y y
y y y
y y
4 4 3 4 6 2 3
2 2 3 1 2 3
2 3 2 1
2 2` + - = + - -
= + - +
= + -
^ ^
^ ^
h h
h h
2y
2y 1-
3
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63Chapter 2Algebra and Surds
Perfect squares
You have looked at some special binomial products, including
2a b a ab b2 2 2+ = + +] g and 2 .a b a ab b2 2 2- = - +] g When factorising, use these results the other way around.
Factorise
1. a a2 11 52 + +
2. 5 7 2y y2 + +
3. x x3 10 72 + +
4. 3 8 4x x2 + +
5. 2 5 3b b2 - +
6. 7 9 2x x2 - +
7. 3 5 2y y2 + -
8. x x2 11 12
2+ +
9. p p5 13 62 + -
10. x x6 13 52 + +
11. y y2 11 62 - -
12. x x10 3 12 + -
13. 8 14 3t t2 - +
14. x x6 122 - -
15. 6 47 8y y2 + -
16. n n4 11 62 +-
17. t t8 18 52 + -
18. q q12 23 102 + +
19. r r8 22 62 + -
20. x x4 4 152 - -
21. y y6 13 22 +-
22. p p6 5 62 - -
23. x x8 31 21
2+ +
24. b b12 43 362 +-
25. x x6 53 92 - -
26. 9 30 25x x2 + +
27. 16 24 9y y2 + +
28. k k25 20 42 +-
29. a a36 12 12 +-
30. m m49 84 362 + +
2.10 Exercises
a ab b a b
a ab b a b
2
2
2 2 2
2 2 2
+ + = +
- + = -
]]
gg
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EXAMPLES
Factorise
1. 8 16x x2 - +
Solution
8 16 2(4) 4x x x x
x 4
2 2 2
2
- + = - +
= -] g
2. 4 20 25a a2 + +
Solution
4 20 25 2(2 )(5) 5a a a a
a
2
2 5
2 2 2
2
+ + = + +
= +
]
]
g
g
Factorise
1. y y2 12 - +
2. 6 9x x2
+ +
3. m m10 252 + +
4. 4 4t t2 - +
5. x x12 362 - +
6. x x4 12 92 + +
7. b b16 8 12 - +
8. a a9 12 42 + +
9. x x25 40 162 - +
10. y y49 14 12 + +
11. y y9 30 252 +-
12. k k16 24 92 +-
13. 25 10 1x x2
+ +
14. a a81 36 42 +-
15. 49 84 36m m2 + +
16. t t412
+ +
17. x x
34
942
- +
18. yy
95
6
2512
+ +
19. xx
2 122
+ +
20. kk
25 04
222
- +
2.11 Exercises
In a perfect square, the
constant term is always a
square number.
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65Chapter 2Algebra and Surds
Difference of 2 squares
A special case of binomial products is a b a b a b2 2+ - = -] ]g g .
a b a ba b2 2 + -- = ] ]g g
EXAMPLES
Factorise
1. 36d2 -
Solution
d d
d d
36 6
6 6
2 2 2=
= +
- -
-] ]g g
2. b9 12 -
Solution
( ) ( )
b b
b b
9 1 3 1
3 1 3 1
2 2 2- = -
= + -
] g
3. ( ) ( )a b3 12 2+ - -
Solution
[( ) ( )][( ) ( )]
( ) ( )
( )( )
a b a b a b
a b a b
a b a b
3 1 3 1 3 1
3 1 3 1
2 4
2 2+ - - = + + - + - -
= + + - + - +
= + + - +
] ]g g
Factorise
1. 4a2 -
2. 9x2 -
3. y 12 -
4. 25x2 -
5. 4 49x2 -
6. 16 9y2 -
7. z1 4 2-
8. t25 12 -
9. 9 4t2 -
10. x9 16 2-
11. 4x y2 2-
12. 36x y2 2-
2.12 Exercises
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13. 4 9a b2 2-
14. x y1002 2-
15. 4 81a b2 2-
16. 2x y2 2
+ -
] g 17. a b1 22 2- - -] ]g g 18. z w12 2- +] g 19. x
412
-
20.y
91
2
-
21. x y2 2 12 2+ - +] ^g h 22. x 14 -
23. 9 4x y6 2-
24. x y164 4-
25. 1a8 -
Sums and differences of 2 cubes
a b a ab ba b3 3 2 2+ - ++ = ] ^g h
a b a b a ab b3 3 2 2- = - + +] ^g h
Proof
( ) ( )a b a ab b a a b ab a b ab b
a b
2 2 3 2 2 2 2 3
3 3
+ - + = - + + - +
= +
Proof
( ) ( )a b a ab b a a b ab a b ab b
a b
2 2 3 2 2 2 2 3
3 3
- + + = + + - - -
= -
EXAMPLES
Factorise
1. 8 1x3 +
Solution
( ) [ ( )( ) ]
( ) ( )
x x
x x x
x x x
8 1 2 1
2 1 2 2 1 1
2 1 4 2 1
3 3 3
2 2
2
+ = +
= + - +
= + - +
]]
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67Chapter 2Algebra and Surds
Factorise
1. b 83 -
2. 27x3
+
3. 1t3 +
4. 64a3 -
5. 1 x3-
6. 8 27y3+
7. 8y z3 3+
8. 125x y3 3-
9. 8 27x y3 3+
10. 1a b3 3 -
11. 1000 8t3+
12.x8
273
-
13.a b
1000 13 3
+
14. x y1 3 3+ -] g 15. x y z216125 3 3 3+
16. 2 1a a3 3- - +] ]g g 17.
