108
EPMATH153 — Extension Mathematics 1 also first half of EPMATH309 — Extension Mathematics by Scott Sciffer
EPMATH153 — Extension Mathematics 1
also first half of
EPMATH309 — Extension Mathematics
by Scott Sciffer
Contents
Topic 1 — Number Systems
Lecture 1 — Natural numbers and integers 3
Lecture 2 — Rational numbers, Pythagoras’ Theorem and surds 8
Lecture 3 — Real and complex numbers 16
Topic 2 — Algebra
Lecture 4 — Basic algebra 24
Lecture 5 — More algebra 30
Lecture 6 — Solving equations 36
Lecture 7 — Quadratic equations 42
Lecture 8 — Inequalities 49
Lecture 9 — Sequences and series 59
Topic 3 — Graphing
Lecture 10 — Functions and graphs 63
Lecture 11 — Graphing straight lines and parabolas 71
Lecture 12 — Graphing higher order polynomials, circles, ellipses,hyperbolae 77
Lecture 13 — Curve sketching 85
Exam 2001 92
Solutions to exercises 96
2
Lecture 1. Natural numbers and integers
§ 1.1 The Natural numbers
The numbers
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, .....
are called the natural numbers. You may also have seen them called the counting numbers,the whole numbers or even the cardinal numbers. The abbreviated notation for the naturalnumbers is a boldface, capital ‘N’— N— but because mathematics was, and still is, mainlytaught with blackboard and chalk, and because it is very difficult to write in boldface withchalk, a special symbol was invented:
There is some controversy about whether or not zero should be included in the naturalnumbers. It is not really worth arguing about — simply be aware that sometimes whenwe talk of the natural numbers we intend zero to be included and sometimes we intend itto be excluded. Hopefully it will be clear which applies from the context!
Notice that the natural numbers do not stop. The line of dots at the end indicatesthey continue forever; there is no largest natural number, for if there was you could simplyadd one to it and get an even bigger number. Secondly we can perform the usual arithmeticoperations on these numbers — addition, multiplication, subtraction and division. Noticethat if we add two natural numbers, or multiply them, the result is another natural num-ber. Anywhere in mathematics we come across this sort of behaviour we call it a closure
property. We say that the natural numbers are closed under addition and multiplication,since using these operations we cannot ‘escape’ from the natural numbers. Subtractionand division are a different matter, however. Subtracting two natural numbers may ormay not result in a natural number, depending on which is bigger. Thus 7− 5 = 2 gives anatural number but 5−7 = −2 does not. Likewise division of natural numbers will usuallyresult in fractions, so the natural numbers are not closed under subtraction and division.This is strong evidence that we will need to find a bigger system of numbers to work with.Before looking for other number systems, however, let’s look at a few important propertiesof the natural numbers.
Prime Numbers
A natural number f is a factor of the natural number n if we can find a naturalnumber g so that f × g = n. (Alternatively, if n÷ f is some natural number g. We wouldsay f divides n.)
3
So f = 7 is a factor of n = 28, since we can find g = 4, and 7×4 = 28. (Alternatively,28 ÷ 7 = 4 which is a natural number.) Notice that this means that 1 is a factor of everynatural number, and every natural number is a factor of itself, since 1 × n = n alwaysworks. Most numbers will have lots of factors besides these, but a few numbers have onlythese trivial factors, and they are a very special class of numbers, the prime numbers.
A natural number p, p �= 1, is called a prime number if its only factors are 1 and p
(that is, itself).
Notice that 1 is specifically ruled out as a prime even though it does satisfy thecondition that its only factors are 1 and itself. There is a very good reason why we don’twant to count 1 as a prime number. The Fundamental Theorem of Arithmetic or Unique
Factorisation Theorem says that every natural number can be written as a product of itsprime factors in only one way. This property explains the importance of the prime numbers— they are in a sense the fundamental building blocks out of which we create the othernatural numbers. For example 12 = 2 × 2 × 3 is the only way to write 12 as a product ofprimes. However if we allowed 1 as a prime number we could write 12 = 1 × 2 × 2 × 3 or12 = 1 × 1 × 2 × 2 × 3 and so on.
Finding the prime factorisation of a number is relatively easy for small numbers (evenup into the thousands). Finding the prime factorisation of big numbers (and I mean BIG)is extremely difficult, and (believe it or not) extremely useful. Most modern codes andcyphers are based on the properties of prime numbers, and the fact that factorising very bignumbers is so time consuming as to be essentially impossible. Anyway, for small numbers,it is simply a case of continually splitting the number down into factors, and splitting thefactors, etc until you can go no further. Thus
780 = 2 × 390 = 2 × 3 × 130 = 22 × 3 × 65 = 22 × 3 × 5 × 13.
The first prime number is 2, and it is the only even prime number since every other evennumber will have a non-trivial factor of 2; 4 = 2 × 2, 6 = 2 × 3, etc. In a similar way,although 3 is prime, all of its multiples; 6,9,12,15,18,..., are not prime. The list of primenumbers begins
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, .....
Once again the line of dots at the end is to indicate that the prime numbers go on forever.This is not so obvious as it was for the natural numbers, and will require a little proof. Itis not essential that you know the proof of any theorems in this course, although it mayhelp you to remember and understand the result.
4
Theorem There are infinitely many prime numbers.
Proof Well suppose there weren’t. Then it would be possible to take all the primenumbers, multiply them together and add 1, and call the result m, say. The FundamentalTheorem of Arithmetic says that m can be expressed (in a unique way) as a product ofits prime factors, but what can these prime factors possibly be? All the primes we startedwith are not factors of m since they leave a remainder of 1 when we divide m. Thereforem must have some other prime factors, but this contradicts our the assumption that wehad included all prime numbers. Therefore the original assumption must be wrong, andthere must be infinitely many primes.
There, you’ve seen your first proof. I hope it wasn’t too painful. In any case don’tworry if you can’t understand the proof of the theorem, the important thing is to rememberwhat it was we proved. You can use the result of a theorem even if you do not understandits proof, although you are more likely to misuse a theorem which you don’t understand.That’s why it is nice if you can follow proofs, but not essential for getting by.
LCM and GCD
The lowest common multiple (LCM) of two numbers is the smallest number whichis divisible by both. For example, the LCM of 6 and 4 is 12, and the LCM of 8 and 2is 8. The LCM may be more familiar to you as the number you choose for the commondenominator when adding or subtracting fractions.
16
+14
=212
+312
=512
Finding the LCM of two numbers is easy when you know their prime factorisations. Foreach prime number, find the maximum number of times it appears in either of the twonumbers, and it must appear that many times in the LCM. For example 24 = 23 × 3 and36 = 22 × 32. That is, the prime 2 divides 24 three times, and divides 36 twice, so it mustdivide the LCM three times. The prime 3 divides 24 once and 36 twice, so it divides theLCM twice. Therefore the LCM of 24 and 36 is 23 × 32 = 72.
The greatest common divisor (GCD) of two numbers is the largest number which isa factor of both numbers. Thus the GCD of 6 and 4 is 2 and the GCD of 8 and 2 is also2. Finding the GCD of two numbers is easy when you know their prime factorisations.For each prime number, find the minimum number of times it appears in either of the twonumbers, and it must appear that many times in the GCD. Again using 24 = 23.3 and
5
36 = 22.32 we find that the GCD should be 22.3 = 12. Once again you may have seenthis notion in fraction work. The GCD is the number by which to divide numerator anddenominator to reduce a fraction to its lowest form.
2436
=24/1236/12
=23
Two numbers with GCD equal to 1 have no factors in common (except 1 of course) andare termed relatively prime or coprime.
Exercises 1.1
1 Find all prime numbers less than 100.2 Decompose the following numbers into their prime factors.
(a) 144 (b) 196 (c) 200 (d) 2753 Find the LCM and GCD of each pair of numbers from exercise 2 above.4 What is the smallest common denominator you could choose to do
1144
+1
196?
§ 1.2 Integers
The positive rational numbers (fractions) are usually taught before the integers sincefractions occur quite natuarlly at an early age. However, mathematically it is neater tointroduce the integers first, so that’s what we’ll do.
The integers are the numbers
.....,−4,−3,−2,−1, 0, 1, 2, 3, 4, .......
That is, the natural numbers (with zero thrown in) and all the negatives of the naturalnumbers. The usual abbreviation for the integers is Z (from a German word I believe),which on a blackboard is written:
The integers can be represented on a number line.
The best way to interpret the minus sign is as a ‘change of direction’ on the numberline. For a moment consider only the natural numbers on the number line, and throwin zero as well. Addition of a number can be thought of as travelling to the right, and
6
subtraction as travelling to the left. Now 2 − 3 is not a natural number, but it is onestep left of zero, and so the notation -1 seems appropriate. In this way we extend thenatural numbers to the integers. Each integer has a ‘size’ (the distance from 0) and adirection (positive or negative, right or left from 0). In order to extend arithmetic fromthe natural numbers to the integers we need to know how to combine the direction partsof the numbers. Each time two negatives are together, it indicates a double change ofdirection on the number line, but that means we are back to the positive direction. Twonegatives really do make a positive! Thus
(−5) − (−3) = (−5) + 3 = −2
(−5) × (−3) = −− 15 = 15
It is very important that you are comfortable and confident with adding and multiplyingnegative numbers together. This is the single biggest cause of ‘silly’ mistakes in mathe-matics. It is all too easy to miss a negative sign in a calculation, so it is as well to workslowly and deliberately when lots of negatives get together in the one place!
One thing we have achieved by extending the natural numbers to the integers is toclose the system under subtraction. If you take any two integers and subtract them theresult is again an integer. However we still have a problem with division, since 2 ÷ 3 stilldoes not have an answer among the integers. We therefore need to invent fractions.
Exercises 1.2
(a) 5 − 7 = (b) 5 ×−7 =(c) −5 −−7 = (d) −5 ÷−7 =(e) 3 − 7 −−5 = (f) −− 3 + −7 =(g) −3 ×−5 ×−7 = (h) −3 − 5 − 7 =
7
Lecture 2 Rational numbers, Pythagoras’ Theorem and surds
§ 2.1 Rational numbers
The need for rational numbers arises quite naturally at a very earlier age. As thefather of twins it is especially important to be able to divide one piece of cake into two(very equal!) halves. From a mathematical point of view we need to invent the rationals toget around the problem that the natural numbers and integers aren’t closed under division.
A rational number is any number which can be written as the ratio of integers (afraction). Thus r is rational if we can find two integers p and q such that
r =p
q.
The rational numbers are abbreviated Q (and I have no idea why ‘Q’ was chosen, unlessit has comes from ‘quotient’) which has the blackboard form:
Notice that I make a subtle distinction between the words rational and fraction. Tome ‘fraction’ refers to the form of the number; it is written as the ratio of integers; whereasrational is a property of the number; it can be written as the ratio of integers. Thus tome 0.5 is rational because it can be written as 1/2, but 0.5 is not itself a fraction. I freelyadmit that this is a pedantic point of view and most people, even many mathematicians,would not make this distinction.
At this point we will pause for a while and remind ourselves of the rules of arithmeticinvolving fractions. An inability to manipulated fractions confidently is a major problemamong students, particularly when algebra is introduced. Personally I believe the littleblack box, the calculator, is in large part to blame. Certainly pre-calculator students seemto have better arithmetic skills. It is imperative that you can manipulate fractions well,especially when we come to algebra.
Equalitya
b=
c
dif and only if ad = bc.
Thus 3/9 = 4/12 since 3 × 12 = 9 × 4. However, you will probably prefer to show twofractions are equal by reducing them to their lowest form, that is by dividing numerator
(top) and denominator (bottom) by their GCD. Doing this, 3/9 and 4/12 both reduce to1/3.
Multiplicationa
b× c
d=
ac
bd.
8
Thus23× 1
4=
212
=16.
Divisiona
b÷ c
d=
a
b× d
c=
ad
bc.
This is often referred to as ‘invert and multiply’. Thus
23÷ 1
4=
23× 4
1=
83.
Addition and subtractiona
b± c
d=
ad
bd± bc
bd=
ad ± bc
bd.
This process is called finding a common denomimator for the fractions. Thus
524
+736
=180864
+168864
=348864
=2972
.
Notice how cumbersome this calculation was because we did not use the smallest commondenominator. As mentioned above, the smallest common denominator is the LCM of 24and 36, which is 72. We could therefore have done the calculation as
524
+736
=1572
+1472
=2972
.
Again I cannot overemphasise the importance of manipulating fractions well. Thisneeds to be second nature to you so that you can concentrate on the problem you aresolving rather than the mechanics of arithmetic. A few other points on handling fractions:
A fraction is called improper if its numerator (top) is bigger than its denominator(bottom). Such a fraction may be broken down into an integer part plus a proper fraction.This is called a mixed fraction. For example
85
= 135.
I recall from my school days that this was encouraged, even enforced, and improper frac-tions were considered wrong. This convention is misguided. There is nothing wrong withusing improper fractions, in fact they have some definite advantages over mixed fractions,so either form is acceptable.
Another convention enforced at school is that all fractions must be reduced to theirlowest form by cancelling factors in the numerator and denominator (dividing by the GCDif you like). This convention is by and large enforced. Thus although 2/4 is the same
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number as 1/2 the former answer is frowned upon since it is not user friendly. Since thenumber has a simpler form you should use it.
Finally we should discuss division by zero. Recall when we do 6 ÷ 2 we ask ourselves‘how many 2’s make 6?’ and so we get the answer 3. Now if we do 6 ÷ 0 we ask ‘howmany 0’s make 6?’ and we get the answer that no matter how many 0’s we have we can’tpossibly get 6, thus 6÷ 0 has no answer. If we ever arrive at 6÷ 0 in a calculation we sayit is undefined. The same is true of virtually any other number divided by zero, but whatabout 0 ÷ 0? How many 0’s make 0? Any number would do, so again we can’t give ananswer, and so division by zero is always undefined.
We still have a bit to talk about with rational numbers — their decimal form, forexample — but let’s pause to see what we have achieved. We have produced a numbersystem in which addition, subtraction, multiplication and division (except by 0) are allpossible, and what is more the system is closed under all these operations, so it wouldappear that we do not have to seek any larger class of numbers. Such a system is so nicethat mathematicians give it a name, it is called a field of numbers. Throughout primaryschool and junior high school these are all the numbers we need; and throughout our dailylives these are the only numbers we commonly use.
Exercises 2.1
1 Reduce the following fractions to their lowest form.(a) 3/15 (b) 12/108 (c) 15/85 (d) 36/84 (e) 144/192
2 Write the following mixed fractions as improper fractions.(a) 12
3 (b) 2 47 (c) 5 3
11 (d) 3 1217
3 Write the following improper fractions as mixed fractions.
(a)73
(b)247
(c)4311
(d)11217
4 Add, subtract, multiply and divide each of the following pairs of fractions.
(a)12
and13
(b)25
and37
(c)2512
and59
§ 2.2 Pythagoras’ Theorem and√
2
Pythagoras’ Theorem is the most famous theorem in all mathematics, and quite im-portant. I introduce it here to demonstrate the limitations of the rational numbers. Thehypotenuse of a right-angled triangle is the side opposite the right-angle: the longest side.
Pythagoras’ Theorem The square of the length of the hypotenuse of a right-angled
triangle equals the sum of the squares of the lengths of the other two sides.
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Proof Consider the following diagrams.
The two big squares with sides of length a + b have identical areas. The four shadedtriangles in each diagram are identical, so the remaining white areas must be identical. Inthe left diagram this is made up of two squares, with areas a2 and b2. In the right diagramthis is one square with area h2, where h is the hypotenuse of the right-angled triangle.Since these areas are equal,
a2 + b2 = h2.
Pythagoras’ theorem was known long before his time, in Babylon, Egypt, and ancientChina; however he is said to have been the first to prove it, although we do not know howhe did it. Pythagoras’ Theorem is one of the very few pieces of geometry we will do in thiscourse, but it was worth doing for many reasons. Firstly it is so famous that you might feelcheated if you didn’t see it, like going to a Don Maclean concert and not hearing AmericanPie. Secondly it is so remarkably useful — just wait until we do trigonometry. But finally,the reason I want it here, is that we can use it to show that
√2 must exist. Imagine a
right-angled triangle with sides of length 1 coming from the right-angle, and ask yourselfwhat is the length of the hypotenuse? Since we aren’t sure, call that length h, and applythe theorem.
12 + 12 = h2
11
So h must be a number whose square is 2. Try as they might, the Greeks could notfind a rational number to do this.
Theorem There is no rational number h whose square equals 2.
Proof Suppose that there is such a rational number h. Since it is rational we couldwrite it as the ratio of whole numbers,
h =p
qwhere q �= 0.
So now if h2 = 2 we getp2
q2= 2
and multiplying both sides by q2 gives
p2 = 2q2.
Now squaring a number doubles up all its prime factors, so any prime number divides asquare an even number of times (maybe zero). We don’t know how many times 2 dividesp2, but it must be an even number of times. Similarly 2 divides q2 an even number of times,so it divides 2q2 once more than that, that is an odd number of times. But if p2 = 2q2
they must have the same prime factorisation by the Fundamental Theorem of Arithmetic,so we have a contradiction; an even number can’t equal an odd number. We conclude thatour original assumption must be wrong, and therefore h cannot be a rational number.
Don’t worry if you can’t follow the details of the proof. The important fact is, theremust be a number whose square is 2, since we can draw a triangle the length of whosehypotenuse has this property — and the theorem says that that length cannot be a rationalnumber.
So our search for the ideal number system must go on. Of course there are many morenumbers which are important and which aren’t rational. You have probably already metπ and later in the year we shall come across another important number which is called e.In fact most numbers aren’t rational, but they are seldom used.
Exercises 2.2
1 Find the hypotenuse of a right-angled triangle if the other two sides are(a) 3 and 4 (b) 5 and 12 (c) 9 and 40.
12
2 Find the length of the side of a triangle if the hypotenuse and its other side are(a) 13 and 12 (b) 41 and 9 (c)
√20 and 4.
3 A swimming pool is 50 metres long and 15 metres wide. How far, to the nearestmetre, must you swim to go diagonally across the pool?
§ 2.3 Surds
We have just seen that√
2 is not a rational number. In a similar way we could showthat
√3,
√5 and so on are also irrational. (Of course
√4,
√9 and so on are rational.)
We could also take cube roots, or even higher roots of numbers, and combine these in anyfashion using addition, multiplication and so on. Thus we can produce a number like√
2 + 3√
3.
Numbers produced in this fashion are called surds. Many of the irrational numbers we useare of this form, however it is a mistake to believe that all irrational numbers are surds.In fact most are not, for example π is not a surd. Once again confidence in manipulatingsurds is important as it allows you to concentrate on the real issues in a problem ratherthan the superficial task of calculating. We will concentrate on squareroots, though similarrules apply for cuberoots etc.
Multiplication of surds
√a ×
√b =
√a × b.
For example√
2 ×√
5 =√
10.
Lowest form
Often surds can be simplified by factoring perfect squares out of the squareroot. Thus
√72 =
√36 ×
√2 = 6
√2.
In general the answer 6√
2 is preferable to√
72 because it is in a more compact, userfriendly form. There may be occasions, however, when you need to go in the reversedirection, moving the 6 back under the squareroot as 36. Similar rules apply to cuberootsand so on.
13
Addition and subtraction of surds.
In general it is only possible to simplify addition or subtraction of surds which havethe same number under the squareroot. Thus
√2 +
√3 cannot be simplified any further.
But √2 +
√8 =
√2 +
√4 × 2 =
√2 + 2
√2 = 3
√2.
