Module 1 - Algebra (self Study)

SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self Study Course

MODULE 1

ALGEBRA

Module Topics 1. Simplifying expressions and algebraic functions 2. Rearranging formulae 3. Indices 4. Rationalising a denominator containing a square root 5. Linear and quadratic equations 6. Simultaneous linear equations 7. Inequalities 8. Partial fractions Most modules are based on the book by James but the first two, on Algebra and Trigonometry, are selfcontained. Facility in carrying out algebraic manipulation is very important in your studies, and Module 1 covers most of the basic algebraic topics. Many of you will know much of the material in the first two modules and so these will be mainly revision. It is important, however, that you are able to carry out mathematical manipulations quickly and so, even if the material is familiar, you are still strongly advised to work through the exercises to improve your speed and accuracy. To some of you the discussed material will not all be revision, in which case you should spend time working carefully through the modules. The book (not the course text) Introduction to Engineering Mathematics by Croft, Davison and Hargreaves, published by Addison-Wesley, contains a large number of chapters on the basic material covered in the first two modules and this book is recommended if you need extra examples. After working through this module please attend one of the testing sessions and then have your test marked by your nominated tutor. Work Scheme Study the following sections, read carefully the worked Examples and do the stated Exercises. Solutions to the Exercises are given towards the end of this module, starting on p.21. 1. Removing brackets

A basic rule in removing brackets from mathematical expressions is a(b + c) = ab + ac.

The quantity a outside the bracket multiplies both the quantities b and c inside. Note that the meaning of an expression depends crucially on including brackets, where appropriate. Omitting them in the above expression, for instance, gives ab + c, which is not the same. You should also recall that (a + b)c = ac + bc,

a(b − c) = ab − ac. –1–

If more than one set of brackets is present the inner ones are removed first. Now go through the following worked Example. Example 1. Remove the brackets in the following and simplify the resulting expressions by combining like terms: (i) a − 3 − 2(4b − 2 − 3(3a − 2b)), (ii) (x − 2)2 . (i)

= a − 3 − 2(4b − 2 − 9a + 6b) = a − 3 − 8b + 4 + 18a − 12b = 19a − 20b + 1.

(ii)

= (x − 2)(x − 2) = x(x − 2) − 2(x − 2) = x2 − 2x − 2x + 4 = x2 − 4x + 4.

In determining the above solution note that it was essential to use the following rules: (positive) × (positive) = positive,

(positive) × (negative) = negative

(negative) × (positive) = negative,

(negative) × (negative) = positive.

The process of removing brackets is also commonly called expanding brackets. ***Do Exercise 1.

Remove brackets and simplify the following

(i)

x − 1 + 2(3x − 8),

(v)

(x − 1)(x + 2)(x − 3),

(ii)

2a + 3(c − a) − 2(b − a),

(iii)

(x + 7)(x − 5),

(iv)

(x + 4)2 ,

(y − 3)3 .

(vi)

2. Factorisation This is the reverse of the process of removing brackets. You are given an algebraic expression and you try to rewrite it as the product of factors. In many cases you may have an additional implicit requirement that the factors contain only integers (positive or negative) - you will not always be able to find factors satisfying this constraint. With very simple expressions the factorisation is straightforward: e.g.

5x − 10 = 5(x − 2),

t2 + t = t(t + 1).

Sometimes you must try to factorise a quadratic expression ax2 + bx + c, where a, b and c are numbers, into a product of linear factors. For example, if you are asked to factorise the quadratic expression x2 − 4x − 21 then you seek integers m and n such that x2 − 4x − 21 = (x − m)(x − n). Since the right-hand side can be expanded to give x(x − n) − m(x − n) = x2 − nx − mx + mn = x2 − (m + n)x + mn, comparison with the original quadratic shows that you require (i)

m + n = 4,

(ii)

mn = −21.

There are eight possible ways of satisfying condition (ii) with integers: m = 1, n = −21; m = 3, n = −7; m = 7, n = −3; m = 21, n = −1; m = −1, n = 21; m = −3, n = 7; m = −7, n = 3; m = −21, n = 1. It is easily seen that only the third pair of numbers satisfies condition (i). Hence, the required factorisation is x2 − 4x − 21 = (x − 7)(x − (−3)) = (x − 7)(x + 3). When the coefficient of x2 is not unity then the above approach must be slightly modified. For the quadratic 2x2 + 5x − 3 you write 2x2 + 5x − 3 = (2x − m)(x − n) –2–

and, after expanding the right-hand side and comparing coefficients, require 2n + m = −5,

mn = −3.

It relatively easy to deduce that the above pair of equations has solution m = 1, n = −3, so the appropriate factorisation is 2x2 + 5x − 3 = (2x − 1)(x + 3). In many situations the quadratic does not factorise into linear factors with integer coefficients. Show, for instance, that x2 + x + 1 cannot be written in the form (x − m)(x − n) for any integers m and n. With practice you will be able to spot the linear factors and write them down, although it is advisable to remove the brackets in your answer to verify that you do indeed get back to the original expression. That is to say, you should verify that your answer is correct. ***Do Exercise 2. (i)

2

x + 7x + 12,

Factorise (ii)

x2 − 2x − 3,

(iii)

x3 − 25x,

(iv)

8 + 2x − x2 ,

(v)

2x2 − 3x − 2.

3. Algebraic fractions The rules for adding, subtracting, multiplying and dividing arithmetic fractions carry over into algebra. It is important you express fractions in their simplest form, by cancelling common 12x3 both have factors of 3 and x, and cancelling factors. The numerator and denominator of the quotient 3x these gives 12x3 = 4x2 . 3x This answer applies only when x 6= 0, since division by zero is not allowable. As a further example of simplifying fractions you can see that 3x2 3x3 = . 3x2 + x 3x + 1 Note that the denominator 3x2 +x = x(3x+1) and so it is not divisible by 3 or x2 , just x or 3x+1. In general a fraction (either arithmetic or algebraic) is expressed in its simplest form by factorising the numerator and denominator separately and then cancelling any common factors. x2 − 2x − 15 x+3 On factorising the numerator the above quotient can be written

Example 2.

Simplify

(x − 5)(x + 3) = x − 5, x+3 provided x 6= −3. The latter restriction on x is necessary because in obtaining the answer x − 5 the common factor x + 3 was cancelled by dividing both numerator and denominator by this factor. Clearly this is only acceptable provided x + 3 6= 0, i.e. x 6= −3, and hence the quotient in Example 2 simplifies to x − 5 only if x 6= −3. When x = −3 the original fraction has the value 0/0 which has no meaning. To add or subtract algebraic fractions you must first write each fraction in its simplest form by cancelling any common factors. Next you have to determine the simplest algebraic expression that has the given denominators as its factors - i.e. the lowest common denominator (LCD). Each fraction is then written with this LCD as its denominator, before combining.

–3–

Example 3.

Express as single fractions

(i)

x y + , y x

(ii)

3 2x + 1 + 2 (x + 1)2 x + 3x + 2

(i) LCD of denominators y and x is xy, so xx yy x2 + y 2 x y + = + = . y x xy xy xy (ii) This time the LCD of denominators (x + 1)2 and x2 + 3x + 2 (= (x + 1)(x + 2)) is (x + 1)2 (x + 2), hence 3(x + 2) (2x + 1)(x + 1) 3 2x + 1 = + + (x + 1)2 (x + 2)(x + 1) (x + 1)2 (x + 2) (x + 2)(x + 1)2 3x + 6 + (2x2 + 2x + x + 1) = (x + 1)2 (x + 2) 2x2 + 6x + 7 . = (x + 1)2 (x + 2) Note that the numerator in the above expression does not factorise into linear terms with integer coefficients. Example 4.

