MATHEMATICS Learner’s Study and Revision Guide for Grade 12 BRUSHING UP ON YOUR BASIC ALGEBRA
Revision Notes and Exercises by
Roseinnes Phahle 1
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Contents Unit 1 Indices or exponents Exponential equations More problems involving indices Multiplying out brackets Surds Factorization Simultaneous linear equations Quadratic equations Completing the square Formula for solving quadratic equations Simultaneous equations: one linear and the other not Equations involving radicals Equations and simplifications involving rational expressions Evaluating limits Algebraic division Inequalities of quadratic and rational expressions What you have achieved in this Unit Answers
3 5 6 7 8 8 10 10 11 12 13 13 14 15 16 23 27 28
About this study guide 1. This Unit revises the algebra you need to know to answer the question on algebra in Paper 1 of the Grade 12 Mathematics examination. Pay much attention to the answers on page 28 because they carry many very useful HINTS. 2. You must follow this Unit with Unit 2 which includes the examination type questions in algebra. 3. Logarithms which also form part of the algebra question are dealt with in Unit 3. 4. You are strongly urged to study Units 1, 2 and 3 together as all three taken together is what you need to know to be good at answering the question on algebra. 5. Working through this Unit will at the same time be preparing you for the Unit 14 on Circle Geometry (“completing the square”) and Unit 9 on Differentiation (“indices” for applying the differentiation rule, “multiplying out brackets” and “evaluating limits” for differentiation from first principles, and “algebraic division” for factorizing and sketching cubic expressions). 6. Just one topic in algebra is not included in this Unit. This is the Factor Theorem dealt with in Unit 10. 7. Finally, don’t be a loner. Work through this guide Unit in a team with your classmates especially as you may need their help because some parts of this Unit don’t show you “how” but assume that you have previously covered the topics in any of your year Grades 10 to 12. However, pay special attention to the answers given at the end of the Unit because many of them carry hints and enrichment notes.
Basic algebra
REVISION UNIT 1: BRUSHING UP ON YOUR BASIC ALGEBRA SKILLS In this chapter we revise most of the algebra you have learnt since Grade 7 and which you will find necessary to know in order to solve problems in the Grade 12 end of year examination. INDICES or EXPONENTS The laws of indices are summarized below. Make sure you know these laws and, working as a group with your class mates, give yourself examples to confirm that you understand. Laws
Examples – make up your own
x m × x n = x m .x n =
xm ÷ xn =
xm = xn
x −m =
1 = x −m
(x )
m n
n
n
=
x=
xm =
3
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EXERCISE 1.1 – In all cases express your answer with positive indices 1. 5 x 3 .x 2 = 4.
4s 3 = 2 s −3
(
7. 3 p −3 10.
x −5 = 3x
5.
)
2
3. 2k 2 .3k −1 =
2. a 0 a 4 a =
( )
=
8. 3 x 3
3 x −3 y 4 yx
11.
−2
6.
7 m −5 m 3 = 14m −1
9. a 3 b −3 .4a 5 b 3 =
=
2 x −4 x 2 y −2
12.
h 3 .3
h =
EXERCISE 1.2 Without using a calculator evaluate the following:
1.
24
5.
41
2.
5 3 + 33
6.
4 −1
3.
(5 + 3)3
7.
2 −2
4.
40
8.
27 3
9.
(2 )
3 2
10. 64
−2
3
1 11. 2 4
1
−3 2
EXERCISE 1.3 Simplify each of the following expressions leaving your answer with positive exponents:
1.
x 5 × x −3
5.
x 5 ÷ x −3
9. 12 x 2 ÷ 6 x 5
2.
(x )
6.
5 x 3 y × 2 xy 3
3.
6x3 ÷ 2x
7.
x −5 × x −2
3x 4 10. 9x 7
4.
4 x × 3x
8.
x× x
4 3
2
3
3
11.
x x
Basic algebra
EXERCISE 1.4 – Simplify and express your answer with only positive indices.
5 4
3 4
1. 3 x .2 x , x
3.
−
5 4
13 32 0 m n .n 1 2
m n
5 3
3 2
=
2. x y .63 x y =
=
− 2 2 12 xy x 4. 2 x3 y2
(
)
3
=
2
2
1 2
53 − 2 x y × x −1 y 0 = 5. 0 xy
6.
x3
( y ) .x 2 2
5 3
y
−
5 4
=
EXPONENTIAL EQUATIONS EXERCISE 1.5 Without using a calculator, determine the value of x in each of the following:
1.
3 x = 27
5.
5 2 x × 5 3 = 58
2.
2 x = 0,25
6.
33 × 35 = 3 x
3.
7x = 1
7.
2 − x = 64
8.
3 ×3 = 81 3x
4.
3x =
1 9
6
9
5
9. 115 ÷ 11x = 121 10.
2 x × 26 = 16 23
11.
35 × 35 = 27 3x
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MORE PROBLEMS INVOLVING INDICES Illustrations leading to exercise in opposite column x −1
Example 1: Simplify
Solution:
x−2
2 .3 6 x−2
2 x −1.3 x − 2 2 x .2 −1.3 x .3 −2 = 6 x .6 − 2 6 x−2 =
(2 × 3)x .2 −1.3−2 6 x .6 − 2
=
6 x .6 2 6 x .21.3 2
=
6×6 2 × 3× 3
=2 Example 2: Simplify
Solution:
4.3 x +1 3 x + 3 x −1
EXERCISE 1.6 Simplify each of the following expressions: 1.
4.3 x .3 3 x 1 + 3 −1 4.3 = 1 1 + 3 4 = 12 ÷ 3
=9
(
(2 )
x 2
2.
+ 2 2 x −1 4x
3.
5 ⋅ 3 x+2 − 4 ⋅ 3 x 5 ⋅ 3 x − 4 ⋅ 3 x−2
4.
2 −2 − 3 −2 2 −2 ⋅ 3 + 2 ⋅ 3−2
5.
6 ⋅ 5 x + 5 x+2 5 x −1
6.
a x − a x −1 a x + a x+2
7.