x1
27
3
-
18. 3y x3 3+ +] g 19. x y1 23 3+ + -] ^g h 20. 8 3a b3 3+ -] g
2.13 Exercises
2. 27 64a b3 3-
Solution
( )[ ( )( ) ]( )( )
a b a b
a b a a b ba b a ab b
27 64 3 4
3 4 3 3 4 43 4 9 12 16
3 3 3 3
2 2
2 2
- = -
= - + +
= - + +
] ]
] ]
g g
g g
Mixed factors
Sometimes more than one method of factorising is needed to completely
factorise an expression.
EXAMPLE
Factorise 5 45.x2 -
Solution
5 45 5( 9) (using simple factors)
5( 3)( 3) (the difference of two squares)
x x
x x
2 2- = -
= + -
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Factorise
1. x2 182 -
2. p p3 3 362 - -
3. y5 53 -
4. 4 8 24a b a b ab a b3 2 2 2 2+ - -
5. a a5 10 52 - +
6. x x2 11 122- + -
7. z z z3 27 603 2+ +
8. ab a b9 43 3
-
9. x x3 -
10. x x6 8 82 + -
11. m n mn3 15 5- - +
12. x x3 42 2- - +] ]g g 13. y y y5 5162 + +-^ ^h h14. x x x8 84 3- + -
15. x 16 -
16. x x x3 103 2- -
17. x x x3 9 273 2- - +
18. 4x y y2 3 -
19. 24 3b3-
20. 18 33 30x x2 + -
21. 3 6 3x x2 - +
22. 2 25 50x x x3 2+ - -
23. 6 9z z z3 2
+ +
24. 4 13 9x x4 2- +
25. 2 2 8 8x x y x y 5 2 3 3 3+ - -
26. 4 36a a3 -
27. 40 5x x4-
28. a a13 364 2 +-
29. k k k4 40 1003 2+ +
30. x x x3 9 3 93 2+ - -
2.14 Exercises
You will study this in
Chapter 12.
DID YOU KNOW?
Long division can be used to nd factors of an expression. For example, 1x - is a factor of
4 5x x+ -3 . We can nd the other factor by dividing 4 5x x+ -3 by 1.x -
-
5
4
5 5
5 5
0
x x
x
x x
x x
x x
x
x
2
3
2
-
+ +
+
-
-
2
3
2
1x - + 4 5x -g
So the other factor of 4 5x x+ -3 is 5x x2 + +
4 5 ( 1) ( 5)x x x x x3
` + - = - + +2
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69Chapter 2Algebra and Surds
Completing the Square
Factorising a perfect square uses the results
a ab b a b22 2 2!! + = ] g
EXAMPLES
1. Complete the square on .x x62 +
Solution
Using 2 :a ab b2 2+ +
a x
ab x2 6
=
=
Substituting :a x=
xb x
b
2 6
3
=
=
To complete the square:
a ab b a b
x x x
x x x
2
2 3 3 3
6 9 3
2 2 2
2 2 2
2 2
+ + = +
+ + = +
+ + = +
]] ]
]
gg g
g
2. Complete the square on .n n102 -
Solution
Using :a ab b22 2+-
a n
ab x2 10
=
=
Substituting :a n=
nb nb
2 10
5
=
=
To complete the square:
a ab b a b
n n n
n n n
2
2 5 5 5
10 25 5
2 2 2
2 2 2
2 2
- + = -
- + = -
- + = -
]] ]
]
gg g
g
Notice that 3 is half of 6.
Notice that 5 is half of 10.
To complete the square on ,a pa2 + dividepby 2 and square it.
2 2a pa
pa
p2
2 2
+ + = +d dn n
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EXAMPLES
1. Complete the square on .x x122 +
Solution
Divide 12 by 2 and square it:
x x x x
x x
x
122
1212 6
12 36
6
22
2 2
2
2
+ + = + +
= + +
= +
c
]
m
g
2. Complete the square on .y y22 -
Solution
Divide 2 by 2 and square it:
y y y y
y y
y
222
2 1
2 1
1
22
2 2
2
2
+ = +
= +
=
- -
-
-
c
^
m
h
Complete the square on
1. x x42 +
2. 6b b2 -
3. 10x x2 -
4. 8y y2 +
5. 14m m2 -
6. 18q q2 +
7. 2x x2 +
8. 16t t2 -
9. 20x x2 -
10. 44w w2 +
11. 32x x2 -
12.y y32 +
13. 7x x2 -
14. a a2 +
15. 9x x2 +
16.yy
2
52
-
17. k k
2112
-
18. 6x xy2 +
19. a ab42 -
20.p pq82 -
2.15 Exercises
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71Chapter 2Algebra and Surds
Simplify
1.a
55 10+
2.t
36 3-
3.y
6
8 2+
4.d4 2
8
-
5.x x
x
5 22
2
-
6.y y
y
8 16
42
- +
-
7.a a
ab a
3
2 42
2
-
-
8. s s
s s
5 6
22
2
+ +
+ -
9.b
b
1
12
3
-
-
10.p
p p
6 9
2 7 152
-
+ -
11.