Express as a single fraction

x + 2 can be expressed

(x + 2)(2x − 1) 3 2x2 − x + 4x − 2 + 3 3 = + = 2x − 1 2x − 1 2x − 1 2x − 1 2x2 + 3x + 1 , = 2x − 1

which can be written

(i)

1 1 + , x x−1

3 2x − 1

x+2 and the LCD of 1 and 2x − 1 is 2x − 1. Hence 1

x+2+

***Do Exercise 3.

x+2+

(2x + 1)(x + 1) . 2x − 1

Express as a single fraction (ii)

1 2 , + (t + 2)2 t+2

(iii)

x+

1 , x2 + 1

(iv)

2x 3x − x2 + x − 2 (x − 1)2

4. Formulae Physical quantities are often related to each other using formulae, e.g. the area of a circle, A, is related to its radius, R, through the formula A = πR2 . To evaluate a formula you must substitute numbers in place of the symbols, remembering that you must be very careful about the units you use. In the formula A = πR2 , A is called the subject of the formula since it appears by itself on one side of the formula, and nowhere else. It is often necessary to rearrange a formula so that a different variable becomes the subject. During this rearrangement, or transposition, you can carry out a number of different operations to the formula provided you do the same thing to both sides. The possible operations are addition, subtraction, multiplication, division (by a non-zero quantity) and taking ’functions’ of both sides (e.g. taking the square or square root). Example 5. An object with initial speed u and constant acceleration a travels a distance s in time t given by s = ut + 12 at2 . Rearrange the formula so that the subject is a. Subtracting ut from both sides gives s − ut = –4–

1 2 at , 2

then multiplying both sides by 2 leads to 2(s − ut) = at2 . Finally dividing both sides by t2 implies a=

***Do Exercise 4. (i) (iii)

(iv)

A = πr

2

2(s − ut) . t2

Rearrange the following formulae to obtain expressions for the stated variable

for r,

(ii)

v 2 = u2 + 2as

1 1 1 = + for R, R R1 R2 s l for l. T = 2π g

for s,

(first express right-hand side as a single fraction),

5. Indices Products of a number can be written compactly using indices, or powers. For example, 81 = 3 × 3 × 3 × 3 and you write 81 = 34 . The basic rules for manipulating indices are am an = am+n , am = am−n , an a−m =

1 am

1 ), a−m

or (am =

(am )n = (an )m = amn , 1

1

m

(am ) n = (a n )m = a n . Note that

(am cn )3 = (am cn )(am cn )(am cn ) = am am am cn cn cn = am+m+m cn+n+n = a3m c3n ,

and in general one obtains the result

Example 6.

(am cn )p = amp cnp .

3

Evaluate 16 2

Using the above rules

3

1

16 2 = (16 2 )3 = 43 = 4 × 4 × 4 = 64.

Check the answer using the xy button on your calculator. Example 7. a2 b 3 c 1

acd 2


12

Simplify 3

=

1

a2 b 2 c 2 1

acd 2

a2 b 3 c


12

1

acd 2 3

1

1

3

1

1

= a2−1 b 2 c 2 −1 d− 2 = ab 2 c− 2 d− 2 = –5–

3

ab 2 1

1

c2 d2

.

***Do Exercise 5.

Simplify 1

(i)

3

(81) 4 , 

(v)

a−4 a−1

(ii)

1

a 2 a2 ,

 13 ,

(vi)

(iii)

3

(a− 3 ) , a

(iv)

1

2

a 3 b 3 c(a2 c4 ),

x(xa )2 .

6. Rationalising a denominator involving a square root This section is concerned with rewriting 1 √ so that no square roots appear in the denominator. The method for achieving an expression such as a− b √
this usesthe result (x + y) (x − y) = x2 − y 2 = x2 − xy + yx − y 2 . Putting x = a and y = b gives   √ √ √ a + b a − b = a2 − b. Hence, multiplying numerator and denominator by a + b you obtain  √  √ 1 a + b a+ b 1    √ = . √ √ = 2 a −b a− b a− b a+ b

1 √ . 1+ 2 √ Multiply the numerator and denominator by 1 − 2 to give Example 8.

Rationalise the denominator in

√
√ √ √ 1 1− 2 1− 2 1− 2 √
= √
= = −1 + 2. 1−2 −1 1+ 2 1− 2 Check your answer again using your calculator. ***Do Exercise 6. Rationalise the denominator in 6 1 √ , √ . (i) (ii) √ 1− 7 2+ 3 7. Linear equations Physical quantities are related by equations. You often need to solve an equation in an unknown quantity, say x. That is to say you need to find the value, or values, of x which will make both sides of the equation equal. The simplest equations to solve are linear equations, in which x appears only to the first power, that is as 1 x, and not as x3 , x 4 , etc. The standard form is ax + b = 0, but linear equations often appear in non-standard form. Example 9.

Solve the equation 3x + 10 = 4(1 − x).

Multiplying out the right-hand side (RHS) 3x + 10 = 4 − 4x. Adding 4x to each side gives 3x + 4x + 10 = 4, –6–

and then subtracting 10 from both sides 7x = 4 − 10 = −6. Dividing both sides by 7 then leads to the solution x = −6/7. 8. Quadratic equations

Quadratic equations have the standard form ax2 + bx + c = 0, where a 6= 0.

If the quadratic expression on the left-hand side can be factorised then it is easy to write down the solution of the quadratic equation. Example 10.

Solve 6x2 − 13x − 5 = 0.

The left-hand side factorises to give (3x + 1)(2x − 5) = 0, and hence either 3x + 1 = 0 or 2x − 5 = 0. Solving these linear equations leads to x=−

5 1 or x = . 3 2

There are two values of x, therefore, which satisfy the above quadratic equation. Sometimes you may be given a quadratic which you find difficult, or impossible, to factorise. In these situations you can always use the result: the solutions of the equation ax2 + bx + c = 0 (with a 6= 0) are √ −b ± b2 − 4ac . x= 2a Comparing the general equation with that considered in Example 10 it is seen that a = 6, b = −13, c = −5. 1 5 Substitute these values into the general formula and verify the solutions x = − , x = . 3 2 Note that if b2 − 4ac > 0 then its square root is also a non-zero real number, with the plus and minus signs in the general formula leading to two distinct real roots. When b2 = 4ac there is only one value of x which satisfies the equation. For instance, the equation x2 − 6x + 9 = 0 can be written (x − 3)2 = 0, with solution x = 3 (twice). x = 3 is known as a repeated root. Finally, if b2 − 4ac < 0 then its square root is a complex number (see a later module). In this situation the quadratic is said to possess complex roots. A third method for solving quadratic equations involves completing the square. Since (x + a)2 = (x + a)(x + a) = x2 + ax + ax + a2 = x2 + 2ax + a2 , by subtracting a2 from both sides it follows that (x + a)2 − a2 = x2 + 2ax. Hence, by choosing a = 4, the expression x2 + 8x can be written x2 + 8x = (x + 4)2 − 42 = (x + 4)2 − 16. –7–

Confirm the above answer by removing the bracket. Completing the square is used in the example below. Example 11.

Solve x2 + 6x + 6 = 0 by completing the square.