3 2 x +1 − 2 ⋅ 3 x +1 − 3 x −1 2 ⋅ 3 2 x − 5 ⋅ 3 2 x +1
8.
x 2 y n +1 x n +1 ÷ ÷ 2 y n x n −1 y 2 n +1
4.3 x +1 4.3 x .31 = 3 x + 3 x −1 3 x + 3 x .3 −1 =
7 x+2 − 7 x 8⋅7x
)
( )
n −1
NOTE: Another instructive example belonging to the same family as the problems in the above exercise is Example 5 on page 16. Have a look at this example now.
Basic algebra
MULTIPLYING OUT BRACKETS The exercise below is given to you to work out together with your study partners to remind you what is involved in multiplying out brackets. What is essential to remember is that a minus outside a bracket changes the signs inside the bracket when it is removed. EXERCISE 1.7 Simplify the following expressions: 1. 5( x + 7 ) − 3(1 − 8 x ) =
3. 8( x − 1) + x(8 x − 3) =
2. − x( x − 4 ) + 7(8 − 2 x ) =
4. − (− 6 x − 8) + 7( x − 8) =
EXERCISE 1.8 Find each product by multiplying out the brackets: 1.
(7 x − 2)(6 x − 3)
10. ( x − h )
2.
(3x − 5)(4 x + 3)
11. ( x − 2 )(2 x + 1)(( x − 3)
3.
(a + 6)(− 5a − 2)
12. ( x − 3)
4.
(2 x − 7 )2
13. (2 x + 3) 3 x 2 − 8 x + 5
5.
(x + 3)(x − 3)
14. (6k − 1) 6k 2 − 6k + 5
6.
(x + 3)(x + 3)
15. ( p − 7 ) 6 p 2 − 3 p − 4
7.
(x − 3)(x − 3)
16.
(3x − 5)(2 x 2 − 2 x + 3)
8.
(3d + 5)(d − 4)
17.
(x + h )3
9.
( x + h )2
18.
(x − h )3
2
3
7
(
)
(
)
(
)
Preparation for the Mathematics examination brought to you by Kagiso Trust
SURDS You can evaluate
3
27 exactly because
But you cannot evaluate
3
1 3
( )
27 = 27 = 3
1 3 3
=3
s×
1 3
= 31 = 3 .
27 exactly. This is an example of a surd and is an irrational number. We can
9 × 3 = 3 3 but still can’t work it out exactly.
simplify it to
Rationalisation of denominators This means having only rational terms in the expression of the denominators. Example 1:
Example 2:
1 2
=
1 3−2
2 2 =
.
1 2
=
3+2 3+2
2 2
.
1 3−2
=
3+2 3+2 = = − 3−2 −1 3−4
EXERCISE 1.9 Simplify: 1.
3.
(− 4 + 3 3 )(3 + 3 )
(− 1 + 3 )(− 5 + 2 3 )
2.
2+ 5
−3+ 2
4.
3− 3
−3+ 5
FACTORISATION Types of factorization Type Extracting a common factor Difference of two squares Trinomial Factorisation by grouping
Example
3 x − 6 = 3( x − 2 )
9 x 2 − 25 = (3 x − 5)(3 x + 5)
2 x 2 + 5 x − 3 = (2 x − 1)(x + 3)
(
)
5 x 3 + 2 x 2 − 15 x − 6 = 5 x 3 + 2 x 2 − (15 x + 6 ) = x 2 (5 x + 2 ) − 3(5 x + 2 ) = (5 x + 2 ) x 2 − 3
(
Cubics
)
These are dealt with in Unit 10.
If you are not sure how to carry out any of the above factorizations ask a friend in your class to help you and also look up your textbook for explanations.
Basic algebra
EXERCISE 1.10 Factorise the following completely:
1.
x 2 + x − 12
10. 9 x 2 − 1
2.
x 2 − 4 x − 21
11. cos 2 x + 2cosxsinx + sin 2 x
3.
x 2 − 5x + 4
4.
x2 −1
12. 3 x 3 + 9 x 2 − 2 x − 6
5.
6 x 2 + 42 + 60
13. 9 x 3 + 3 x 2 − 6 x − 2
6.
4 x 2 − 25
14. 35 x 3 − 21x 2 + 20 x − 12
7.
9 x 2 + 30 x + 25
15. 35 x 3 − 20 x 2 + 56 x − 32
8.
cos 2 x − sin 2 x
16. 42 x 3 + 48 x 2 − 49 x − 56
9.
cos 2 x − 2cosxsinx + sin 2 x
17. 35 x 3 − 7 x 2 + 15 x − 3
Factorise the following by grouping:
EXERCISE 1.11 Factorise the following completely: 1. 9 x 2 + 6 x + 1
7. 81x 2 − 126 x + 49
13. 25 + 40 x + 16 x 2
2. 36 x 2 − 400
8. 147 x 2 + 210 x + 75
14.
64 x 2 + 32 x + 4
3. x 2 + 14 x + 49
9.
15.
2 x 2 − 7 x − 15
4. 4 x 2 − 12 x + 9
10. 8 x 2 − 392
16. 2 x 3 − 18 x
5. 5 x 2 − 10 x + 5
11. 25 x 2 + 20 x + 4
17. 6 x 2 − x − 2
6. 1 − x 2
12.
x 2 − 8 x + 16
500 x 2 + 900 x + 405
9
18. ax + 4 x − 2a − 8
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SIMULTANEOUS LINEAR EQUATIONS EXERCISE 1.12 Use the elimination method to solve the following system of simultaneous equations:
1.
x − 7 y = 26 7 x − y = −10
3.
2.
5 x − 4 y = −30 − 7 x − 8 y = 30
4.
− 7 x − 4 y = −10 8 x − 8 y = 24 6 x − 3 y = −21 − 12 x + 2 y = 6
EXERCISE 1.13 Use the substitution method to solve the following system of simultaneous equations:
1.
x + 5 y = −15 4 x − 8 y = −4
3.
7 x − 5 y = 17 − 4 x + y = −6
2.
x + 3 y = −21 −x− y =7
4.
− 5 x − y = −15 − 3x = y − 1
QUADRATIC EQUATIONS EXERCISE 1.14 Solve each of the following factored quadratic equations: 1.
(3x + 1)(x − 8) = 0
2.