a a
a
2 3
12
2
+ -
-
12.x
x xy
8
2 233
-
- -+] ]g g
13.x x
x x x
6 9
3 9 272
3 2
+ +
+ - -
14.p
p p
8 1
2 3 23
2
+
- -
15. 2 2ay by ax bx
ay ax by bx
- - +
- + -
2.16 Exercises
Algebraic Fractions
Simplifying fractions
EXAMPLES
Simplify
1.2
4 2x +
Solution
x x
x2
4 22
2 1
2 1
2+=
+
= +
] g
2.8
2 3 2
x
x x3
2
-
- -
Solution
x
x x
x x x
x x
x x
x
8
2 3 2
2 2 4
2 1 2
2 4
2 1
3
2
2
2
-
- -=
- + +
+ -
=
+ +
+
] ^] ]
g hg g
Factorise rst, then cancel.
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Operations with algebraic fractions
EXAMPLES
Simplify
1.x x
51
43-
- +
Solution
x x x x
x x
x
51
43
20
4 1 3
204 4 5 15
2019
5--
+=
- +
= - - -
= - -
-] ]g g
2.b
a b ab
b
a
27
2 10
4 12
253
2 2
'+
+
+
-
Solution
b
a b ab
b
a
b
a b ab
a
b
b b b
ab a
a a
b
a b b
ab
27
2 10
4 12
25
27
2 10
25
4 12
3 3 9
2 5
5 5
4 3
5 3 9
8
3
2 2
3
2
2
2
2
' #
#
+
+
+
-=
+
+
-
+
=
+ - +
+
+ -
+
=- - +
] ^]
] ]]
] ^
g hg
g gg
g h
3.5
22
1x x-
++
Solution
x x x x
x x
x x
x x
x xx
52
21
5 2
2 5
5 2
2 4 5
5 23 1
2 1
-+
+=
- +
+ -
=
- +
+ + -
=
- +
-
+
] ]] ]
] ]] ]
g gg g
g gg g
Do algebraic fractions
the same way as ordinary
fractions.
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73Chapter 2Algebra and Surds
1. Simplify
(a)2 4
3x x+
(b)5
1
3
2y y++
(c)3
24
a a+-
(d)6
3
2
2p p-+
+
(e)2
53
1x x--
-
2. Simplify
(a) 2
3
6 3
2
b a
b b2#
+ -
+
(b)2 1
4
2
1
q q
p
p
q2
2 3
#
+ +
-
+
+
(c)xyab
x y xy
ab a53
2
12 62
2 2'
+
-
(d)x y
ax ay bx by
ab a b
x y2 2 2 2
3 3
#
-
- + -
+
+
(e)x
x x
x x
x x
25
6 9
4 5
5 62
2
2
2
'-
- +
+ -
- +
3. Simplify
(a)2 3x x+
(b)1
1 2x x-
-
(c) 13
a b+
+
(d)2
xx
x2-
+
(e)1
p q
p q
- ++
(f)1
13
1x x+
+-
(g) 42 23x x2 --
+
(h)2 1
11
1
a a a2 + ++
+
(i)2
23
11
5y y y+
-+
+-
(j)16
2
12
7
x x x2 2--
- -
4. Simplify
(a)y
x
x
y
y
x x
4 12
3
6 24
9
27
2 822
3
2
# #
- -
-
+
- -
(b)y y
a a
y
aay
y y
4 4
5
4
3 155
22
2
2
2
' #- +
-
-
- - -
(c)x x
x
x
x x
33
9
2 84 16
32
2
#-
+
-
+
-
+
(d)b
b
b b
b
b
b
2 6
5
6 12
2
'+ + -
-+
(e)x x
x x
x
xx
x x
5 10
8 15
10
92 10
5 62
2
2
2 2
' #+
- + -
-
+ +
5. Simplify
(a)7 10
1
2 15
2
6
4
x x x x x x2 2 2- +-
- -
+
+ -
(b)4
52
32
2
x x x2 --
--
+
(c)2 3
p pq pq q2 2++
-
(d)1
a b
a
a b
b
a b2 2+-
-+
-
(e) x y
x y
y x
x
y x
y
2 2-
+
+-
-
-
2.17 Exercises
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 land breadth b. We
can substitute any values for land bto nd the area of different rectangles.
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EXAMPLES
1. P l b2 2= + is the formula for nding the perimeter of a rectangle
with length land breadth b. FindPwhen .l 1 3= and . .b 3 2=
Solution
. .
. .
