A-Level Notes CORE 1

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1 A-Level Notes CORE 1

2 Basic algebra Glossary Coefficient For example, in the expression x³ 3x² x + 4, the coefficient of x³ is, the coefficient of x² is 3, and the coefficient of x is 1. (The final 4 is referred to as the constant term.) Completing the square Writing a quadratic in the form A(x + B)² + C, where A, B and C are constants. Discriminant For the quadratic equation ax² + bx + c = 0, the discriminant is the expression b² 4ac. If the discriminant is positive then there are two real solutions to the equation; if it is zero then there is one real solution to the equation (the equation is a perfect square); and if it is negative then there are no real solutions to the equation. Elimination method A method of solving linear simultaneous equations. See the Notes and Examples. Linear equation An equation involving only terms in x (or any variable), no terms in x², x³ etc. Linear expression An expression in which the highest term is a term in x. A linear expression may also include a constant term. A linear expression is of the form ax + b, and all graphs of the form y = ax + b are straight lines, hence the name linear. Parabola The graph of a quadratic is an example of a parabola. You may learn more about parabolas if you study Further Maths. Parabolas are a type of conic section. MEI, 0/03/05 1/3

3 Perfect square A quadratic expression is a perfect square if it can be written as the square of a linear expression. E.g. x + 8x+ 16is a perfect square because it may be written as ( x + 4) 4x 4x 1 5x + + is a perfect square because it may be written as ( x + 1) 1 x + is a perfect square because it may be written as ( 5x ) However, x 3x 1 ( x ) + = is not a perfect square Quadratic equations that are perfect squares have a discriminant of 0 and have only one solution. Quadratic equation An equation of the form ax² + bx + c = 0, where a, b and c are constants and a 0. Quadratic expression A polynomial expression in which the highest term is a term in x². A quadratic expression may also have a term in x and a constant term. Quadratic formula The formula for the solutions of the quadratic equation ax² + bx + c = 0. The formula is given by b± b² 4ac x = a Real Numbers All of the numbers you have met up to now, or will meet on the standard Maths AS/A level course are real numbers. A real number is a number that may be represented on a number line. The real numbers are made up the rational numbers and the irrational numbers. Unless you study beyond the standard Maths A level, the only type of numbers you will meet are the real numbers. If you do Further Mathematics AS/A level, or study a strongly Maths-related subject at university, you will meet imaginary numbers (which enable you to give square roots for negative numbers) and complex numbers (which have a real part and an imaginary part). If complex numbers are allowed, ALL quadratic equations have roots. Quadratic equations with a negative discriminant have no real roots, but have two complex roots. MEI, 0/03/05 /3

4 Simultaneous equations A set of two or more equations involving two or more variables. There should be the same number of equations as there are variables. Substitution method A method of solving simultaneous equations. See the Notes and Examples. Turning Point A turning point on a curve is a point where the gradient of the curve is zero. On the graph of a quadratic, there is always one turning point. On one side of this point the gradient of the graph is positive, on the other side of this point the gradient of the graph is negative. Maximum Minimum Inflection Variable An unknown quantity represented by a letter, such as x. Vertex The turning point of a quadratic is sometimes called the vertex. MEI, 0/03/05 3/3

5 Basic Algebra Section 1: Basic algebraic manipulation and solving simple equations Notes and Examples These notes contain subsections on Manipulating algebraic expressions Linear equations Changing the subject of a formula Manipulating algebraic expressions Throughout your mathematics you will need to be able to manipulate algebraic expressions confidently. The examples below remind you of the important techniques of collecting like terms, removing brackets, factorising, multiplying, and adding, subtracting and simplifying algebraic fractions. You may also like to try the Algebra Puzzle, either on your own or in a group, in which you need to be able to recognise equivalent algebraic expressions. Collecting like terms Example 1 Simplify the expression 3a + b a + 3b ab + a There are three different types of like term in this expression. There are terms in a, terms in b, and a term in ab. Notice that the term in ab cannot be combined with either the terms in a or the terms in b, but remains as a term on its own. 3a + b a + 3b ab + a = 3a a + a + b + 3b ab = 4a + 5b ab In the example the expression has been rewritten with each set of like terms grouped together, before simplifying by adding / subtracting the like terms. You may well not need to write down this intermediate stage. For practice in examples like this one, try the interactive resource Collecting terms. Removing brackets Example Simplify the expressions (i) 3(p q) + (3p + q) (ii) x(x + 3y) y(x 5y) MEI, 01/03/05 1/7