(n + 5)(2n − 6) = 0
EXERCISE 1.15 Solve each of the following quadratic equations by factorization: 1. 2. 3. 4. 5.
x2 x2 x2 x2 x2
− x − 42 = 0 − 5x − 6 = 0 − 6x − 7 = 0 + 6x + 5 = 0 + 35 = 12 x
6. 7. 8. 9. 10.
x2 x2 x2 x2 x2
− 5 = −4 x − 8 x = −7 = 21 − 4 x = −x = 5x
Basic algebra
COMPLETING THE SQUARE An example is used to demonstrate what is known as the completion of a square. Example:
x 2 + 6x − 1 = 0 x 2 + 6x = 1
Add the square of half the coefficient of x to each side of the equation: 2
6 6 x 2 + 6x + = 1 + 2 2
2
x 2 + 6x + 9 = 1 + 9
(x + 3)(x + 3) = 10 ( x + 3) 2
= 10
x + 3 = ± 10
So
x = −3 + 10
or
x = −3 − 10
Method: The technique used in the above method is called “completing the square”. The method uses the expansion of
( x + b )2
= x 2 + 2bx + b 2
where the last term b 2 is the square of half of the coefficient of the x term. NOTE: To complete the square, the coefficient of x 2 must be +1. EXERCISE 1.16 In the following find the value that completes the square and then rewrite as a perfect square: 1. x 2 + 24 x + ____ 2. x 2 − 12 x + ____
3.
x 2 − 8 x + ____
5.
x 2 − 9 x + ___
4.
x 2 + 4 x + ____
6.
x 2 + 3 x ____
7. Write each of the following expressions in the form ( x − a ) + ( y − b ) = r 2 : 2
(i) x 2 − 4 x + y 2 − 2 y − 4 = 0
2
(ii) x 2 + y 2 − 6 x − 2 y − 15 = 0 11
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FORMULA FOR SOLVING QUADRATIC EQUATIONS Completing the square on the quadratic equation ax 2 + bx + c = 0 leads to the formula:
x=
− b ± b 2 − 4ac 2a
See whether you can derive this formula. HINT: You must make the coefficient of x 2 equal to +1 by placing a outside a bracket. EXERCISE 1.17 Solve each of the following quadratic equations by completing the square:
x 2 − 5 = −4 x
1.
x 2 − x − 42 = 0
9.
2.
x 2 − 5x − 6 = 0
10. x 2 − 8 x = −7
3.
x 2 − 6x − 7 = 0
11. x 2 = 21 − 4 x
4.
x 2 + 6x + 5 = 0
12. x 2 = − x
5.
x 2 + 35 = 12 x
13. x 2 = 5 x
6. 5 x 2 − x = 84
14. 5 x 2 = 12 − 7 x
7.
x 2 = 25
15. 4 x 2 = 8 x + 5
8.
x 2 − 126 = 11x
16. 3 x 2 − x + 42 = 0
EXERCISE 1.18 Solve each of the following quadratic equations by formula: 1. 5 x 2 − x = 84
4. 5 x 2 = 12 − 7 x
2.
x 2 = 25
5.
4 x 2 = 8x + 5
3.
x 2 − 126 = 11x
6.
3 x 2 − x + 42 = 0
Basic algebra
EXERCISE 1.19 Solve each of the following quadratic equations by taking square roots
1.
2 x 2 + 4 = 176
3.
5 x 2 − 6 = 83
2.
7 x 2 + 3 = 143
4.
− 4 x 2 − 6 = −45
SIMULTANEOUS EQUATIONS: ONE LINEAR AND THE OTHER NOT EXERCISE 1.20 Solve each of the following system of simultaneous equations:
1.
x 2 + y 2 − 3 x + 20 y + 36 = 0 x + 2 y = −1
5.
x 2 + y 2 − 9 x − y + 18 = 0 x + 3y = 1
2.
4 x 2 + 2 y 2 − 3 x + 13 y − 64 = 0 x+ y =2
6.
x 2 + y 2 + 9x − 3y = 0 3x − y = 0
3 x 2 + 2 y 2 + 2 x + 115 y + 116 = 0 3. x + 3 y = −2 4.
3 x 2 + y 2 + x + 63 y − 78 = 0 7. x + 2y − 4 = 0
5 x 2 + y 2 + 29 x − y − 6 = 0 3x + y = 3
8.
2x 2 + 3y 2 − 2x − 2 y = 0 x+ y =0
EQUATIONS INVOLVING RADICALS EXERCISE 1.21 Solve each of the following equations and state the values of x for which each is undefined:
1. 2. 3.
− 3 − 2 x − 8 = −5
4.
6 + x − 9 = 13
5.
4x − 8 = 8x + 8
7 x − 8 = 14 24 − 4 x = x − 5
6. 15 = 10 + 4 x − 3
13
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EQUATIONS AND SIMPLIFICATIONS INVOLVING RATIONAL EXPRESSIONS EXERCISE 1.22 Simplify each of the following expressions:
1.
3x − 5 x + 4 + 4 3
3.
4 x 3x + 4 + 5 4x − 3
2.
3 4 + x − 2 5x
4.
2 3x + 4 + 3x − 5 6
EXERCISE 1.23 Solve each of the following equations:
1.
x+4 1 5 = − 2 2 x 2x 2x
3.
3 x 2 + 9 x + 20 + = x−2 2 x
2.
1 x−3 −1 = 3x x
4.
x − 1 5 3 x + 12 + 2 = 5x x2 x
EXERCISE 1.24 Simplify each of the following expressions:
1.
x+4 2 x + 9 x + 20
6.
x 2 − 3 x − 4 4 x − 16 ÷ x+3 x+3
2.
x 2 + 8 x + 15 x+5
7.
7k 3k 2 ÷ 5k − 15 15k − 45
3.
8 x 2 + 72 x 1 × 8 x+9
8.
8 j + 28 4 ÷ 4 j + 20 j + 5
4.
1 x 2 + 6 x − 40 × 8 x + 80 x+7
9.
x +1 1 ÷ x − 9 x − 10 x − 10
5.
x 2 − 2 x − 15 4x × 2 x + 3 x + 4 x − 45
10.