P l b2 2
2 1 3 2 3 2
2 6 6 4
9
= +
= +
= +
=
] ]g g
2. V r h2r= is the formula for nding the volume of a cylinder with
radius rand height h. Find V(correct to 1 decimal place) when 2.1r=
and 8.7.h =
Solution
. ( . )
120.5
V r h
2 1 8 7
correct to1 decimal place
2
2
r
r
=
=
=
] g
3. IfF C
59
32= + is the formula for changing degrees Celsius C] gintodegrees Fahrenheit F] gndFwhen 25.C = Solution
F C
59
32
5
2532
5225
32
5225 160
5385
77
9
= +
= +
= +
= +
=
=
] g
This means that 25 Cis the same as .77 F
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75Chapter 2Algebra and Surds
1. Given 3.1a = and 2.3b = - nd,
correct to 1 decimal place.
(a) ab3(b) b
(c) a5 2
(d) ab3
(e) a b 2+] g (f) a b-
(g) b2-
2. T a n d1= + -] g is the formulafor nding the term of an
arithmetic series. Find Twhen
,a n4 18= - = and .d 3=
3. Given ,y mx b= + the equation
of a straight line, ndyif
,m x3 2= = - and 1.b = -
4. If 100 5h t t2= - is the height of
a particle at time t, nd hwhen
5.t=
5. Given vertical velocity ,v gt= -
nd v when 9.8g= and 20.t=
6. If 2 3y x= + is the equation of
a function, ndywhen 1.3,x =
correct to 1 decimal place.
7. S r r h2r= +] gis the formula forthe surface area of a cylinder.
Find S when 5r= and 7,h =
correct to the nearest whole
number.
8. A r2r= is the area of a circle with
radius r. FindAwhen 9.5,r=
correct to 3 signicant gures.
9. Given u ar 1n
n=
- is the nth term
of a geometric series, nd unif
5,a = 2r= - and 4.n =
10. Given3
V lbh= 1 is the volume
formula for a rectangular
pyramid, nd Vif . , .l b4 7 5 1= =
and 6.5.h =
11. The gradient of a straight line is
given by .m x x
y y
2 1
2 1=
-
-
Find m
if , ,x x y3 1 21 2 1
= = - = - and
5.y2
=
12. If2
A h a b= +1 ] ggives the areaof a trapezium, ndAwhen
, .h a7 2 5= = and 3.9.b =
13. Find Vif3
V r3r= 4 is the volume
formula for a sphere with radius r
and 7.6,r= to 1 decimal place.
14. The velocity of an object at a
certain time tis given by the
formula .v u at = + Find vwhen
4 5,u a= =1 3and
6.t= 5
15. Given1
,Sr
a=
-nd Sif 5a =
and3
.r= 2 Sis the sum to innity
of a geometric series.
16. ,c a b2 2= + according to
Pythagoras theorem. Find thevalue of cif 6a = and 8.b =
17. Given 16y x2= - is the
equation of a semicircle, nd the
exact value ofywhen 2.x =
2.18 Exercises
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18. Find the value ofEin the energy
equationE mc2= if 8.3m = and
1.7.c=
19. 1100
A P r n
= +c m is the formulafor nding compound interest.
FindAwhen ,P r200 12= = and
5,n = correct to 2 decimal places.
20. If Sr
ra
1
1=
-
-n^ h
is the sum of
a geometric series, nd Sif
,a r3 2= = and 5.n =
21. Find the value of ca b2
3 2
if
4 3
,a b2 3
= =3 2c cm m and .c
21 4
= c m
Surds
An irrational numberis a number that cannot be written as a ratio or fraction
(rational). Surdsare special types of irrational numbers, such as 2, 3and 5 .
Some surds give rational values: for example, 9 3.=
Others, like 2,donot 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
a b ab
a bb
a
b
a
#
'
=
= =
Class Investigations
Is there an exact decimal equivalent for1. 2?
Can you draw a line of length exactly2. 2?
Do these calculations give the same results?3.
(a) 9 4# and 9 4#
(b)9
4and
94
(c) 9 4+ and 9 4+
(d) 9 4- and 9 4-
Here are some basic properties of surds.
x x x2 2
= =^ h
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EXAMPLES
1. Express in simplest surd form 45 .
Solution
45 9 5
9 5
3 5
3 5
#
#
#
=
=
=
=
2. Simplify 3 40 .
Solution
3 33
3 2
6
40 4 104 10
10
10
#
# #
# #
=
=
=
=
3. Write 5 2as a single surd.
Solution
5 2 25 2
50
#=
=
54 also equals
3 15# but this will
not simplify. We look
for a number that is a
perfect square.
Find a factor of 40 that
is a perfect square.
1. Express these surds in simplest
surd form.