The method consists of rewriting the first two terms of this quadratic by completing the square. Using the result proved above it follows that x2 + 6x = (x + 3)2 − 32 = (x + 3)2 − 9. Hence

x2 + 6x + 6 = (x + 3)2 − 9 + 6 = (x + 3)2 − 3,

and so the given quadratic equation can also be expressed (x + 3)2 − 3 = 0. Adding 3 to both sides gives

(x + 3)2 = 3

which, after taking square roots, becomes

√ x + 3 = ± 3. √ √ √ The two solutions, therefore, are x = −3 ± 3 (i.e. x = −3 + 3 and x = −3 − 3). You can verify the above solutions by using the general formula for the solution of a quadratic.

It is important to note that the technique of completing the square is also very useful in other areas. For instance, you will certainly use the technique for some integration problems. 9. Solution of simultaneous linear equations A linear equation in two unknowns x and y has the form ax + by = c, where a, b, c are constants and a and b are non-zero. If you have two such equations then you have two simultaneous linear equations in two unknowns. A set of values for x and y which satisfies both equations is called a solution of the pair of equations. To solve the system of equations it is necessary to eliminate one of the unknowns by adding or subtracting appropriate multiples of the equations (see below). Example 12.

Solve the set of equations 4x + 2y = 5 5x − 3y = −2.

First you must decide which unknown to eliminate. Suppose the variable y is chosen, then it is necessary to ensure the coefficients of the y terms in both equations have equal magnitude. An obvious way to achieve this is to multiply the first equation by 3 and the second equation by 2: 12x + 6y = 15 10x − 6y = −4. Adding the above equations leads to 12x + 10x = 15 − 4, which gives 22x = 11, or x = 12 . Substituting back into the first given equation leads to   1 + 2y = 5. 4 2 This implies 2 + 2y = 5, i.e. 2y = 3, y = 32 . Hence, the solution set is x = 12 , y = 32 . You should always verify your answer by substituting back into both equations stated in the question.

–8–

10. Solving equations using graphs A variety of different types of equation can be solved using graphs. By drawing a graph (or graphs) you can see whether the given equation has a solution and, if so, obtain approximate values for each solution. Note that a graphical method may not give you precise values for these solutions although the accuracy of the answer can usually be improved by drawing better graphs, using the zoom facility on graphical calculators to home in on the points of intersection or by algebraic means. Example 13.

Solve, by using graphs, (i) 3x + 10 = 4(1 − x), (ii) 6x2 − 13x − 5 = 0,

(iii) 4x + 2y = 5, 5x − 3y = −2, (the equations solved algebraically in Examples 9, 10 and 12). (i) In this case you can plot y = 3x + 10 and y = 4(1 − x) and then look for the point of intersection. Alternatively, you rearrange the terms as in Example 9 to obtain 7x + 6 = 0, plot y = 7x + 6 and look for the value (or values) of x where it intersects the x-axis (where y = 0). y

.. ...... ....... ... .... ... . ... ... .... ... ... .. ... . ... . .. ... ... ... ... ....... ... .. ...... ... . ....... ....... . ...... ....... .. . . ...... . . . . . . . ...... .. ... ....... ...... ... ....... ... ...... ... ...... ...... ... ....... ... ...... ... ............ . ... . ...... .. ...... ...... .... ... ....... ...... .. ........... ...... ...... ...... ... ............ ...... . . . . . . . ...... ... . ... ......... ... ...... ...... ........ ..... ...... . ....... .. ...... ............ .... .... . ............ . . . ... .. ......... ...... ........... . . . . . . . . . . . ...... .... ....... ........ .. ...... .......... ....... ... .............. ....... ...... ... . . . . . . ... .......... . . .. ..... . ...... . . . . .. . . .... ... .. . ................................................................................................................................................................................................................................................................................. . . . . ....... . . . . . .... . ... . . ...... .... . . . . . . . . . . . ...... ... . .. . . . . . . ...... .. ...... .... ... ...... ... ... ...... ... ...... ... ... ...... . . . . ...... . .. . . . . ...... .. .. . ...... . . ...... .. .. . . . ...... ... .. . . . ...... . . .. ...... . . . . ..... .. .. . . . .. .. . . . . .. . . . . . ... ..... .. .... .. ... . ......

y = 7x + 6

The lines y = 3x + 10 and y = 4(1 − x) intersect near x = −0.8.

20

A more accurate graph would provide a better solution.

y = 3x + 10

10

In the alternative method the graph of y = 7x + 6 is again a straight line, and it is easily seen that y = 7x + 6 = 0 when x is approximately −0.8.

x

−3 −2 −1

1

2

3

y = 4(1 − x)

−10

−20

(ii) y

.. ...... ........ ... .... ... ... .... .. ... ... ... ... .. .. . . .. ... .. .. .. .. .... .. .. .. . . . .. .. .. .... .. .. ... .. .. .. .. ... .. .. . . . . .. . .. .... .. ... ... ... ... . .. ... ..... ... . ... ... ... .... ... ... ... ... ... ... ... .. . . . . .. .. .... .. .. .. ... .. .. .. .. .. .. .. ... . . .. ... . .... . .. ...................................................................................................................................................................................................................................................................... . . . . . . .. ... .... .. ... .. .. . ... ... ... ..... ... ..... .... ... .. . ....... ... ... ... ... .... ... ... .... ... ... ... ... . . .... .... ..... .... ......... ............. .... ...... ... ... ... ... ... ... .... .. . .....

20

10

x

−3 −2 −1

1

2

3

−10

−20

–9–

To find a solution of 6x2 − 13x − 5 = 0 draw a graph of y = 6x2 − 13x − 5 and look for the values of x, if any, for which y = 0 (i.e. where the graph intersects the x-axis). The graph suggests that approximate solutions for x are −0.3 and 2.5. By zooming in and expanding the graph near these points it is possible to achieve more accurate solutions.

(You should observe from the graph that y never becomes −20, for instance, and hence the equation 6x2 − 13x − 5 = −20, which can be written 6x2 − 13x + 15 = 0, has no real solutions (or roots).) (iii) In this case you look for the point of intersection of the two graphs 4x + 2y = 5 and 5x − 3y = −2. First draw the graphs of both these equations. .. ...... .......... ... ... .... ... ... .. ... ... ... ... .... .. . ... ..... . .. ... .. ... ... ... ... ..... ... .... .. . . ..... .. ... .... ... .. ... ... .. ....... .... ... . . .... .... ... ..... .. ... ... ... ... ... ...... ... ... ... ... ..... ..... . . ... . . ..... .. ... .. ... ... ...... ... . . . ... ....... . ... . .. . ... . .. ... ... . . . .. ... . . . . . . ............................................................................................................................................................................................................................................. . . . ... . . . . ... . . . .. . . . . . . ... .. .... ... ... ... ... ... ... ... ... ... .. . . . . ... ..... .. . . . ... . .. . . ... . . .. ... ... . . . . ... . . . . ... . .. . . . . ... .. ... ... . . . . . . ... . . ...... .. ... . . ... . .. . . . . ... .. ... . ... . . . . . ... . . . .. ... . . . . ... . .. . . . . ... .. ... .. . . . ...... .... ... ..

y

3

5x − 3y = −2

2

Clearly the lines intersect at approximately x = 0.5, y = 1.5.