1 x+4 ÷ 2 x + 9 x + 20 x − 25
2
2
Basic algebra
EVALUATING LIMITS EXERCISE 1.25 Show that 1. Lim x→2
x2 − 4 =4 x−2
x 2 − 2x + 1 =0 x →1 x −1 tan θ 3. Lim =1 θ →0 sin θ 2. Lim
EXERCISE 1.26 Determine the following: 1. Lim
x 2 − 16 x−4
2. Lim
4 x 2 − 3x 3− x
3. Lim
x 2 − x − 12 x−4
x →4
x →2
x→4
x →1
4. Lim 3 x +
1 x
EXERCISE 1.27 If f ( x ) = 2 x 2 − 5 x + 7 , evaluate (i) f ( x + h ) ;
(ii) f ( x + h ) − f ( x ) ;
(iii)
f (x + h ) − f (x ) h
(iv) Lim
f (x + h ) − f (x ) h
(iii) Lim
f (x + h ) − f (x ) h
h →0
EXERCISE 1.28 Given that f ( x ) = 3 x 2 − 5 x + 7 determine the expressions for (i) f ( x + h ) − f ( x )
(ii)
f (x + h ) − f (x ) h
15
h →0
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ALGEBRAIC DIVISION Some algebraic division amounts to simplification and is easy to perform. Example 1:
3x 3 3x 2 = 2x 2
Example 2:
3x − 6 3x 6 = − = x−2 3 3 3
or use factorization:
3 x − 6 3(x − 2 ) = = x−2 3 3 Example 3:
x 3 + 3x 2 − 5 x x 3 3x 2 5 x = + − = x 2 + 3x − 5 x x x x
or use factorization:
(
)
x 3 + 3x 2 − 5 x x x 2 + 3x − 5 = = x 2 + 3x − 5 x x Example 4:
x 2 − 5 x + 6 (x − 2 )(x − 3) = = x−2 x−3 x−3
In some cases, as you have already seen in Exercise 1.6 on page 6, you may need to apply the laws of indices to simplify and carry out the division. Here is another example. Example 5:
(
2 x 4.2 −1 − 6.21 = 2 x 4.2 −1 − 1
(
)
)
21 2.2 −1 2 − 3. −1 2 = 1 2 −1 4 − −1 2 =
(
2 2 − 3.21+1 4 − 21
(
2 2 − 3.2 2 = 4−2
)
)
Basic algebra
=
2(2 − 3.4 ) 2
= 2 − 12
= −10 Example 6: However, dividing something like x 3 + 3 x 2 − 2 x + 7 by x − 2 , say, is somewhat more complicated and requires a different procedure which is very much like dividing plain numbers by long division. But first let’s sort out some terminology. the binomial x − 2 is known as the divisor the polynomial x 3 + 3 x 2 − 2 x + 7 is known as the dividend the answer you get on division is known as the quotient plus a remainder (if there is one). As in the method of long division, we set up the question of dividing in the form:
x − 2 x 3 + 3x 2 − 2 x + 7 Note that the terms of both the dividend and divisor are arranged in a descending order of the powers of x . This arrangement you must always have when you do polynomial division which is shown below. Explanation of what you do
What you must write down
Divide x of the divisor into x 3 of the dividend. This gives x 2 . Write the x 2 above the x 3 .
x2
x − 2 x 3 + 3x 2 − 2 x + 7
Multiply x − 2 by x 2 . That is, x 2 ( x − 2 ) = x 3 − 2 x 2
x2
x − 2 x 3 + 3x 2 − 2 x + 7
Write x 3 − 2x 2 below x 3 + 3x 2 and draw a line under.
x 3 − 2x 2
17
Preparation for the Mathematics examination brought to you by Kagiso Trust
Explanation of what you do
Subtract x 3 − 2x 2 from x 3 + 3x 2 . Because you are doing a subtraction, the signs of x 3 − 2x 2 will change to − x 3 + 2x 2 as shown by left superscripts in the box opposite. This allows you to simply add the like terms.
What you must write down
x2
x − 2 x 3 + 3x 2 − 2 x + 7 −
x 3 − + 2x 2
•
5x 2
x 3 − x 3 is nothing and this is indicated by the dot. Bring down − 2 x .
x2
x − 2 x 3 + 3x 2 − 2 x + 7 x3 − 2x 2 •
Divide x of the divisor into 5x 2 . This gives + 5 x . Write + 5 x above the 3x 2 .
5x 2 − 2 x
x 2 + 5x
x − 2 x 3 + 3x 2 − 2 x + 7 x3 − 2x 2 •
Multiply x − 2 by + 5 x . That is, + 5 x( x − 2 ) = 5 x 2 − 10 x . Write 5 x − 10 x under 5 x − 2 x and draw a line under. 2
5x 2 − 2 x
x 2 + 5x
x − 2 x 3 + 3x 2 − 2 x + 7
2
x3 − 2x 2 •
5x 2 − 2 x 5 x 2 − 10 x
Basic algebra
Explanation of what you do
What you must write down
Subtract now, taking care to change signs because of the subtraction (shown by left superscrpts).
x 2 + 5x
x − 2 x 3 + 3x 2 − 2 x + 7 x3 − 2x 2 •
5x 2 − 2 x − 5 x 2 − + 10 x •
Bring down the +7 which is the last term in the dividend.
8x
x 2 + 5x
x − 2 x 3 + 3x 2 − 2 x + 7 x3 − 2x 2 •
5x 2 − 2 x − 5 x 2 − + 10 x •
Divide x into 8 x . This gives + 8 . Write the + 8 above the − 2 x .
8x + 7
x 2 + 5x + 8
x − 2 x 3 + 3x 2 − 2 x + 7 x3 − 2x 2 •
5x 2 − 2 x − 5 x 2 − + 10 x •
19
8x + 7
Preparation for the Mathematics examination brought to you by Kagiso Trust
Explanation of what you do Multiply x − 2 by 8 . That is, 8( x − 2 ) = 8 x − 16 Write 8 x − 16 under 8 x + 7 and draw a line under.
What you must write down
x 2 + 5x + 8
x − 2 x 3 + 3x 2 − 2 x + 7 x3 − 2x 2 •
5x 2 − 2 x − 5 x 2 − + 10 x •
Subtract 8 x − 16 from 8 x + 7 and be mindful of the change of signs as shown by the left superscripts.