Here, although the two surds appeared to have different squareroot parts, by factoringout the perfect square 4, we see they share the same squareroot part,
√2, and so we can
combine them.
Division of surds - rationalising the denominator
Much the same as for improper fractions, the High School system has decreed thatsurds in the denominator of a quotient are unsightly, and should be removed. In fact thereis little to choose between 1/
√2 and
√2/2, and I believe the former is perhaps the nicer
form. Nonetheless there certainly are occasions when it is useful to remove a surd fromthe denominator of a quotient, so we should know how to do it. In fact, it is a good chanceto trot out a much used mathematical trick — doing absolutely nothing in a clever way!
Yes, the trick of doing nothing is much favoured by all mathematicians I know. Onecommon form it takes is to add zero (in disguise) to a number. This does not change thenumber one iota, but it can simplify things if the disguise is wisely chosen. The form weuse here is to multiply the number by 1 (in disguise), which of course doesn’t change thenumber, but can remove a surd from the denominator if we chose the correct disguise.
Given a surd a + b√
c we call the number a − b√
c its conjugate surd. To removea + b
√c from a denominator we multiply by 1, disguised as the conjugate surd over itself.
Thus, for example,
12 +
√3
=1
2 +√
3× 2 −
√3
2 −√
3=
2 −√
34 − 2
√3 + 2
√3 − 3
= 2 −√
3.
Exercises 2.3
1 Write the following surds in their ‘lowest form’.(a)
√27 (b)
√32 (c)
√75 (d)
√108
(e)√
125 (f)√
192 (g) 3√
16 (h) 3√
1922 Write the following surds in the form n
√x.
(a) 2√
3 (b) 5√
2 (c) 3√
7 (d) 3 3√
3
14
3 Simplify.(a) 4
√3 − 2
√3 + 5
√3 (b)
√8 +
√32 (c)
√75 −
√48 (d)
√27 −
√108 −
√12
(e)√
128 −√
98 +√
72 (f) 4√
48 − 5√
27 (g)√
x3 +√
x −√
4x
4 Simplify.(a)
√3 ×
√3 (b) 3
√3 × 5
√3 (c) 4
√5 × 2
√2 (d)
√2(√
8 −√
2)(e) (3
√2)2 (f) 2
√3(4
√2−5
√3) (g) 3
√3(√
8+2√
2) (h) (√
7−2√
3)(√
5−3√
2)5 Multiply each of the follwowing by its conjugate surd.
(a) 3 +√
2 (b) 7 −√
3 (c) 4√
3 − 5 (d) 2√
7 + 16 Rationalise the denominator of the following surds.
(a)3√2
(b)√
3√6
(c)√
5 −√
2√3
(d)1√
2 + 1
(e)3√
3 − 1(f)
5√5 +
√7
(g)3 −
√2
3 +√
2(h)
3 + 2√
52√
5 − 1
15
Lecture 3 Real and complex numbers
§ 3.1 Real numbers — decimal representation
We return to building up our ideal number system, having progressed from naturalnumbers to integers then rationals. Through Pythagoras’ Theorem we saw that
√2 must
exist, since it is the length of the side of a triangle, and we believe in the concept oflength. However from the Fundamental Theorem of Arithmetic we saw that
√2 could not
be rational. This number, and others like it, are called surds — and we have learnt alittle about calculating with surds — but still it is not enough. There are other numbers,the most notable being π, which do not fall into any of these categories. These, togetherwith all the rest, are the real numbers. In many ways they are very intuitive, and yetto describe them fully is beyond the scope of this course. Paradoxically, however, theyare the numbers which we shall use more than any others. To understand them better,we will investigate the decimal system of representing numbers. This will take us on twodigressions, but fortunately they are fairly important digressions which we would have hadto study at some stage anyway.
Place-value representation of natural numbers
Our way of writing natural numbers involves only ten symbols, the digits 0, 1, 2, 3,4, 5, 6, 7, 8, 9. We are able to use just ten symbols to write natural numbers becausethe meaning of a symbol is dependent on its position within the number. Thus a 9 in thefirst place (rightmost) means 9 lots of one, whereas a 9 in the second place means 9 lotsof ten, and a 9 in the third place means 9 lots of one hundred, and so on. This is calledthe place-value system. The significance of the numbers 1, 10, 100, ... are that they are thepowers of 10, respectively 100, 101, 102, .... (We’ll discuss why 100 should equal 1 shortly.)Thus the number 6853 quite literally means
(6 × 103) + (8 × 102) + (5 × 101) + (3 × 100).
We should mention briefly that there is nothing magical about the number ten. Bysome quirk of fate mankind was given ten digits on his hands, which he used for counting,and so he invented a place-value system based upon the number ten. The ancient Babylo-nians based their system on the number sixty for some strange reason. In modern times,the advent of the computer has made base 2 very important. In base 2 there are only twodigits, which we’ll write as 0, 1, but which to a computer can be on, off. The principle isthe same. The number 100112 (the subscript indicates the base) means literally
(1 × 24) + (0 × 23) + (0 × 22) + (1 × 21) + (1 × 20)
16
which is equal to 1910 (I won’t bother with subscripting the 10 from now on).
The index laws for natural numbers
Whenever we write a number raised to some power, the power to which it is raised iscalled the index or exponent. Thus in 102 the index of 10 is 2.
Taking powers of a number is an easily defined operation when the exponent is anatural number. The operation an means multiply a by itself n times. So now if weconsider am × an, it means multiply a by itself m times, and n times, and multiply the
two results. Well the net result of that is that you have multiplied a by itself m+n times.We’ve just proved the first index law! Similar reasoning gives us the other laws.
am × an = a(m+n)
am ÷ an = a(m−n)
(am)n = amn
Notice that our reasoning depended on our definition of taking powers, which only makessense for natural numbers. In the second index law, if n > m, the right-hand side givesus a negative index, but we don’t know what that means. After all can you multiply a
by itself a negative number of times? What we do now is another common way in whichmathematics proceeds. We take the index laws, proven for natural indices, and demandthat they hold in a wider setting, effectively defining what we mean by negative indices.Now we see why 100 must equal 1, because we can put a = 10 and m = n into the secondindex law. Now it is possible to write down two more index laws based on these ideas.
a0 = 1 provided a �= 0
a−n =1an
The index laws now apply to all integers, not just to natural numbers. That’s all I reallywanted in order to introduce decimal notation, but while we’re here we might as wellconsider fractional indices.
Consider what raising a number to the power 1/2 might mean. If we assume the firstindex law holds for fractional powers then
a1/2 × a1/2 = a1/2+1/2 = a1 = a,
or again by the third index law
(a1/2)2 = a(1/2)×2 = a1 = a.
17
So a1/2 would have to be a number which squares to give a. Therefore an index of 1/2means squareroot. Similarly an index of 1/3 means cuberoot, and so forth. In this way wecan apply the index laws to all rational indices (however there is the slight problem thatsquareroots of negative numbers can’t be done).
In fact the index laws can be applied to any real numbers, including surds and numberslike π. At the moment there is no way for me to explain what raising a number to thepower π means. This is a detail which is usually overlooked! We shall simply note thatthe same index laws work for all real numbers, and leave it at that.
Place-value representation of real numbers
The main reason for introducing the index laws at this time was to clarify what ismeant by negative powers. The place-value system can now be extended, adding a decimalpoint to mark where the negative powers of 10 begin, and continuing the digits beyondthe decimal point. Thus 681.25 means, literally
6 × 102 + 8 × 101 + 1 × 100 + 2 × 10−1 + 5 × 10−2.
Of course the digits past the decimal point, 0.25, would be 1/4 in fractional notation. Wearrive at this through a division of 1 by 4, writing 1 as 1.00000... . After a couple of stepsof the division there is no remainder, so the division stops and we have our answer. Whathappens when we try to write 1/3 in decimal form? (Try it!)
Now 3 into 1 doesn’t go, but 3 into 10 goes 3 times with remainder 1. So far we havethe answer as 0.3. Then 3 into 10 goes 3 times with remainder 1. So far the answer is 0.33.But we must continue. We notice that it gets a bit repetitive, in fact very repetitive. Nomatter what we do the division is never completed, and the ‘answer so far’ is a long stringof 3’s past the decimal point. Of course we cannot write down 3’s forever, but once thepattern is established we can write a ‘dot’ over the three, 0.3̇, to indicate that it continuesforever, and now we have written 1/3 exactly in decimal notation.
Now it is an interesting fact that every fraction may be written in a repeating decimalformat, although it may involve not just one number repeating, but a whole block ofnumbers repeating. We find that 1/7 is given by 0.1̇42857̇, where the two ‘dots’ abovethe number indicate that the whole block of numbers repeats. Fractions (in their lowestform) will only give terminating decimals if their denominators have only 2 or 5 as primesfactors (can you think why?), so most fractions lead to repeating decimals. They mustrepeat since in any division there are only a finite number of different remainders possible.Once a remainder crops up which has appeared before the pattern must start to repeat.
18
Now we should see how a mathematician thinks. If every fraction gives rise to arepeating (or terminating) decimal, must every repeating decimal equal some fraction?The answer is YES! I don’t want to give a full proof, but let’s consider an example whichcaptures all the necessary ideas. We will show that x = 0.7̇32̇ is a rational number; thatis, it can be written as a fraction.
1000x = 732.7̇32̇−x = 0.7̇32̇
999x = 732
Therefore x = 732/999 = 244/333. So far we have outlined an important fact, or Theorem,which you should remember.
Theorem A number is rational if and only if its decimal representation is repeating (or
terminating).
Of course a terminating decimal can always be thought of as having repeated 0’s (orrepeated 9’s!).
Now above we accepted the idea that once a pattern was established in the deci-mal representation of a number we could consider it to continue forever. Without ac-cepting this we could not even write 1/3 as a decimal. But what about a pattern like0.101001000100001000001.... ? It is quite clear what the pattern is, and how it should beextended forever, nevertheless this is not a rational number because it does not containa repeating block of digits. Why shouldn’t this be just as valid a number as 1/3? Andfor that matter, why does there have to be a predetermined pattern to the digits. If wewere to write down the decimal representation of
√2, for as many digits as we could be
bothered, we would see no obvious pattern, and yet√
2 is a real number.And so we come to the definition of the real numbers, as best we can in this course.
The real numbers are all the possible numbers we can obtain be allowing neverendingstrings of decimal digits.
This will include all the surds like√
2 = 1.14142135... as well as even stranger numberslike π = 3.1415926... . You should always remember that whenever you use decimalsto represent numbers like these you are using an approximation because their decimalexpansion continues forever. It is always good policy to leave numbers in their exact form(√
2 or π) until the very end of a calculation. This prevent inaccuracies and, believe it ornot, makes algebra easier because converting everything into long strings of digits tendsto hide patterns which would be obvious otherwise. Even at the end of a calculation there
19
is no need to give a decimal approximation unless you are asked for it. An answer of√
2π
is more accurate, more succinct, and more easily marked then an answer of 2.5066283.
Exercises 3.1
1 Write 26 in base 2, base 5 and base 7.
2 What number (base 10) does 111b represent if b = 2, 5 or 7?
3 Evaluate.
(a) 2−1 (b) 3−3 (c) 912 (d) 4−
12 (e) 70
4 Write using index notation.
(a)1x
(b)1x3
(c)√
x (d)2
(2x + 5)4(e)
13√
x5 Write without negative indices.
(a) x−5 (b) 2x−3 (c) (2x)−3 (d)(
1x + y
)−2
(e)(
x + y
x − y
)−1
6 Use the index laws to simplify the following.
(a) 26 × 29 × 22 (b) 33 ÷ 3−7 (c) (25)2 (d)(32 × (33)4)
39
(e)x2
x−3(f) (3y−2)4 (g)
a2(b2)6
a4b9(h)
(x2)−3(y3)2
x−1y4
7 Write the following fractions as repeating decimals.
(a) 2/9 (b) 3/11 (c) 3/7 (d) 7/13 (e) 1 211
8 Write the following repeating decimals as fractions.
(a) 0.4̇ (b) 0.2̇3̇ (c) 0.23̇45̇ (d) 1.23̇ (e) 0.9̇
9 Find√
2 from your calculator, and write down all the digits it displays. Nowsubtract these digits away from
√2. Why isn’t the answer zero? Why doesn’t the
answer have more digits?
§ 3.2 Scientific notation and significant figures
So far we have been looking at numbers the way a mathematician thinks of them.However in more practical disciplines, like the sciences and engineering, numbers are oftentreated somewhat differently.
Because the magnitude of numbers used in the sciences can vary widely, dependingon the units used, it is common to report numbers in scientific notation. This consists ofwriting the number with just one digit left of the decimal point, multiplied by a power of10, to indicate the order of magnitude. So for example the number 625 would be written6.25 × 102, and the number 0.625 would be written as 6.25 × 10−1. Very large or verysmall numbers are reported this way on hand calculators.
20
Since in the sciences most numbers represent a measurement of some quantity, andthat measurement will not be exact, it is conventional to indicate the accuracy of themeasurement by reporting the value to a certain number of significant figures. After all,when you listen to the weather at night, you don’t expect to hear that the daily maximumwas 25.3678◦C! It is extremely doubtful that the instruments used would be accuratebeyond the first decimal place, so the remaining digits are meaningless. To indicate this,the temperature should be reported as 25.4◦C (three significant figures). In fact, even if thetemperature could be measured that accurately, for the purposes of reporting the weatherthe decimal digit is not relevant. Thus the temperature would probably be given as 25◦C,which is two significant figures. As a general rule, the last digit reported in the numbershould be the only one which is not certainly correct. Therefore a temperature of 25◦Cindicates that the temperature lies between 24.5◦C and 25.5◦C, whereas a temperatureof 25.3◦C indicates that the temperature lies between 25.25◦C and 25.35◦C. Note thatleading zeroes in a decimal do not count towards significant figures, since they indicatethe order of magnitude of the number, not its accuracy. On the other hand, final zeroespast the decimal point do count towards significant figures, since they indicate improvedaccuracy.
Exercises 3.2
1 Write the following numbers in scientific notation.
(a) 7892 (b) 0.007654 (c) 760 (d) 7.765
2 How many significant figures have each of the following.
(a) 987 (b) 9.87 (c) 9.870 (d) 0.0987
3 Three students working together on an experiment each wrote their lab reportsseparately, giving their results as 1
2 , 0.5 and 0.50. Explain why none of these answersare the same.
§ 3.3 Epilogue — Complex numbers
Let’s now review how we stand with our number systems. We began with the naturalnumbers, then the integers, then the rational numbers. At this stage we had a systemwhich was closed under the usual operations (we called it a field), and was sufficient foreveryday use. We were still missing important numbers we knew existed, however, like√
2 and π, so we invented decimal representation and allowed neverending strings of digitsto represent numbers. This captures all the numbers we wanted, and is still a field. Weshould be happy with the number system we have arrived at, and indeed for the purposes
21
of this course we are content. However, since there is but one step to go in this process,let’s just mention it in passing.
Recall that we used Pythagoras’ Theorem to justify that there must be a number x
such that x2 = 2, and we called that number√
2 because there was no convenient wayto write it as a fraction or as a decimal. Now I can’t draw a triangle with hypotenuseof length -1, but still I might wish to solve x2 = −1. Now the real numbers are no useto me here because anything squared is greater than (or equal to) zero. Pragmatically Icould blunder forward and invent the number
√−1, just as we invented the number
√2.
This is what mathematicians do, and the resulting numbers are called complex numbers.For some reason people baulked at this notation, and decided instead to give it the specialname i, but what’s in a name? For whatever reason, this number is termed imaginary,and unfortunately this has led to the belief that it is not a ‘true’ number. Initially youwill probably also hold this belief. Nonetheless, you can add it, subtract it, multiply anddivide it, and do algebraic operations with it — so if its not a number, at least it behaveslike one.
A complex number is a number which can be written as a + ib where a and b are realnumbers and i is the squareroot of -1. Most often a complex variable is represented bythe letter z, just as most often real variables are represented by an x (as first preference).The real and imaginary parts of the complex number z = a + ib are
Re(z) = a Im(z) = b.
The complex conjugate of a number is the number obtained by changing the sign of itsimaginary part, and is denoted by a bar over the variable. Thus if z = a + ib the complexconjugate of z is z̄ = a − ib.
Performing arithmetic with complex numbers is just like doing algebra, except thathigh powers of i can be simplified using the property that i2 = −1. Addition and subtrac-tion are simple.
(2 + 3i) + (7 − 6i) = 9 − 3i
(2 + 3i) − (7 − 6i) = −5 + 9i
Multiplication is nearly as easy.
(2 + 3i)(7 − 6i) = 14 − 12i + 21i − 18i2 = 32 + 9i (since i2 = −1)
Division is performed in much the same way as rationalising the denominator of surds —multiplying by 1 in disguise. The particular disguise of 1 which works is to divide the
22
conjugate of the denominator by itself, thus
2 + 3i
7 − 6i=
2 + 3i
7 − 6i× 7 + 6i
7 + 6i
=14 + 12i + 21i + 18i2
49 − 36i2
=−4 + 33i
85.
Exercises 3.3
1. Find the real and imaginary parts, and the complex conjugate of the complexnumbers 5 + 3i and −2 −
√2i.
2. Add, subtract, multiply and divide the numbers 5 + 3i and −3 + 5i.
23
Lecture 4. Basic algebra
This lecture is intended to develop a smorgasbord of algebraic skills which will beneeded throughout the rest of the course. Algebra is, unfortunately, one of those areas inwhich there is no substitute for experience. It is always helpful to know why things aretrue, but in algebra it is not just knowing the rules, it is knowing when and how to applythem. This usually requires lots of exercises. By doing many examples in which you arriveat similar situations you learn to recognise algebraic patterns, and how to deal with them.
Algebra is the manipulation of symbols (usually we use letters, sometimes Greekletters) using the rules of arithmetic. In effect the pronumeral takes the place of a number,any number, and so by performing an algebraic calculation we are finding an answer whichis valid for any choice of numbers. A consequence of this is that we cannot use anyproperties which are specific to particular numbers. For example, x/y cannot be reducedto a fraction of lower form algebraicly, despite the fact that for some choices of x and y
(say x = 3, y = 6) there is a lower form.
§ 4.1 The Laws of Arithmetic
In this section we will note some well known facts about the way arithmetic operationscan be manipulated. Most of these facts are so well known we won’t need to commentabout them any further.
The Commutative Law
Both addition and multiplication are commutative; that is, the numbers can be inter-changed without affecting the result.
a + b = b + a
a × b = b × a
For example 2 + 3 = 3 + 2 and 2 × 3 = 3 × 2. Of course subtraction and division are notcommutative; 2 − 3 �= 3 − 2 and 2 ÷ 3 �= 3 ÷ 2.
The Associative Law
Both addition and multiplication are associative; that is, the order in which the opera-tion is performed does not matter. (Bracketed expressions are calculated first — discussedbelow in § 4.2)
(a + b) + c = a + (b + c)
(a × b) × c = a × (b × c)
24
For example (2+3)+4 = 5+4 = 9 and 2+(3+4) = 2+7 = 9. Similarly (2×3)×4 = 6×4 =24 and 2 × (3 × 4) = 2 × 12 = 24. Of course subtraction and division are not associative;(2− 3)− 4 = −1− 4 = −5 but 2− (3− 4) = 2− (−1) = 3 and (2÷ 3)÷ 4 = 2
3 ÷ 4 = 212 = 1
6
but 2 ÷ (3 ÷ 4) = 2 ÷ 34 = 8
3 .