1

These values can again be confirmed by drawing more accurate graphs

x

−2

−1

1

2

−2 −2

4x + 2y = 5

−3

It is worth noting that if the two graphs are parallel then they do not intersect and the system has no solution. This situation would arise if you were asked to solve the equations x + y = 1 and 2x + 2y = 5, for instance. Graphical methods can easily be extended to more complicated equations. ***Do Exercise 7. (i)

4 − 2x = 3,

***Do Exercise 8. (i)

(ii)

2x + y = 3,

(ii)

(i)

x3 = 2 − x,

4(1 − x) = −3(2x − 1),

(iii)

(iv)

2 1 = . x x+1

x2 − 9x + 14 = 0,

(iii)

x2 − 9x − 8 = 0,

(iv)

8 = 6 − x. x+3

Solve the following sets of equations (ii)

4x − y = 3, ***Do Exercise 10.

2 = 5, x+1

Solve the equations

x2 − x − 12 = 0,

***Do Exercise 9. (i)

Solve the equations

3x + 4y = 12, 9x + 2y = −9,

(iii)

4x − 2y = 5 5x + 3y = −2

Solve the following equations (using graphs in both cases to illustrate your results): (ii)

x3 = x.

11. Inequalities The statement that the number a is less than the number b is written a < b. This means the same as the statement b is greater than a, which is written b > a. Every inequality, that is a statement that one number is less than another, may be written using either the symbol < or the symbol >. An inequality a < b has a geometric meaning; if a and b are represented by points on a number line with – 10 –

positive numbers to the right of the origin, then the point with coordinate a lies to the left of the point with coordinate b. . . . ......................................................................................................................................................................................................................................................................................................

0

a

b

The symbol < in a < b is often called the sign (or direction or sense) of the inequality - reversing the sign leads to a > b. The statement a ≤ b means that either a < b or a = b. Similarly a ≥ b means that either a > b or a = b. (You read the symbols ≤ and ≥ as “less than or equal to” and “greater than or equal to” respectively.) The statement “a is positive and b is negative” is equivalent to a > 0 and b < 0; whereas “a non-negative” and “b non-positive” may be written a ≥ 0 and b ≤ 0 respectively. If a and b are two distinct numbers then one of them must be greater than the other; this is the first basic rule for inequalities, rule I below. If a and b are both positive and given by non-terminating decimals you pick the greater number by successively comparing digits. For example, if a = 12.47325 and b = 12.475111, then a < b since the first digit in which they differ is the third digit after the decimal point and that digit is greater for b. A negative number is less than any positive number. If a and b are both negative it is useful to refer to the number line if you are in any doubt about which is the greater. For example, −3.18 > −7.23. . . . ................................................................................................................................................................................................................................................................................................................................................................................................................................................. . . .

−7.23

−3.18

0

When working with inequalities there are a number of basic rules. I. If a and b are two numbers, then one of the three statements a = b, a < b, a > b is true and the other two are false. II. If a < b and b < c then a < c. III. If a < b then a + c < b + c (i.e. adding or subtracting the same quantity to both sides of an inequality leaves the inequality sign unchanged). IV. If a < b and c > 0 then ac < bc (multiplying or dividing both sides of an inequality by the same positive quantity does not change the sign of the inequality). V. If a < b and c < 0 then ac > bc (multiplying or dividing both sides of an inequality by the same negative quantity reverses the sign of the inequality). The above rules can be illustrated using the number line. Rule III is equivalent to translating the figure to the right or left, depending on the sign of c - translation preserves the order. Rule IV is equivalent to magnifying (or shrinking) the number line - again the order is unchanged. However, you can easily show that Rule V implies a magnification and reflection, with the order (and hence sign of inequality) reversed. 12. Solution of inequalities To solve an inequality you must find all numbers x for which the inequality is true. The actual steps taken in determining the solution are similar to those used in solving the corresponding equation, but you must be careful in dealing with the sign of the inequality. Example 14.

Solve 3 − 2x < 4x − 5.

First you must move all terms involving x to the left-hand side and all numbers to the other side. To achieve this you must remove all x terms from the right-hand side (by subtracting 4x) and all numbers – 11 –

from the left-hand side (by subtracting 3). Hence (3 − 2x) − 4x − 3 < (4x − 5) − 4x − 3 i.e. − 6x < −8. 1 To obtain the inequality for x you must multiply by − (or divide by −6), remembering from Rule 6 V that you must also reverse the inequality −8 −6x > −6 −6 4 i.e. x > . 3 The solution is a set of numbers (called the solution set) - it consists of all numbers greater than

4 . 3

If a and b are numbers such that a < b then the statement a < x < b is equivalent to the two statements x > a and x < b. In writing the double inequality it is essential that a < b. Thus, the inequalities x > 1 and x < 2 can be expressed by the double inequality 1 < x < 2, but it is not possible to write x > 2 and x < −1 as the double inequality 2 < x < −1 (since 2 is not less than −1). You may find it helpful to observe that acceptable double inequalities, such as 1 < x < 2, represent a single region on the number line, whereas the inequalities x > 2 and x < −1 give two distinct sections of the line. The solution of a double inequality is equivalent to the solution of two simultaneous inequalities, each of which is solved using the method discussed above. Example 15.

Solve x − 6 < 2x − 5 ≤ x − 3.

This means that you must find the values of x which satisfy the inequalities x − 6 < 2x − 5 and 2x − 5 ≤ x − 3. Each of these is treated separately: (x − 6) − x < (2x − 5) − x (x − 6) − x + 5 < (2x − 5) − x + 5 i.e. − 1 < x. (2x − 5) − x ≤ (x − 3) − x (2x − 5) − x + 5 ≤ (x − 3) − x + 5 x ≤ 2. Hence the double inequality implies x > −1 and x ≤ 2. The solution set therefore consists of all numbers x such that −1 < x ≤ 2. The absolute value | a | of a number a is defined by  a if a ≥ 0 |a|= −a if a < 0. Thus | 8 |= 8, | −8 |= −(−8) = 8. Note therefore that | a |> 0 if a 6= 0, and | a |= 0 if a = 0. It also follows that |ab| = |a| |b|, 1 1 , = a |a| a |a| . = b |b| |a| is also often called the modulus of a or the magnitude of a. – 12 –

The absolute value has a geometric interpretation and it is the distance from the origin to the corresponding point on the number line. Hence |a| < 4 means −4 < a < 4, i.e. ←−− |a| −−→

... ... ... ... ... ... ... ... . ................................................................................................................................................................................................................................................................................................................................................................................

−4

a

0

4

The quantity |a − b| is the distance between the points a and b on the number line, whatever the numerical values of a and b. The corresponding graphs, of course, may vary: e.g.

if a < 0, b > 0

if a > 0, b > 0

←−−−−−−− |a − b| −−−−−−−→

←−−− |a − b| −−−→

.. ... ... ... .. ... ... ... . . . .............................................................................................................................................................................................................................

a

Example 16.

0

.. .. .. ... .. .. ... ... . . .............................................................................................................................................................................................................................

a

0

b

b

Solve |x − 2| < 5.

From the comments above the inequality means −5 < x − 2 < 5. This is a double inequality which is equivalent to −5 < x − 2 and x − 2 < 5. Adding 2 to both sides of the first of these inequalities gives −3 < x whereas adding 2 to the second inequality leads to x < 7. Hence the solution set consists of all numbers greater than −3 and less than 7; i.e. −3 < x < 7. As discussed above the result has a geometrical interpretation. The quantity |x − 2| is the distance of the point x on the number line from the point 2, and this distance is less than 5. ←−−−−−− 5 −−−−−−→←−−−−−− 5 −−−−−−→

... ... ... ... ... .. ... ... ... .. .. .. .............................................................................................................................................................................................................................................................................................................