8x + 7 8 x − 16
x 2 + 5x + 8
x − 2 x 3 + 3x 2 − 2 x + 7 x3 − 2x 2 •
5x 2 − 2 x − 5 x 2 − + 10 x • −
8x + 7 8 x − + 16 •
23
The division is now completed and 23 is the remainder after x 3 + 3 x 2 − 2 x + 7 is divided by x − 2 . Note that the division procedure can be summarized by a repetition of cycles which entail the following steps: DIVIDE (always by the leading term of the divisor) MULTIPLY (the divisor by what you obtained in the above step) SUBTRACT (remembering that this implies a change of signs) BRING DOWN (the next term in the dividend) DIVIDE (repeating the above steps until the last term in the dividend is brought down and subtraction is done to complete the cycle).
Basic algebra
Thus x 3 + 3 x 2 − 2 x + 7 is divided by x − 2 gives the answer x 2 + 5 x + 8 plus remainder 23. This may be written as
x 3 + 3x 2 − 2 x + 7 23 = x 2 + 5x + 8 + x−2 x−2 In much the same way as we can write 75 divided by 11 as
75 9 6 = 6 = 6+ . 11 11 11
The above procedure for carrying out polynomial division seemed very long but that’s only because it was being explained to you. All there is to it and is necessary to write down is what you see in the last box which is
x 2 + 5x + 8
x − 2 x 3 + 3x 2 − 2 x + 7 x3 − 2x 2 •
5x 2 − 2 x 5 x 2 − 10 x •
8x + 7 8 x − 16 •
23
Easy! But as with all of your Mathematics, practice will make it easier and faster. by x − 2 . Notice that the x 2 term is now
Example 7: Suppose we had to divide x 3 − 2 x + 7
missing in the polynomial. Account for it by inserting 0x 2 in its place and carry out the division as explained above.
x − 2 x3 + 0x 2 − 2x + 7
Its your turn to complete the division inside this box. The answer is
x3 − 2x + 7 11 . = x 2 + 2x + 2 + x−2 x−2 21
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EXERCISE 1.29 Carry out the following divisions:
(8 x 3 − 2 x 2 + 4 x − 7) ÷ ( x − 1) 2. ( x 3 − 3 x 2 − 8 x − 11) ÷ ( x − 5) 3. (2 x 3 + 9 x 2 − 78 x + 17) ÷ ( x + 9 ) 4. ( x 3 − 4 x 2 − 19 x − 6) ÷ ( x − 7 ) 1.
( x 3 + 17 x 2 + 71x + 2) ÷ (x + 7 ) 6. ( x 3 + 14 x 2 + 50 x + 27) ÷ ( x + 5) 7. (8 x 3 + 4 x − 7) ÷ ( x − 1) 8. (8 x 3 − 2 x 2 − 7) ÷ ( x − 1) 5.
EXERCISE 1.30 1. Evaluate f (1) given that f ( x ) = 8 x 3 − 2 x 2 + 4 x − 7 . Compare your answer with No 1 in the above exercise. What do you notice? 2. Repeat with No 2 above by evaluating f (5) given that f ( x ) = x 3 − 3 x 2 − 8 x − 11 . What do you notice in comparing the answers? Following what you have just been doing, complete the following for the rest of the questions in the above exercise: 3. f ( x ) =
f (− 9) =
6.
What do you notice?
4.
f (x ) = f (7 ) =
What do you notice?
7.
What do you notice?
5.
f (x ) = f (− 7 ) = What do you notice?
f (x ) = f (− 5) =
f (x ) = f (1) = What do you notice?
8.
f (x ) = f (1) = What do you notice?
Can we draw conclusions from what we have observed in comparing the two exercises above? If we divide f ( x ) by x − a then f (a ) is what? If we divide f ( x ) by x + a then f (− a ) is what?
Basic algebra
INEQUALITIES OF QUADRATIC AND RATIONAL EXPRESSIONS We will demonstrate how you must tackle inequalities of especially quadratic and rational expressions by examining the signs of the factors in the expressions or sketching the graphs of the expressions. Either method will only be possible if the expressions can be expressed in terms of linear factors. Linear inequalities A solution to an inequality is given by a range or ranges of values. Example 1: If − 22 ≤ 3 x − 4 ≤ 8 then
− 22 + 4 ≤ 3x ≤ 8 + 4 − 18 ≤ 3 x ≤ 12
−6≤ x ≤ 4 Quadratic inequalities Given a quadratic inequality, it is possible to find a range or ranges of values that satisfy the inequality. The values may be found by any of the following methods: 1. Sketching a curve 2. Examining the signs of the factors 3. Completing the square Example 1: Find by sketching the range or ranges of values taken by x if 3 x 2 + 4 x − 4 ≤ 0 . Solution:
Let y = 3 x 2 + 4 x − 4
which on factorization becomes y = (3 x − 2 )( x + 2 ) y
The graph of y = 3 x 2 + 4 x − 4 will thus cut
f(x)=3x^2+4x-4
9 8 7
2 the x -axis at ;0 and (− 2;0 ) , and the 3 y -axis at (0;−4 ) .
6 5 4 3 2 1
x -13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
10
-1 -2 -3 -4
This is information that allows us to make the following sketch:
-5 -6 -7 -8 -9
From the sketch, it can be seen that y ≤ 0 ,meaning that the curve is below the x -axis for
−2≤ x≤
2 3 23
11
12
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Preparation for the Mathematics examination brought to you by Kagiso Trust
Example 2: Finding by examining the signs of the factors the range or ranges of values x can take if
− 2 x + 15 ≤ x 2 Solution: First, re-arrange the inequality so that all the terms of the quadratic are on the left hand side:
− x 2 − 2 x + 15 ≤ 0 You are advised always to make the coefficient of x 2 positive so the sketch of your curve will be concave upwards (“smiley”) as in the above one. Thus
x 2 + 2 x − 15 ≥ 0
(x − 3)(x + 5) ≥ 0 Set up a table of values as shown below:
(x − 3) ( x + 5) (x − 3)(x + 5)
x < −5
−5< x < 3
x>3
+
To complete the above table, take for example the column under x < −5 . Think of any value of x that
is less than − 5 ; for example, − 6 . Ask yourself whether ( x − 3) is positive or negative for this value.
Ask likewise for ( x + 5) . Enter the appropriate signs in the column as illustrated above. Then in the last row, multiply the signs of ( x − 3) and ( x + 5) which is also shown above.