The Distributive Law
Multiplication is said to be distributive over addition. This means that when wemultiply the sum of two numbers, it is the same as multiplying each number in turn, andadding the results.
a × (b + c) = (a × b) + (a × c)
For example 2× (3 + 4) = 2× 7 = 14 which is the same as (2 × 3) + (2× 4) = 6 + 8 = 14.When the distributive law is used in the direction given above we would say we expanded
the expression. If it is used is the reverse direction it is called factoring the expression. Useof the distributive law in either direction is the standard move in most algebraic problems.One of the most common situations encountered is
(a + b)(c + d) = (a + b)c + (a + b)d = ac + bc + ad + bd.
Probably the best way to do this is to begin with a term from the first set of brackets, andmultiply it by each term from the second set; then take the next term from the first set ofbrackets and repeat the process.
Factoring expressions (that is, using the distributive law in reverse) is a more difficulttask as it sometimes requires a ‘good eye’ to recognise the potential factors. This onlycomes from experience. An easy case is when each term of an expression has the samevariable occuring in it. For example
a3 − 3ab2 + 2a2b
has an a in each term, and so we can factor this to give
a(a2 − 3b2 + 2ab).
25
What is less easy to see is that a2−3b2 +2ab can also be factored. Check that (a+3b)(a−b) = a2 − 3b2 + 2ab! So now the original expression can be written as
a3 − 3ab2 + 2a2b = a(a + 3b)(a − b).
We’ll develop some algebraic ‘tricks of the trade’ to try and make factoring easier.
Exercises 4.1
1 Use the distributive law to expand the following expressions.(a) 2y(4 − y) (b) 2(3t − 4) − (t + 1) (c) (p + 3)(p + q − 4)(d) 3x2 + 2x(x − 4) (e) 6a3 + 3a(a2 − 2a + 6) (f) 2y(6 + 4y) − 6y(2 − 3y)(g) (x + 3)(x + 2) (h) (m + 1)(m − 6) (i) (3c + 4)(5c − 3)(j) (n − 3)(n + 3) (k) (n + 3)2 (l) (a − 7)(3a2 − a)(m) (a + b)(a2 + b2) (n) (a + 1)(a2 + 3a + 3) (o) (2x2 − 3x)(x3 − x2 + 3x − 4)
2 Use the distributive law in reverse to factorise the following expressions.(a) ax + ay (b) y2 − 4y (c) 5a2 − 15a
(d) 7ab + 14bc (e) −3 + 6x (f) x3 + 7ax2
(g) at2 + 2at + 5t (h) 8m2n + 6mn2 + 10mn (i) (a + b)2 + x(a + b)(j) uv + 5u + 3v + 15 (k) 2xy − 3yz + 4x − 6z (l) x(a − b) + y(b − a)(m) xy + y2 − x − y (n) 4 − 4x + cx − c (o) x3 − 3x2 + 2x − 6
§ 4.2 Precedence of operations
As we saw in the last section, there are times when the order in which operations areperformed affects the result. For example 2÷3÷4 could equal 1
6 or 83 depending on which
of the divisions is performed first. We therefore need a convention to determine which ofthese answers we mean.
The difficulty is with the notation, and the fact that it is ambiguous; that is, thereare two possible interpretations of the expression 2 ÷ 3 ÷ 4. The convention which isapplied to remove this ambiguity is known as the precedence of operations, or the order of
operations, and it merely tells us which operations recieve priority over others. In this way,any mathematical expression will have one and only one meaning to everybody, so therecan be no ambiguity. I stress that the rules we are about to write down are a convention.We could have chosen other rules, but those given here are universally accepted.
26
Precedence rules
(i) Perform all operations on bracketed groups of objects first. (Note that the numer-ator and denominator of a fraction are considered to be grouped even if the bracketsare not specifically written. The same applies to expressions under squareroot signs.)
(ii) Perform all exponentiation (taking powers) next.
(iii) Then all multiplications and divisions, working left to right.
(iv) Finally all additions and subtractions, working left to right.
Under these rules we see that 2÷3÷4 = 16 , because there are no groupings, or powers,
and by rule three we evaluate the divisions working from left to right.
Exercise 4.2
Evaluate the following expressions.(a) 6 ÷ 32 − 2 × 3 (b) (6 ÷ 3)2 − 2 × 3 (c) 6 ÷ 3(2−2) × 3 (d) 6 ÷ (32 − 2) × 3(e) 15 + 5 × 23 (f) 33 + 3 ×
√64 (g)
√81 ÷ 9 (h)
√81 ÷ 9
§ 4.3 Squaring
Some algebraic operations occur so frequently that it is worthwhile treating them asspecial, and remembering specific tricks to solve them. One such example is squaring asum.
(a + b)2 = (a + b) × (a + b)
= a2 + ab + ba + b2
= a2 + 2ab + b2.
This formula is beautifully illustrated by drawing a square whose side has length a + b!
27
In words you would say, the square of a sum, is the sum of the squares plus twice
the product. It is an extremely common (and unforgivable) mistake to forget that lastbit!
As with a lot of algebra, this rule is simply an application of the distributive law, andcan therefore be applied in reverse; and as usual this is more difficult since you need to beable to ‘spot’ the factor to use. For example
a2 + 6a + 9 = (a + 3)2
or
4x2 + 4xy + y2 = (2x + y)2.
Not so easy, but it comes with practice. In the last case, for example, we just look forterms which are obviously squares. This gives us (2x)2 and y2. A quick check shows theremaining term is twice the product; 4xy = 2(2x)(y).
A very similar result is
(a − b)2 = (a − b) × (a − b)
= a2 − ab − ba + (−b)2
= a2 − 2ab + b2.
In fact this can be thought of as (a + (−b))2, and then it is seen to be just a special caseof the previous rule.
Exercises 4.3
1 Expand the following.(a) (x + 5)2 (b) (y − 2)2 (c) (x + 2y)2
(d) (3a − 2b)2 (e) (m2 + n2)2 (f) (3xy − z)2
2 Write the following as perfect squares.(a) x2 + 4x + 4 (b) t2 − 8t + 16 (c) 9a2 + 4b2 + 12ab
(d) s2 + 4t2 − 4st (e) s2 + s +14
(f) x2 +1x2
+ 2
§ 4.4 Completing the square
Consider an expression like x2 + 12x. We could factorise this as x(x + 12), and inmany circumstances that is what we would want to do, but suppose for some reason wewanted to write it in such a way that a square appeared. How could we do that?
28
Comparing x2 + 12x with a2 + 2ab + b2 we see that x2 could play the role of one ofthe squared terms, and 12x could play the role of ‘twice the product’. This gives us x = a
and 12x = 2ab, so b must be 6. Unfortunately we don’t have anything to play the role ofb2. All is not lost. We simply do nothing by adding and subtracting the term we want, inthis case b2 = 36. Thus
x2 + 12x = (x2 + 12x + 36) − 36
= (x + 6)2 − 36.
This process is called completing the square. The number to be added and subtractedis always the ‘twice the product’ term (2ab), halved and divided by the first square((2ab)/(2a) = b), then squared (b2). One more example,
t2 − 4√
2t = (t2 − 4√
2t + 8) − 8
= (t − 2√
2)2 − 8.
The reason we added and subtracted 8 was that 8 =(−4
√2t
2t
)2
.
Exercises 4.4
1 Complete the square on the following.(a) x2 + 8x (b) y2 − 16y (c) m2 − 14m (d) y2 + 3y
(e) a2 + a (f) x2 + 2√
7x (g) x2 + 8xy (h) 4a2 − 4ab + 4
29
Lecture 5. More algebra
§ 5.1 Factorising quadratic expressions (monic)
An expression of the formax2 + bx + c
where a, b, c are numbers and x is the variable, is said to be quadratic in x. We would liketo be able to factorise such an expression into the product of linear expressions; that is,expressions of the form ax+b. It is not always possible to do this, and even if it is possibleit may be too difficult to do easily, but we wish to develop some method to try to factorisequadratic expressions if possible.
The first case we will consider is that of a monic quadratic, which just means that thecoefficient of the x2 term is 1. Consider
(x + a)(x + b) = x2 + (a + b)x + ab.
By comparing our monic quadratic with the right-hand side of this equation we may beable to find two numbers a and b which will work. For example, given the quadraticx2 + 4x + 3, and comparing it with the above, we need to find two numbers a and b suchthat
a + b = 4 and ab = 3.
The two numbers which do this are 1 and 3, so we have found that x2 + 4x + 3 factors togive (x+1)(x+3). If you have any doubts you can always check your answer by expandingit; (x + 1)(x + 3) = x2 + 3x + x + 3 = x2 + 4x + 3.
One more example. To factorise m2 − 2m − 15 we would need to find two numberswhich add to give −2 and multiply to give −15. In my head I would think of factors of 15which differ by 2, and come up with 3 and 5. Since the product should be negative I needto make one of the numbers negative, and since the sum is also to be negative, it shouldbe the bigger one that is negative. In this way I arrive at 3 and -5. Let’s check it,
(m + 3)(m − 5) = m2 − 5m + 3m − 15 = m2 − 2m − 15
so it worked.
30
Exercises 5.1
1 By inspection, find values of a and b satisfying(a)
{a + b = 10ab = 21
(b){
a + b = 8ab = 7
(c){
a + b = -10ab = 24
(d){
a + b = 8ab = 15
(e){
a + b = -3ab = -28
(f){
a + b = 5ab = -6
2 Factorise the following monic quadratics. (Refer to question 1)(a) x2 + 10x + 21 (b) x2 + 8x + 7 (c) x2 − 10x + 24(d) x2 + 8x + 15 (e) x2 − 3x − 28 (f) x2 + 5x − 6
3 Factorise the following monic quadratics.(a) x2 − 2x − 63 (b) y2 + 6y + 5 (c) z2 − z − 110(d) a2 + 14a + 45 (e) b2 − 29b + 100 (f) m2 + 4m − 77
§ 5.2 Factorising quadratic expressions (non-monic)
Now we have to deal with the case where the coefficient of the squared term is not1. One solution is the ‘do nothing’ option; dividing and multiplying by the coefficient tocreate a monic quadratic. For example we could write 4y2 +4y− 3 as 4× (y2 + y− 3
4 ) andthen factor (y2 + y − 3
4 ) by the method above. This is fine, but tends to introduce messyfractions. I’ll outline another method, whose proof may seem a little confusing, but whichis easy enough to use.
We wish to factor a quadratic into linear factors, so
ax2 + bx + c = (dx + e)(fx + g).
Expanding these we find thata = df
b = ef + dg and
c = eg
This means that ac = (df)(eg) = (ef)(dg). That is, there are factors of ac whose sum is b.We can then write
ax2 + bx + c = ax2 + efx + dgx + c
= dfx2 + efx + dgx + eg
= fx(dx + e) + g(dx + e)
= (dx + e)(fx + g)
31
Perhaps we need to demonstrate this with an example! Let’s try to factorise 4y2 + 4y − 3.Then ac = −12 and b = 4. Factors of −12 whose sum is 4 are 6 and −2. So
4y2 + 4y − 3 = 4y2 + 6y − 2y − 3
= 2y(2y + 3) − (2y + 3)
= (2y − 1)(2y + 3).
Yes, I agree that factoring quadratics can be a bit tricky, and it is something which I amnot too good at either. Usually if it is easy, a bit of trial and error can work. A harderexample can be done by the methods above, but you are not guaranteed that there willbe a solution. In a short while we will introduce the quadratic formula, which will givea foolproof means of factoring. Nevertheless, the methods here are quicker and easier onsimple problems which work out to have neat solutions.
Exercises 5.2
1. Factorise the following non-monic quadratics.(a) 3x2 + 5x + 2 (b) 2x2 + 5x + 2 (c) 2x2 + 7x + 3(d) 6x2 − 7x − 3 (e) 3x2 − 13x + 14 (f) −10x2 − 7x + 12
§ 5.3 Differences of squares
Another situation which arises very frequently is the difference between two squares.Observe that by the distributive law
(a + b)(a − b) = a2 − ab + ba − b2
= a2 − b2
This tells us that the difference of two squares can always be factored into the sum of thetwo numbers times the difference of the numbers. (It was this rule that was at the heartof rationalising denominators of surds.)
Exercises 5.3
1 Factorise the following.(a) x2 − 16 (b) y2 − x2 (c) 4a2 − 9b2 (d) 4a2 − 2b2
(e) 25 − n2 (f) a2 − b2c2 (g) x4 − y4 (h) (y + 4)2 − (y − 2)2
32
§ 5.4 Sums and differences of higher powers
There are a few more algebraic tricks which, although they are not needed as often asthose above, are still handy to know. Recall that we have already considered the differenceof two squares.
x2 − y2 = (x − y)(x + y).
We might well ask about the difference of two cubes. Is there a similar formula? Yes thereis!
x3 − y3 = (x − y)(x2 + xy + y2)
To prove it we simply expand the right-hand side of the equation and observe all the termswhich cancel.
(x − y)(x2 + xy + y2) = x3 − x2y + x2y − xy2 + xy2 − y3
= x3 − y3
Now if it works for squares and cubes, perhaps it should work for higher powers. Indeedit does. Following the pattern for x4 − y4 gives us
x4 − y4 = (x − y)(x3 + x2y + xy2 + y3).
(In fact in this case we can do better. Treating it as a difference of squares, x4 − y4 =(x2)2 − (y2)2, we can factor it as (x2 − y2)(x2 + y2) and then the first term factors againto give (x − y)(x + y)(x2 + y2).)
Since such a nice pattern exists for the difference of powers, we might expect a nicepattern for the sum of two powers. However, there is no such pattern for squares, sincex2 + y2 cannot be factored. Our hopes are not dashed yet though. If we experiment withthe sum of two cubes we find
x3 + y3 = (x + y)(x2 − xy + y2).
Once again to prove this we simply expand
(x + y)(x2 − xy + y2) = x3 + x2y − x2y + xy2 − xy2 + y3
= x3 + y3.
More experimenting would show that for x4 + y4 we again can’t do anything, but that
x5 + y5 = (x + y)(x4 − x3y + x2y2 − xy3 + y4).
33
So there seems to be a pattern for the sum of odd powers but not for the sum of evenpowers.
Exercises 5.4
1. Factorise x6 − y6.2. Factorise x7 − y7 and x7 + y7.
§ 5.5 Pascal’s Triangle
Earlier we saw how to square a sum, (x + y)2 = x2 + 2xy + y2, but often we maywish to cube a sum, or take even higher powers. Of course this can be done by repeatedapplications of the distributive law, expanding and expanding again. This process is notonly time consuming, it is likely to result in a ‘silly’ mistake. It would be nice to havea simple method for doing these higher powers. Like most of algebra, it comes down tospotting the right pattern. Consider
(x + y)3 = (x + y)(x + y)2
= (x + y)(x2 + 2xy + y2)
= x3 + 2x2y + xy2 + x2y + 2xy2 + y3
= x3 + 3x2y + 3xy2 + y3.
For fourth powers we have
(x + y)4 = (x + y)(x + y)3
= (x + y)(x3 + 3x2y + 3xy2 + y3)
= x4 + 3x3y + 3x2y2 + xy3 + x3y + 3x2y2 + 3xy3 + y4
= x4 + 4x3y + 6x2y2 + 4xy3 + y4.
Now if we write these out, along with the rules for lower powers, a pattern starts to emerge.
(x + y)0 = 1
(x + y)1 = x + y
(x + y)2 = x2 + 2xy + y2
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
Observe how across each row we run through terms in x and y which diminish by onepower of x each time, and increase by one power of y. Can you also spot the pattern
34
among the coefficients? If we write them separately, dropping all the variables it may bemore apparent.
11 1
1 2 11 3 3 1
1 4 6 4 1
This pattern is very famous, and is known as Pascal’s Triangle. Notice how each termcan be found by adding the two terms above it, to left and right. So the next row of thetriangle would be
1 5 10 10 5 1.
If our hunch is correct we should now be able to write down the formula for (x + y)5.Firstly we get the pattern of the variables correct.
x5 + x4y + x3y2 + x2y3 + xy4 + y5
Then we add the coefficients from the fifth row of Pascal’s triangle, (I count the startingrow of ‘1’ as the zeroth row!).
1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5
As one of your exercises you can check that this is correct.Why does it work? I don’t want to give a full proof, but some explanation is in
order. Let’s take (x + y)4 as our example, and look at the coefficient of x2y2. Writing(x + y)4 = (x + y)(x + y)3, think about how we could get terms involving x2y2. Thesecan arise in two ways; either we multiply x from the first set of brackets by xy2 from thesecond set, or y from the first set of brackets by x2y from the second set. So the coefficientof x2y2 in (x + y)4 comes from the sum of the coefficients of xy2 and x2y in (x + y)3; thatis, the two numbers above to the left and right in Pascal’s triangle!
Exercises 5.5
1. Extend Pascal’s triangle to the sixth row.2. Expand (x + y)5 using the distributive law, and check the result against that pre-
dicted above by Pascal’s triangle.3. Use Pascal’s triangle to expand the following.
(a) (a + b)6 (b) (x − y)4 (c) (x + 2y)3
35
Lecture 6. Solving equations
You guessed it, more algebra! In the last two lectures we looked at ways of manipu-lating expressions into different forms without changing their substance. To this end wefound the mathematical principle of doing nothing (in a clever way) very useful. In thischapter we will be concerned with equality between two mathematical expressions, andways of manipulating these into different forms without changing their substance. To thisend the mathematical principle of do the same thing to both expressions (in a clever way)will be the unifying idea.
§ 6.1 Changing the subject of a formula
The words formula and equation are virtually synonymous. Perhaps the word ‘for-mula’ suggests a mystical power, a magical way to solve a problem, whereas a mere ‘equa-tion’ is just a true mathematical statement. Tied up with this perception is the idea thata formula must have a subject whereas an equation need not. The subject of a formula isthe variable which the formula is ‘about’, and so it takes the form variable = expression.To demonstrate this distinction, compare the ‘formula’
s = ut +12at2
whose subject is s, with the ‘equation’
2(s − ut) = at2
which has no subject.Often we will be confronted with an equation without a subject, or with the wrong
variable (from our point of view) as the subject. It is then our task to manipulate theequation so as to make the desired variable the subject, and guarantee that the equality isstill true. The only legitimate moves in this game are to perform the same operations on
both sides of the equality, since doing the same thing to two equal quantities keeps them
equal. Consider the following example.
m
3− 4 =
12
We wish to make m the subject of the equation, so we would like to get rid of everythingbut the m from the left. To get rid of the −4 we add 4 to both sides, thus
m
3− 4 + 4 =
12
+ 4
m
3=
12
+ 4 =92.
36
Notice that it appears that the 4 shifts to the other side of the equation, with a change ofsign; and I suppose it does, but the reason it does is that we performed the same operationon two equal things, so they remained equal.
Next we multiply both sides of the equation by 3, thus
m
3× 3 =
92× 3
m =272
.
We have finished, since m is now the subject. Again it appears that a 3 ‘underneath’ onone side of the equation shifts to be a 3 ‘on top’ of the other side. This is true enough,but don’t think of it as a special rule — its just another case of doing the same thing toboth sides.
We could also have done this another way. Firstly we could have multiplied both sidesby 3.
3 ×(
m
3− 4
)= 3 ×
(12
)Notice that the entire expression on both sides is multiplied by 3, so this becomes
m − 12 =32.
Now adding 12 to both sides gives the answer
m − 12 + 12 =32
+ 12
m =272
.
One more example. Find the value of x if
x + 13
+x
4= 5.
The first step towards simplifying this is to get rid of all fractions. This can be done bymultiplying both sides by 12. (Why 12? Because it is the LCM of 3 and 4!)