(a) 12
(b) 63
(c) 24(d) 50
(e) 72
(f) 200
(g) 48
(h) 75
(i) 32
(j) 54
(k) 112
(l) 300
(m) 128
(n) 243
(o) 245
(p) 108
(q) 99
(r) 125
2. Simplify
(a) 2 27
(b) 5 80
2.19 Exercises
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(c) 4 98
(d) 2 28
(e) 8 20
(f) 4 56
(g) 8 405(h) 15 8
(i) 7 40
(j) 8 45
3. Write as a single surd.
(a) 3 2
(b) 2 5
(c) 4 11
(d) 8 2
(e) 5 3
(f) 4 10
(g) 3 13
(h) 7 2
(i) 11 3
(j) 12 7
4. Evaluate xif
(a) 3 5x =
(b) 2 3 x=
(c) 3 7 x=
(d) 5 2 x=
(e) 2 11 x=
(f) 7 3x =
(g) 4 19 x=
(h) 6 23x =
(i) 5 31 x=
(j) 8 15x =
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 4 72 2 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- + = +
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79Chapter 2Algebra and Surds
Multiplication and division
Simplify
1. 5 2 5+
2. 3 2 2 2-
3. 3 5 3+
4. 7 3 4 3-
5. 5 4 5-
6. 4 6 6-
7. 2 8 2-
8. 5 4 5 3 5+ +
9. 2 2 2 3 2- -
10. 5 45+
11. 8 2-
12. 3 48+
13. 12 27-
14. 50 32-
15. 28 63+
16. 2 8 18-
17. 3 54 2 24+
18. 90 5 40 2 10- -
19. 4 48 3 147 5 12+ +
20. 3 2 8 12+ -
21. 2863 50--
22. 12 45 48 5-- -
23. 150 45 24+ +
24. 32 243 50 147-- +
25. 80 3 245 2 50- +
2.20 Exercises
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 a2
#
#
#
=
=
= =
EXAMPLES
Simplify
1. 2 2 5 7# -
Solution
2 2 5 7 10 14# - = -
b
a
b
a=
CONTINUED
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2. 4 2 5 18#
Solution
4 2 5 18 20 36
20 6
120
#
#
=
=
=
3.4 2
2 14
Solution
4 2
2 14
4 2
2 2 7
27
#=
=
4.15 2
3 10
Solution
15 2
3 10
15 2
3 5 2
55
# #=
=
5.310
2d n
Solution
33
310
3
10
3
10
2
2
2
=
=
=1
d ^^n h
h
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81Chapter 2Algebra and Surds
Simplify
1. 7 3#
2. 3 5#
3. 2 3 3#
4. 5 7 2 2#
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. 28 6#
12. 3 2 5 14#
13. 10 2 2#
14. 2 6 7 6# -
15. 2 2^ h 16. 2 7
2^ h 17. 3 5 2# #
18. 2 3 7 5# # -
19. 2 6 3 3# #
20. 2 5 3 2 5 5# #- -
21. 2 2
4 12
22.3 6
12 18
23.10 2
5 8
24.2 12
16 2
25.5 10
10 30
26.6 20
2 2
27.8 10
4 2
28.3 15
3
29.8
2
30.6 10
3 15
31.5 8
5 12
32.10 10
15 18
33.2 6
15
34.32
2d n
35.
7
52
d n
2.21 Exercises
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.
b c ab ac a + = +^ h
Proofa b c a b a c
ab ac
# #+ = +
= +
^ h
Binomial product:
a b c d ad bd ac bc + + = + + +^ ^h h Proof
a b c d a c a d b c b d
ac ad bc bd
# # # #+ + = + + +
= + + +
^ ^h h
Perfect squares:
a b a ab b22
+ = + +^ h Proof
a b a b a b
a ab ab b
a ab b2
2
2 2
+ = + +
= + + +
= + +
^ ^ ^h h h
a b a ab b22
- = - +^ h Proof
a b a b a b
a ab ab b
a ab b2
2
2 2
- = - -
= - - +
= - +
^ ^ ^h h h
Difference of two squares:
a b a b a b+ - = -^ ^h h
Proof
a b a b a ab ab b
a b
2 2+ - = - + -
= -
^ ^h h
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83Chapter 2Algebra and Surds
EXAMPLES
Expand and simplify
1. 2 5 2+^ hSolution
( )2 5 2 2 5 2 2
10 4
10 2
# #+ = +
= +
= +
2. 3 7 2 3 3 2-^ hSolution
( )3 7 2 3 3 2 3 7 2 3 3 7 3 2
6 21 9 14
# #- = -
= -
3. 2 3 5 3 2+ -^ ^h hSolution
( ) ( )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 5 2 3+ -^ ^h hSolution
( 2 )( 2 ) 2 2 2 2
5 2 2 4 3
5 12
7
5 3 5 3 5 5 5 3 3 5 3 3
15 15
# # # #
#
+ - = - + -
= - + -
= -
= -
Another way to do this question is by using the difference of two squares.
( ) ( )5 2 3 5 2 3 5 2 35 4 3
7
2 2
#
+ - = -
= -
= -
^ ^h h
Notice that using the
difference of two
squares gives a rational
answer.