It’s left to you as an exercise to fill in the rest of the signs in the above table. Answer:
x ≤ −5 and x ≥ 3
Example 3: Use the method of completing the square to find the range (or ranges) of values of x can take if 3 x + 2 ≤ 2 x 2 Solution:
First, place all the terms on the left hand side:
− 2 x 2 + 3x + 2 ≤ 0 Then, make the coefficient of x 2 positive:
2 x 2 − 3x − 2 ≥ 0 Now start the process of completing the square:
Basic algebra
x2 −
3 x −1 ≥ 0 2 2
2
2
2
3 3 3 x − x + − −1 ≥ 0 2 4 4 2
3 9 16 25 3 3 + = x − x + ≥ +1 = 2 16 16 16 4 4 2
2
3 25 x − ≥ 4 16 Therefore,
x−
3 5 ≤− 4 4
or
x−
3 5 ≥ 4 4
x≤
3 5 2 − =− 4 4 4
or
x≥
3 5 8 + = 4 4 4
x≤−
1 2
NOTE: Supposing we had x −
Then The symbol
x−
or
x≥2
2
3 25 instead. ≤ 4 16
3 5 ≤ 4 4
is a mathematical way of saying that
− simplifying to −
5 3 5 ≤ x− ≤ 4 4 4 1 ≤ x ≤ 2. 2
Inequalities of rational expressions A rational expression is of the form
g (x ) or what we may think of as an algebraic fraction. f (x )
The method of examining the signs of the factors is applicable to inequalities involving rational expressions such as in the example that follows.
25
Preparation for the Mathematics examination brought to you by Kagiso Trust
Example 4: Solve for x if
3x + 1 ≥0 1− x
Solution: Critical x -values are given by
3x + 1 = 0 x=−
that is,
1 3
and
1− x = 0
and
x =1
Set up a table of signs:
x<−
3x + 1 1− x 3x + 1 1− x
1 3
−
1 < x <1 3
x >1
+ -
How the signs are worked out was explained in the Example 2 above. It’s left to you to complete the table.
−
Answer:
1 ≤ x ≤1 3
EXERCISE 1.31 Solve the following inequalities by the method of examining the signs of the factors of each of the expressions. Illustrate your solutions with rough sketches.
1.
x 2 + 2x < 0
2.
x 2 + 6x + 7 ≥ 0
Basic algebra
3.
x 2 + 3 ≥ 4x
4. 8 x − 15 > x 2
5.
x 2 + 4 x ≥ −3
6.
x 2 + 8 x + 14 > 0
WHAT YOU HAVE ACHIEVED IN THIS UNIT This UNIT has given you the necessary tools you need to be able to answer the algebra question in your examination at the end of Grade 12 year. But it has also made extensions by introducing the algebra you will need to factorise cubic expressions using the remainder or factor theorem, what you will need to do to find the radius and centre of a circle, and the working out you will need to do to evaluate limits and carry out differentiation from first principles. A topic which is usually part of the algebra question but applies to financial mathematics is on logarithms is also covered later in this booklet. So if you have thoroughly studied this UNIT on algebra you will have prepared yourself for answering much more than just the algebra question in the examination. It is a UNIT you will need to study over and over again until you feel confident that you can do the algebra. 27
Preparation for the Mathematics examination brought to you by Kagiso Trust
ANSWERS EXERCISE 1.1 1. 5x 5 2. a 5 3. 6k 4. 4s 6 1 5. 3x 6 1 6. 2m 9 7. 6 p 1 8. 9x 6 9. 4a 8 3y3 10. 4 x 2y2 11. 6 x 12. 3h
7 4
EXERCISE 1.4 3
1. 2. 3. 4. 5.
16 152 512 1 4
6.
1 4 1 4 1 3
8.
5
8. x 6 2 9. x3 1 10. 3 3x 1 11. x
1. 6x 4
EXERCISE 1.2
7.
EXERCISE 1.3 1. x 2 2. x 12 3. 3x 2 4. 12x 5 5. x 8 6. 10 x 4 y 4 1 7. x7
2. 6 x 2 y 2
1 16 8 11. 27
1 3
1
m n2
4.
x y
5.
11 2 26 3
1 2
x3 y
9. 64 10.
1
3.
6.
1 xy
11 4
Basic algebra
EXERCISE 1.5 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.
6. 7. 8. 9. 10. 11. 12. 13. 14. 15.
x=3 x = −2 x=0 x = −2 x = 2,5 x=8 x = −6 x = 11 x=3 x =1 x=7
16. 6 x 3 − 16 x 2 + 19 x − 15 17. x 3 + 3 x 2 h + 3 xh 2 + h 3 18. x 3 − 3 x 2 h + 3 xh 2 − h 3
EXERCISE 1.6
EXERCISE 1.9
1. 6 2.
−3−5 3 2. 11 − 7 3 1 3. 2+ 5 3+ 3 6 1 4. 3− 2 3+ 5 4 1.
3 2
( (
3. 9 4.
1 7
5. 155 6.
x 2 + 6x + 9 x 2 − 6x + 9 3d 2 − 7 d − 20 x 2 + 2 xh + h 2 x 2 − 2 xh + h 2 2x3 − 9x 2 + 7x + 6 x 3 − 9 x 2 + 27 x − 27 6 x 3 − 7 x 2 − 14 x + 15 36k 3 − 42k 2 + 36k − 5 6 p 3 − 45 p 2 + 17 p + 28
a −1 a a2 +1
(
)
19 − 3 x + 2 3 x +1 ⋅ 13 8. x 2
7.
EXERCISE 1.7 1. 10(3 + 5 x ) 2. − 6(9 y + 4 ) 2 3. 3 x + 7 x − 56 4. r (7 r − 6 ) EXERCISE 1.8 1. 42 x 2 − 23 x + 6 2. 12 x 2 − 11x − 15 3. − 5a 2 − 32a − 12 4. 4 x 2 − 28 x + 49 5. x 2 − 9
)( )(
EXERCISE 1.10 1. ( x + 3)( x − 4 ) 2. ( x + 3)( x − 7 ) 3. ( x − 1)( x − 4 ) 4. ( x − 1)( x + 1) 5. 6( x + 2 )( x + 5) 6. (2 x − 5)(2 x + 5)
) )
7. (3 x + 5) Note how some of the following expressions with trigonometric ratios can be factorised by using the methods of factorisation you use in algebra. For example, this next one is a difference of two squares. 8. (cos x − sin x )(cos x + sin x ) 2
9.
