SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING. Self-paced Course

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1 SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self-paced Course MODULE ALGEBRA Module Topics Simplifying expressions and algebraic functions Rearranging formulae 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 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 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 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 If more than one set of brackets is present the inner ones are removed first Now go through the following worked Example

2 Example Remove the brackets in the following and simplify the resulting expressions by combining like terms: (i) a (4b (a b)), (ii) (x ) (i) = a (4b 9a +6b) =a 8b a b = 9a 0b + (ii) = (x )(x ) = x(x ) (x ) = x x x +4=x 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 Remove brackets and simplify the following (i) x + (x 8), (ii) a + (c a) (b a), (iii) (x + 7)(x 5), (iv) (x + 4), (v) (x )(x + )(x ), (vi) (y ) 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: eg 5x 0 = 5(x ), t + t = t(t + ) Sometimes you must try to factorise a quadratic expression ax + bx + c, wherea, b and c are numbers, into a product of linear factors For example, if you are asked to factorise the quadratic expression x 4x then you seek integers m and n such that Since the right-hand side can be expanded to give x 4x = (x m)(x n) x(x n) m(x n) =x nx mx + mn = x (m + n)x + mn, comparison with the original quadratic shows that you require (i) m + n =4, (ii) mn = There are eight possible ways of satisfying condition (ii) with integers: m =, n = ; m =, n = 7; m = 7, n = ; m =, n = ; m =, n = ; m =, n = 7; m = 7, n = ; m =, n = It is easily seen that only the third pair of numbers satisfies condition (i) Hence, the required factorisation is x 4x = (x 7)(x ( )) = (x 7)(x + ) When the coe cient of x is not unity then the above approach must be slightly modified For the quadratic x +5x you write x +5x =(x m)(x n) and, after expanding the right-hand side and comparing coe cients, require n + m = 5, mn =

3 It relatively easy to deduce that the above pair of equations has solution m =, n = factorisation is x +5x =(x )(x + ), so the appropriate In many situations the quadratic does not factorise into linear factors with integer coe cients Show, for instance, that x + x + 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 Factorise (i) x +7x+, (ii) x x, (iii) x 5x, (iv) 8+x x, (v) x x 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 factors The numerator and denominator of the quotient x x these gives x x =4x 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 x x + x = x x + both have factors of and x, and cancelling Note that the denominator x +x = x(x+) and so it is not divisible by or x,justx or x+ 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 Example Simplify x x 5 x + On factorising the numerator the above quotient can be written (x 5)(x + ) x + = x 5, provided x 6= The latter restriction on x is necessary because in obtaining the answer x 5 the common factor x + was cancelled by dividing both numerator and denominator by this factor Clearly this is only acceptable provided x +6= 0, ie x 6=, and hence the quotient in Example simplifies to x 5 only if x 6= When x = 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 - ie the lowest common denominator (LCD) Each fraction is then written with this LCD as its denominator, before combining

4 Example Express as single fractions (i) (i) LCD of denominators y and x is xy, so x y + y x, (ii) x + + (x + ) x +x + x y + y x = xx xy + yy xy = x + y xy (ii) This time the LCD of denominators (x + ) and x +x + (= (x + )(x + )) is (x + ) (x + ), hence (x + ) + x + (x + )(x + ) = (x + ) (x + )(x + ) (x + ) + (x + ) (x + )(x + ) = x +6+(x +x + x + ) (x + ) (x + ) = x +6x +7 (x + ) (x + ) Note that the numerator in the above expression does not factorise into linear terms with integer coe Example 4 Express as a single fraction x ++ x x + can be expressed x + and the LCD of and x isx Hence which can be written x ++ (x + )(x ) = + x x x = x x +4x + x = x +x +, x (x + )(x + ) x cients ***Do Exercise (i) Express as a single fraction x + x, (ii) (t + ) + t +, (iii) x + x +, (iv) x x + x x (x ) 4 Formulae Physical quantities are often related to each other using formulae, eg the area of a circle, A, is related to its radius, R, through the formula A = R 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 = R, 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 di erent variable becomes the subject During this rearrangement, or transposition, you can carry out a number of di erent 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 (eg 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 + at Rearrange the formula so that the subject is a Subtracting ut from both sides gives s ut = at, 4