12 ×(
x + 13
+x
4
)= 12 × 5
4(x + 1) + 3x = 60
7x + 4 = 60
Now we subtract 4 from both sides to get
7x = 56
37
and divide both sides by 7 to arrive at x = 8.
Exercises 6.1
1 Find the value of the variable in each of the following equations.(a) y − 7 = 2 (b) 5x = 40 (c) −a = 8(d) 42 = 7x (e) 7m = 5m + 16 (f) 11x + 17 = 39(g) 3(x + 6) = 2(x + 40) (h) 4(x − 3) = 12 (i) 172 − (7y + 18) = 5(y + 20)(j) 9a + 8 = 3a + 44 (k) 9y = 5(y + 8) (l) 3(3y + 1) = 4(y + 8)
(m)x + 2
5= 40 (n)
12x
= 3 (o)92x
= 4 − 3x
(p)x
3− 2 =
x
6+
152
(q)3x + 4
x= 2 (r)
x
x + 2+
4x + 6
= 1
2 Make x the subject of each of the following equations.
(a) y = 3x − 4 (b) y =1x
(c) y =√
x − 9
(d) y =x + 12x + 1
(e) y =1
x + 3+ 3 (f) xy = x + y
§ 6.2 Simultaneous equations - by substitution
So far we have dealt with individual equations involving just one unknown. We nowwish to consider the problem of satisfying two (or more) equations, involving two (or more)unknowns. These are refered to as simultaneous equations, since we seek a solution thatnot only works for one of the equations, but satisfies all of them simultaneously. We havealready encountered this type of problem before. Remember when we learned to factorthe quadratic x2 + 4x + 3, we sought two numbers a and b which added to give 4 andmultiplied to give 3. That is,
a + b = 4 and ab = 3.
These are simultaneous equations, and by the ‘staring into space’ method we found thesolution a = 1 and b = 3.
For most simultaneous equations the ‘staring into space’ method fails, so we may needto do some algebra. The principle of the method of substitution is to choose one of theequations to be solved, and rearrange it to make one of the variables its subject. Theresulting formula is then substituted for that variable in the other equations, effectivelyremoving one variable (and one equation) from the system. This process is repeated untilonly one equation with one variable remains, and can be solved as usual. Working backthrough the previous equations yields the values of all the other variables. In principle
38
it doesn’t matter which equations or variables are removed, or in what order, though inpractise some options will be easier than others, and only experience (that is, doing theexercises!) can help you decide what is the easiest path to the answer.
As our first example consider the following system of two simultaneous equations.
3a + 2b = 5
2a − b = −6
Rearranging the first of these to make a the subject leads to
a =5 − 2b
3.
This formula for a is substituted into the second equation, which therefore becomes
2(
5 − 2b
3
)− b = −6.
This is a single equation, with only one unknown, b, so we can solve it as usual. Multiplyingboth sides by 3 gives us
2(5 − 2b) − 3b = −18
10 − 7b = −18
−7b = −28
b = 4.
Once we have found the value for b we can substitute it into the formula for a, to finda = (5−8)/3 = −1. So the solution to the problem is a = −1 and b = 4. Check that theseactually do satisfy the original two equations!
Now a harder example. Find all possible solution of the pair of simultaneous equations,
x2 + y2 = 16
3x − 4y − 20 = 0.
In this case we will rearrange the second equation, since the presence of squares in the firstmakes it a little more difficult to manipulate. From the second equation we find
y =3x − 20
4.
Substituting this into the first equation gives
x2 +(
3x − 204
)2
= 16
x2 +(
9x2 − 120x + 40016
)= 16
16x2 + (9x2 − 120x + 400) = 256
25x2 − 120x + 144 = 0.
39
We now need to be able to solve this quadratic equation in order to obtain the answer.Can you see that this factors? In fact its a perfect square, (5x−12)2 = 0 and so x = 12/5,which gives y = −16/5.
Exercises 6.2
Solve the following systems of simultaneous equations by substitution.
(a){
a − b = 2a + b = 4
(b){
5x + 2y = 123x − 2y = 4 (c)
{4p − 3q = 115p + 3q = 7
(d){
2x + 3y = −14x + 3y = −4 (e)
{ 7t + v = 124t + v = 13
(f){
4x + 5y + 2 = 04x + y + 10 = 0
(g){ 4w1 + 3w2 = 11
3w1 + w2 = 2 (h){
y = x2
y = x(i)
{y = x2
2x + y = 0
(j){
x2 + y2 = 9x + y = 3
(k){
xy = 14x − y + 3 = 0 (l)
{a + b = 2b + c = 3a + c = 4
§ 6.3 Simultaneous equations - by elimination
The substitution method can be applied to a wide range of simultaneous equations,which is a great advantage, but it is not always the neatest approach. The method ofelimination is often less cumbersome than substitution, but it can only be applied to someproblems. The most common problem to which the method of elimination can be appliedis simultaneous linear equations. An equation is linear if none of the variables are raisedto any power (other than 1);
ax + by + c = 0
is the general form of a linear equation with two variables x and y.The principle of elimination is to add (or subtract) the equations from one another so
as to cancel one of the variables, hence the term elimination. In fact this is just a specialcase of ‘do the same thing to both sides of the equation’. In order to get the desiredcancellation it may be necessary to multiply one or all equations by some constants. Forexample, consider the system of equations we did above by substitution.
3a + 2b = 5
2a − b = −6
If we decided to eliminate the variable b, we would double the second equation to get
3a + 2b = 5
4a − 2b = −12.
40
Now when we add the equations all the b’s disappear, leaving
7a = −7.
So a = −1, and substituting this back into either equation gives us b = 4. (Checking thisin both equations is a good means of ensuring you didn’t make a mistake!)
Once again we’ll do the same problem, but this time eliminating the a variable. Inthis case I will multiply the first equation by 2 and the second equation by 3, so that thecoefficients of a are the same in both equations.
6a + 4b = 10
6a − 3b = −18
Now we subtract these equations from one another to eliminate a,
7b = 28
and once again we arrive at the same answer, b = 4 and a = −1, which is comforting.Now although this method is often used on simultaneous linear equations, the idea
can be used in other cases. Consider the equations
x2 + y2 = 16
y = x2 − 14.
If we write the second equation in reverse order,
x2 + y2 = 16
x2 − 14 = y
and subtract, we can eliminate the x2 term.
y2 + 14 = 16 − y
This is again a quadratic equation y2+y−2 = 0, which we can factor to give (y+2)(y−1) =0. So there are two possible answers for y, namely y = −2 and y = 1. The correspondingvalues of x are obtained by back substitution, to give x2 = 12 and x2 = 15 respectively.So there are in fact four different pairs of values which work;
x = 2√
3 y = −2
x = −2√
3 y = −2
x =√
15 y = 1
x = −√
15 y = 1.
Exercises 6.3
Repeat Exercise 6.2 using the method of elimination, if possible.
41
Lecture 7. Quadratic equations
§ 7.1 Solving quadratic equations by factoring
The simplest of all algebraic expressions is the linear expression ax + b, where a, b arenumbers and x is the variable. We would say this expression is linear in x. It contains x
only to the powers 0 and 1 (the constant term can be thought of as bx0). A linear equationis one which can be put in the form ax + b = 0, and this is very easy to solve (x = −b/a)so we shall move on to the next most common expression.
We saw before that a quadratic expression is one in the form
ax2 + bx + c
where a, b, c are numbers and x is the variable. We say that the expression is quadratic in
x. It contains x only to the powers 0, 1 and 2. A quadratic equation is one which can beput in the form
ax2 + bx + c = 0
and we would like to develop a general technique to solve such equations.Mathematics is built on a few very simple principles which are often so obvious you
would wonder that they are useful at all. Here is another one.
If the product of two (or more) numbers is zero, then one of the numbers is itself zero.
Thusif xy = 0 then x = 0 or y = 0
andif xyz = 0 then x = 0 or y = 0 or z = 0
and so on. (Notice that here I am using the mathematical ‘or’, the so called inclusive or,which means one or the other or both. The usual ‘or’ of common speech is the exclusive
or, which means one or the other but not both. )Consider the quadratic equation
x2 + 4x + 3 = 0.
We have seen earlier how to factor the quadratic expression on the left of this equation,
(x + 1)(x + 3) = 0.
42
Now this is in the form of a product of two numbers — the numbers (x + 1) and (x + 3)— equals zero. By the principle outlined above this means that one of the two numbersmust itself be zero. We therefore conclude that
x + 1 = 0 or x + 3 = 0.
This gives us two possible solutions to the equation, either x = −1 or x = −3.On more example along these lines; 4y2 + 4y− 3 = 0. We saw earlier that this factors
to give (2y − 1)(2y + 3) = 0. Then either (2y − 1) = 0 or (2y + 3) = 0. Thus the twopossible solutions are y = 1
2 or y = −32 .
Exercises 7.1
1 Write down the solutions of the following equations.(a) (x − 4)(x − 6) = 0 (b) (y + 2)(y − 2) = 0 (c) x(x + 5) = 0(d) (3x − 2)(x + 3) = 0 (e) 4b(2b − 3) = 0 (f) (x + 1)(x + 2)(x − 3) = 0
2 Solve the following equations by factoring.(a) x2 + 7x + 12 = 0 (b) y2 + 8y + 7 = 0 (c) z2 + 13z + 40 = 0(d) x2 − 8x + 7 = 0 (e) b2 − 10b + 25 = 0 (f) c2 + 15c + 36 = 0(g) x2 − 2x − 8 = 0 (h) d2 − 10d − 11 = 0 (i) h2 − 4h + 4 = 0(j) 3p2 + 5p + 2 = 0 (k) 2q2 + 5q + 2 = 0 (l) 2r2 − 11r + 12 = 0(m) 35x2 = 12x − 1 (n) 2 + x − 10x2 = 0 (o) 41x = 3 + 26x2
(p) x3 + 3x2 + 2x = 0 (q) x3 + 6x2 + 9x = 0 (r) 3x3 − 13x2 + 14x = 0
§ 7.2 The quadratic formula
The factorising method is very neat and tidy; when it works. However, it may notbe easy to spot the factorisation of a quadratic expression and worse still this might bebecause there is no such factorisation!
How can we decide this? There is a formula which we can use, the quadratic formula.It looks complicated, and most students who use it just learn that it works — anotherpiece of abracadabra mathematical magic — but really we already have all the skills itneeds to prove the formula.
Theorem The quadratic formula.
The two possible solutions of the quadratic equation
ax2 + bx + c = 0
43
are
x =−b ±
√b2 − 4ac
2a.
Proof Our first step is to make the task easy by working with a monic quadratic, sowe write the equation as
a
(x2 +
b
ax +
c
a
)= 0.
Now a �= 0, since if it were we would only have a linear equation, not a quadratric one,and the solution would be trivial. Since a �= 0 then the other number must be equal tozero,
x2 +b
ax +
c
a= 0.
We now complete the square on the quadratic expression by adding and subtracting thenumber ( b
2a )2 — half the coefficient of the x, squared — to the left hand side. (Equivalentlyyou could add this number to both sides of the equation.)
x2 +b
ax +
c
a+ (
b
2a)2 − (
b
2a)2 = 0
The reason we chose to add this strange number was to get a perfect square involving thevariable x, (
x +b
2a
)2
+c
a− (
b
2a)2 = 0.
Now all we need to do is make x the subject of the equation.(x +
b
2a
)2
= (b
2a)2 − c
a
=b2
4a2− 4ac
4a2
=b2 − 4ac
4a2
Now we wish to take the squareroot of both sides of the equation; but first a word ofwarning. Although the squareroot of a number is always considered positive, so
√4 = 2,
nevertheless the equation a2 = 4 has two valid answers, 2 and −2. So although squareroot-ing both sides of an equation maintains the equality, it does not give all possible solutions.For that we need to consider the negative as well as the positive squareroot. So
x +b
2a= ±
√b2 − 4ac
4a2
= ±√
b2 − 4ac
2a
44
Finally we arrive at the famous formula, which is one of the few which I will ask you tocommit to memory:
x =−b ±
√b2 − 4ac
2a.
To see how this formula works, let’s do the two examples above using it. For x2 +4x + 3 = 0 we have a = 1, b = 4 and c = 3, so the formula gives the two solutions as
x =−4 ±
√42 − 4 × 1 × 32 × 1
=−4 ±
√16 − 122
=−4 ± 2
2
That is, the two solutions are x = −1 or x = −3, as we found before. Notice that thisgives us the factorisation, (x − (−1))(x − (−3)) = (x + 1)(x + 3). For the other example,4y2 + 4y − 3 = 0, we have a = 4, b = 4 and c = −3, so the formula gives the two solutionsas
y =−4 ±
√42 − 4 × 4 × (−3)
2 × 4
=−4 ±
√16 + 488
=−4 ± 8
8
That is, the two solutions are y = 12 or y = −3
2 , as we found before. Notice that this givesus the monic factorisation, (y − 1
2 )(y − (− 32 )) = (y − 1
2 )(y + 32 ). If we now multiply by
the coefficient of y2, namely 4, we arrive at the correct factorisation 4(y − 12 )(y + 3
2 ) =(2y − 1)(2y + 3).
Now let’s look at two more examples which cannot be done by factoring. Considerthe equations x2 + 3x + 1 = 0 and x2 + x + 1 = 0. For the first of these we have
x =−3 ±
√32 − 4 × 1 × 12 × 1
=−3 ±
√5
2.
This does allow us to solve the quadratic equation, and obtain a factorisation of sorts, butnot in a way we were likely to find by trial and error, requiring, as it does, surds. In thesecond case we have
x =−1 ±
√12 − 4 × 1 × 12 × 1
=−1 ±
√−3
2.
45
Now any number squared is positive (or zero), so there is no way to squareroot −3, (unlesswe use the complex numbers, see lecture 3). This tells us that there are no real solutionsto the equation x2 + x + 1 = 0, and the expression x2 + x + 1 cannot be factored in anyway using real numbers.
One more example, similar to the last two. Consider the quadratic equation x2 +2x + 1 = 0. You may be able to see how to factor this, but let’s suppose not and use theformula.
x =−2 ±
√22 − 4 × 1 × 12 × 1
Now because the expression within the squareroot is zero, the ‘two’ answers we get are x =−1 and x = −1, so in fact there is only one answer to the equation — although it is oftenuseful to think of this as getting the same answer twice. This example, and the precedingtwo examples, demonstrate the importance of the term within the squareroot. This termis considered so important by some that they give it a special name, the discriminant,and a special symbol, ∆, the capital Greek letter delta. The discriminant of the quadraticax2 + bx + c is ∆ = b2 − 4ac, and the equation ax2 + bx + c = 0 has
2 distinct real solutions if ∆ > 0
1 repeated real solution if ∆ = 0
no real solutions if ∆ < 0
Exercises 7.2
1 Solve the following equations using the quadratic formula, if a solution exists.(a) x2 − 4x − 1 = 0 (b) 2x2 + 7x + 2 = 0 (c) 7x2 = 4x + 1(d) x2 + 3x + 5 = 0 (e) 3x2 + 9x + 5 = 0 (f) 9x2 − 6x + 1 = 0(g) 4x2 − 36x = 0 (h) 6a2 + 5a = 6 (i) 7b2 − 3b − 1 = 0(j) 4x2 − 12x + 9 = 0 (k) x2 = 2x + 48 (l) 18x2 − 24x − 3x3 = 0
2 Without solving the equations, use the discriminant to determine the number ofdistinct real solutions to the following quadratic equations.(a) 3x2 + 5x + 2 = 0 (b) 3x2 − 4x + 2 = 0 (c) 2x2 + x − 2 = 0(d) 9x2 − 30x + 25 = 0 (e) 7x2 − 5x + 4 = 0 (f) x2 + x + 1
4 = 0
§ 7.3 Solving cubic, quartic and higher order equations
Since we have solved linear and quadratic equations, it is natural to go one stepfurther and ask about the solution of cubic (order 3), quartic (order 4) or even higher order
46
equations. It turns out that the solution of these is surprisingly difficult. So difficult, infact, that for orders five and above, it can be proven that there is no general procedure orformula which can ever solve the equation! However in special cases it may still be possibleto find solutions. We’ll consider some examples.
Consider the cubic equation x3 − x2 − 6x = 0. Although this is a cubic, and we don’tknow how to solve a cubic in general, this particular cubic has a factor of x, so we canwrite the equation as x(x2 − x − 6) = 0. For the product of two things to equal zero, oneor other of them must be zero. We conclude that either x = 0 or x2 − x − 6 = 0. So theproblem has been reduced from a cubic equation to a quadratic equation, which we cando. Factoring the quadratic we find x2 − x − 6 = (x − 3)(x + 2), and so the cubic factorsas x3 − x2 − 6x = x(x − 3)(x + 2) and the solution of the equation is x = 0, 3 or − 2.
A similar, though harder problem, is to solve x3 − x2 − x + 1 = 0. Perhaps you cansee at a glance that x = 1 works? This means that (x − 1) must be a factor of this cubic.Having spotted this, you can factorise the expression through grouping pairs. (A morecertain method is polynomial division.) With some experimentation we find
x3 − x2 − x + 1 = x2(x − 1) − (x − 1) = (x2 − 1)(x − 1).
Once again we have reduced the problem to a quadratic, but in fact its now very easy,since the factor x2−1 is the difference of two squares, and so can be written (x+1)(x−1).We have now factorised the cubic completely as x3 − x2 − x + 1 = (x + 1)(x− 1)2, and sothe solution of the equation is x = −1 or x = 1 (twice).
Another trick is to disguise a quadratic equation as something more difficult. Considerthe equation x4 −3x2 −4 = 0. At first glance this is a quartic in x, and so we have no wayto solve it, but look more closely. It involves only even powers of x, and so can be thoughtof as a quadratic in x2. If that’s still not clear, try writing it as
(x2)2 − 3(x2) − 4 = 0
and introduce a new variable m = x2. Now the equation becomes
m2 − 3m − 4 = 0
which we can factor as (m − 4)(m + 1) = 0. It follows that the solutions are m = 4 andm = −1. But wait! We can’t stop there because the question was to find x, not m. So infact we get x2 = 4 or x2 = −1. The former tells us that x = ±2 while the latter has noreal solutions. Therefore the only values of x that work are ±2. In effect we have factoredthe quartic as
x4 − 3x2 − 4 = (x2 − 4)(x2 + 1) = (x + 2)(x − 2)(x2 + 1).
47
Finally, while on the subject of ‘quadratics in disguise’, consider the following equationwhich appears on the surface to have nothing to do with quadratics.
x +1x
= −3
To solve this we could multiply both sides by x to get
x2 + 1 = −3x,
then add 3x to both sides of the equation to arrive at
x2 + 3x + 1 = 0
which we solved earlier. You will find that solving quadratic equations pops up again andagain, when you least expect it.
Exercises 7.3
1 Solve the following cubic equations.(a) x3 + 5x2 + 6x = 0 (b) x3 + 2x2 − x − 2 = 0
2 Solve the following quartic equations.(a) x4 − 5x2 + 4 = 0 (b) x4 − 8x3 + 15x2 = 0
3 Solve the following equations.
(a) x − 1 =1
x + 1(b) x2 − 8
x2= 2
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Lecture 8. Inequalities
Dealing with inequalities is similar in many respects to dealing with equalities, how-ever there are some complications. Once again the only legitimate move in simplifyinginequalities is to do the same thing to both sides of the inequality, but unlike equalities,not all moves are valid. To understand what is permitted when solving inequalities, andwhat is not, we need to have an understanding of the ordering of the real numbers, andhow the different operations affect that ordering.