6 Each term in the bracket must be multiplied by the number or expression outside the bracket. (i) 3(p q) + (3p + q) = 3p 6q + 6p + q = 9p 4q (ii) x(x + 3y) y(x 5y) = x² + 6xy xy + 10y² = x² + 4xy + 10y² Factorising To factorise an expression, look for numbers and/or letters which are common factors of each term. We often talk about taking out a factor this can cause confusion as it tends to make you think that subtraction is involved. In fact you are, of course, dividing each term by the common factor which you are taking out. Example 3 Factorise the following expressions. (i) 6a + 1b + 3c (ii) 6x²y 10xy² + xy (i) 3 is a factor of each term. 6a + 1b + 3c = 3(a + 4b + c) (ii) xy is a factor of each term. 6x²y 10xy² + xy = xy(3x 5y + ) Check your answers by multiplying out the brackets. Multiplication Example 4 Simplify the expression xy 3yz 4x z. xy 3yz 4x z = 3 4 x x y y z z 3 3 = 4xyz You may be happy to do this in your head, without writing out the intermediate line of working. Adding and subtracting algebraic fractions Algebraic fractions follow the same rules as numerical fractions. When adding or subtracting, you need to find the common denominator, which may be a number or an algebraic expression. MEI, 01/03/05 /7

7 Example 5 Simplify x x 5x (i) (ii) 1 1 x x Core 1 (i) The common denominator is 1, as 3, 4 and 6 are all factors of 1. x x 5x 8x 3x 10x + = x + 3x 10x = 1 x = 1 (ii) The common denominator is x². 1 1 x = x x x x x = x Simplifying fractions You are familiar with the idea of cancelling to simplify numerical fractions: for example, 9 can be simplified to by dividing both the numerator and the denominator by 3. You can also cancel before carrying out a multiplication, to make the numbers simpler: e.g =. The same technique can be used in algebra. As with 3 4 factorising, remember that cancelling involves dividing, not subtracting. Example 6 Simplify 3 6xy + x y (i) 10x y 3a a+ (ii) a+ 1 a+ (i) It is very important to remember that you can only cancel if you can divide each term in both the numerator and denominator by the same expression. In this case, don t be tempted to divide by x²y although this is a factor of both x²y and 10x²y, it is not a factor of 6xy³. In a case like this, it may be best to factorise the top first, so that it is easier to see the factors. 3 6xy + x y xy(3 y + x) = 10xy 10xy 3y + x = 5x xy is a common factor of both top and bottom MEI, 01/03/05 3/7

8 (ii) Again, factorise where possible first. 3a a+ 3a ( a+ 1) = a+ 1 a+ a+ 1 a+ 6a = a + Core 1 (a + 1) is a common factor of both top and Notice that you cannot cancel a here, as it is not a factor of a +. Linear equations A linear equation involves only terms in x (or whatever variable is being used) and numbers. So it has no terms involving x², x³ etc. Equations like these are called linear because the graph of an expression involving only terms in x and numbers (e.g. y = x + 1) is always a straight line. Solving a linear equation may involve simple algebraic techniques such as gathering like terms and multiplying out brackets. Example 7 shows a variety of techniques that you might need to use. Example 7 Solve these equations. (i) 5x = 3x + 8 (ii) 3( y 1) = 4 ( y 3) (iii) a 1 = a (i) 5x = 3x+ 8 5x= 3x+ 8+ 5x= 3x+ 10 5x 3x= 10 x = 10 x = 5 Add to each side Subtract 3x from each side Divide each side by (ii) 3(y 1) = 4 ( y 3) 6y 3= 4 y+ 6 6y 3= 10 y 6y = 13 y 8y = 13 y = 13 8 Multiply out the brackets Add 3 to each side Add y to each side Divide each side by 8 MEI, 01/03/05 4/7