5 then multiplying both sides by leads to (s ut) =at Finally dividing both sides by t implies a = (s ut) t ***Do Exercise 4 Rearrange the following formulae to obtain expressions for the stated variable (i) A = r for r, (ii) v = u +as for s, (iii) R = + for R, (first express right-hand side as a single fraction), R R s l (iv) T = for l g 5 Indices Products of a number can be written compactly using indices, or powers For example, 8 = and you write 8 = 4 The basic rules for manipulating indices are Note that and in general one obtains the result a m a n = a m+n, a m a n = am n, a m = a m or (am = a m ), (a m ) n =(a n ) m = a mn, (a m ) n =(a n ) m = a m n (a m c n ) =(a m c n )(a m c n )(a m c n )=a m a m a m c n c n c n = a m+m+m c n+n+n = a m c n, (a m c n ) p = a mp c np Example 6 Evaluate 6 Using the above rules 6 = (6 ) =4 =4 4 4 = 64 Check the answer using the x y button on your calculator Example 7 Simplify a b c acd a b c acd = a b c acd = a b c d ab = ab c d = c d 5

6 ***Do Exercise 5 Simplify (i) (8) 4, (ii) a a, (iii) (v) a 4 a, (vi) x(x a ) (a ), (iv) a b c(a c 4 ), a 6 Rationalising a denominator involving a square root This section is concerned with rewriting an expression such as p so that no square roots appear in the denominator The method for achieving a b this uses the result (x + y)(x y) =x y = x xy + yx y Putting x = a and y = p b gives a + p p b a b = a b Hence, multiplying numerator and denominator by a + p b you obtain a p = b a a + p b p b a + p = a + p b b a b Example 8 Rationalise the denominator in Multiply the numerator and denominator by + p p to give + p p p = p = p = + p Check your answer again using your calculator ***Do Exercise 6 6 (i) p, (ii) 7 Rationalise the denominator in p p + 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 x, and not as x,x 4, etc The standard form is ax + b =0, but linear equations often appear in non-standard form Example 9 Solve the equation x + 0 = 4( x) Multiplying out the right-hand side (RHS) x + 0 = 4 4x Adding 4x to each side gives x +4x + 0 = 4, 6

7 and then subtracting 0 from both sides 7x =4 0 = 6 Dividing both sides by 7 then leads to the solution x = 6/7 8 Quadratic equations Quadratic equations have the standard form ax + 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 0 Solve 6x x 5 = 0 The left-hand side factorises to give (x + )(x 5) = 0, and hence either x + = 0 or x 5 = 0 Solving these linear equations leads to x = or x = 5 There are two values of x, therefore, which satisfy the above quadratic equation Sometimes you may be given a quadratic which you find di cult, or impossible, to factorise In these situations you can always use the result: the solutions of the equation ax + bx + c =0(witha 6= 0) are x = b ± p b 4ac a Comparing the general equation with that considered in Example 0 it is seen that a =6,b=,c= 5 Substitute these values into the general formula and verify the solutions x =, x = 5 Note that if b 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 b =4ac there is only one value of x which satisfies the equation For instance, the equation x 6x + 9 = 0 can be written (x ) = 0, with solution x =(twice) x = is known as a repeated root Finally, if b 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) =(x + a)(x + a) =x + ax + ax + a = x +ax + a, by subtracting a from both sides it follows that (x + a) a = x +ax Hence, by choosing a = 4, the expression x +8x can be written x +8x =(x + 4) 4 =(x + 4) 6 7