§ 8.1 The ordering of the real numbers
There are certain properties of the real numbers which are generally taken for grantedsince they are ‘obvious’. These fundamental properties are called axioms. We have al-ready encountered some axioms: the associative, commutative and distributive laws ofarithmetic. Another of these fundamental properties is that the real numbers have a natu-ral ordering; that is, there is a way of comparing numbers which arranges them in a strictorder. This ordering is usually referred to as “greater than” (or “less than”). Given anytwo numbers, then either they are equal or one is greater than the other. We’ll visualisethe order on the real numbers by thinking of the number line. Travelling to the righton the number line is getting bigger, while travelling left is getting smaller. A number is‘greater than’ another number if it is further to the right on the number line, and ‘lessthan’ the other number if it is further to the left. The symbols for ‘greater than’ and ‘lessthan’ are, respectively, > and <. In addition, if we wish to allow the chance of equality aswell, we have ‘greater than or equal to’, ≥, and ‘less than or equal to’, ≤. So to describeall numbers greater than 2 but less than or equal to 4 we write
2 < x ≤ 4.
An alternative way of describing these same numbers is via interval notation. (An intervalis just a single, connected chunk out of the number line.) We would write
x ∈ (2, 4].
Here the symbol ‘∈’ means ‘is an element of’ or ‘is contained in’, and the notation (2, 4]indicates all numbers between 2 and 4, excluding the 2 because of the rounded bracket,but including the 4 because of the square bracket. In order to describe the numbers whichare greater than or equal to 1 we need a new symbol, ∞, to represent infinity. Thus wewould write
x ∈ [1,∞).
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(Note: Infinity is not a number. It cannot be used as a number. It cannot be added,subtracted, multiplied or divided. You cannot perform arithmetic with ∞, it is merely anotational convenience introduced because the real numbers do not have an end. Noticethe interval notation always uses a rounded bracket next to ∞ to emphasise that it is notincluded in the real numbers.)
If we wished to also include those numbers less than or equal to -1 we would write
x ∈ (−∞,−1] ∪ [1,∞)
where the symbol ‘∪’ means the union of the two pieces; that is, x is in one interval or theother.
A third way to represent the same information is graphically on a number line. Therelevant regions on the number line are shaded darker, with an open circle to indicate aboundary of the region which is excluded and a closed circle to indicate a boundary whichis included. Thus the numbers 2 < x ≤ 4 would be indicated graphically by
Exercises 8.1
1 Decide whether each of the following statements is true or false.
(a) −5 > 3 (b) 17 ≤ 1
5 (c) 17 < −1
5 (d) π ≤ 227 (e) 1
2−√
2≥ 1 +
√2
2
2 Express each of the following sets of numbers using interval notation and graphically.
(a) −2 ≤ x < 1 (b) x > 3 or x ≤ −2 (c) x < 3 and x ≥ −2
§ 8.2 Effect of operations on inequalities
In order to understand how and why various operations affect inequalities, we willconsider their effect on the number line.
Addition and subtraction
Addition of a number a has the effect of translation (sliding) by a units to the righton the number line. (Of course a translation to the right by a negative amount actuallytakes you to the left.)
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Similarly subtraction of a number a has the effect of translation by a units to the left.(Once again a left translation by a negative amount actually takes you to the right.) As aresult of this, adding or subtracting the same number to both sides of an inequality simplyleads to both numbers sliding along the number line by the same amount. Their relativepositions, with regard to rightness and leftness, do not change, and so the inequality isunaffected.
Addition or subtraction of a number from both sides of an inequality preserves the
inequality.
For example, to solve
7x − 13 > 6x + 22
we add 13 to both sides of the inequality
7x > 6x + 35
and subtract 6x from both sides
x > 35.
Multiplication and division
To visualise the effect of multiplication on the number line, imagine it as a rubberband, nailed down at zero. Multiplication by a positive number has the effect of stretchingthe rubber band by that factor. (Of course for a number less than 1 this is really a‘shrinking’ rather than a ‘stretching’.)
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In a similar way division by a positive number is a shrinking of the rubber band by thatfactor (or stretching if the number you divide by is less than 1). The important point is,that while undergoing stretching or shrinking, numbers maintain their relative position onthe band. Bigger numbers stay bigger, and smaller numbers stay smaller.
Multiplication or division by a positive number on both sides of an inequality preserves
the inequality.
The effect of multiplication or division by a negative number is slightly different. Therubber band which is the number line is still stretched or shrunk by a factor equal to thesize of the negative number, but the negative sign indicates a ‘flipping’ of the number line.
Thus the positive arm gets stretched in the negative direction and the negative arm getsstretched in the positive direction. Due to this flipping of the numberline, the relativepositions of all numbers (with respect to rightness or leftness) is reversed. The stretchingor shrinking aspect does not change relative positions, thus
52
Multiplication or division by a negative number on both sides of an inequality reverses
the inequality.
For example to solve6 − 3x > 18
we subtract 6 to both sides of the inequality
−3x > 12
and divide both sides by −3, reversing the inequality since the division is by a negative
x < −4.
A far trickier problem, which continues to stump many students, is to solve the followinginequality.
3 − 1x≥ 2
If we subtract 2 from both sides and add 1/x to both sides we arrive at
1 ≥ 1x
.
The trap is then to multiply both sides by x and claim the answer is x ≥ 1. Why is thiswrong (or at least only partly right)? When multiplying an inequality by a number weneed to reverse the inequality if the number is negative, and leave the inequality unchangedif the number is positive. In this case the number we are multiplying by is x, a variable,so we don’t know if it is positive or negative. We therefore have to solve this problem intwo parts. If x is positive; that is x > 0, we proceed as above to the solution x ≥ 1. Thecombination of these two conditions is just x ≥ 1. Alternatively, if x < 0 (negative), weproceed as above but reverse the inequality, to arrive at x ≤ 1. The combination of theseconditions is x < 0. So in fact the full answer to the question is
x ≥ 1 or x < 0.
Other operations
One has to be very wary about applying any operations other than those above toinequalities. Unlike equality, where if two things are equal, performing the same operationon both always leaves them equal - inequalities are much more fickle. For example 2 > −3,
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but if we squared both sides we would be claiming that 4 > 9! By the same token 2 > 1and if we square both sides we get 4 > 1, which is correct this time. To understand why, wehave to understand what squaring does to the number line. Firstly squaring the positivenumbers just ‘stretches’ them out, although not uniformly. But squaring negative numbersflips them over to be positive, as well as stretching them.
Thus squaring ‘folds’ the numberline over itself at zero, and the sense of order is lost bythis folding.
It is the same for other operations too. Some books may suggest that you remember aspecial rule for taking reciprocals, that the inequality reverses. This rule is fine for numbersof the same sign, for example
2 < 3 implies that12
>13
However if you stuck to this rule always you would say
2 > −3 implies that12
< −13
which is obviously wrong. The correct way to view this is simply to perform divisions,reversing the sign if the division is by a negative. Thus
2 < 3 implies 1 <32
implies13
<12
and2 > −3 implies 1 > −3
2implies − 1
3<
12.
The moral of this story is that the only operations which can be performed on inequalitiessafely are the standard operations of addition, subtraction, multiplication and division(with appropriate reversal of the inequality if necessary). Using any other operation isfraught with peril, and requires justification.
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Exercises 8.2
Solve the following inequalities.(a) x + 4 > 7 (b)
x
3− 4 ≤ 6 (c) 3 + 4y ≥ −2(1 − y)
(d)r − 3
2> −6 (e) −5x < 20 + x (f) −8 < 3x − 2 ≤ 4
(g)x + 6
2≤ x (h)
x
4≤ x
6+
112
(i)2b − 5
8+ 3 ≤ b + 6
12(j)
1x + 2
≤ 1 (k)1
x + 3> 1 (l)
x
x + 1>
12
§ 8.3 Quadratic inequalities
Naturally enough these are inequalities involving a quadratic expression, which canbe written in the form
ax2 + bx + c > 0
or a comparable expression involving <, ≥ or ≤. I will describe what I hope is the easiestmethod of solving these inequalities — precisely why it works may be easier to follow oncewe have learnt to graph quadratics.
Firstly we momentarily ignore the inequality and imagine it as equality. We then solvethe quadratic equation, either by factoring or via the quadratic formula. The solutionsare called the roots of the equation. Now the solution of the original inequality is eitherthe region between the roots, or the two regions outside the roots. Knowing this, we onlyhave to test a single point from one region to see if it satisfies the inequality or not. We’lldiscuss why this works in a minute, but for now let’s do an example.
To solve x2 + 4x + 3 < 0, we first solve the equation x2 + 4x + 3 = 0. We have donethis before (factoring is easy — (x + 1)(x + 3) = 0) and the roots are −1 and −3. Nowchoose a number to test — x = 0 is an easy option. At x = 0, a point outside (that is,not between) the two roots the inequality fails. Therefore the solution must be the regionbetween the roots, −3 < x < −1.
For our second example we’ll solve x2 + 3x + 1 ≥ 0. First, replace the inequalityby equality. You may recall we have tried to solve this quadratic equation before and weneeded to use the quadratic formula. We found
x =−3 ±
√5
2.
Once again I will use x = 0 as my test point, and this is outside the two roots. It is easyto confirm that the inequality is true for x = 0, and so it is true for all numbers outside
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the roots. Also, in this case equality is allowed, and so the roots themselves are part ofthe solution. Therefore x ≤ −3−
√5
2 or x ≥ −3+√
52 .
A few words now on why it works. Suppose the two roots of the quadratic are a andb, so that the quadratic factors as
(x − a)(x − b).
Then for any choice of x bigger than both a and b, we have (x−a) and (x−b) both positive,and so the quadratic is positive. Similarly, for a choice of x less than than both a and b,we have (x−a) and (x− b) both negative, and so the quadratic is positive (product of twonegatives is positive). That is, anywhere outside the roots the quadratic (x − a)(x − b) ispositive. Anywhere between the roots, however, one of (x − a) or (x − b) is positive, andthe other is negative, and so the quadratic (x − a)(x − b) is negative. It therefore sufficesto test one point from a region to determine if that region satisfies the inequality or not.
Exercises 8.3
Solve the following quadratic inequalities. Graph your solutions on the number line.(a) x2 + 7x + 12 > 0 (b) y2 + 8y + 7 ≤ 0 (c) z2 + 13z + 40 ≥ 0(d) 9 − x2 > 0 (e) x2 + x − 6 > 0 (f) 2x2 − 15x + 25 ≤ 0(g) 6x2 + x ≥ 2 (h) x2 ≥ 16 (i) x2 < x + 6
§ 8.4 Absolute value
A real number has both a magnitude and a direction (sign). The technical jargonfor the size or magnitude of a number is its absolute value or modulus. The notation torepresent this size is vertical lines either side of the number, thus |−2| means ‘the absolutevalue of −2’, and |x| means ‘the absolute value of x’. In practise we find the absolute valueof a number by throwing away its sign (if it is negative). The formal definition of absolutevalue is a little different. For a variable it is not clear whether it is negative or positive,and so we must treat each case separately.
|x| ={
x if x ≥ 0−x if x < 0
This means that for a positive number (or zero) the absolute value or size is equal to itself,and for a negative number the absolute value is minus itself. For example, |2| = 2 and| − 2| = −(−2) = 2. The effect for negative numbers is the same as if we had droppedthe negative sign. As you can see, this means that an absolute value is always positive (orzero).
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In some books you may see an alternative definition, that |x| =√
x2. This is in factidentical to our definition above, since for positive numbers the squareroot of the squareis the number itself, whereas for negative numbers we have a change of sign becausesquareroots are always taken to be positive. For example, |2| =
√22 =
√4 = 2 and
| − 2| =√
(−2)2 =√
4 = 2. However I don’t believe this definition is as easy to use, or asintuitive.
So much for the definition, how are we to interpret or visualise this concept? Themost valuable way of viewing the absolute value of a number is as its distance from 0 onthe number line. Of course distances are always positive (or zero), so that fits in well.Following through our example, |2| can be interpreted as ‘the distance of 2 from 0’, so theanswer is 2. Similarly |−2| can be interpreted as ‘the distance of −2 from 0’, so the answeris again 2.
Continuing in this fashion, the distance between any two numbers can be written asthe absolute value of their difference — say, distance from 3 to 5 is |3 − 5|. Notice itwouldn’t matter if I wrote it the other way around (|5 − 3|) because distances don’t havea direction.
Exercises 8.4
Find the value of the following expressions involving absolute values.(a) |17| (b) | − π| (c) |2 − 3| (d) 3 − | − 4| (e) |3 − | − 4|| (f) | − x|
§ 8.5 Equations and inequalities involving absolute values
There are generally two approaches to solving equations/inequalities involving abso-lute values. One is the formal algebraic approach, which is tedious and confusing to moststudents, yet is rigorous and logical. The other approach is more intuitive and geometric,and allows us to ‘see’ the solution to simple problems quickly. It is this second approachI wish to develop here.
Let’s solve the equation |x−2| = 1 by thinking in terms of distances. The term |x−2|can be thought of as meaning the distance between x and 2, and so the equation readsgeometrically as
the distance from x to 2 equals 1.
Draw a number line (or imagine it). Which numbers are a distance 1 away from thenumber 2? Answers 3 or 1.
We’ll try an inequality the same way. Consider the inequality |x−3| < |x−1|. Readingthe equation in terms of distances, it says
57
the distance from x to 1 is greater than the distance from x to 3.
Another way of saying this isx is closer to 3 than it is to 1.
Draw or imagine a number line again. Any number right of the midpoint is closer to3 than to 1. Therefore x > 2.
Exercises 8.5
1 Solve the following equations involving absolute values.(a) |x| = 6 (b) | − x| = 6 (c) −| − 4x| = 20(d) |x − 2| = 4 (e) |5 − x| = 8 (f) |2x − 4| = 12(g) |x + 1| = x (h) |2x| = 9 − x (i) |x + 3| = |x − 1|(j) |x + 1| + |x − 1| = 4 (k) |x − 2| = |x − 4| + 1 (l) |x − 2| + |x + 1| = 3
2 Solve the following inequalities involving absolute values.(a) |x| > 6 (b) | − x| > 6 (c) −| − 4x| < 20(d) |x − 2| ≤ 4 (e) |5 − x| ≥ 8 (f) |2x − 4| > 12(g) |x + 1| > x (h) |2x| ≥ 9 − x (i) |x + 3| ≤ |x − 1|(j) |x + 1| + |x − 1| < 4 (k) |x − 2| ≥ |x − 4| + 1 (l) |x − 2| + |x + 1| ≤ 3
58
Lecture 9. Arithmetic and geometric sequences and series
§ 9.1 Sequences in general
A sequence is an ordered list of numbers, as opposed to a set which is just a collectionof numbers whose order does not particularly matter. Thus
{1, 4, 3, 5, 2} and {1, 2, 3, 4, 5}
are the same as sets, since the numbers involved are identical, but different as sequencesbecause of the different order in which the numbers appear. In general there need not beany discernable pattern in the numbers in order to justify the title of ‘sequence’, howeversequences in which the terms follow a set pattern are the more interesting and useful.Infinite sequences are also generally more fun then finite ones!
A sequence which follows a set pattern may be defined in several ways. For example
1, 4, 9, 16, 25, 36, .....
is a sequence defined by writing down the first few terms until the pattern becomes obvious.The series of dots at the end — ...... — indicates that the sequence extends forever.Alternatively there might be a direct formula for describing the value of each term, forexample
xn = n2
describes the same sequence as above. As n takes the values 1, 2, 3... the formula computesthe first, second, third ... terms of the sequence. Finally the sequence could be describedby an indirect formula, such as
xn+1 = xn + 2n + 1 starting from x1 = 1.
Such a formula is called an iterative formula. It expresses the terms of the sequence as afunction of the previous terms. In order to find the 10th term of the sequence the formulamust be iterated to find all the intervening terms. This can be painful if we want to knowthe 573rd term of the sequence. A direct formula allows us to jump straight to the valueof the 573rd term without needing to know any of the other terms. It is not always easy towrite down a direct formula for a sequence, even if the pattern is a relatively simple one.
59
Exercises 9.1
1. Find the next three terms of the following sequences, and write down a formula todescribe the sequences (if you can).
1 3 5 7
2 6 18 54
1 8 27 64
2 3 5 7 11 13 17
1 1 2 3 5 8 13 21
§ 9.2 Arithmetic progressions
A commonly occuring type of sequence with a simple pattern is the arithmetic se-
quence or arithmetic progression. In an arithmetic progression each term is obtained fromthe last by adding a constant number, called the common difference, d. Thus
1, 4, 7, 10, 13, 16, 19, ....
is an A.P. with a common difference of d = 3. In fact if we know the first term of thesequence, x1 = a, and we know the common difference d, the rest of the sequence is easyto find. As an iterative formula we could write
xn+1 = xn + d starting from x1 = a,
but simpler still is the direct formula
xn = a + (n − 1)d.
A series is the sum of the terms of a sequence, thus 1, 4, 7, 10, 13, 16, 19 is an arithmeticsequence while 1 + 4 + 7 + 10 + 13 + 16 + 19 is an arithmetic series. Finding the sum of anarithmetic series can be accomplished by a neat trick. Write the terms of the series downtwice, once in its usual order and once in reverse, thus
1 + 4 + 7 + 10 + 13 + 16 + 19 +
19 + 16 + 13 + 10 + 7 + 4 + 1.
Now when we pair each number with the one below we see that the pairs all add up to thesame value, 20, while the number of pairs is 7. The total is therefore 7 × 20 = 140, but
60
that’s the result of adding up the series twice. Therefore the sum of the series (once) is140/2 = 70. This process is the basis of the general formula for the sum of an arithmeticseries
sn =n
2(a + l)
where n is the number of terms to be added, a is the first term and l = a + (n− 1)d is thelast term to be added.
Exercises 9.2
1. Find the sum of the numbers from 1 to 1000.2. What is the 27th term of the arithmetic sequence 5, 8, 11, 14...? What is the sum
of the first 27 terms?3. Find the sum of the odd numbers from 1 to 2n+1.
§ 9.3 Geometric progressions
The geometric sequence or geometric progression is one in which each term is obtainedfrom the last by multiplying by some number, called the common ratio, r. If in addition weknow the first term of the sequence, a, then we can find any term easily. For example, thegeometric progression with first term a = 1 and common ratio r = 2 is 1, 2, 4, 8, 16, 32, ......The general formula for any term of a G.P. is
xn = arn−1.
The ultimate behaviour of a G.P. depends on the value of r. There are a number of cases.
Case 1: r > 1 — In this case the value of rn−1 grows larger and larger, and the sequencegrows towards ∞ (if a > 0) or −∞ (if a < 0).
Case 2: r = 1— This results in the rather dull constant sequence, a, a, a, a, .....
Case 3: 1 > r > 0 — The value of rn−1 shrinks away to nothing as n gets larger, so thesequence gets closer and closer to zero.
Case 4: r = 0 — Even more dull than r = 1. The sequence goes a, 0, 0, 0, .... This probablyshouldn’t even be considered a G.P.
Case 5: 0 > r > −1 — As in Case 3 rn−1 shrinks in size, but it alternates in sign as itshrinks. The sequence therefore bounces either side of zero as it shrinks away.
Case 6: r = −1 — The sequence a,−a, a,−a, a,−a, ....
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Case 7: r < −1 — The value of rn−1 grows larger and alternates in sign as n gets larger,so every second term of the sequence is growing towards ∞ while every other term growstowards −∞.