−3

0

2

7

Hence the point x must satisfy −3 < x < 7. Inequalities containing a quotient must be solved with care as shown below. 2−x < 4. 3+x The denominator can be removed by multiplying both sides by 3 + x. However, the quantity 3 + x can be positive or negative and the sign of the inequality depends on which of those values it takes. Both situations must be treated separately. Example 17.

Solve

Case (i) 3 + x > 0. In this case multiplying both sides by 3 + x gives 2 − x < 4(3 + x) 2 − x < 12 + 4x 2 − 12 < 4x + x i.e.

− 10 < 5x x > −2. – 13 –

It is important to note that when x > −2 it follows that x + 2 > 0, (or x + 3 > 1) and hence the necessary condition 3 + x > 0 is always satisfied. Case (ii) 3 + x < 0. Now you can again multiply both sides by 3 + x, but after doing so the inequality must be reversed. This time, therefore, you obtain 2 − x > 4(3 + x) 2 − x > 12 + 4x −10 > 5x i.e. x < −2. However, case (ii) requires 3 + x < 0, i.e. x < −3, and the latter inequality and x < −2 are both satisfied only if x < −3. The final solution to the original inequality, therefore, is the set of values of x satisfying x < −3 and x > −2. 13. Solving inequalities using graphs To solve an inequality of the form f (x) < g(x) you can draw graphs of f (x) and g(x) and investigate where the first lies below the second. Generally this will be achieved by finding the points of intersection, which involves solving the equation f (x) = g(x) (which has been considered in previous sections). Example 18.

Solve graphically the inequality x2 − 4x + 1 < 3.

The graphs of y = x2 − 4x + 1 and y = 3 are shown on the diagram and you can see that the solution of the inequality consists of all values of x between the points P and Q. y

.. . ..... .. .. ....... .. .. .. .... ... .. .. .. . . . . . . . . . . . .. .. .. .... .. .. .. ... .. .. .. ... .. .. ... ... .. ................................................................................................................................................................................................................................................................... ... .. .. ....... ..... ... .. .. . .. ... ... ... .... .... . .. . .. ..... .... .. .. . ........ ... . ... ..... ... ... .. .... ... .. ..... ..... .. .. ... .. . ... ... .. ... .. .. ... .. .. .............. .. .. . . . ... . ........ ... .. ... ... .. .. ... .. .. .. . . ... ... ... .... .......................................................................................................................................................................................................................................................................................... . . . . . . . .... ... ... . .. . ... .... ... ... ... ... ... ... ... .. .. ... .. . ............ . ... ... ... .. .. ... ... .. ... ... ... .. ... . ... . .. ... .......... ... ... ... ... ... .... ... .... ... . ... . .... . ... ..... .... .... ...... ... ...... .......... ....................... ........... ... .. ..

3

x

−1 P

Q 5

−3

P and Q are determined by solving x2 − 4x + 1 = 3, which reduces to x2 − 4x − 2 = 0. Using the formula gives p √ √ √ 4 ± ((−4)2 − 4(1)(−2)) 4 ± 24 4±2 6 = = = 2 ± 6, x= 2(1) 2 2 √ √ and so the solution of the inequality is the set of all x satisfying 2 − 6 < x < 2 + 6. Example 19.

Use graphical means to solve the inequality in Example 17.

2−x and y = 4. The first graph is not easy to obtain 3+x but from plotting points, using a graphics calculator or using methods discussed in a later module on graph plotting you can find the graph is in two parts as shown below:

Here you need to draw the graphs of y =

– 14 –

y

... .. ..... ....... .. .. .... ... ... .... . ... .... .......... .. .. .. .... ... .. ... ... ... ... ... .. ... ... ... ... .. . ... . ... ... ... .. .. ... .. .. ... ... ... ... ... . . ................................................................................................................................................................................................................................................... . .. .. ... ... ... ..... .. ... ... ........ . . ... ...... ....... ... ... ... ......... . .. . . ................ ... .... .. .................... . ............................................................................................................................................................................................................................................................................................................. . . ... .... ... . . ... .... ... .... .. ............. ... ......... ... ........ . . .... ...... . .. ..... .. .... . .... .... . .. . ... . ... ... .. ... . ... ... . . ... ... . .... ... ... .. ... ... ... ... ... .... .. ... . .. . ... ... . . . ....... ... .. ... .... ... . ...

10

4

y=4

x

−6

−3

P

3

−10

The straight line y = 4 has also been added to the figure. You can easily observe that the solution of the inequality consists of all values of x to the right of P and to the left of the asymptote x = −3. 2−x From solving = 4, (i.e. 2 − x = 4(3 + x) = 12 + 4x, 5x = −10, x = −2), you know that P is 3+x situated at x = −2. Hence the solution set is all values of x satisfying x < −3 and x > −2 (which agrees with the earlier result!). ***Do Exercise 11. (i) (iv)

6x + 5 ≥ x − 5, 2x2 − 2 < x2 − x,

14. Partial fractions degree n has the form

Solve the inequalities (ii)

−5 < x − 4 < 2 − x, (v)

1<

3x − 1 < 2, x−3

(iii) (vi)

x + 10 < 2x − 5 ≤ x − 3, |2x − 5| < 9,

(vii)

|4 − 3x| ≤ 3.

Polynomial functions arise in many engineering situations. A polynomial of an xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 ,

where an , · · · a0 are all constants. The degree is the highest power occurring in the polynomial. Hence, x2 + 1 is a polynomial of degree 2 and x4 + x2 + 1 is a polynomial of degree 4. In solving problems it is sometimes necessary to consider a quotient of polynomials (i.e. one polynomial divided by another). Such quotients can be split into a sum of much simpler fractions, called partial fractions, and this splitting is very important, for instance, in determining solutions to some integration problems and in applying Laplace transform methods to differential equations. Given a quotient of polynomials in which the denominator has degree d and the numerator has degree n then the quotient, or fraction, is proper if n < d and is said to be improper if n ≥ d. It can be shown that any polynomial with real cofficients can be factorised into a product of linear and quadratic factors (with all coefficients real). In particular, therefore, the denominator can be expressed as a product of such factors. For the moment consider only proper fractions. Next, factorise the denominator, if it is not already in the required form, so that it consists of a product of linear and quadratic factors. Each factor produces a partial fraction of a particular form according to the following rules:

– 15 –

factor in denominator

partial fraction

ax + b

(linear)

A ax + b

(ax + b)2

(repeated linear)

B A + ax + b (ax + b)2

ax2 + bx + c

(quadratic)

Ax + B ax2 + bx + c

(ax2 + bx + c)2

(repeated quadratic)

Cx + D Ax + B + 2 2 ax + bx + c (ax + bx + c)2

In all cases the unknown constants A, B, C... are determined by evaluating an identity at some chosen values of x or by equating coefficients, or from using some combination of the two methods (as shown in the examples below). Example 20.

Express as partial fractions

x−5 x2 + 2x − 3

The denominator can be factorised to give (x + 3)(x − 1), and then there is a partial fraction corresponding to each linear factor. Hence, A B x−5 = + , (x + 3)(x − 1) x+3 x−1 where A and B are constants. Rewriting the right-hand side using a common denominator gives A(x − 1) + B(x + 3) x−5 = , (x + 3)(x − 1) (x + 3)(x − 1) and multiplying both sides by (x + 3)(x − 1), or by equating numerators (since the denominators are identical), leads to the equation x − 5 = A(x − 1) + B(x + 3). The latter equation must hold for all values of x, and in these situations is often called an identity. Substituting particular values for x into the equation gives linear equations in A and B (most of which will be related to the others). Obviously you can save a lot of effort by choosing suitable values for x and with linear factors it is usually best to choose values which ensure the linear factors are zero, in turn. For the above, therefore, the convenient choices are x = 1 and x = −3. Consider these in turn: x=1

1 − 5 = A(0) + B(4) −4 = 4B,

x = −3

i.e.