8 Confirm the above answer by removing the bracket Completing the square is used in the example below Example Solve x +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 Hence x +6x =(x + ) =(x + ) 9 x +6x +6=(x + ) 9+6=(x + ), and so the given quadratic equation can also be expressed (x + ) =0 Adding to both sides gives which, after taking square roots, becomes (x + ) = x +=± p The two solutions, therefore, are x = ± p (iex = + p and x = 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 p ) 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 Solve the set of equations 4x +y =5 5x y = First you must decide which unknown to eliminate Suppose the variable y is chosen, then it is necessary to ensure the coe cients of the y terms in both equations have equal magnitude An obvious way to achieve this is to multiply the first equation by and the second equation by : x +6y = 5 0x 6y = 4 Adding the above equations leads to x + 0x = 5 4, which gives x =, or x = Substituting back into the first given equation leads to 4 +y =5 This implies + y =5, ie y =,y = Hence, the solution set is x =,y = You should always verify your answer by substituting back into both equations stated in the question 8

9 0 Solving equations using graphs A variety of di erent 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 Solve, by using graphs, (i) x + 0 = 4( x), (ii) 6x x 5 = 0, (iii) 4x +y =5, 5x y =, (the equations solved algebraically in Examples 9, 0 and ) (i) In this case you can plot y =x + 0 and y = 4( 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 y =x + 0 x The lines y = x + 0 and y = 4( x) intersect near x = 08 A more accurate graph would provide a better solution 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 08 0 y = 4( x) 0 (ii) y x To find a solution of 6x x 5=0 draw a graph of y =6x x 5 and look for the values of x, if any, for which y = 0 (ie where the graph intersects the x-axis) The graph suggests that approximate solutions for x are 0 and 5 By zooming in and expanding the graph near these points it is possible to achieve more accurate solutions 0 9

10 (You should observe from the graph that y never becomes 0, for instance, and hence the equation 6x x 5= 0, which can be written 6x x + 5 = 0, has no real solutions (or roots)) (iii) In this case you look for the point of intersection of the two graphs 4x +y = 5 and 5x y = First draw the graphs of both these equations y 5x y = x Clearly the lines intersect at approximately x = 05, y = 5 These values can again be confirmed by drawing more accurate graphs 4x +y =5 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 = and x+y = 5, for instance Graphical methods can easily be extended to more complicated equations ***Do Exercise 7 (i) 4 x =, (ii) Solve the equations x + =5, (iii) 4( x) = (x ), (iv) x = x + ***Do Exercise 8 Solve the equations (i) x x = 0, (ii) x 9x + 4 = 0, (iii) x 9x 8=0, (iv) 8 =6 x x + ***Do Exercise 9 Solve the following sets of equations (i) x + y =, (ii) x +4y =, (iii) 4x y =5 4x y =, 9x +y = 9, 5x +y = ***Do Exercise 0 Solve the following equations (using graphs in both cases to illustrate your results): (i) x = x, (ii) x = x 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, whichiswrittenb>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<bhas a geometric meaning; if a and b are represented by points on a number line with 0

11 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<bis often called the sign (or direction or sense) of the inequality - reversing the sign leads to a>b The statement a apple b means that either a<bor a = b Similarly a b means that either a>bor a = b (You read the symbols apple 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 apple 0respectively 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 = 475 and b = 475, then a<bsince the first digit in which they di er 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, 8 > 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>bis true and the other two are false II If a<band b<cthen a<c III If a<bthen a + c<b+ c (ie adding or subtracting the same quantity to both sides of an inequality leaves the inequality sign unchanged) IV If a<band c>0thenac < 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<band c<0thenac > 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 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 4 Solve x <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

12 from the left-hand side (by subtracting ) Hence ( x) 4x < (4x 5) 4x ie 6x < 8 To obtain the inequality for x you must multiply by V that you must also reverse the inequality 6 6x 6 > 8 6 ie x> 4 (or divide by 6), remembering from Rule The solution is a set of numbers (called the solution set) - it consists of all numbers greater than 4 If a and b are numbers such that a<bthen the statement a<x<bis equivalent to the two statements x>aand x<b In writing the double inequality it is essential that a<b Thus, the inequalities x> and x< can be expressed by the double inequality <x<, but it is not possible to write x> and x< as the double inequality <x< (since is not less than ) You may find it helpful to observe that acceptable double inequalities, such as <x<, represent a single region on the number line, whereas the inequalities x> and x< 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 5 Solve x 6 < x 5 apple x This means that you must find the values of x which satisfy the inequalities x 6 < x 5 and x 5 apple x Each of these is treated separately: (x 6) x<(x 5) x (x 6) x +5< (x 5) x +5 ie <x (x 5) x apple (x ) x (x 5) x +5apple (x ) x +5 x apple Hence the double inequality implies x> numbers x such that <xapple and x apple The solution set therefore consists of all 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 > 0ifa6= 0, and a = 0ifa = 0 It also follows that ab = a b, a = a, a b = a b a is also often called the modulus of a or the magnitude of a