A geometric series is the sum of a geometric sequence. Just as in the case of anarithmetic series, a neat trick allows us to quickly and easily find the sum of a geometricseries no matter how many terms are involved. Let us write sn for the sum of n terms ofthe series, thus
sn = a + ar + ar2 + ar3 + ..... + arn−1.
Multiplying this expression by r and subtracting gives
sn = a + ar + ar2 + ar3 + ..... + arn−1 −rsn = ar + ar2 + ar3 + ..... + arn−1 + arn
—————-———————————————————–
(1 − r)sn = a − arn
Rearranging gives us a simple formula for the sum,
sn =a(1 − rn)(1 − r)
.
Note that the formula fails if r = 1 since we would be dividing by zero; but if r = 1 thenit is easy to see that sn = na.
An interesting and useful consideration is what happens if we add more and moreterms of our geometric series. If |r| > 1 (that is, if r > 1 or r < −1) then rn grows verylarge in size (and alternates if r is negative). However if |r| < 1 (that is, if −1 < r < 1)then rn shrinks away to nothing the bigger we make n. This means that sn gets closerand closer to a/(1 − r) as n gets bigger. Under these circumstances we say that we cansum the series ‘to infinity’, and the result is
s∞ =a
1 − rfor − 1 < r < 1.
Exercises 9.3
1. What is the 12th term of the geometric sequence 8, 4, 2, 1...? What is the sum ofthe first 12 terms? What is the sum of the entire series (i.e. infinitely many terms)?
2. Prove that 0.9̇ = 1.
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Lecture 10. Functions and graphs
The preceding lectures have concentrated on algebra and algebraic methods of solvingproblems. Algebra is extremely important because it is the tool by which we prove, justifyor calculate. There are, however, few people who gain insight through algebra. More oftenit is through geometry - the representation of mathematical ideas pictorially - that we seewhat is true and what is not. The subsequent proof may require an algebraic approach,and this may prove to be illuminating, but the initial spark is usually a geometric one.
§ 10.1 The cartesian plane
We have already found it useful to imagine the real numbers pictorially by viewingthem as points on the number line. The number line is useful for displaying informationabout a single variable, as we did when we solved inequalities, however most often we willbe interested in displaying information about two variables, and their relationship to eachother. To do this we will need two number lines. This could be done simply by drawingtwo copies of the numberline; one for each variable. Thus we could represent x = 2 andy = 3 by
but this is a little awkward. Instead we draw two copies of the number line, crossing eachother at rightangles at the number 0. To represent x = 2 and y = 3 we find the pointdirectly above the 2 on the horizontal number line and to the right of the 3 on the verticalnumber line.
By utilising the entire plane in this way, the information x = 2 and y = 3 can berepresented by a single point. Whenever this method is used for labelling points it isreferred to as the cartesian plane.
Now we come to some of the terminology and notation used for the cartesian plane.The number lines themselves are referred to as the axes (plural of axis), and the point
63
where they cross each other is known as the origin. Points in the cartesian plane areusually labelled by a capital letter, and the numbers describing its position relative to theaxes are called its coordinates. The coordinates are usually written as an ordered pair ofnumbers, in parentheses, with the coordinate of the horizontal axis first, followed by thecoordinate of the vertical axis. Thus the point x = 2 and y = 3 would be written as (2, 3).The jargon for the first coordinate is the absissca and for the second coordinate is theordinate. Since most of the time the variables we use will be named x and y, it is morecommon to hear the absissca referred to as the x-coordinate or x-value, and the ordinatereferred to as the y-coordinate or y-value. There are a few simple quantities which can beeasily calculated from the coordinates of points in the cartesian plane.
Distance between two points in the cartesian plane
Given any two points in the plane a simple application of Pythagoras’ Theorem allowsus to calculate the distance between them. If we label the two points P and Q, and theircoordinates (x1, y1) and (x2, y2) respectively, then we can produce a right-angled trianglewhose hypotenuse is the line joining P and Q, as shown below.
Now the length of the horizontal side of the triangle is |x1 − x2| and the length of thevertical side of the triangle is |y1 − y2|, and so by Pythagoras’ Theorem the distance fromP to Q is √
(x1 − x2)2 + (y1 − y2)2.
Notice that we have dropped the absolute value signs from the formula. Since we aresquaring the numbers (x1 − x2) and (y1 − y2) anyway, the absolute value signs are notimportant. This also means that it does not matter which point you consider to be thefirst point and which is the second, the distance turns out the same.
For example, calculate the distance from the point P (1,−3) to Q(−3, 0). By thedistance formula this is√
(1 − (−3))2 + (−3 − 0)2 =√
42 + (−3)2 =√
16 + 9 =√
25 = 5.
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If we use the points in the reverse order we get√(−3 − 1)2 + (0 − (−3))2 =
√(−4)2 + 32 =
√16 + 9 =
√25 = 5
so the order of the points did not matter.
Midpoint between two points in the cartesian plane
Given two points P (x1, y1) and Q(x2, y2) in the cartesian plane, the point which liesexactly halfway between these two is simply the point whose x and y coordinates arehalfway between those of P and Q. Therefore we just average the coordinates of P and Q.
The midpoint between P and Q is therefore the point(x1 + x2
2,y1 + y2
2
).
So, for example, the midpoint between (1,−3) and (−3, 0) is(1 + (−3)
2,−3 + 0
2
)= (−1,
−32
).
Exercises 10.1
For each of the following pairs of points, find their midpoint and the distance betweenthem.
(a) (1,2) and (4,6) (b) (2,-3) and (-4,-1) (c) (3,7) and (3,-2)
§ 10.2 What is a function?
A point in the cartesian plane is described by an ordered pair of numbers, the abscissa(x-value) and the ordinate (y-value). You may consider this point as ‘associating’ the two
65
numbers with each other. For example, if the x-value represents temperature in oF andthe y-value is temperature in oC, then the point (32, 0) would indicate the associationbetween these different temperature scales at the freezing point of water. Similarly thepoint (212, 100) would give the relationship at the boiling point of water, and it would bepossible to describe all the points relating temperature in Fahrenheit to temperature inCelcius. This gives us an intuitive idea of what a function is. It is a means of associatingpairs of numbers through some relationship.
Another feature of the intuitive idea of a function is that of dependence. We usuallyconsider that the y-value is dependent on the x-value, meaning that if the x-value is knownthen the y-value can be determined according to some rule. This brings us to our definitionof a function.
A function is a rule associating to each x-value exactly one y-value.
Now we may have to clarify and elaborate this definition somewhat, but it is a goodstarting point. If we drop the idea that each x-value can only be associated with oney-value, then we get what is called a relation.
A relation is a rule which associates x-values with y-values.
These definitions are a little vague because they do not describe what constitutes arule. Most commonly the rule takes the form of a formula, for example
y = 3x2 − 2x + 7.
This is clearly a function. Given any x-value we can use the formula to calculate theassociated y-value. Since y is the subject of the formula it is dependent on x, and x
is described as the independent variable. The rule, however, may be more complicated,involving several different formulae and conditional statements describing when each is tobe applied. We have already seen a function described in just this way — the absolutevalue function. Another example is,
y =
{x2 if x > 013 if x = 0−3x + 2 if x < 0.
Although more complicated, we can see each value of x satisfies only one of the conditionalcriteria, and so determines only one value of y. Now we could in principle extend this ideaof creating a rule through a sequence of different cases. Taken to its illogical conclusion,
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a rule could in principle consist of a list of all the possible different x-values, and a list ofall the associated y-values. Such a thing would then be a function, though not the sort offunction which is ever likely to be useful.
An example of a formula without a subject is
x2 + y2 = 25.
In this case we can’t identify one variable as being dependent on the other. Neverthelessthis equation does determine some points in the cartesian plane. For example the points(0,5), (3,4) and (-3,4) all satisfy the equation. However the points (3,-4) and (-3,-4) alsowork, so for x = 3 and x = −3 there are two values of y satisfying the equation. Thismeans that a given value of x does not determine exactly one value of y, so the equationis a relation, not a function.
The domain of a function is the set of all x-values for which the function is defined,while the range is the set of all y-values which can result from the function. For example,for the function
f(x) =√
9 − x2
the domain is x ∈ [−3, 3], since any value outside this range results in the squareroot of anegative number. The range is y ∈ [0, 3].
§ 10.3 Graphs
A function is a rule which associates y-values with x-values to create ordered pairs ofnumbers. When we plot these ordered pairs on the cartesian plane we create the graph
of the function. For most functions which we will encounter the function is given by asingle formula, and in general the graph of the function will be a continuous, smooth curveof points in the cartesian plane. However we should not assume that all functions giverise to smooth continuous curves. Some graphs may have sharp corners, and others maybe made up of several pieces which are not joined to each other. We will be continuallydeveloping methods to help in drawing the graphs of functions, and in this section we willonly consider the most primitive of methods.
The most obvious way of graphing a function is simply to start finding the orderedpairs defined by the function, and plotting them one by one. For example, consider thefunction
y =5(x − 32)
9.
We could simply start choosing some values for x and calculate the corresponding valuesfor y. We could start by putting x = 32 and finding that y = 0, so we plot the point
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(32, 0). Then we could try x = 100 and find y ≈ 37.8, and so plot (100, 37.8). Finally wemight try x = 212 and find that y = 100, and so plot (212, 100). We would end up withthe graph below.
Now we see the inherent weakness of this method. When its all said and done, we onlyhave three points on the graph. Now they appear to lie along a straight line, so perhapsthe entire graph is just a straight line, but we can’t be sure. We could test this hypothesisby plotting some more points. Try x = 149 and find y = 65, and the point (149, 65) doesappear to lie along the same straight line. Therefore it seems reasonable to extrapolate
from these four points to conclude that the graph is a straight line.
Essentially this method requires that we plot enough points to detect a pattern ofsome sort, and then extend this pattern (extrapolation) to draw the entire graph. Thisextension procedure is an assumption, and on more complicated examples may misleadyou. For this reason I never recommend that you apply this method in isolation, and wewill develop other methods to be used in conjunction with plotting points to avoid thecommon traps. Eventually I hope you will not need to plot any individual points in order
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to graph functions, but for the moment, just to get a feel for functions and graphs, this isour only method.
A more complicated example would be to graph the function
y = x3 − x.
In order to plot some points it may be useful to draw up a table of x-value along withtheir associated y-values.
Plotting these points we find
If you now tried to extrapolate these points to obtain the entire graph you may bemisled into believing the graph is flat between -1 and 1, whereas in fact the graph lookslike
69
If you are in doubt about what the graph is doing in some region, try plotting a fewmore points in the doubtful area. In the example above plotting x = 1
2 , y = − 38 and
x = −12 , y = 3
8 would help enormously in finding the correct shape of the graph.
Exercises 10.3
Use graph paper to plot some points on the graphs of the following functions orrelations. Attempt to extrapolate to obtain a complete graph of the function. What is thedomain and range of each function?
(a) y = 2x (b) y = x2 (c) y =√
16 − x2 (d) y =1x
(e) y = x3 (f) y = x2 + 4x − 21 (g) y =x
x2 + 1(h) y2 = x3 + x
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Lecture 11. Graphing straight lines and parabolas
In the last lecture we investigated graphing functions by plotting points and extrap-olating between them. This method is primitive but effective in many circumstances,although it can sometimes be misleading if used without adequate caution. A more com-plete approach to graphing, or curve sketching, involves knowing the graphs of standardfunctions ‘by heart’, and using this information to predict the graphs of more complicatedfunctions. This lecture begins the process of developing a repetoire of graphs of commoncurves, starting with straight lines and parabolas.
§ 11.1 The straight line
We have already noted that an expression of the form ax + b, where x is a variableand a and b are constants, is called a linear expression. This is because the graph of theexpression
y = ax + b
is a straight line. In order to graph this function we could plot some points, and the easiestpoint to plot is when x = 0 — then y = b. Thus the graph of y = ax + b cuts acrossthe y-axis at height b. This is called the y-intercept of the curve, so the number b has animmediate meaning on the graph of the function. We wish to establish a meaning for thenumber a. We could plot the points corresponding to x = 1, 2, 3 etc, and we would findy = a + b, 2a + b, 3a + b etc.
So we see that the number a is the amount by which y changes whenever x changesby the amount 1. This means that the number a determines the steepness of the graph.This quantity is usually referred to as the gradient or slope of the line, and is usuallyrepresented by the symbol m rather than a. (Why you would choose the letter m I do notknow!) The gradient of the line may be calculated from any two points on the line via theformula
71
m =riserun
=y2 − y1
x2 − x1.
Now the straight line y = mx + b is easy to graph since it has y-intercept b and gradientm. Note that if m = 0 then the line is horizontal (y = b) and that if m is negative then y
is decreasing from left to right.
Given any two points there is only one straight line which goes through these points,and this line can be found by determining the values of m and b. For example to find theequation of the line through the points (1, 3) and (2, 5) we could substitute these valuesinto the equation y = mx + b, thus
3 = m + b
5 = 2m + b.
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We can now solve these equations, either by elimination or substitution. Let’s eliminateb by subtracting the first equation from the second. This gives m = 2. Substituting thisvalue back shows b = 1. Thus the equation of the line through the two points is y = 2x+1.
Another approach to this problem is to use the formula above to find the gradient ofthe line,
m =y2 − y1
x2 − x1=
5 − 32 − 1
= 2.
Notice that it doesn’t matter in which order we use the points, since swapping the orderchanges the sign of the numerator and the denominator, and so the ratio is unchanged.Now since we know m = 2 the equation of the line must be
y = 2x + b
and we can substitute either point into the equation to find b. Substituting the point (2, 5)gives
5 = 2 × 2 + b
and so we find b = 1, and the line is y = 2x + 1. Finally we could streamline this processeven further by noting that if (x1, y1) is any point on the line, then
b = y1 − mx1
and so the equation of the line can be written as
y = mx + b
y = mx + (y1 − mx1)
(y − y1) = m(x − x1).
With this formula we can do the example again. As before we can find that m = 2, andusing either point (say (1, 3) this time), we instantly get the equation of the line
(y − 3) = 2(x − 1).
Of course this can be rearranged to be y = 2x + 1, as we had before.In summary then, there are three formulae you should know for straight lines.
y = mx + b gradient-intercept form
(y − y1) = m(x − x1) point-gradient form and
m =riserun
=y2 − y1
x2 − x1the gradient.
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There is an added complication with these forms of a straight line. We cannot describe avertical line using an equation involving the gradient (why not?). However the equationof a vertical line is quite simple, x = c where c is the x-intercept. If you wish to use ageneral equation which can incorporate all straight lines, including vertical lines, then itcan easily be done, but you can’t make y the subject of the equation. The general form ofa straight line is
ax + by + c = 0.
(Note that a vertical line x = c is not a function, it is a relation, since there are manyvalues of y for the one value of x.)
Exercise 11.1
1 Find the equation of each of the following straight lines and graph them.(a) The line with slope 2 cutting the y-axis at 3.(b) The line with slope -1 through the point (2,−5).(c) The line through the two points (1,−1) and (−3, 5).(d) The line through the points (2,−7) and (5,−7).(e) The line through the points (2,−7) and (2,−6).
2 Solve the equations y = 2x + 1 and y = 3x − 1 simultaneously, and graph the twolines on the same graph. At which point do the two lines intersect?
§ 11.2 Parabolas
The simplest of mathematical expressions, linear expressions, give rise to straight linegraphs. The next simplest of expressions is the quadratic, and it gives rise to parabolic
graphs. The parabola is a very famous shape, one of a family of shapes known as conicsections since it can be obtained by slicing through a cone at an angle. The parabola, likeall conic sections, appears frequently in nature. For example the path of an object thrownthrough the air is parabolic. It is also a shape rich in geometrical properties which canbe exploited. For example, the curved reflective dish in torches or headlights is parabolicsince this shape reflects light into parallel beams, focussing it in the forward direction. Forthe moment we will not be concerned with the geometric properties of the parabola, butwill simply focus on graphing quadratic functions.
Now might be a convenient time to introduce another notation. In order to indicatethat the y variable is a function of x, the notation f(x) is often used synonymously withy. Thus if we write
y = ax2 + bx + c or f(x) = ax2 + bx + c
74
we mean the same thing. The first thing to notice when graphing a quadratic is the sign ofits leading coefficient. (Recall the leading coefficient is the number in front of the highestpower of x. For a quadratic, this is the coefficient of x2.) If the leading coefficient ispositive the graph is an ‘up-the-right-way’ parabola, and if it’s negative the graph is an‘upside-down’ parabola.
The next features you can easily find are the roots of the quadratic, and its axis ofsymmetry. The roots of any expression are the points where it equals zero, so these are thex-intercepts on the graph. For a quadratic the roots can be found by factoring, or by usingthe quadratic formula. However recall that a quadratic may have no real roots, in whichcase the graph does not cross the x-axis. A graph of a quadratic always has a vertical axisof symmetry along the line x = −b/(2a). This is easy to remember because it is half-waybetween the roots (if it has any) and in any case can be obtained just by dropping thesquareroot term from the quadratic formula. The point on the parabola along this axisof symmetry is called the vertex, and is the maximum (a < 0) or minimum (a > 0) valuethe function takes. This value can be found by evaluating the function at x = −b/(2a).Finally you can plot the y-intercept, which is c, obtained by putting x=0.
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As an example, let’s graph f(x) = x2 − x − 2. Observe that the leading coefficient ispositive, so this is an ‘up-the-right-way’ parabola. It factors to be f(x) = (x−2)(x+1), soits roots are at x = 2 and x = −1. The axis of symmetry is halfway between these roots, atx = 1
2 , and at that point f(x) attains its minimum value, which is f( 12 ) = (1
2 −2)( 12 +1) =
− 94 . The y-intercept is at 2. With this information we draw the graph.
Notice now that we have drawn the graph how easy it is to solve the quadratic in-equality
x2 − x − 2 < 0.
On the graph this represents the region where the graph dips below the x-axis, and this isclearly between the roots, since it is an ‘up-the-right-way’ parabola.
Exercises 11.2
1 Sketch the graphs of the following quadratics.(a) y = (x−1)(x−3) (b) f(x) = x(x+4) (c) y = x2+3x+1 (d) f(x) = x2+x+1
2 Using your sketches from question 1 solve the following quadratic inequalities.(a) (x − 1)(x − 3) ≥ 0 (b) x(x + 4) < 0 (c) x2 + 3x + 1 > 0 (d) x2 + x + 1 ≤ 0
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Lecture 12. Graphing higher order polynomials, circles,ellipses,hyperbolae
§ 12.1 Higher order polynomials
Having considered the graphs of linear and quadratic functions it is natural (at leastto a mathematician) to progress to cubic, quartic and higher order polynomials. In factmany of the ideas we used to sketch quadratics still apply, but as so often happens thereare added complications, some of which we are not yet ready to deal with. Still we can goa long way towards sketching any polynomial with a few simple rules.
The first thing we need to consider is what is the ‘right-way-up’ for higher orderpolynomials. Once again we shall say that the curve is the ‘right-way-up’ if the leadingcoefficient is positive, and it is ‘upside-down’ if the leading coefficient is negative. Theoverall behaviour for large values (positive or negative) of x is completely determined bythe leading term; that is, the term involving the largest power of x. This is because forlarge values of x the leading term is so much bigger than all the other terms combinedthat they have little influence on the function. For example, in the cubic
f(x) = x3 + 10x2 + 25x + 357
if we consider the size of each term for x = 100 say, we find that the terms are respectively1000000, 1000, 2500 and 357. The overall answer is 1003857, so the leading term totallydominates the contributions of the other terms. This will always be the case, provided wetake a large enough value of x. The behaviour splits into two classes, according to whetherthe highest power of x is odd or even. For odd powers n of x, when x is large and positive,xn is extremely large and positive; and when x is large and negative, xn is extremely largeand negative. For even powers n of x, when x is large (either positive or negative) thenxn is extremely large and positive. From this we can sketch the two possibilities for largevalues of x, and for ‘up-the right-way’ polynomials.