(cos x − sin x )2
10. (3 x − 1)(3 x + 1)
11. (cos x + sin x )
29
2
Preparation for the Mathematics examination brought to you by Kagiso Trust
12. (3 x − 1)(3 x + 1)
13. (cos x + sin x )
2
( ) 15. (3 x + 1)(3 x − 2 ) 16. (5 x − 3)(7 x + 4 ) 17. (7 x − 4 )(5 x + 8) 18. (7 x + 8)(6 x − 7 ) 19. (5 x − 1)(7 x + 3) 14. ( x + 3) 3 x 2 − 2 2
2
2
2
2
EXERCISE 1.11 2 1. (3 x + 1) 2. 4(3 x − 10 )(3 x + 10 ) 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.
( x + 7 )2 (2 x − 3)2 2 5( x − 1) (1 − x )(1 + x ) (9 x − 7 )2 2 3(7 x + 5) ( x − 4 )2 8(x − 7 )(x + 7 ) (5 x + 2)2 2 5(10 x + 9 ) (5 x + 4)2 (8 x + 2)2 (2 x + 3)(x − 5) 2 x(x − 3)(x + 3) (3x − 2)(2 x + 1) (a + 4)(x − 2)
EXERCISE 1.12 1. 2. 3. 4.
(x; y ) = (5;−3) (x; y ) = (− 10;5) (x; y ) = (2;−1) (x; y ) = (1;9)
Another way of solving two simultaneous linear equations is by drawing their graphs which will be straight lines. The solution is given by the coordinates of the point of intersection of the two lines. All you need to draw a straight line are at least two points on the line. The easiest way is to find the intercepts on the axes. Put x = 0 and find y so as to get the y - intercept. Similarly, find the x intercept by putting y = 0. Join the two points to get the straight line. Now solve Exercise 1.6 graphically and see if you get the same answers as the above solutions obtained by the elimination method.
EXERCISE 1.13 1. ( x; y ) = (− 5;−2 ) 2. ( x; y ) = (0;−7 ) 3. ( x; y ) = (1;−2 ) 4. ( x; y ) = (7;−20 ) EXERCISE 1.14 1 1. x = − or x = 8 3 2. n = −5 or n = 3 EXERCISE 1.15 1. x = −6 or x = 7 2. x = −1 or x = 6 3. x = −1 or x = 7 4. x = −1or x = −5 5. x = 5 or x = 7 6. x = 1 or x = −5 7. x = −1or x = 7 8. x = 3 or x = −7 9. x = 0 or x = −1 10. x = 0 or x = 5 The solutions above are also known as the roots of the equations you have been solving. By putting y equals to each of the equations you could also solve them graphically. The roots or solutions
Basic algebra
would be given by the x -intercepts of the graphs. The roots are also known as the zeroes of the expressions because y is zero at the x -intercepts. The y -intercept is of course found by putting x =0 in each equation. Extended Exercise: Use the intercepts to draw rough sketches of equations.
EXERCISE 1.16 2 1. 144; ( x + 12 ) 2.
36;
3. 16;
( x − 6 )2 ( x − 4 )2 ( x + 2 )2
4.
4;
5.
81 9 ; x − 4 2
6.
9 ; 4
7. (i) (ii)
3 x + 2
(x − 3)
EXERCISE 1.19 1. x = ±9,27 2. x = ±4,47 3. x = ±4,42 4. x = ±3,12
(x; y ) = (7;−4) or (x; y ) = (3;−2) (x; y ) = (− 1;3) or (x; y ) = (5;−3) (x; y ) = (1,29;−1,10) or (x; y ) = (− 9,72;−3,91) (x; y ) = (0;3) or (x; y ) = (− 1;6) (x; y ) = (4;−1) (x; y ) = (0;0) (x; y ) = (2;1) or (x; y ) = (8;−2) (x; y ) = (0;0)
1.
2
2. 3.
2
4. 5.
= 32
6.
+ ( y − 1) = 5 2
8.
EXERCISE 1.17 1. x = 7 or x = −6 2. x = 6 or x = −1 3. x = 7 or x = −1 4. x = −1 or x = −5 5. x = 1 or x = −5 6. x = 7 or x = 1 7. x = 7 or x = −3 8. x = 1or x = 0 9. x = 5 or x = 0 10. x = 4,2 or x = −4,0 11. x = ±5 12. x = 7 or x = −18 13. x = 1 or x = −2,4 14. x = 2,5 or x = −0,5 15. No solution
x = 4,20 or x = −4,00 x = ±5 x = 18 or x = −7 4. x = −2,4 or x = 1 1. 2. 3.
EXERCISE 1.20
( x − 2 )2 + ( y − 2 )2 2
EXERCISE 1.18
2
7.
All of the problems in this exercise were to be solved in an analytic way. In the case of this particular exercise, that means using the linear or straight line equation to make a substitution in the other equation. Now this other equation is either an equation of a circle or a general conic section (which is not studied at Grade 12). The way you tell that an equation is an equation of a circle is two-fold: • •
First, the coefficients x 2 and y 2 terms must be the same; and Second, the equation must not have an xy term.
Can you now identify the equations of circles in this exercise?
31
Preparation for the Mathematics examination brought to you by Kagiso Trust
Can you tell from the solution whether the line cuts the curve (circle or general conic) at two distinct points or whether it is a tangent to the curve? Remember that a solution to two simultaneous equations in two unknowns x and y can also be obtained by drawing the graphs of the two equations. The solution would then be given by the coordinates of the points of intersection of the two graphs. In some cases you have two distinct solutions. This means that if you were to solve the equations graphically you would see that the line and the curve intersect at two points. In other cases there is just one solution. This means that if you were to solve the equations graphically you would see the line being a tangent to the curve.
EXERCISE 1.21
x = −6 and undefined for x ≥ −1,5 x = 58 and undefined for x ≤ 9 x = −4 and undefined for x ≤ −1 x = 12 and undefined for x ≤ 2 5. x = 3 ± 2 2 and undefined for x ≥ 6 6. x = 5,5 and undefined for x ≥ −0,75 1. 2. 3. 4.