13 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, ie 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: eg if a<0,b>0 ifa>0,b>0 a b! a b! a 0 b 0 a b Example 6 Solve x < 5 From the comments above the inequality means 5 <x < 5 This is a double inequality which is equivalent to 5 <x and x < 5 Adding to both sides of the first of these inequalities gives <xwhereas adding to the second inequality leads to x<7 Hence the solution set consists of all numbers greater than and less than 7; ie <x<7 As discussed above the result has a geometrical interpretation The quantity x is the distance of the point x on the number line from the point, and this distance is less than 5 5! 5! 0 7 Hence the point x must satisfy <x<7 Inequalities containing a quotient must be solved with care as shown below Example 7 Solve x +x < 4 The denominator can be removed by multiplying both sides by + x However, the quantity + 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 Case (i) + x>0 In this case multiplying both sides by + x gives x < 4( + x) x < + 4x < 4x + x 0 < 5x ie x>

14 It is important to note that when x> it follows that x +> 0, (or x +> ) and hence the necessary condition + x>0 is always satisfied Case (ii) + x<0 Now you can again multiply both sides by + x, but after doing so the inequality must be reversed This time, therefore, you obtain x>4( + x) x> + 4x 0 > 5x ie x< However, case (ii) requires + x<0, ie x<, and the latter inequality and x< are both satisfied only if x< The final solution to the original inequality, therefore, is the set of values of x satisfying x< and x> 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 8 Solve graphically the inequality x 4x +< The graphs of y = x 4x + and y = 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 P Q 5 x P and Q are determined by solving x 4x + =, which reduces to x 4x =0 Using the formula gives x = 4 ± p (( 4) 4()( )) = 4 ± p 4 = 4 ± p 6 =± p 6, () and so the solution of the inequality is the set of all x satisfying p 6 <x<+ p 6 Example 9 Use graphical means to solve the inequality in Example 7 Here you need to draw the graphs of y = x and y = 4 The first graph is not easy to obtain +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: 4

15 6 0 4 y y =4 P 0 x 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 = From solving x = 4, (ie x = 4( + x) = + 4x, 5x = 0,x = ), you know that P is +x situated at x = Hence the solution set is all values of x satisfying x< and x> (which agrees with the earlier result!) ***Do Exercise Solve the inequalities (i) 6x +5 x 5, (ii) 5 <x 4 < x, (iii) x + 0 < x 5 apple x, (iv) x <x x, (v) < x x <, (vi) x 5 < 9, (vii) 4 x apple 4 Partial fractions Polynomial functions arise in many engineering situations A polynomial of degree n has the form a n x n + a n x n + + a x + a x + a 0, where a n, a 0 are all constants The degree is the highest power occurring in the polynomial Hence, x + is a polynomial of degree and x 4 + x + is a polynomial of degree 4 In solving problems it is sometimes necessary to consider a quotient of polynomials (ie 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 di erential 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<dand is said to be improper if n d It can be shown that any polynomial with real co cients can be factorised into a product of linear and quadratic factors (with all coe cients 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: 5