B = −1.

− 3 − 5 = A(−3 − 1) + B(0) −8 = −4A,

i.e.

Thus the partial fractions are x2

A = 2. 2 1 x−5 = − . + 2x − 3 x+3 x−1

– 16 –

Example 21.

Express as partial fractions

The denominator factorises again:

x2

x + 6x + 9

x2 + 6x + 9 = (x + 3)2 , so A B x = + , x2 + 6x + 9 x + 3 (x + 3)2

and multiplying both sides by (x + 3)2 leads to the identity x = A(x + 3) + B. With a linear factor it is convenient to choose x = −3, in which case −3 = A(0) + B,

i e. B = −3.

The remaining coefficient could be found by substituting a different value for x (for example, with x = 0, 0 = A(3) + B, 3A = −B = −(−3) = 3, A = 1). Alternatively you can compare coefficients of powers of x on both sides of the identity x = A(x + 3) + B. For the x term, equating coefficients gives 1 = A, as above. From the constant terms (i.e. x0 terms) it follows that 0 = 3A + B, which is identically satisfied by A = 1, B = −3, as expected. The final result, therefore, is 1 3 x = − . x2 + 6x + 9 x + 3 (x + 3)2 Let us now consider a more complicated example. Example 22.

Express as partial fractions

(x2

3x2 + 5x + 2 + 4x + 5)(x − 1)

The quadratic x2 + 4x + 5 does not split into a product of linear factors with real coefficients, so the denominator is already in its simplest form. The rules stated earlier tell you that for this example the appropriate partial fractions are A Bx + C 3x2 + 5x + 2 = + . (x2 + 4x + 5)(x − 1) x − 1 x2 + 4x + 5 Writing the right-hand side in terms of the common denominator (x−1)(x2 +4x+5) and then equating numerators, or multiplying throughout by the product (x − 1)(x2 + 4x + 5), gives 3x2 + 5x + 2 = A(x2 + 4x + 5) + (Bx + C)(x − 1). Choose x = 1 (to make the linear factor zero) 3 + 5 + 2 = A(1 + 4 + 5) + (Bx + C)(0), 10 = 10A,

A = 1.

Before equating coefficients, multiplying out the right-hand side leads to 3x2 + 5x + 2 = A(x2 + 4x + 5) + Bx(x − 1) + C(x − 1) = A(x2 + 4x + 5) + Bx2 − Bx + Cx − C. – 17 –

Equating coefficients now gives x2

3=A+B =1+B

(since A = 1)

i.e. B = 2. x

5 = 4A − B + C = 4(1) − 2 + C C = 5 − 4 + 2 = 3.

All three constants have now been calculated but comparing constant terms on both sides of the equation provides a check on your answers: 2 = 5A − C,

constant

and this equation is satisfied by A = 1, C = 3. Examples 20-22 have involved partial fractions for proper fractions. Attention is now fixed on improper fractions. Recall that the fraction (or quotient) is improper if the degree of the polynomial in the numerator, n say, is greater than the degree of the polynomial in the denominator, x4 + 1 is improper, since the numerator has degree 4 and the denominator d. For example, the fraction 2 x +1 degree 2. It can be shown that

2 x4 + 1 = x2 − 1 + 2 . x2 + 1 x +1

The corresponding general result can be written polynomial degree ≤ (d − 1) polynomial degree n = polynomial degree (n − d) + . polynomial degree d orig. denominator degree d Note that for the specific example and the general result the final term on the right-hand side is a proper fraction. Considering the right-hand side of the general expression, both the first term and the numerator in the final term can be determined by long division, but partial fractions would still usually be needed for this final term. An alternative, and often simpler, approach is not to use long division but determine the polynomial of degree (n − d) as part of the partial fraction process. This second method is illustrated below. (2x + 1)(x + 1) in terms of partial fractions. 2x − 1 Here the numerator is quadratic, the denominator is linear and hence the general result above gives

Example 23.

Express

const (2x + 1)(x + 1) = linear + . 2x − 1 2x − 1 Note that the denominator of the final term above is linear, and hence its numerator (which is of lower order) must be a constant. To be more precise C (2x + 1)(x + 1) = Ax + B + , 2x − 1 2x − 1 and, after multiplying both sides by 2x − 1, the identity obtained is (2x + 1)(x + 1) = (Ax + B)(2x − 1) + C. Choosing x = 12 ,

     1 1 +1 + 1 = (Ax + B)(0) + C 2 2 2 – 18 –

  3 = C, 2 2

C = 3.

Equating coefficients in the identity: x2

2 = 2A,

constant

1 = −B + C,

A = 1. B = C − 1 = 2.

2 + 1 = −A + 2B,

[check term in x : Hence

satisfied.]

3 (2x + 1)(x + 1) =x+2+ , 2x − 1 2x − 1

which is the reverse of Example 4. ***Do Exercise 12. (i) (iv)

1 , (x − 1)(x − 2) x2 , (x − 1)2

Express in partial fractions (ii)

(v)

x , (x2 + 5x + 4)

(iii)

x3 − 1 . (x + 1)(x + 3)

– 19 –

1 , (x2 + 1)(x2 + 4)

You have now worked through the Module, so please attend a testing session as soon as possible. The following questions are similar to ones that you will be asked in the Test.

Specimen Test 1 1.

Expand (x + 1)(3 − 2x)

2.

Simplify

3.

Express as a single fraction

4.

Solve

5.

Solve x2 + 3x + 1 = 0

6.

Solve the set of equations

1

x(x 3 )4 1

x2 3 2 − x+1 x+2

3 =2 x+1

4x − y = 9 x + 3y = −1 7.

Solve 5 − x ≤ 3(x − 1)

8.

Solve |2 + x| < 3

9.

Express in partial fractions

x2 + 1 x(x − 1)

(Solutions to the Specimen Test can be found on p.27)

– 20 –

Worked solutions to Exercises 1. (i)

= x − 1 + 6x − 16 = 7x − 17

(ii)

= 2a + 3c − 3a − 2b + 2a = a − 2b + 3c

(iii)

= x(x − 5) + 7(x − 5) = x2 + 2x − 35

(iv)

= (x + 4)(x + 4) = x(x + 4) + 4(x + 4) = x2 + 8x + 16

(v)

= (x − 1)(x(x − 3) + 2(x − 3)) = (x − 1)(x2 − x − 6) = x(x2 − x − 6) − 1(x2 − x − 6) = x3 − x2 − 6x − x2 + x + 6 = x3 − 2x2 − 5x + 6

(vi)

= (y − 3)(y − 3)(y − 3) = (y − 3)(y(y − 3) − 3(y − 3)) = y(y 2 − 6y + 9) − 3(y 2 − 6y + 9) = y 3 − 9y 2 + 27y − 27

x2 + 7x + 12 = (x + 4)(x + 3) [ (x − m)(x − n) = x2 − (m + n)x + mn requires mn = 12, m + n = −7. Solution is m = −4, n = −3 or equivalently m = −3, n = −4.]

2. (i)

(ii)

x2 − 2x − 3 = (x − 3)(x + 1) [as above : m = 3, n = −1].