9 (iii) a 1 = a a 1= 3(a+ 3) a 1= 6a+ 9 a = 6a+ 10 4a = 10 a =.5 Core 1 Multiply both sides by 3 Multiply out the brackets Add 1 to each side Subtract 6a from each side Divide both sides by -4 In Example 8, the problem is given in words and you need to express this algebraically before solving the equation. Example 8 Sarah has a choice of tariffs for text messages on her mobile phone. Tariff A: 10p for the first 5 messages each day, p for all others Tariff B: 4p per message How many messages would Sarah need to send each day for the two tariffs to cost the same? (She always sends at least 5!) Let the number of messages Sarah sends per day be n. Under Tariff A, she has to pay 10p for each of 5 messages and p for each of n - 5 messages. Cost = 50 + ( n 5) Under Tariff B, she has to pay 4p for each of n messages. Cost = 4n For the cost to be the same 50 + ( n 5) = 4n 50 + n 10 = 4n 40 + n= 4n 40 = n 0 = n She needs to send 0 messages per day for the two tariffs to cost the same. For practice in examples like this one, try the interactive resource Forming and solving linear equations. Changing the subject of a formula Changing the subject of a formula is similar to solving an equation, but you are working with letters rather than numbers. Example 9 The volume V of a cylinder with radius r and height h is given by V = π r h. Make r the subject of this formula. MEI, 01/03/05 5/7

10 V = π r h V = r π h V = r π h V r = π h Divide both sides by πh Square root both sides Finish by writing the equation with r on the left side. In the next example, two different methods are shown. The answers look a bit different but they are equivalent. Make sure that you can see how you could rewrite one solution to give the other. Example 10 The surface area A of a cylinder with radius r and height h is given by A= π r( h+ r). Make h the subject of this formula. (1) A= π r( h+ r) A= πrh+ π r A r rh π = π A πr = h πr A πr h = πr () A= π r( h+ r) A = h+ r πr A r = h πr A h= r πr Multiply out the brackets Subtract πr² from each side Divide each side by πr Rewrite with h on the left Divide each side by πr Subtract r from each side Rewrite with h on the left side In the next example, the new subject appears more than once. You need to collect the terms involving the new subject together and then factorise to isolate the new subject. MEI, 01/03/05 6/7

11 Example 11 Make x the subject of the formula cx a = a( b + x). cx a = a( b + x) cx a = ab + ax cx ax a = ab cx ax = ab + a ( c a) x= a( b+ 1) ab ( + 1) x = c a Multiply out the brackets Subtract ax from each side to collect the terms in x together Add a to each side Factorise Divide both sides by c - a MEI, 01/03/05 7/7

12 Basic Algebra Section 1: Basic algebraic manipulation and solving simple equations Crucial points 1. Make sure that you are not making basic errors Errors in algebra are very common. Sometimes these are just careless mistakes, but sometimes you may make errors because you have not understood a technique correctly. If you have problems with any technique in this unit, read the worked examples very carefully and make sure that you understand each step. If you are not sure, make sure that you consult a teacher.. Make sure you know what is meant by taking out a factor When factorising, make sure that you understand that taking out a factor means dividing each term by that factor, NOT subtracting. 3. Algebraic fractions If you are having trouble with algebraic fractions, you may find it helps to practise some numerical fractions first, so that you can be sure that you remember the techniques involved. 4. Cancelling fractions Remember that when you cancel fractions, you are dividing the numerator and denominator by the same thing. You have to divide each term by the same thing, and it may help to factorise numerator and denominator if possible so that you can see any factors. p A fraction like cannot be simplified, as p is not a factor of the p + q whole denominator. 5. Check your answer when solving equations When you solve an equation, always check your solution by substituting it into the original equation to make sure that it fits. 6. Rearranging formulae If you are stuck on rearranging a formula, try writing in numbers instead of all the letters except for the new subject, so that you have an equation with one unknown. Then think what you would do to solve the equation, and do the same thing to the formula that you are rearranging. 7. Changing the subject of a formula When changing the subject of a formula, make sure the new subject does not appear on both sides of the equals sign in the rearranged formula. MEI, 01/03/05 1/