16 factor in denominator ax + b (linear) partial fraction A ax + b (ax + b) ax + bx + c (ax + bx + c) (repeated linear) (quadratic) (repeated quadratic) A ax + b + B (ax + b) Ax + B ax + bx + c Ax + B ax + bx + c + Cx + D (ax + bx + c) In all cases the unknown constants A, B, C are determined by evaluating an identity at some chosen values of x or by equating coe cients, or from using some combination of the two methods (as shown in the examples below) Example 0 Express as partial fractions ), and then there is a partial fraction corre- The denominator can be factorised to give (x + )(x sponding to each linear factor Hence, x 5 x +x x 5 (x + )(x ) = A x + + B x, where A and B are constants Rewriting the right-hand side using a common denominator gives x 5 (x + )(x ) A(x ) + B(x + ) =, (x + )(x ) and multiplying both sides by (x + )(x identical), leads to the equation ), or by equating numerators (since the denominators are x 5=A(x ) + B(x + ) 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 e ort 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 = and x = Consider these in turn: x = 5=A(0) + B(4) 4=4B, ie B = x = 5=A( ) + B(0) 8= 4A, ie A = Thus the partial fractions are x 5 x +x = x + x 6

17 Example Express as partial fractions The denominator factorises again: x x +6x +9 x +6x +9=(x + ), so x x +6x +9 = A x + + and multiplying both sides by (x + ) leads to the identity x = A(x + ) + B B (x + ), With a linear factor it is convenient to choose x =, in which case =A(0) + B, ie B = The remaining coe cient could be found by substituting a di erent value for x (for example, with x = 0, 0 = A() + B, A = B = ( ) =, A = ) Alternatively you can compare coe cients of powers of x on both sides of the identity x = A(x + ) + B For the x term, equating coe cients gives = A, as above From the constant terms (ie x 0 terms) it follows that 0 = A + B, which is identically satisfied by A =,B =, as expected The final result, therefore, is Let us now consider a more complicated example Example x x +6x +9 = x + Express as partial fractions (x + ) x +5x + (x +4x + 5)(x ) The quadratic x +4x + 5 does not split into a product of linear factors with real coe cients, so the denominator is already in its simplest form The rules stated earlier tell you that for this example the appropriate partial fractions are x +5x + (x +4x + 5)(x ) = A x + Bx + C x +4x +5 Writing the right-hand side in terms of the common denominator (x )(x +4x+5) and then equating numerators, or multiplying throughout by the product (x )(x +4x + 5), gives x +5x +=A(x +4x + 5) + (Bx + C)(x ) Choose x = (to make the linear factor zero) +5+=A( ) + (Bx + C)(0), 0 = 0A, A = Before equating coe cients, multiplying out the right-hand side leads to x +5x +=A(x +4x + 5) + Bx(x ) + C(x ) = A(x +4x + 5) + Bx Bx + Cx C 7

18 Equating coe cients now gives x =A + B =+B (since A = ) ie B = x 5=4A B + C = 4() +C C =5 4+= All three constants have now been calculated but comparing constant terms on both sides of the equation provides a check on your answers: constant = 5A C, and this equation is satisfied by A =,C = Examples 0- 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, d For example, the fraction x4 + x is improper, since the numerator has degree 4 and the denominator + degree It can be shown that The corresponding general result can be written x 4 + x + = x + x + polynomial degree n degree apple (d ) = polynomial degree (n d)+polynomial 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 Example Express (x + )(x + ) x in terms of partial fractions Here the numerator is quadratic, the denominator is linear and hence the general result above gives (x + )(x + ) x = linear + const x 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 (x + )(x + ) x = Ax + B + C x, and, after multiplying both sides by x, the identity obtained is (x + )(x + ) = (Ax + B)(x ) + C Choosing x =, + + =(Ax + B)(0) + C 8

19 Equating coe cients in the identity: x =A, A = = C, C = constant = B + C, B = C = [check term in x : + = A +B, satisfied] Hence which is the reverse of Example 4 ***Do Exercise (i) (iv) (x + )(x + ) x Express in partial fractions = x ++ x, (x )(x ), (ii) x (x +5x + 4), (iii) (x + )(x + 4), x (x ), (v) x (x + )(x + ) 9

20 Specimen Test Expand (x + )( x) Simplify x(x ) 4 x Express as a single fraction x + x + 4 Solve x + = 5 Solve x +x +=0 6 Solve the set of equations 4x y = 9 x +y = 7 Solve 5 x apple (x ) 8 Solve +x < 9 Express in partial fractions x + x(x ) 0