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1. Expand and simplify(a) 5 32 +^ h(b) 2 2 53 -^ h(c) 3 3 2 54 +^ h(d) 5 2 2 37 -^ h(e) 3 2 4 6-- ^ h(f) 3 5 11 3 7+^ h(g) 3 2 2 4 3- +^ h(h) 5 5 35 -^ h(i) 3 12 10+^ h(j) 2 3 18 3+^ h(k) 4 2 3 62- -^ h(l) 7 3 20 2 35- - +^ h(m)10 3 2 2 12-^ h(n) 5 22 +- ^ h(o) 2 3 2 12-^ h
2. Expand and simplify(a) 2 3 5 3 3+ +^ ^h h(b) 5 2 2 7- -^ ^h h(c) 2 5 3 2 5 3 2+ -^ ^h h(d) 3 10 2 5 4 2 6 6- +^ ^h h(e) 2 5 7 2 5 3 2- -^ ^h h(f) 5 6 2 3 5 3+ -^ ^h h(g) 7 3 7 3+ -^ ^h h(h) 2 3 2 3- +^ ^h h(i) 6 3 2 6 3 2+ -^ ^h h(j) 3 5 2 3 5 2+ -^ ^h h(k) 8 5 8 5- +
^ ^h h
(l) 2 9 3 2 9 3+ -^ ^h h
(m) 2 11 5 2 2 11 5 2+ -^ ^h h(n) 5 2
2+^ h
(o) 2 2 3 2-^ h (p) 3 2 7
2+^ h
(q) 2 3 3 52
+^ h (r) 7 2 5
2-^ h
(s) 2 8 3 52
-^ h (t) 3 5 2 2
2+^ h
3. If 3 2a = , simplify
(a) a2
2(b) a3
(2(c) a)3
(d) 1a 2+] g (e) a a3 3+] ]g g
4. Evaluate aand bif
(a) 2 5 1 a b2
+ = +^ h (b) 2 2 5 2 3 5- -^ ^h h
a b 10= +
5. Expand and simplify
(a) a a3 2 3 2+ - + +^ ^h h(b) 1p p 2- -_ i
6. Evaluate kif
.k2 7 3 2 7 3- + =^ ^h h 7. Simplify .x y x y 2 3+ -_ _i i 8. If 2 3 5 a b
2- = -^ h , evaluate
aand b.
9. Evaluate aand bif
.a b7 2 3 22
- = +^ h 10. A rectangle has sides 5 1+ and
2 5 1- . Find its exact area.
2.22 Exercises
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
rationalising the denominator,5
3becomes
5
3 5.
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85Chapter 2Algebra and Surds
Squaring a surd in the denominator will rationalise it since .x x2
=^ h
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,
5
3 3 2.236'=
5
3 53 2.236 5# '=
This is hard to do
without a calculator.
This is easier to calculate.
b
abb ba b
# =
Multiplying by
b
b
is the same as
multiplying by 1.
Proof
b
a
b
b
b
a b
b
a b
2# =
=
EXAMPLES
1. Rationalise the denominator of5
3.
Solution
5
3
5
5
5
3 5# =
2. Rationalise the denominator of
5 3
2.
Solution
5 3
2
3
3
5 9
2 3
5 3
2 3
15
2 3
#
#
=
=
=
Dont multiply by
5 3
5 3as it takes
longer to simplify.
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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 ofc d
a b
+
+, multiply by
c d
c d
-
-
Proof
c d
a b
c d
c d
c d
a b c d
c d
a b c d
c d
a b c d
c d
2 2
#
+
+
-
-=
+ -
+ -
=
-
+ -
=-
+ -
^ ^^ ^
^ ^^ ^
^ ^
h hh h
h hh h
h h
EXAMPLES
1. Write with a rational denominator
.2 3
5
-
Solution
2 35
2 32 3
2 35 2 3
2 9
10 3 5
7
10 3 5
7
10 3 5
2 2#
- +
+ =
-
+
=-
+
=-
+
= -+
^ ^ h h
2. Write with a rational denominator
3 4 2
2 3 5.
+
+
Solution
3 4 2
2 3 5
3 4 2
3 4 2
3 4 2
2 3 5 3 4 2
3 16 2
2 3 8 6 15 4 10
2 2#
#
#
+
+
-
-=
-
+ -
=-
- + -
^ ^^ ^
h hh h
Multiply by the conjugate
surd 2 3.+
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87Chapter 2Algebra and Surds
29
6 8 6 15 4 10
29
6 8 6 15 4 10
=-
- + -
=- + - +
3. Evaluate aand bif .a b3 2
3 3
-
= +
Solution
3 2
3 3
3 2
3 2
3 2 3 2
3 3 3 2
3 2
3 9 3 6
3 2
3 3 3 6
1
9 3 6
9 3 6
9 9 6
9 54
2 2
#
#
#
- +
+=
- +
+
=
-
+
=
-
+
=+
= +
= +
= +
^ ^^
^ ^
h hh
h h
.a b9 54So and= =
4. Evaluate as a fraction with rational denominator
3 2
2
3 2
5.