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If the polynomial is ‘upside-down’ these sketches are inverted.
Getting this behaviour correct sets the shape of the whole graph. Now we just plotthe roots of the polynomial, and the x-intercept, as we did for the quadratic. The latteris easily found since it is just the constant term (put x = 0 into the equation). Findingthe roots is, however, a difficult job in general. We need to be able to factorise theploynomial to find the roots. For example, here are the graphs of y = x(x− 1)(x + 2) andy = (4 − x2)(x2 + x).
Take note of the number of ‘wiggles’ each function has. Linear functions have nowiggless, quadratics have one, cubics have two and quartics have three. In general thenumbers of ‘wiggles’ on a polynomial graph is one less than the degree (the highest powerof x) of the polynomial. We will have to discuss adjustments to this rule later, but it willremain true provided we are prepared to double count wiggless which occur at the samespot, just as we double count repeated roots to an equation.
A word of warning! Unlike the parabola, higher order polynomials do not in generalhave any symmetry. We therefore cannot assume that the maximum and minimum valuesof the polynomial occur midway between adjacent roots. We will have to wait for calculusbefore we can determine these values exactly.
Exercises 12.1
1 Sketch the graphs of the following polynomials.
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(a) y = (x−1)(x−3)(x+2) (b) f(x) = x(x+5)(x−1)(3−x) (c) p(x) = x2(x−1)(d) y = 4x − x3 (e) q(x) = x3 + 3x2 − x − 3 (f) f(x) = x4 + 2x3 − x2 − 2x
§ 12.2 Circles
A circle is the set of all points which are equidistant from a given point; the centre.So to specify a circle completely we only need to know the position of its centre, and itsradius. Let’s say the centre of the circle is to be the point (x0, y0) and the radius is to ber. Then the point (x, y) is on the circumference of the circle if and only if its distance tothe centre equals r.
That is, using the distance formula,√(x − x0)2 + (y − y0)2 = r.
Squaring both sides gives us the standard form of the equation of a circle.
(x − x0)2 + (y − y0)2 = r2
Given an equation in this form it is an easy task to graph it. For example the equation(x−3)2 +(y +1)2 = 25 is a circle, centred at the point (3,−1) (since (y +1) = (y− (−1)))with radius 5.
Notice that this is a relation, not a function, since vertical lines through the graphcut it at two points.
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If the equation of the circle is given in standard form it is easy to graph, but it couldarise in non-standard form, in which case it needs to be rearranged into standard form bycompleting the square. For example, the equation
2x2 + 12x + 2y2 − 8y = 24
is a circle. In order to graph it we rearrange it into standard form, by dividing by 2, andgrouping terms involving x and y together. Thus
(x2 + 6x) + (y2 − 4y) = 12.
Now to complete the square on the x we need to add (6/2)2 = 9 to both sides, and tocomplete the square on y we add (−4/2)2 = 4 to both sides. This gives
(x2 + 6x + 9) + (y2 − 4y + 4) = 12 + 9 + 4
(x + 3)2 + (y − 2)2 = 25.
So this is a circle centred at (−3, 2) with radius 5.
Exercises 12.2
1 State the centre and radius of the following circles, and hence sketch their graphs.
(a) x2 + y2 = 4 (b) (x − 2)2 + (y + 5)2 = 14 (c) x2 + (y + 1)2 = 3
2 Write the equation of the circle x2 + 2x + y2 − 10y + 10 = 0 in standard form, andhence draw its graph.
§ 12.3 Ellipses
The ellipse (or oval) is another famous shape among the family known as conic sections.It arises when a cone or a cylinder is cut with an inclined plane. In nature its most famousrole is as the path traced out by planets orbiting the sun. Like all conic sections it is richin geometrical properties which we won’t consider here at this time. Perhaps you haveseen one of these properties, however. If you pin the two ends of a piece of string to thepage, hook a pencil in the string, pull it tight and draw, then you will trace out the shapeof an ellipse. What this means is that the sum of the distances from the pencil tip to thepins is always the same, since it equals the length of the string. From this property, andthe distance formula, it is possible to derive the equation of the ellipse, however we shan’tbother since it is an uninspiring and painful piece of algebra.
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What we will say about the equation of an ellipse is that it is very similar to that ofa circle. Indeed an ellipse itself looks like a flattened circle. If we take the equation of acircle, centred at the origin and radius r,
x2 + y2 = r2
we can divide both sides by r2, and write this as(x
r
)2
+(
y
r
)2
= 1.
Now if we think of an ellipse as being a ‘squished’ circle, with different ‘radii’ in eachdirection; say ‘radius’ a in the x-direction and ‘radius’ b in the y-direction; then we havethe equation of an ellipse. (
x
a
)2
+(
y
b
)2
= 1.
The numbers a and b are not called radii anymore, their correct names are the major and
minor semi-axes, depending on which is bigger. However the analogy to radius is a goodone. Even the area of an ellipse (πab) is similar in form to the area of a circle (πr2). Thesemi-axes tell us the distance from the centre to the ellipse in each direction.
If the ellipse is to be centred at the point (x0, y0) then the equation simply becomes(x − x0
a
)2
+(
y − y0
b
)2
= 1.
The points which are furthest from and closest to the centre are called the vertices. Theseare (x0 ± a, y0) and (x0, y0 ± b). So, to plot the ellipse(
x − 13
)2
+(
y + 24
)2
= 1
we simply find the centre, (1,−2), and the vertices (−2,−2), (4,−2), (1, 2) and (1,−6).Now connect these with a squashed circle and you’re done.
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If an ellipse is given in non-standard form we need to put it into standard form bycompleting the square. For example, the equation 2x2 + y2 − 4x + 2y + 1 = 0 is an ellipse.Writing this as
2(x2 − 2x) + (y2 + 2y) = −1
2(x2 − 2x + 1) + (y2 + 2y + 1) = 1 + 2 + 1
2(x − 1)2 + (y + 1)2 = 2
(x − 1)2 +(
y + 1√2
)2
= 1
we see that it is centred at the point (1,−1) and has a = 1 and b =√
2. It is thereforeelongated in the y-direction.
Exercises 12.3
1 State the centre and the semi-minor axes of the following ellipses. Hence sketchtheir graphs.
(a)(
x
3
)2
+(
y
2
)2
= 1 (b)(
x − 13
)2
+(
y + 24
)2
= 1 (c)x2
3+
(y − 1)2
4= 1
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2 Write the equation of the ellipse 9x2 + 36x + 4y2 − 8y + 4 = 0 in standard form,and hence draw its graph.
§ 12.4 Hyperbolae
This is the last of that famous family of curves, the conic sections, and it too canbe obtained by cutting across a cone (an infinite cone in both directions, that is!) with aplane. The most common form in which we are likely to meet an hyperbola is
xy = c
where c is a constant. This can also be arranged with y (or x) as the subject
y =c
x
(or x =
c
y
).
This equation has the form of an inverse proportionality law which occurs occasionally inscience. For example, Boyle’s Law relating the pressure and volume of a gas.
To graph the hyperbola the most interesting behaviour happens near x = 0 or y = 0.We’ll consider the case of an ‘up-the-right-way’ hyperbola, with c > 0. Imagine x veryclose to zero, but still positive. Then y = c/x will be very large, and positive. So as x getsclose to zero from the positive side y zooms up towards infinity. The function does nothave a value at x = 0 because you can’t divide by zero. For x close to zero but negativethe function is very large, but negative. So as x gets close to zero from the negative side y
zooms down towards minus infinity. Similar reasoning can be applied to the case where y isclose to zero. The arguments are the same for an ‘upside-down’ hyperbola (c < 0), exceptthat all the signs are reversed. We can therefore draw a rough sketch of the hyperbola.
In fact the dotted lines on the rough sketches are really quite accurate. About all weneed is to plot one point on each branch of the hyperbola, just to pin it down. The obviouspoint to plot is when |x| = |y|. So the complete graph of the hyperbola xy = c is
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Exercises 12.4
1 Sketch the graphs of the following hyperbolae.
(a) xy = 1 (b) y =−5x
(c) y =1
x + 1
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Lecture 13. Curve sketching
In the preceding lectures we examined a number of commonly occuring functionsand relations and saw how to sketch their graphs. The techniques used there were oftenspecific to the function in question, though sometimes we applied quite general principlesto the problem. This lecture concentrates on general principles which are applicable to thesketching of any graph. The aim of this section is that using the ideas here you should beable to quickly draw a rough sketch of virtually any function.
§ 13.1 The translation principle
The translation principle says thatIf a new function is created by translating the graph of an old function by an amount
(x0, y0) then the equation of the new function is the same as that of the old function with
(x, y) replaced by (x − x0, y − y0) throughout.
This sounds complicated, but it isn’t really. In fact we’ve already seen this idea atwork. For example, a straight line of slope 2 through the origin has equation
y = 2x.
If we translate the line so that it passes through the point (3, 5) it’s equation becomes
(y − 5) = 2(x − 3);
(replace x by (x− 3) and y by (y − 5)). This is the idea behind the point-gradient form ofa straight line.
Another example we have already seen is that of a circle. A circle radius 1 centred atthe origin has equation
x2 + y2 = 1.
If we move the circle to be centred at the point (−2, 3) the equation becomes
(x + 2)2 + (y − 3)2 = 1.
(Note that x is replaced by (x −−2) = (x + 2).)So far we have seen examples of functions we could probably have graphed easily by
other means. Now consider a more difficult example,
y =x
x + 1.
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This does not appear to be any of the functions we are familiar with. Let’s begin by ‘doingnothing’ to the numerator.
y =(x + 1) − 1
x + 1= 1 − 1
x + 1.
A little rearranging gets us
(y − 1) = − 1x + 1
.
Now this is just the equation of an (upside-down) hyperbola (y = −1/x) with (x + 1)taking the place of x and (y − 1) in place of y. Therefore the graph of this function is justan upside-down hyperbola ‘centred’ about the point (−1, 1).
-5 -4 -3 -2 -1 1 2 3
-4
-2
2
4
6
Exercises 13.1
1 The parabola y = x2 is translated so that its vertex is at (−2, 1). Sketch the newparabola and write down its equation.
2 Draw the graph of y = |x|. Using the translation principle deduce the graphs ofy = |x + 2| and y = |x| + 2.
§ 13.2 Asymptotic behaviour
The asymptotic behaviour of a function is the behaviour as one of the variables growslarge. In some instances we can completely determine the graph of a function from itsasymptotics; for example, the hyperbola from last lecture.
When graphing any curve at all one of the first considerations should be ‘what happensfor large (positive and negative) values of x?’. The jargon for this is ‘as x tends to infinity’and as ‘x tends to minus infinity’; and these are written symbolically as x → ∞ andx → −∞. Let’s continue with the example y = x/(x + 1).
Consider what happens as x → ∞. The ratio of x to x + 1 becomes very close to 1,although it is always slightly less than 1. We can write this as y → 1−, which reads ’y
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tends to 1 from below’. If this is difficult to see at first, try a sequence of values for x
getting larger, say x = 10, 100, 1000....., and see what happens to y. In a similar way wecan argue that as x → −∞ then y → 1+ (y tends to 1 from above). Once again, if indoubt, try putting x = −10,−100,−1000....
So far we have considered x → ±∞. What about y → ±∞? A common way thisarises is if the formula for the function indicates division by zero. Of course at a pointwhere divison by zero occurs the function is not defined, but near such a point the functionwill be defined, and division by a number near zero typically (but not always) results inanswers ‘near’ ±∞. The most crucial thing to decide is whether the trend is towards +∞or −∞. Let’s continue with the example y = x/(x + 1).
At the point x = −1 we have division by zero and so the function is not defined.However as we approach x = −1 from above (x → −1+) the value of y is very largeand negative (y → −∞). Again, if unsure, you could try x = −0.9,−0.99,−0.999... andobserve the trend. Another way is to try some ‘dodgy’ arithmetic
y =−1+
−1+ + 1=
−1+
0+→ −∞.
That is, the numerator is negative and the denominator is ‘positive zero’ (a number veryclose to zero but positive), so the result is negative but large. In a similar way as weapproach x = −1 from below (x → −1−) we get
y =−1−
−1− + 1=
−1−
0−→ ∞.
That is, the numerator is negative and the denominator is ‘negative zero’ (a number veryclose to zero but negative), so the result is positive but large.
Piecing this information together on a graph gives
-5 -4 -3 -2 -1 1 2 3
-4
-2
2
4
6
from which it is easy to interpolate to obtain the full graph.
87
Let’s try a more complicated example!
y =2x2 + 1
x2 − x − 2
First let’s consider x → ∞. In this case the highest power of x dominates all other terms,top and bottom, so we can write
y =2x2 + 1
x2 − x − 2→ 2x2
x2= 2 as x → ∞.
In fact, with a little more care we could see that y → 2+, since the numerator is a littlemore than twice the denominator. Perhaps dividing the top and bottom by x2 makes thisa little clearer.
y =2x2 + 1
x2 − x − 2=
2 + 1/x2
1 − 1/x − 2/x2→ 2+ as x → ∞.
In a similar fashion
y =2x2 + 1
x2 − x − 2=
2 + 1/x2
1 − 1/x − 2/x2→ 2− as x → −∞.
We should now look for any potential asymptotes caused by a division by zero. Since thedenominator factors, x2 − x − 2 = (x + 1)(x − 2), divisions by zero arise at x = −1 orx = 2. We must consider what happens to the function near these points.
As x → 2+, y → 2.(2+)2 + 1(2+ + 1)(2+ − 2)
→ 93.0+
→ ∞
As x → 2−, y → 2.(2−)2 + 1(2− + 1)(2− − 2)
→ 93.0−
→ −∞
As x → −1+, y → 2.(−1+)2 + 1(−1+ + 1)(−1+ − 2)
→ 30+.(−3)
→ −∞
As x → −1−, y → 2.(−1−)2 + 1(−1− + 1)(−1− − 2)
→ 30−.(−3)
→ ∞
Plotting this information on a graph gives us
88
-4 -2 2 4
-4
-2
2
4
6
Since elsewhere we expect the graph to be a smoothly flowing curve, we can join thesepieces together in the obvious way with confidence.
-4 -2 2 4
-4
-2
2
4
6
Note that the left-most piece of the graph must necessarily drop below the asymptotey = 2, and therefore there is a dip (local minimum) in this section of the graph. Finding itsexact location will have to wait for calculus, but we know it is there simply by examiningthe asymptotic behaviour.
Exercises 13.2
1 By considering the asymptotic behaviour, sketch the graph of y = 1/x2.
2 Use the translation principle, and your answer to 1, to sketch y = 1/(x + 1)2.
3 By examining the asymptotic behaviour, sketch the graphs of
(a)1
x2 + 1(b)
x
x2 + 1(c)
x2
x2 + 14 By examining the asymptotic behaviour, sketch the graphs of
(a)1
x2 − 1(b)
x
x2 − 1(c)
x2
x2 − 1
5 By examining the asymptotic behaviour, sketch the graph of y = x +1x
.
89
§ 13.3 Graphing reciprocal functions
An example where the ideas of the previous section are particularly easy to applyis when we desire to graph the reciprocal of a familiar function. For example, considergraphing
1x2 + 4x + 3
.
Now the denominator here is a parabola, with roots at -1 and -3. This is easily graphed.
-5 -4 -3 -2 -1 1
2
4
6
8
From this easy graph, the appropriate asymptotic behaviour of the reciprocal is easyto determine.
As x → −∞, y → 0+
As x → −3−, y → ∞As x → −3+, y → −∞As x → −1−, y → −∞As x → −1+, y → ∞
As x → ∞, y → 0+
These can be drawn directly onto the graph, thus
-5 -4 -3 -2 -1 1
-4
-2
2
4
90
Exercises 13.3
1 By considering graph of the denominator, sketch the graphs of
(a) y =1x2
(b) y =1
x2 − 1(c) y =
1√1 − x2
(d) y =1
|x| − 2
91
Exam 2001
Question 1 [25 marks] Number Systems
(a)(i) Find the prime factorisations of 36 and 96.(ii) What is the lowest common multiple (LCM) of 36 and 96?(iii) What is the greatest common divisor (GCD) of 36 and 96?
(b)(i) Write
√96 as a surd in its lowest form.
(ii) Simplify 3√
27 − 2√
108 −√
12
(iii) Rationalise the denominator of3√
3 + 1.
(iv) The size of a television screen is reported as the diagonal distance across thescreen. For a screen 25 cm wide and 20 cm high, what would be the screen sizeto the nearest centimetre?
(c)(i) What (base 10) number is represented in base 5 by 43215?(ii) How would the (base 10) number 78 be represented in base 3?
(d) Express 0.6̇7̇ as a fraction in its lowest form.(e) Express 0.01234567
(i) correct to 5 decimal places(ii) in scientific notation using 5 significant figures
(f) Find the exact values (i.e. using surds) of(i) 8
56
(ii) (√
3)−3
(g) If i2 = −1 simplify(i) (3 + 2i)(4 − 5i)
(ii)2
3 + i
92
Question 2 [25 marks] Basic Algebra
(a) Simplify(a2)3b−1c
a3b−2a2√
c(b) Expand and simplify the following
(i) (2x − y)(2x + y)(ii) (x − 3y)(2 − (x + y))(iii) (3x − y)2
(iv) (x − 2)4
(c) Fully factorise the following(i) x3 + 5x2 − 36x
(ii) 6x2 + 5x − 4(iii) 6xy − 15x + 2y − 5(iv) x6 − y6
(d) Simplifyx
x + 1+
2x − 1x2 − 1
.
(e)(i) Sum the odd numbers less than 100.
(ii) Express 1 +13
+19
+127
+181
as a fraction in its lowest form.
(iii) Express 1 +13
+19
+127
+181
+ ....... as a fraction in its lowest form.
93
Question 3 [25 marks] Solving Equations
(a) Solvex + 3
5− x + 5
3= 1.
(b) Show thatx + 1x + 2
+x + 3x + 4
= 5
can be arranged to form a quadratic equation. Hence solve.(c)
(i) Find the distance between the points (−4, 1) and (2,−0.5).(ii) What is the midpoint between the points (−4, 1) and (2,−0.5)?(iii) What is the equation of the line joining the points (−4, 1) and (2,−0.5)?
(d)(i) On the same axes, graph the straight lines y = 1 − x and x + 4y = 0.(ii) Using the method of elimination, or otherwise, find the point of intersection of
these lines.(e)
(i) Graph y =12x2 − 2x− 6 showing clearly the roots, axis of symmetry, vertex and
y-intercept.(ii) By completing the square, on the same axes graph x2 − 4x + y2 + 6y − 12 = 0.(iii) Using the method of substitution, or otherwise, find the points of intersection of
the line and the circle.
94
Question 4 [25 marks] Inequalities and Graphing
(a) Indicate the set of numbers which are less than -1 or greater than or equal to 2(i) graphically on a number line(ii) using interval notation.
(b) Solve 3 − 4x ≤ −2(1 − x)(c)
(i) Solve 3x2 + 7x + 4 = 0(ii) Draw a rough sketch of y = 3x2 + 7x + 4(iii) Solve 3x2 + 7x + 4 > 0(iv) Draw a rough sketch of y = 3x3 + 7x2 + 4x
(v) Solve 3x3 + 7x2 + 4x > 0(d)
(i) Solve |x + 2| + |x| = 4(ii) Solve |x + 2| + |x| < 4
(e)
(i) Sketch y =1
x − 2(ii) Solve
1x − 2
< 1.