(iii)

x3 − 25x = x(x2 − 25) = x(x − 5)(x + 5)

(iv)

8 + 2x − x2 = −(x2 − 2x − 8) = −(x − 4)(x + 2) = (4 − x)(x + 2)

(v)

2x2 − 3x − 2 = (2x + 1)(x − 2)

[(2x − m)(x − n) = 2x2 − (2n + m)x + mn ⇒ mn = −2, 2n + m = 3, leading to the solution n = 2, m = −1.] 3. (i)

(ii)

(iii)

(iv)

LCD of x and x − 1 is x(x − 1) 1(x) x−1+x 2x − 1 1(x − 1) + = = Solution = x(x − 1) (x − 1)(x) x(x − 1) x(x − 1) LCD of (t + 2)2 and (t + 2) is (t + 2)2 1 + 2t + 4 1 2(t + 2) 2t + 5 = Solution = + = (t + 2)2 (t + 2)(t + 2) (t + 2)2 (t + 2)2 x 1 1 = + 2 +1 1 x +1 LCD of 1 and x2 + 1 is x2 + 1 1 x3 + x + 1 x(x2 + 1) + = Solution = 1(x2 + 1) x2 + 1 x2 + 1 x+

x2

x2 + x − 2 = (x + 2)(x − 1) LCD of (x + 2)(x − 1) and (x − 1)2 is (x + 2)(x − 1)2 3x(x + 2) x(2(x − 1) − 3(x + 2)) 2x(x − 1) − = 2 (x + 2)(x − 1)(x − 1) (x − 1) (x + 2) (x + 2)(x − 1)2 x(−x − 8) x(x + 8) = =− 2 (x + 2)(x − 1) (x + 2)(x − 1)2

Solution =

– 21 –

4. (i) (ii) (iii)

(iv)

A = r2 , r = ± π

A π

(positive if distance)

v 2 − u2 2a LCD of R1 and R2 is R1 R2 1(R2 ) 1(R1 ) R2 + R1 1 = + = R R1 (R2 ) R2 (R1 ) R1 R2 R1 R2 R= R1 + R2 s T2 T l l = , = 2π g 4π 2 g

v 2 − u2 = 2as,

l=

5. (i)

r

s=

gT 2 4π 2 3

3

(81) 4 = (34 ) 4 = 33 = 27 1

5

(ii)

a 2 a2 = a 2

(iii)

a−1 (a− 3 )3 1 = = a−2 = 2 a a a

1

(iv) (v) (vi)

6. (i) (ii)

7. (i)

1

2

7

2

= a 3 b 3 c(a2 c4 ) = a 3 b 3 c5  −4  13 1 1 a = (a−3 ) 3 = a−1 = a−1 a x(xa )2 = x(x2a ) = x2a+1 √ √ √ √ 6(1 + 7) 6(1 + 7) 6 (1 + 7) √ √ = = = −(1 + 7) = 1−7 −6 (1 − 7) (1 + 7) √ √ √ √ √ √ 1 2− 3 ( 2 − 3) √ √ √ = =− 2+ 3 = √ 2−3 ( 2 + 3) ( 2 − 3) 2x = 4 − 3 = 1 ⇒ x =

1 2

(ii)

2 = 5(x + 1) = 5x + 5, 5x = −3, x = −

(iii)

4 − 4x = −6x + 3, 2x = −1, x = −

(iv)

2=

8. (i) (ii) (iii) (iv)

3 5

1 2

x , 2(x + 1) = x, 2x + 2 = x, x = −2 x+1

x2 − x − 12 = (x − 4)(x + 3) = 0 ⇒ x = 4 or − 3 x2 − 9x + 14 = (x − 7)(x − 2) = 0 ⇒ x = 7 or 2 p √ −(−9) ± ((−9)2 − 4(1)(−8)) 9 ± 113 = = 9.8151; or − 0.8151. x= 2 2 8 = (6 − x)(x + 3) = 6x + 18 − x2 − 3x = 18 + 3x − x2 ⇒ x2 − 3x − 10 = (x − 5)(x + 2) = 0 ⇒ x = 5 or − 2 – 22 –

9. (i) Adding the given equations leads to 6x = 6, x = 1. Substituting this value into the first given equation then implies 2(1) + y = 3, so y = 1. Hence solution is x = 1, y = 1. (ii) Eliminate y 3x + 4y = 12 18x + 4y = −18

(first equation) (second equation × 2)

Subtracting second equation from first, −15x = 12 − (−18) = 30, which implies x = −2. Substituting this solution back into the original first equation gives 3(−2) + 4y = 12,

4y = 12 − (−6) = 18,

y=

9 18 = . 4 2

9 . 2 (iii) This time we choose to eliminate x Hence solution is x = −2, y =

(first equation × 5)

20x − 10y = 25

(second equation × 4) 20x + 12y = −8 (subtracting) − 22y = 25 − (−8) = 33 3 y=− 2 3 Substituting back into first equation in question 4x − 2(− ) = 5, 2 3 1 Thus solution is x = , y = − . 2 2

4x = 5 − 3 = 2, x =

1 . 2

10. (i) The graphs of y = 2 − x and y = x3 are shown below. Clearly they intersect at x = 1 approximately. In fact this is a solution since when x = 1 both 2 − x and x3 are equal (= 1). No other solution is possible. .. y ... y . ...... .. .. ...... ... .. ......... ... .... ... ... .... ... ... ... .... .... ... ... .. ... .... ... ... ... .... .. ... .. .... . . . . ... . .... .... .. ... ... ... .... ... .. ... .. .... ... . ... ... ... .... .. ..... ... ... ........ .... ... ..... ... ..... .. .... ... ... ... ... ... .... ... .... ... .... .. ... .... . ... ... ... ...... ... ... ... ... .... .. ... . ... .... .. ... ... ... . . ... .... .. . . .. . .. . . . . ........................................................................................................................................................................................................... ..... .. . . . . . . . ... . .. . . .... . . . ... ... .... .. . . . .. .. . . .. .. . . ... .... ... ... ... ... .... ... .... .... . .. ... ... .. .... ... .. ... .... .. ... ... .... .. ... .... .. .

. .. .......... ... .. ... ... ... .. . ... ... .. ... .. ... ... ... ... . ... .. ... .. ... ... .. .... ... . ... ... ... .... ... ... .... ... .. . ... . . .... ... ... ... ... ... .... ... ... .... .. ..... ... . . ... ... ..... ... ....... ..... ... ....... . . ... . . ... .... ... .... .. ... ... .. ... .... .. ... ...... .... .. ... .... . .......................................................................................................................................................................................................... . . . .... .... .... ...... .... . . ... ...... ... . . .. ..... ..... .. ... .... .. ... ... ........ . . ... ...... . . ... ... ... . . ... ... ... . ... . .. ... . . ... . . .. . . . ... . . . ... . .... . . . .. .... ... ... .. .... .. ... .... .. ... ... ... ... .... .

Graph for Problem 10(i)

Graph for Problem 10(ii)

x

−1

x

−1

+1

+1

(ii) Graphs of y = x3 and y = x are displayed above. These graphs intersect at x = 0, x = 1, x = −1 approximately. It is obvious by substitution that all three of the latter values are the exact points of intersection. – 23 –

Algebraically, at the points of intersection x3 = x. Hence x3 − x = x(x2 − 1) = x(x − 1)(x + 1) = 0 with solutions x = 0, +1, −1. 11. (i) 6x + 5 ≥ x − 5 6x + 5 − x − 5 ≥ x − 5 − x − 5 5x ≥ −10, ⇒ x ≥ −2 (ii) Equivalent to

−5 < x − 4

and

x − 4 < 2 − x.

