Ch2 Algebra and Surdskjtfc

8/13/2019 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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45Chapter 2Algebra and Surds

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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46 Maths In Focus Mathematics Extension 1 Preliminary Course

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


20/50


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


21/50


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.


22/50


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


23/50


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

+ = +

= + - +

= + - +

]]

gg


24/50


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- = -

= + -


25/50


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


26/50


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


27/50


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


28/50


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.


29/50


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.


30/50


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.


31/50


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


32/50


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


33/50


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


34/50


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


35/50


(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- + = +


36/50


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


37/50


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


38/50


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.


39/50


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


40/50


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.


41/50


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.


42/50


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.


43/50


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.+


44/50


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


45/50


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


46/50


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.


2b

b 2

-

+ b 4!] g


47/50


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-


(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=


(a)5 3

2

(b)2

1 3+


(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


48/50


14. Simplify

(a)93

xyx y

5

4

(b)15 5

5x -

15. Simplify

(a) 3 112^ h

(b) 2 33^ h


(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 - - -


(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


(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+ =


49/50


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


50/50



(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

Documents

Ch2 Algebra and Surdskjtfc