21 Worked solutions to Exercises (i) = x +6x 6 = 7x 7 (ii) = a +c a b +a = a b +c (iii) = x(x 5) + 7(x 5) = x +x 5 (iv) = (x + 4)(x + 4) = x(x + 4) + 4(x + 4) = x +8x + 6 (v) = (x )(x(x ) + (x )) =(x )(x x 6) = x(x x 6) (x x 6) = x x 6x x + x +6 = x x 5x +6 (vi) = (y )(y )(y ) = (y )(y(y ) (y )) = y(y 6y + 9) (y 6y + 9) = y 9y + 7y 7 (i) x +7x + = (x + 4)(x + ) [(x m)(x n) =x (m + n)x + mn requires mn =, m + n = 7 Solution is m = 4, n = or equivalently m =, n = 4] (ii) x x =(x )(x + ) [as above : m =, n = ] (iii) x 5x = x(x 5) = x(x 5)(x + 5) (iv) 8 + x x = (x x 8) = (x 4)(x + ) = (4 x)(x + ) (v) x x =(x + )(x ) [(x m)(x n) =x (n + m)x + mn ) mn =, n + m =, leading to the solution n =, m = ] (i) LCD of x and x isx(x ) (x ) Solution = x(x ) + (x) (x )(x) = x +x x(x ) = x x(x ) (ii) LCD of (t + ) and (t + ) is (t + ) Solution = (iii) x + x + = x + x + LCD of and x +isx + (t + ) +t +4 t +5 + = (t + ) (t + )(t + ) (t + ) = (t + ) Solution = x(x + ) (x + ) + x + = x + x + x + (iv) x + x =(x + )(x ) LCD of (x + )(x ) and (x ) is (x + )(x ) x(x ) Solution = (x + )(x )(x ) = x( x 8) (x + )(x ) = x(x + 8) (x + )(x ) x(x + ) x((x ) (x + )) (x ) = (x + ) (x + )(x )

22 4 (i) r A A = r, r = ± (ii) v u =as, s = v u (positive if distance) a (iii) LCD of R and R is R R R = (R ) R (R ) + (R ) R (R ) = R + R R R (iv) R = R R R + R T = l = gt 4 s l g, T 4 = l g 5 (i) (8) 4 =( 4 ) 4 = = 7 (ii) a a = a 5 (iii) (a ) a = a a = a = a (iv) = a b c(a c 4 )=a 7 b c 5 (v) a 4 a =(a ) = a = a (vi) x(x a ) = x(x a )=x a+ 6 (i) = ( ( + p 7) = 6( + p 7) 7 ( p p ) 6 ( + p 7) p 7) (ii) = ( p + p ) ( p = 6( + p 7) 6 p p p = = p + p ) = ( + p 7) 7 (i) x =4 =) x = (ii) = 5(x + ) = 5x +5, 5x =, x= 5 (iii) 4 4x = 6x +, x =, x= (iv) = x, (x + ) = x, x +=x, x = x + 8 (i) x x = (x 4)(x + ) = 0 ) x = 4 or (ii) x 9x + 4 = (x 7)(x ) = 0 ) x = 7 or (iii) x = ( 9) ± p (( 9) 4()( 8)) = 9 ± p (iv) 8 = (6 x)(x + ) = 6x + 8 x x = 8 + x x ) x x 0 = (x 5)(x + ) = 0 ) x = 5 or =985; or 085

23 9 (i) Adding the given equations leads to 6x = 6, x = Substituting this value into the first given equation then implies () + y =, so y = Hence solution is x =, y = (ii) Eliminate y x +4y = (first equation) 8x +4y = 8 (second equation ) Subtracting second equation from first, 5x = ( 8) = 0, which implies x = Substituting this solution back into the original first equation gives ( ) + 4y =, 4y = ( 6) = 8, y = 8 4 = 9 Hence solution is x =, y = 9 (iii) This time we choose to eliminate x (first equation 5) 0x 0y = 5 (second equation 4) 0x + y = 8 (subtracting) y = 5 ( 8) = y = Substituting back into first equation in question 4x ( Thus solution is x =, y = ) = 5, 4x =5 =, x = 0 (i) The graphs of y = x and y = x are shown below Clearly they intersect at x = approximately In fact this is a solution since when x = both x and x are equal (= ) No other solution is possible y y + x + x Graph for Problem 0(i) Graph for Problem 0(ii) (ii) Graphs of y = x and y = x are displayed above These graphs intersect at x = 0, x =, x = approximately It is obvious by substitution that all three of the latter values are the exact points of intersection