First inequality (add 4 to both sides)

− 5 + 4 < x − 4 + 4,

i.e. − 1 < x.

Second inequality (add x + 4 to both sides) (x − 4) + x + 4 < (2 − x) + x + 4 which implies 2x < 6, i.e. x < 3. For this example, therefore, the solution set is −1 < x < 3. (iii) Equivalent to x + 10 < 2x − 5 and 2x − 5 ≤ x − 3. First inequality (x + 10) − x + 5 < (2x − 5) − x + 5, i.e. 15 < x. Second inequality (2x − 5) − x + 5 ≤ (x − 3) − x + 5, i.e. x ≤ 2. A solution must satisfy both inequalities simultaneously and this is not possible. Therefore, there is NO solution. (iv)

(2x2 − 2) − x2 + x + 2 < (x2 − x) − x2 + x + 2 x2 + x < 2 x2 + x − 2 < 0 (x + 2)(x − 1) < 0 For the product of brackets to be negative, the quantities in the two brackets must have opposite signs. Hence x + 2 > 0 and x − 1 < 0, or x + 2 < 0 and x − 1 > 0. Thus x > −2 and x < 1, or x < −2 and x > 1. The second pair of inequalities is not possible, so the required solution is −2 < x < 1. 3x − 1 3x − 1 3x − 1 and < 2. The graphs of y = , y = 1 and y = 2 are shown on x−3 x−3 x−3 the next page. The graphs show that the double inequality is never satisfied when x > 3. At points of intersection 1(x − 3) = 3x − 1, 2x = −2, x = −1 3x − 1 = 2(x − 3) = 2x − 6, x = −5

(v) Equivalent to 1 <

Therefore the solution set is −5 < x < −1.

– 24 –

y

. ........ ......... ... ... ... ... ... ... .. ... ... .... ... .. ..... ... ..... ... ... ..... . ...... ... . . ...... ... ....... ....... ... ... ........ . ... . ......... . ...... ... ... ... . ... . . ... ... ... ... ... ... ... ... . . .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... ......... .... .... .... .... .... ........... .... .... .... .... .... .... .... .... .... .... .... .... .... .... . ... ... ... ... ... ... ............................. . ... . ....................... .. ................... . ... . .. ....................................................................................................................................................................................................................................................................................................................................................................... ........... . . . .......... . . ... ......... ....... .... ....... . ... . ...... .. ...... . ... ...... .... ..... .. . . . . ....................................................................................................................................................................................................................................................................................................................................................... .... .. . ... ... ... .. ... ... .. ... .. .. ...... .. .... . .... . . . . . . . . . . . . . . . . . . . . . . . . . ............................................................................................................................................................................................................................................................................................................................................................................................................... . . .. ... . . . . . . . . . .. . . . . . . ... ... ... .... .. ... ... .. ... . .... .. .. .. .. .. .. ... .. . ... .. .. .. ... .. .. ... .. . .. ... .. .. ... ... ... .. ... . ... ... .. ... ... .. .. ... ... . ... ..

3 2 1

x

−8

−6

−4

−2

2

4

6

8

Graph for Problem 11(v) (vi) The stated inequality implies −9 < 2x − 5 < 9 which is equivalent to − 9 < 2x − 5 and 2x − 5 < 9. First inequality above gives − 9 + 5 < 2x, x > −2. Second inequality implies 2x < 9 + 5, x < 7. −2 < x < 7.

Thus the solution set is

[Alternatively you could divide the original inequality by 2 to give x − must be less than

5 5 9 < . Distance of x from 2 2 2

9 leading to the same answer]. 2

(vii) In this case the given inequality leads to the double inequality −3 ≤ 4 − 3x ≤ 3. The above is equivalent to

− 3 ≤ 4 − 3x

and

4 − 3x ≤ 3.

The first inequality gives 3x ≤ 4 + 3,

x≤

7 3

Second inequality implies 3x ≥ 4 − 3,

x≥

1 . 3

7 1 Hence the solution set is ≤ x ≤ . 3 3 4 4 4 [Alternatively − x ≤ 1, or x − ≤ 1, which means that the distance of x from must be less 3 3 3 than or equal to 1, giving the earlier answer). – 25 –

12. (i)

A B A(x − 2) + B(x − 1) 1 = + = (x − 1)(x − 2) x−1 x−2 (x − 1)(x − 2) comparing numerators gives the identity choosing x = 1, choosing x = 2, Hence the answer is

1 = A(x − 2) + B(x − 1) 1 = A(−1) ⇒ A = −1 1 = B(1) ⇒ B = 1

1 1 − . x−2 x−1

(ii) Denominator factorises to (x + 4)(x + 1), hence A B A(x + 1) + B(x + 4) x = + = (x + 4)(x + 1) x+4 x+1 (x + 4)(x + 1) comparing numerators leads to identity x = A(x + 1) + B(x + 4) 1 choosing x = −1, − 1 = B(3), B = − 3 4 choosing x = −4, − 4 = A(−3), A = 3 Answer is

4 3

x+4

(iii) (x2

−

1 3

x+1

.

Ax + B Cx + D (Ax + B)(x2 + 4) + (Cx + D)(x2 + 1) 1 = 2 + 2 = 2 + 1)(x + 4) x +1 x +4 (x2 + 1)(x2 + 4)

Equating numerators here leads to the identity 1 = (Ax + B)(x2 + 4) + (Cx + D)(x2 + 1) Equating co-efficients of powers of x :x3 , 2

x , x,

A+C =0 B+D =0 4A + C = 0

const.

1 = 4B + D

The only solution of the first and third equations is A = C = 0. From the second equation it follows 1 that D = −B, so substituting into the fourth equation leads to 3B = 1, B = . 3 Hence the solution is

1 3

x2 + 1

−

1 3

x2 + 4

.

(iv) Numerator has degree 2, denominator has degree 2; hence division gives constant plus remainder. That is to say E Bx + C D x2 + =A+ =A+ (x − 1)2 (x − 1)2 x − 1 (x − 1)2 [Note that you do not need to calculate the intermediate constants B and C.] Thus

A(x − 1)2 + D(x − 1) + E x2 = (x − 1)2 (x − 1)2 which gives the identity x2 = A(x − 1)2 + D(x − 1) + E – 26 –

choose x = 1, coefficients of x2 ,

1=E 1=A 0 = A − D + E,

constant terms, Hence the solution is

1+

D=2

1 2 + . x − 1 (x − 1)2

(v) Numerator has degree 3, denominator degree 2, hence Cx + D x3 − 1 = Ax + B + (x + 1)(x + 3) (x + 1)(x + 3) F E + = Ax + B + x+1 x+3 LCD of right hand side is (x + 1)(x + 3) hence (Ax + B)(x + 1)(x + 3) + E(x + 3) + F (x + 1) x3 − 1 = (x + 1)(x + 3) (x + 1)(x + 3) identity x3 − 1 = (Ax + B)(x + 1)(x + 3) + E(x + 3) + F (x + 1) x = −1, − 1 − 1 = E(−1 + 3) ⇒ E = −1 x = −3, − 27 − 1 = F (−2) ⇒ F = 14 1=A coefficient of x3 , x = 0, Hence solution is

x−4−

− 1 = 3B + 3E + F ⇒ −1 = 3B + 3(−1) + 14 ⇒ B = −4 14 1 + . x+1 x+3

– 27 –