EXERCISE 1.22 1.
(x − 2)(3x − 5) 6( x − 5)
2.
19 x − 8 5 x( x − 2 )
3.
16 x + 3 x + 20 5(4 x − 3)
4.
9 x 2 − 3x − 8 5(4 x − 3)
2
EXERCISE 1.23 1.
x=9
2.
2 3 x = −1,6 1 x = or x = −4 3
3. 4.
x =1
EXERCISE 1.24 1. 2. 3. 4. 5. 6. 7. 8.
1 x+5 x+3 x x−4 8( x + 7 ) 4x x+9 x +1 4 7 k j+7 4
9. 1 10. x − 5
EXERCISE 1.25 HINT: Remember you cannot divide by zero. So in
evaluating limits avoid dividing by zero. How you can do this is shown below: 1.
Lim x→2
(x − 2)(x + 2) x2 − 4 = Lim x → 2 x−2 x−2 = Lim( x + 2) x→2
= (2 + 2 ) =4
Basic algebra
2. HINT: Factorise the numerator as was done above. 3. HINT: Replace tan θ by what?
EXERCISE 1.26 HINT: How you evaluate limits is hinted in the exercise
above. 1. 2. 3. 4.
8 15 8 4
EXERCISE 1.27
This is a straightforward substitution exercise in algebra. But it is a very important exercise because it prepares you for the revision of differentiation from first principles.
f (− 3) = 40 2. f ( x + h ) = 2 x 2 + 4 xh + 2h 2 + 5 x + 5h + 7 3. f ( x + h ) − f ( x ) = 4 xh + 2h 2 − 5h f (x + h ) − f (x ) 4. = 4 x + 2h − 5 h Given f ( x ) then f ( x + h ) simply means that wherever an x occurs in f ( x ) it must be replaced by x + h in much the same way as f (− 3) is evaluated by replacing x by -3. 1.
1. 8 x 2 + 6 x + 10 +
This is a significant exercise preparing you for the revision of differentiation from first principles. Rehearse all of it over and over again.
3 x −1
An explanation of the way this answer is written is as follows: What you divide into is called the dividend. What you divide by is called the divisor. The exact number of times a divisor goes in into a dividend is called the quotient. Then what is left after you cannot divide any more is called the remainder. Take as a numerical example 17 ÷ 5 . The answer is 3 remainder 2 but we write it in the form 3
2 2 which is in fact 3 + . 5 5
In the same way, the answer to No 1 above is 8 x 2 + 6 x + 10 plus remainder 3 which we wrote in the form 8 x 2 + 6 x + 10 + 2. 3. 4. 5. 6.
EXERCISE 1.28
1.
EXERCISE 1.29
7. 8.
1 x−5 10 2x 2 − 9x + 3 − x+9 8 2 x 2 + 3x + 2 + x−7 5 x 2 + 10 x + 1 − x+7 2 x 2 + 9x + 5 + x+5 5 8 x 2 + 8 x + 12 + x −1 1 8x 2 + 6 x + 6 x − x −1
3 . x −1
x 2 + 2x + 2 −
What you must also notice is that when you divide a trinomial by a linear expression you get a quotient that is a quadratic. The remainder can be a constant as in all of the above examples. It can also be zero. It will be zero when the divisor divides exactly into the dividend as in the numerical example 12 divided by 4. This is what explains the factor theorem.
f ( x + h ) = 3( x + h ) − 5( x + h ) + 7 2
f ( x + h ) − f ( x ) = 6 xh + 3h 2 − 5h f (x + h ) − f (x ) 3. = 6 x + 3h − 5 h f (x + h ) − f (x ) 4. Lim = 6x − 5 h →0 h 2.
33
Preparation for the Mathematics examination brought to you by Kagiso Trust
EXERCISE 1.30 Your answers in this exercise are to be compared with your answers in the previous exercise. What do you notice, you are asked? What you should notice are that the remainders in the previous exercise are respectively given by the answers to this exercise:
f (1) = 3 f (5) = −1 f (− 9 ) = −10 f (7 ) = 8 f (− 7 ) = −5 f (− 5) = 2 f (1) = 5 f (1) = −1
1. 2. 3. 4. 5. 6. 7. 8.
EXERCISE 1.31 1.
2. y
y
f(x)=x^2+2x x^2+2x<0
9
x^2+6x+7>0 f(x)=x^2+6x+7
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
x -13
-12
-11
-10
-9
-8
-7
Answer: −2< x<0
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
10
11
12
13
x -13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
1
-1
-1
-2
-2
-3
-3
-4
-4
-5
-5
-6
-6
-7
-7
-8
-8
-9
-9
Answer: x ≤ −7 and x ≥ −1
2
3
4
5
6
7
8
9
10
11
12
13
Basic algebra
4.
3. y
y
f(x)=x^2-4x+3 x^2-4x+3>0
9
f(x)=x^2-8x+15 x^2-8x+15<0
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
x -13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
10
11
12
x
13
-13
-3
-2
-1
1
-4 -5
-6
-6
-7
-7
-8
-8
-9
-9
2
3
4
5
6
7
8
9
10
11
12
13
Answer: 3< x<5
f(x)=x^2+4x+3 x^2+4x+3>0
4 3 2 1
x -5
-4
-5
5
-6
-5
-4
6
-7
-6
-3
7
-8
-7
-3
8
-9
-8
-2
y
-10
-9
-2
9
-11
-10
-1
5.
-12
-11
-1
Answer: x ≤ 1 and x ≥ 3
-13
-12
-4
-3
-2
-1
1 -1 -2 -3
2
3
4
5
6
7
8
9
10
11
12
13
6. Expression does not factorise so complete the square or find the x -intercepts by applying the quadratic formula and then making a rough sketch to see where the graph is positive.
-4 -5 -6
y
-7
f(x)=x^2+8x+14 x^2+8x+14>0
9
-8
8
-9
7 6 5 4 3 2 1
x -13
-12
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
1 -1 -2 -3 -4 -5 -6 -7 -8 -9
Answer: x ≤ −3 and x ≥ −1
Answer: x > −2,6 and x < −5,4
35
2
3
4
5
6
7
8
9
10
11
12
13