24 Algebraically, at the points of intersection x = x Hence with solutions x = 0, +, x x = x(x ) = x(x )(x + ) = 0 (i) 6x +5 x 5 6x +5 x 5 x 5 x 5 5x 0, ) x (ii) Equivalent to 5 < x 4 and x 4 < x First inequality (add 4 to both sides) 5+4<x 4+4, ie <x Second inequality (add x + 4 to both sides) (x 4) + x +4< ( x)+x +4 For this example, therefore, the solution set is (iii) Equivalent to x + 0 < x 5 and x 5 apple x (iv) which implies x <6, ie x< <x< First inequality (x + 0) x +5< (x 5) x +5, ie 5 <x Second inequality (x 5) x + 5 apple (x ) x + 5, ie x apple A solution must satisfy both inequalities simultaneously and this is not possible Therefore, there is NO solution (x ) x + x +< (x x) x + x + x + x< x + x < 0 (x + )(x ) < 0 For the product of brackets to be negative, the quantities in the two brackets must have opposite signs Hence x +> 0 and x < 0, or x +< 0 and x > 0 Thus x> and x<, or x< and x> The second pair of inequalities is not possible, so the required solution is <x< (v) Equivalent to < x x x and < The graphs of y =, y = and y = are shown on x x x the next page The graphs show that the double inequality is never satisfied when x> At points of intersection (x ) = x, x =, x= Therefore the solution set is 5 <x< x = (x ) = x 6, x= 5 4

25 8 6 4 Graph for Problem (v) y x (vi) The stated inequality implies 9 < x 5 < 9 which is equivalent to 9 < x 5 and x 5 < 9 First inequality above gives < x, x > Second inequality implies x <9 + 5, x < 7 Thus the solution set is <x<7 [Alternatively you could divide the original inequality by to give x must be less than 9 leading to the same answer] 5 < 9 Distance of x from 5 (vii) In this case the given inequality leads to the double inequality apple 4 x apple The above is equivalent to apple 4 x and 4 x apple The first inequality gives x apple 4+, x apple 7 Second inequality implies x 4, x Hence the solution set is apple x apple 7 [Alternatively 4 4 x apple, or x apple, which means that the distance of x from 4 than or equal to, giving the earlier answer) must be less 5

26 (i) Hence the answer is (x )(x ) = A x + B A(x ) + B(x ) = x (x )(x ) comparing numerators gives the identity = A(x ) + B(x ) x x (ii) Denominator factorises to (x + 4)(x + ), hence choosing x =, =A( ) ) A = choosing x =, =B() ) B = x (x + 4)(x + ) = A x +4 + B A(x + ) + B(x + 4) = x + (x + 4)(x + ) comparing numerators leads to identity x = A(x + ) + B(x + 4) choosing x =, =B(), B= choosing x = 4, 4=A( ), A= 4 (iii) Answer is 4 x +4 x + (x + )(x + 4) = Ax + B x + + Cx + D x +4 = (Ax + B)(x + 4) + (Cx + D)(x + ) (x + )(x + 4) Equating numerators here leads to the identity Equating co-e cients of powers of x :- =(Ax + B)(x + 4) + (Cx + D)(x + ) x, A + C =0 x, B + D =0 x, 4A + C =0 const = 4B + D The only solution of the first and third equations is A = C = 0 From the second equation it follows that D = B, so substituting into the fourth equation leads to B =, B = Hence the solution is x + x +4 (iv) Numerator has degree, denominator has degree ; hence division gives constant plus remainder That is to say x (x ) = A + Bx + C (x ) = A + D x + E (x ) [Note that you do not need to calculate the intermediate constants B and C] Thus x (x ) = A(x ) + D(x ) + E (x ) which gives the identity x = A(x ) + D(x ) + E 6

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