+
+
-
Solution
3 22
3 2
2
3 2
5
3 2 3 2
3 2
3 2
2 3 4 15 2 5
3 4
2 3 4 15 2 5
1
2 3 4 15 2 5
2 3 4 15 2 5
5
2 2
+
+
-
=
+ -
- +
=
-
- + +
=-
- + +
=-
- + +
= - + - -
+
^ ^^ ^
^
h hh h
h
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1. Express with rational
denominator
(a)7
1
(b)2 2
3
(c)5
2 3
(d)5 2
6 7
(e)3
1 2+
(f)2
6 5-
(g)5
5 2 2+
(h)2 7
3 2 4-
(i)4 5
8 3 2+
(j)7 5
4 3 2 2-
2. Express with rational
denominator
(a)3 2
4
+
(b)2 7
3
-
(c)5 2 6
2 3
+
(d)3 4
3 4
+
-
(e)3 2
2 5
-
+
(f)2 5 3 2
3 3 2
+
+
3. Express as a single fraction with
rational denominator
(a)2 1
12 1
1+
+-
(b)2 3
2
2 3
3
-
-
+
(c)5 2
1
3 2 5
3
+
+
-
(d)2 3
2 7
2 3 2
2#
+
-
+
(e) tt1
+ where t 3 2= -
(f) zz12 2- where z 1 2= +
(g)6 3
3 2 4
6 3
2 1
6 1
2
-
++
+
--
-
(h)2
2 3
3
1++
(i)2 3
3
3
2
+
+
(j)6 2
5
5 3
2
+
-
(k)4 3
2 7
4 3
2
+
+-
-
(l)3 2
5 2
3 1
2 3
-
--
+
+
4. Find aand bif
(a)b
a
2 5
3=
(b)b
a
4 2
3 6=
(c) a b5 1
25
+
= +
(d) a b7 4
2 77
-
= +
(e) a b2 1
2 3
-
+= +
2.23 Exercises
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89Chapter 2Algebra and Surds
5. Show that2 1
2 1
2
4
+
-+ is
rational.
6. If x 3 2= + , simplify
(a) 1x x+
(b)1
xx
2
2+
(c)1
x x
2
+b l
7. Write5 2
2
5 2
1
+
+
-
-
3
5 1+as a single fraction with
rational denominator.
8. Show that3 2 2
22
8+
+ is
rational.
9. If x21
3+ = , where ,x 0!
nd xas a surd with rational
denominator.
10. Rationalise the denominator of
2b
b 2
-
+ b 4!] g
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1. Simplify
(a) y y5 7-
(b)a
33 12+
(c) k k2 33 2#-
(d)x y
3 5+
(e) 4 3 5a b a b- - -
(f) 8 32+
(g) 3 5 20 45- +
2. Factorise
(a) 36x2 -
(b) 2 3a a2 + -
(c) 4 8ab ab2 -
(d) y xy x5 15 3- + -
(e) 4 2 6n p- +
(f) 8 x3-
3. Expand and simplify
(a) bb 23 -+
] g(b) x x2 1 3- +] ]g g(c) m m5 3 2+ --] ]g g(d) 4 3x 2-] g (e) 5 5p p- +^ ^h h(f) a a47 2 5+- -] g (g) 2 2 53 -^ h(h) 3 7 3 2+ -^ ^h h
4. Simplify
(a)
b
a
a
b
5
4 12
27
103 3
#-
-
(b)m m
mm
m
2
5 103 3
42
2
'- -
+
+
-
5. The volume of a cube is .V s3=
Evaluate Vwhen 5.4.s =
6. (a) Expand and simplify
.2 5 3 2 5 3+ -^ ^h h Rationalise the denominator of(b)
.2 5 3
3 3
+
7. Simplify .x x x x2
33
1
6
22
-+
+-
+ -
8. If ,a b4 3= = - and ,c 2= - nd the
value of(a) ab2
(b) a bc-
(c) a
(d) bc3] g (e) c a b2 3+] g
9. Simplify
(a)6 15
3 12
(b)2 2
4 32
10. The formula for the distance an object
falls is given by 5 .d t2= Find dwhen
1.5.t=
11. Rationalise the denominator of
(a)5 3
2
(b)2
1 3+
12. Expand and simplify
(a) 3 2 4 3 2- -^ ^h h(b) 7 2
2+^ h
13.Factorise fully
(a) 3 27x2 -
(b) x x6 12 182 - -
(c) 5 40y3 +
Test Yourself 2
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91Chapter 2Algebra and Surds
14. Simplify
(a)93
xyx y
5
4
(b)15 5
5x -
15. Simplify
(a) 3 112^ h
(b) 2 33^ h
16. Expand and simplify
(a) a b a b+ -] ]g g(b) a b 2+] g (c) a b 2-] g
17. Factorise
(a) 2a ab b2 2- +
(b) a b3 3-
18. If 3 1,x = + simplify1
x x+ and
give your answer with a rational
denominator.
19. Simplify
(a)
4 3
a b+
(b)2
35
2x x--
-
20. Simplify5 2
3
2 2 1
2,
+
-
-
writing
your answer with a rational denominator.