13 ! " ax e.g. Make x the subject of b = x ax WRONG b = x bx = ax ax x = b ax RIGHT b = x bx = ax bx + = ax = ax bx = x a b ( ) = x a b x = a b! " x is still present on the RHS of the equation Get the terms involving x on the same side of the equation Isolate x by factorising MEI, 01/03/05 /

14 Notes and Examples Core 1 Basic Algebra Section : Quadratics These notes contain subsections on Multiplying out two brackets Factorising quadratic expressions Graphs of quadratic functions Solving quadratic equations by factorisation Solving quadratic equations using the formula Problem solving In this section of work you will be studying quadratic functions and their graphs. You probably already know something about this topic, but you will now be taking it a little further. Multiplying out two brackets Multiplying out two brackets of the form (ax + b)(cx + d) gives a quadratic function. This should be revision of GCSE work. Make sure that you are confident in this. Example 1 Multiply out ( x+ )(3x 4) ( x + )(3x 4) = 3x 4x+ 6x 8 = 3x + x 8 You need to multiply each term in the first bracket by each term in the second. Use FOIL First, Outer, Inner, Last. You can see a step-by-step version of this example in this PowerPoint presentation. Factorising quadratic expressions Again, this should be revision of GCSE work. It is essential that you are confident in factorisation. Example Factorise the expressions (i) x² + 4x + 3 (ii) x² 4x 1 (iii) x² 7x + 6 MEI, 0/03/05 1/7

15 (i) x² + 4x + 3 = (x..)(x...) x² + 4x + 3 = (x + 1)(x + 3) Start with an x in each bracket You need two numbers whose sum is 4 and whose product is 3. These are +1 and +3. (ii) x² 4x 1 = (x..)(x..) x² 4x 1 = (x 6)(x + ) Start with an x in each bracket You need two numbers whose sum is 4 and whose product is 1. These are 6 and +. (iii) x² 7x + 6 = (x..)(x..) In this case you need to start with x in one bracket and x in the other. x² 7x + 6 = (x 3)(x ) It is not so straightforward to find the two numbers in this case, because of the x in one bracket. The two numbers must have a product of +6, and as the coefficient of x is negative, they must both be negative. Try the different possibilities ( 1 and 6, or and 3, in either order), until you find the correct one. You can see step-by-step examples of factorising quadratics in this PowerPoint presentation. For practice in examples like the ones above, try the interactive resource Factorising quadratics. Sometimes algebraic expressions which look quite complicated can be simplified by factorising. Example 3 x 1 Simplify x + x 3. x 1 ( x+ 1)( x 1) = x + x 3 ( x+ 3)( x 1) = x + 1 x + 3 It can be tempting to try to cancel x² from the top and the bottom. Don t! You can only cancel something which is a factor of the top and the bottom. You can now cancel the factor (x 1) from the top and the bottom. MEI, 0/03/05 /7

16 Graphs of quadratic functions Factorising a quadratic expression gives you information about the graph of a quadratic function. Do not think of this work as just algebraic manipulation, think about it also in terms of the graph of the function. Linking algebra and graphs is a very important mathematical skill; the good news is that being able to consider problems both algebraically and graphically usually makes them easier! A graphic calculator or computer package will be very useful. You may already be familiar with the graph of the simplest quadratic function, y = x². y = x² The curve given by graphs of quadratics is called a parabola. Notice that all quadratic graphs have reflection symmetry. The mirror line is always a vertical line through the turning point or vertex of the curve (shown in yellow on these graphs). All other quadratic graphs have basically the same shape, but they may be stretched, squashed, shifted or inverted. y = x² x y = 3 x x² Factorise the equations of these graphs. What is the relationship between the factorised form and the graph? Can you explain this? Notice that the graphs of functions with a negative x² term are inverted (upside down). You will look at the graphs of quadratic expressions in more detail in Polynomials Section 3. MEI, 0/03/05 3/7