(f)(i) Sketch the graph of y = x3 − x.
(i) Sketch the graph of y =1
x3 − x.
95
Solutions to exercises
Solutions — Lecture 1
Exercise 1.1
1 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,972 Decompose the following numbers into their prime factors.
(a) 2432 (b) 2272 (c) 2352 (d) 52113
LCM(144, 196) = 243272 GCD(144, 196) = 22
LCM(144, 200) = 243252 GCD(144, 200) = 23
LCM(144, 275) = 24325211 GCD(144, 275) = 1LCM(196, 200) = 235272 GCD(196, 200) = 22
LCM(196, 275) = 22527211 GCD(196, 275) = 1LCM(200, 275) = 235211 GCD(200, 275) = 52
4 LCM(144, 196) = 243272 = 7056
Exercises 1.2
(a) −2 (b) −35 (c) 2 (d) 5/7 (e) 1 (f) −4 (g) −105 (h) −15
Solutions — Lecture 2
Exercises 2.1
1 (a) 1/5 (b) 1/9 (c) 3/17 (d) 3/7 (e) 3/42 (a) 5
3 (b) 187 (c) 58
11 (d) 6317
3 (a) 2 13 (b) 3 3
7 (c) 3 1011 (d) 6 10
17
4 Add, subtract, multiply and divide each of the following pairs of fractions.(a) 5
6 , 16 , 1
6 , 32 (b) 29
35 ,− 135 , 6
35 , 1415 (c) 95
36 , 5536 , 125
108 , 154
Exercises 2.2
1 (a) 5 (b) 13 (c) 412 (a) 5 (b) 40 (c) 23
√502 + 152 ≈ 52m
96
Exercises 2.3
1 (a) 3√
3 (b) 4√
2 (c) 5√
3 (d) 6√
3 (e) 5√
5 (f) 8√
3 (g) 2 3√
2 (h) 4 3√
32 (a)
√12 (b)
√50 (c)
√63 (d) 3
√81
3 (a) 7√
3 (b) 6√
2 (c)√
3 (d) −5√
3 (e) 7√
2 (f)√
3 (g) (x − 1)√
x
4 (a) 3 (b) 45 (c) 8√
10 (d) 2 (e) 18 (f) 8√
6 − 30 (g) 12√
6(h)
√35 − 3
√14 − 2
√15 + 6
√6
5 (a) (3 +√
2)(3 −√
2) = 7 (b) (7 −√
3)(7 +√
3) = 46(c) (4
√3 − 5)(4
√3 + 5) = 23 (d) (2
√7 + 1)(2
√7 − 1) = 27
6(a) 3
√2
2 (b)√
22 (c)
√15−
√6
3 (d)√
2 − 1(e) 3(
√3+1)2 (f) 5(
√7−
√5)
2 (g) 11−6√
27 (h) 23+8
√5
19
Solutions — Lecture 3
Exercises 3.1
1 26 = 110102,26 = 1015,26 = 357 .2 1112 = 7,1115 = 31,1117 = 57.
3 (a)12
(b)127
(c) 3 (d)12
(e) 1
4 (a) x−1 (b) x−3 (c) x12 (d) 2(2x + 5)−4 (e) x− 1
3
5 (a)1x5
(b)2x3
(c)1
(2x)3(d) (x + y)2 (e)
x − y
x + y6 (a) 217 (b) 310 (c) 210 (d) 35 (e) x5 (f) 81y−8 (g) a−2b3 (h) x−5y2
7 (a) 0.2̇ (b) 0.2̇7̇ (c) 0.4̇28571̇ (d) 0.5̇38461̇ (e) 1.1̇8̇
8 (a)49
(b)2399
(c)23439990
=7813330
(d)11190
=3730
(e) 1
Exercises 3.2
1 (a) 7.892 × 103 (b) 7.654 × 10−3 (c) 7.6 × 102 (d) 7.765 × 100
2 (a) 3 (b) 3 (c) 4 (d) 33 Each indicates a different level of accuracy in the answer.
Exercises 3.3
1 Re(5 + 3i) = 5, Im(5 + 3i) = 3, 5 + 3i = 5 − 3i
Re(−2 −√
2i) = −2, Im(−2 −√
2i) = −√
2, −2 −√
2i = −2 +√
2i
97
2 (5 + 3i) + (−3 + 5i) = 2 + 8i(5 + 3i) − (−3 + 5i) = 8 − 2i
(5 + 3i)(−3 + 5i) = −15 + 25i − 9i + 15i2 = −30 + 16i
5 + 3i
−3 + 5i=
5 + 3i
−3 + 5i× −3 − 5i
−3 − 5i=
−15 − 25i − 9i − 15i2
9 + 15i − 15i − 25i2=
−34i
34= −i
Solutions — Lecture 4
Exercises 4.1
1(a) 8y − 2y2 (b) 5t − 9 (c) p2 − p + pq + 3q − 12(d) 5x2 − 8x (e) 9a3 − 6a2 + 18a (f) 26y2
(g) x2 + 5x + 6 (h) m2 − 5m − 6 (i) 15c2 + 11c − 12(j) n2 − 9 (k) n2 + 6n + 9 (l) 3a3 − 22a2 + 7a
(m) a3 + a2b + ab2 + b3 (n) a3 + 4a2 + 6a + 3 (o) 2x5 − 5x4 + 9x3 − 17x2 + 12x
2(a) a(x + y) (b) y(y − 4) (c) 5a(a − 3)(d) 7b(a + 2c) (e) 3(2x − 1) (f) x2(x + 7a)(g) t(at + 2a + 5) (h) 2mn(4m + 3n + 5) (i) (a + b)(a + b + x)(j) (u + 3)(v + 5) (k) (2x − 3z)(y + 2) (l) (x − y)(a − b)(m) (y − 1)(x + y) (n) (c − 4)(x − 1) (o) (x2 + 2)(x − 3)
Exercise 4.2
(a) −163 (b) −2 (c) 18 (d) 18
7
(e) 55 (f) 51 (g) 1 (h) 3
Exercises 4.3
1(a) a2 + 10x + 25 (b) y2 − 4y + 4 (c) x2 + 4xy + 4y2
(d) 9a2 − 12ab + 4b2 (e) m4 + 2m2n2 + n4 (f) 9x2y2 − 6xyz + z2
2(a) (x + 2)2 (b) (t − 4)2 (c) (3a + 2b)2
(d) (s − 2t)2 (e) (s + 12 )2 (f) (x + 1
x )2
98
Exercises 4.4
1 (a) (x + 4)2 − 16 (b) (y − 8)2 − 64 (c) (m − 7)2 − 49 (d) (y + 32 )2 − 9
4 (e)(a + 1
2 )2 − 14 (f) (x +
√7)2 − 7 (g) (x + 4y)2 − 16y2 (h) (2a − b)2 − (b2 − 4)
Solutions — Lecture 5
Exercises 5.1
1 (a) 3,7 (b) 1,7 (c) -4,-6 (d) 3,5 (e) -7,4 (f) 6,-12
(a) (x + 3)(x + 7) (b) (x + 1)(x + 7) (c) (x − 4)(x − 6)(d) (x + 3)(x + 5) (e) (x − 7)(x + 4) (f) (x − 6)(x + 1)
3(a) (x − 9)(x + 7) (b) (y + 1)(y + 5) (c) (x − 11)(x + 10)(d) (a + 5)(a + 9) (e) (b − 4)(b − 25) (f) (m + 11)(m − 7)
Exercises 5.2
1.(a) (3x + 2)(x + 1) (b) (2x + 1)(x + 2) (c) (2x + 1)(x + 3)(d) (2x − 3)(3x + 1) (e) (3x − 7)(x − 2) (f) (4 − 5x)(2x + 3)
Exercises 5.3
1 (a) (x + 4)(x − 4) (b) (y + x)(y − x) (c) (2a + 3b)(2a − 3b) (d) 4a2 − 2b2 =(2a +
√2b)(2a −
√2b)
(e) (5 + n)(5− n) (f) (a + bc)(a− bc) (g) (x2 + y2)(x + y)(x− y) (h) 12(y + 1)
Exercises 5.4
1. x6 − y6 = (x − y)(x5 + x4y + x3y2 + x2y3 + xy4 + y5).2. x7 − y7 = (x − y)(x6 + x5y + x4y2 + x3y3 + x2y4 + xy5 + y6)
x7 + y7 = (x + y)(x6 − x5y + x4y2 − x3y3 + x2y4 − xy5 + y6).
Exercises 5.5
1. Fifth row is 1 5 quad 10 10 5 1Sixth row is 1 6 quad 15 20 15 6 1
99
2. (x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
3. (a) (a + b)6 = x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + y6
(b) (x − y)4 = x4 − 4x3y + 6x2y2 − 4xy3 + y4
(c) (x + 2y)3 = x3 + 6x2y + 12xy2 + 8y3
Solutions — Lecture 6
Exercises 6.1
1(a) y = 9 (b) x = 8 (c) a = −8(d) x = 6 (e) m = 8 (f) x = 2(g) x = 62 (h) x = 6 (i) y = 9
2
(j) a = 6 (k) y = 10 (l) y = 295
(m) x = 198 (n) x = 4 (o) x = 158
(p) x = 57 (q) x = −4 (r) x = 2
Exercises 6.2 and Exercises 6.3
(a) a = 3, b = 1 (b) x = 2, y = 1(c) p = 2, q = −1 (d) x = −10, y = 2(e) t = −1
3 , v = 433 (f) x = −3, y = 2
(g) w1 = −1, w2 = 5 (h) x = 0, y = 0 or x = 1, y = 1(i) x = 0, y = 0 or x = −2, y = 4 (j) x = 0, y = 3 or x = 3, y = 0(k) x = −1, y = −1 or x = 1
4 , y = 4 (l) a = 32 , b = 1
2 , c = 52
Solutions — Lecture 7
Exercises 7.1
1(a) x = 4, 6 (b) y = ±2 (c) x = 0,−5(d) x = −3, 2
3 (e) b = 0, 32 (f) x = −1,−2, 3
2(a) x = −3,−4 (b) y = −1,−7 (c) z = −5,−8(d) x = 1, 7 (e) b = 5 (f) c = −3,−12(g) x = −2, 4 (h) d = −1, 11 (i) h = 2(j) p = −1,−2
3 (k) q = −2,−12 (l) r = 4, 3
2
100
(m) x = 17 , 1
5 (n) x = 12 ,−2
5 (o) x = 113 , 3
2
(p) x = 0,−1,−2 (q) x = 0,−3,−3 (r) x = 0, 2, 73
Exercises 7.2
1(a) x = 2 ±
√5 (b) x = −7±
√33
4 (c) x = 2±√
117
(d) No solution (e) x = −9±√
216 (f) x = 1
3
(g) x = 0, 9 (h) a = 23 ,− 3
2 (i) b = 3±√
3714
(j) x = 32 (k) x = −6, 8 (l) x = 0, 2, 4
2(a) 2 (b) 0 (c) 2 (d) 1 (e) 0 (f) 1
Exercises 7.3
1 (a) x = 0,−2,−3 (b) x = −2,−1, 1
2 (a) x = ±1,±2 (b) x = 0, 0, 3, 5
3 (a) x = ±√
2 (b) x = ±2
Solutions — Lecture 8
Exercises 8.1
1 (a) False (b) True (c) False (d) True (e) True
2 (a) x ∈ [−2, 1) (b) x ∈ (−∞,−2] ∪ (3,∞) (c) x ∈ [−2, 3)
Exercises 8.2
(a) x > 3 (b) x ≤ 30 (c) y ≥ − 52
(d) r > −9 (e) x > 103 (f) −2 < x ≤ 2
(g) x ≥ 6 (h) x ≤ 1 (i) b ≤ −454
(j) x < −2, x ≥ −1 (k) −3 < x < −2 (l) x > 1, x < −1
Exercises 8.3
(a) x < −4, x > −3 (b) −7 ≤ y ≤ 1 (c) z ≤ −8, z ≥ −5(d) −3 < x < 3 (e) x < −3, x > 2 (f) 5
2 ≤ x ≤ 5(g) x ≤ −2
3 , x ≥ 12 (h) x ≥ 4, x ≤ −4 (i) −2 < x < 3
101
Exercises 8.4
(a) 17 (b) π (c) 1 (d) -1 (e) 1 (f) |x|
Exercises 8.5
1(a) x = ±6 (b) x = ±6 (c) No solution(d) x = −2, 6 (e) x = −3, 13 (f) x = −4, 8(g) No solution (h) x = 3,−9 (i) x = −1(j) x = ±2 (k) x = 7
2 (l) x ∈ [−1, 2]2
(a) x > 6, x < −6 (b) x > 6, x < −6 (c) x ∈ R
(d) 2 ≤ x ≤ 6 (e) x ≤ −3, x ≥ 13 (f) x < 4, x > 8(g) x ∈ R (h) x ≤ −9, x ≥ 3 (i) x ≤ −1(j) −2 < x < 2 (k) x ≥ 7
2 (l) x ∈ [−1, 2]
Solutions — Lecture 9
Exercise 9.1
1 3 5 7 9 11 13 ... xn = 2n − 1 or xn+1 = xn + 2 , x1 = 1.2 6 16 54 162 486 1558 ... xn = 2.3n−1 or xn+1 = 3xn, x1 = 2.1 8 27 64 125 216 343 ... xn = n3 (or xn+1 = xn + 3n2 + 3n + 1, x1 = 1).2 3 5 7 11 13 17 19 23 29 ... No formula possible. Prime numbers.1 1 2 3 5 8 13 21 ... xn+2 = xn+1 + xn, x1 = 1, x2 = 1. Fibonacchi numbers.
Exercise 9.2
1. 1000.(1 + 1000)/2 = 5005002. x27 = 5 + (27 − 1).3 = 83, s27 = 27.(5 + 83)/2 = 11883. There are n + 1 odd numbers on the list. Thus the sum is (n + 1)(1 + (2n + 1))/2 =
(n + 1)2.
Exercise 9.3
1. x12 = 8.(1/2)11 = 1/256, s12 = 8.(1 − (1/2)12)/(1 − 1/2) = 15 255256
2. 0.9̇ = 0.9 + 0.09 + 0.009 + ... which is a GP with a = 9/10 and r = 1/10. Thens∞ = a/(1 − r) = 1.
102
Solutions — Lecture 10
Exercises 10.1
(a) Midpoint is (52 , 4), Distance is
√(4 − 1)2 + (6 − 2)2 = 5
(b) Midpoint is (−1,−2), Distance is√
62 + 22 = 2√
10(c) Midpoint is (3, 5
2 ), Distance is√
02 + 92 = 9
Exercises 10.3
(a)
-2 -1 1 2
-4
-2
2
4
(b) -2 -1 1 2
1
2
3
4
(c) -4 -2 2 4
1
2
3
4
(d)
-2 -1.5 -1 -0.5 0.5 1 1.5 2
-4
-3
-2
-1
1
2
3
4
(e)
-2 -1.5 -1 -0.5 0.5 1 1.5 2
-3
-2
-1
1
2
3
(f)
-20 -15 -10 -5 5 10 15 20
-400
-200
200
400
(g)
-4 -3 -2 -1 1 2 3 4
-1
-0.75
-0.5
-0.25
0.25
0.5
0.75
1
(h)
2 4 6 8 10
-30
-20
-10
10
20
30
103
Solutions — Lecture 11
Exercises 11.1
-3 -2 -1 1 2 3
-1
1
2
3
4
5
-5 -4 -3 -2 -1 1
-5
-4
-3
-2
-1
1
1(a) y = 2x + 3 1(b) y = −x − 3
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
-4 -2 2 4
-8
-6
-4
-2
1(c) y = −32x + 1
2 1(d) y = −7
-1 -0.5 0.5 1 1.5 2 2.5 3
-2
-1.5
-1
-0.5
0.5
1
1.5
2
0.5 1 1.5 2 2.5 3
2
4
6
8
1(e) x = 2 2. x = 2, y = 5
104
Exercises 11.2
1 2 3 4
-1
1
2
3
-5 -4 -3 -2 -1 1
-4
-2
2
4
(a) x ≥ 3, x ≤ 1 (b) −4 < x < 0
-5 -4 -3 -2 -1 1 2
2
4
6
8
10
-3 -2 -1 1 2
1
2
3
4
5
6
7
(c) x > −3+√
52 ,x < −3−
√5
2 (d) No such x
Solutions — Lecture 12
Exercises 12.1
-3 -2 -1 1 2 3 4
-20
-10
10
-4 -2 2 4
-100
-50
50
100
150
1(a) y = (x − 1)(x − 3)(x + 2) 1(b) y = x(x + 5)(x − 1)(3 − x)
-1 -0.5 0.5 1 1.5 2
-0.3
-0.2
-0.1
0.1
0.2
-3 -2 -1 1 2 3
-15
-10
-5
5
10
15
1(c) y = x2(x − 1) 1(d) y = 4x − x3
105
-4 -3 -2 -1 1 2
-15
-10
-5
5
10
15
-3 -2 -1 1 2
-1
1
2
1(e) y = x3 + 3x2 − x − 3 1(f) y = x4 + 2x3 − x2 − 2x
Exercises 12.2
-2 -1 1 2
-2
-1
1
2 -2 -1 1 2 3 4
-6
-5
-4
-3
-2
-1
1(a) Centre at (0,0), Radius 2 1(b) Centre at (2,-5), Radius 12
-2 -1.5 -1 -0.5 0.5 1 1.5 2
-3
-2.5
-2
-1.5
-1
-0.5
0.5
1
-6 -4 -2 2 4
2
4
6
8
10
1(c) Centre at (0,-1), Radius√
3 2. (x + 1)2 + (y − 5)2 = 16
106
Solutions — Lecture 13
Exercise 13.1
-5 -4 -3 -2 -1 1
2
4
6
8
10
-3 -2 -1 1 2 3
0.5
1
1.5
2
2.5
3
1 (y − 1) = (x + 2)2 2. y = |x|
-5 -4 -3 -2 -1 1
0.5
1
1.5
2
2.5
3
-3 -2 -1 1 2 3
1
2
3
4
5
2 y = |x + 2| 2. y = |x| + 2Exercise 13.2
-3 -2 -1 1 2 3
1
2
3
4
5
-4 -3 -2 -1 1 2
1
2
3
4
5
1 y = 1/x2 2. y = 1/(x + 1)2
-3 -2 -1 1 2 3
0.2
0.4
0.6
0.8
1
-3 -2 -1 1 2 3
-1
-0.75
-0.5
-0.25
0.25
0.5
0.75
1
3(a) y = 1/(x2 + 1) 3(b) y = x/(x2 + 1)
107
-3 -2 -1 1 2 3
0.2
0.4
0.6
0.8
1
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
3(c) y = x2/(x2 + 1) 4(a) y = 1/(x2 − 1)
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
4(b) y = x/(x2 − 1) 4(c) y = x2/(x2 − 1)
5 y = x + 1/x
-4 -2 2 4
-4
-2
2
4
Exercise 13.3
(a) see Exercise 13.2 Question 1 (b) see Exercise 13.2 Question 4(a)
-1 -0.5 0.5 1
0.5
1
1.5
2
2.5
3
-4 -2 2 4
-3
-2
-1
1
2
3
1(c) y = 1/√
1 − x2 1(d) y = 1/(|x| − 2)
108
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