21. Simplify
(a) 3 8
(b) 2 2 4 3#-
(c) 108 48-
(d)2 18
8 6
(e) 5 3 2a b a# #- -
(f)6
2
m n
m n2 5
3
(g) 3 2x y x y - - -
22. Expand and simplify
(a) 2 3 22 +^ h(b) 5 7 3 5 2 2 3- -^ ^h h(c) 3 2 3 2+ -^ ^h h(d) 4 3 5 4 3 5- +^ ^h h(e) 3 7 2
2-^ h
23. Rationalise the denominator of
(a)7
3
(b)5 3
2
(c)5 1
2-
(d)3 2 3
2 2
+
(e)4 5 3 3
5 2
-
+
24. Simplify
(a)x x
53
22
- -
(b)a a
72
32 3+
+
-
(c)x x1
11
22
-
-+
(d)2 3
4
3
1
k k k2 + -+
+
(e)2 5
3
3 2
5
+
-
-
25. Evaluate nif
(a) 108 12 n- =
(b) 112 7 n+ = (c) 2 8 200 n+ =
(d) 4 147 3 75 n+ =
(e) n2 2452
180+ =
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26. Evaluate xx
12
2
+ if x1 2 3
1 2 3=
-
+
27. Rationalise the denominator of2 7
3
(there may be more than one answer).
(a)28
21
(b)28
2 21
(c)14
21
(d) 721
28. Simplify .x x
53
41-
- +
(a)x
20
7+- ] g
(b)20
7x +
(c)x
2017+
(d)x
20
17- +] g
29. Factorise 4 4x x x3 2- - + (there may be
more than one answer).
(a) x x1 42 - -^ ]h g(b) x x1 42 + -^ ]h g(c) x x 42 -] g(d) x x x4 1 1- + -] ] ]g g g
30. Simplify .3 2 2 98+
(a) 5 2
(b) 5 10
(c) 17 2
(d) 10 2
31. Simplify .
x x x4
3
2
2
2
12
-
+-
-+
(a)2 2
5
x x
x
+ -
+
] ]g g(b)
2 2
1
x x
x
+ -
+
] ]g g(c)
2 2
9
x x
x
+ -
+
] ]g g(d)
2 2
3
x x
x
+ -
-
] ]g g
32. Simplify .ab a ab a5 2 7 32 2- - -
(a) 2ab a2+
(b) ab a2 5 2- -
(c) a b13 3-
(d) 2 5ab a2- +
33. Simplify .2780
(a)3 3
4 5
(b)9 3
4 5
(c)9 38 5
(d)3 3
8 5
34. Expand and simplify .x y3 2 2-^ h (a) x xy y 3 12 22 2- -
(b) x xy y 9 12 42 2- -
(c) 3 6 2x xy y 2 2- +
(d) x xy y 9 12 42 2- +
35. Complete the square on .a a162
- (a) a a a16 16 42 2- + = -^ h (b) a a a16 64 82 2- + = -^ h (c) a a a16 8 42 2- + = -^ h (d) a a a16 4 22 2- + = -^ h
-
8/13/2019 Ch2 Algebra and Surdskjtfc
50/50
93Chapter 2Algebra and Surds
1. Expand and simplify
(a) ab a b b aa4 2 32 2- --] ]g g(b) 2 2y y2 2- +_ _i i(c) x2 5 3-] g
2. Find the value of x y+ with rational
denominator if x 3 1= + and
2 5 3
1.y=
-
3. Simplify .7 6 54
2 3
-
4. Complete the square on .x ab
x2 +
5. Factorise
(a) ( ) ( )x x4 5 42+ + +
(b) 6x x y y 4 2 2- -
(c) 125 343x3 +
(d) 2 4 8a b a b2 2- - +
6. Complete the square on .x x4 122 +
7. Simplify .x x
xy x y
4 16 12
2 2 6 62
- +
+ - -
8.| |
da b
ax by c
2 2
1 1=
+
+ +
is the formula for
the perpendicular distance from a
point to a line. Find the exact value
of dwith a rational denominator if
, , ,a b c x2 1 3 41= = - = = - and 5.y1 =
9. Simplify1
.a
a
13
3
+
+^ h
10. Factorise .b
a42 2
2
-
11. Simplify .x
x y
x
x y
x x
x y
3
2
3 6
3 22
-
+
++
-
-
+ -
+
12. (a) Expand .x2 1 3-^ h Simplify(b) .
x x x
x x
8 12 6 1
6 5 43 2
2
- + -
+ -
13. Expand and simplify 3x x1 2- -] ^g h .14. Simplify and express with rational
denominator .3 4
2 5
2 1
5 3
+
+-
-
15. Complete the square on3
.x x2 + 2
16. If ,xk l
lx kx1 2
=+
+
nd the value of xwhen
, ,k l x3 2 51
= = - = and 4.x2
=
17. Find the exact value with rational
denominator of x x x2 3 12 - + if 2 5 .x =
18. Find the exact value of
(a) xx
122
+ if x1 2 3
1 2 3=
-
+
(b) aand bif a b2 3 3
3 43
+
-= +
19.2
A r2 i= 1 is the area of a sector of a
circle. Find the value of iwhenA 12=
and 4.r=
20. If V r h2r= is the volume of a cylinder,
nd the exact value of rwhen 9V= and
16.h =
21. If2
,s u at 2= + 1 nd the exact value of s
Challenge Exercise 2