17 Solving quadratic equations by factorisation Solving quadratic equations is important not just from the algebraic point of view, but because it gives you information about the graph of a quadratic function. The solutions of the equation ax² + bx + c = 0 tells you where the graph of the function y = ax² + bx + c crosses the x-axis, since these are the points where y = 0. Some quadratic equations can be solved by factorising. Example 4 Solve these quadratic equations by factorising. (a) x² + x 8 = 0 (b) x² + 11x + 1 = 0 (a) x² + x 8 = 0 (x + 4)(x ) = 0 x + 4 = 0 or x = 0 x = 4 or For this expression to be zero, one or other of the factors must be zero. (b) x² + 11x + 1 = 0 (x + 3)(x + 4) = 0 x + 3 = 0 or x + 4 = 0 3 x = or 4 For practice in examples like the ones above, try the interactive resource Solving quadratics by factorisation. Solving quadratic equations using the formula It is possible to solve quadratic equations using the method of completing the square (see pages 19 0 of the textbook); however the quadratic formula is just a generalisation of this method. Pages 0 1 of the textbook show how the quadratic formula is derived from the technique of completing the square. Completing the square is covered in greater detail in Polynomials Section 3. The quadratic formula for the solutions of the equation ax² + bx + c = 0 is b± b² 4ac x =. a The expression b² 4ac is called the discriminant. This is very important as it tells you something about the nature of the solutions. In each case the solution(s) correspond to the points where the graph meets the x-axis. MEI, 0/03/05 4/7

18 If the discriminant is positive, then there are two real solutions. (If the discriminant is a positive square number, then the two real solutions are rational and it is possible to solve the equation by factorisation; otherwise the solutions are irrational and you must use the quadratic formula.) y = x² + x 3 Discriminant = 16 Two rational solutions y = x² + x 3 Discriminant = 13 Two real, irrational solutions If the discriminant is zero, then the quadratic is a perfect square and there is one real solution, which can be found by factorisation. y = x² + x + 1 Discriminant = 0 One real solution If the discriminant is negative, then there are no real solutions. y = x² + x + Discriminant = -4 No real solutions As the graph does not meet the x-axis, there cannot be any real solutions. When you need to solve a quadratic equation, it is useful to quickly work out the discriminant before you start, so that you know whether there are real solutions, and whether the equation can be solved by factorisation. MEI, 0/03/05 5/7

19 Example 5 For each of the following quadratic equations, find the discriminant and solve the equation, where possible, by a suitable method (i) x ² 5x + 1 = 0 (ii) 6 x ² + 11x 10 = 0 (iii) 3 x ² x + 4 = 0 (iv) 4 x ² + 1x + 9 = 0 (i) a =, b = -5, c = 1 Discriminant = ( 5)² 4 1 = 5 8 = 17 Since the discriminant is positive, there are two real solutions. As it is not a square number, the equation must be solved using the quadratic formula. b ± b² 4ac x = a 5 ± 17 = 5 ± 17 = 4 =.81or 0.19 (3 s.f.) (ii) a = 6, b = 11, c = -10 Discriminant = 11 ² = = 361 Since the discriminant is positive, there are two real solutions. As it is a square number (19²), the equation can be solved by factorisation. 6x² + 11x 10 = 0 (3x )(x + 5) = 0 x = 3 or x = 5 (iii) a = 3, b = -, c = 4 Discriminant = ( )² = 4 48 = 44 Since the discriminant is negative, there are no real solutions. (iv) a = 4, b = 1, c = 9 Discriminant = 1 ² = = 0 Since the discriminant is zero, there is one solution and the equation can be solved by factorisation into a perfect square. 4x² + 1x + 9 = 0 (x + 3)² = 0 x = 3 Problem solving Some problems, when translated into algebra, involve quadratic equations. Example 6 A rectangular box has width cm greater than its length, and height 3 cm less than its length. The total surface area of the box is 548 cm². What are the dimensions of the box? MEI, 0/03/05 6/7

20 Let the length of the box be x cm. The width of the box is x + cm, and the height is x 3 cm. The surface are of the box is given by x(x + ) + x(x 3) + (x + )(x 3) x(x + ) + x(x 3) + (x + )(x 3) = 548 x(x + ) + x(x 3) + (x + )(x 3) = 74 x² + x + x² 3x + x² x 6 = 74 3x² x 80 = 0 (3x + 8)(x 10) = 0 x = 10 Divide through by The discriminant is 3364, which is 58², so this must factorise 3x + 8 = 0 gives a negative value of x, which does not make sense in this context. So the solution must be x 10 = 0. The length of the box is 10 cm, the width is 1 cm and the height is 7 cm. Notice that in Example 6, you could discard one of the possible solutions as a negative solution did not make sense in the context. This is not always the case. In some situations, a negative solution can have a practical meaning. For example if the height of a stone thrown from the edge of a cliff is negative, this simply means that the stone is below the level of the cliff at that point. However, if the stone was thrown from level ground, then a negative height does not make sense. Some problems leading to quadratic equations do have two possible solutions. Always consider whether your solution(s) make sense in the context. For practice in examples like the ones above, try the interactive resource Forming and solving quadratics. MEI, 0/03/05 7/7

21 Basic Algebra Section : Quadratics Crucial points 1. Make sure that you can multiply out and factorise confidently These algebraic skills are vital in many areas of mathematics at this level. Practise them until you can do them confidently.. Remember the relationship between the solutions of a quadratic equation and the corresponding quadratic graph The solutions of a quadratic equation tell you where the corresponding quadratic graph crosses the x-axis. This is very useful in sketching quadratic graphs. 3. Remember how to calculate the discriminant and what information it gives you Remember that the discriminant of a quadratic equation gives you useful information about the nature of its solutions, so it is often useful to work out the discriminant before you try to solve the equation. 4. When solving problems, make sure your answer makes sense Always look at your answers to problems in the light of the original question. If there are two solutions, do they both make sense, or should one be discarded? Think carefully about the meaning of a negative solution, since this may or may not be a valid solution. MEI, 0/03/05 1/1

22 Notes and Examples Core 1 Basic Algebra Section 3: Simultaneous equations These notes contain subsections on Linear simultaneous equations One linear and one quadratic equation Linear simultaneous equations Simultaneous equations involve more than one equation and more than one unknown. To solve them you need the same number of equations as there are unknowns. One method of solving simultaneous equations involves adding or subtracting multiples of the two equations so that one unknown disappears. This method is called elimination, and is shown in the next example. Example 1 Solve the simultaneous equations 3p+ q = 5 p q = 4 Adding or subtracting these equations will not eliminate either p or q. However, you can multiply the first equation by, and then add. This will eliminate p p+ q = 5 p q = 4 6p+ q = 10 p q = 4 Adding: 7p = 14 p = 3 + q = 5 6+ q = 5 q = 1 The solution is p =, q = -1. Now substitute this value for p into one of the original equations you can use either, but in this example, 1 is used. MEI, 03/03/05 1/4

23 Notice that, in Example 1, you could have multiplied equation by 3 and then subtracted. This would give the same answer. Sometimes you need to multiply each equation by a different number before you can add or subtract. This is the case in the next example. Example 5 pencils and rubbers cost pencils and 3 rubbers cost.35 Find the cost of a pencil and the cost of a rubber. 1 5 p+ r = p+ 3r = 35 Let p represent the cost of a pencil and r represent the cost of a rubber. It is easier to work in pence. The easiest method is to multiply equation 1 by 3 and equation by. (You could of course multiply 1 by 8 and by 5) p+ 6r = p+ 6r = 470 Subtracting: p = 0 p = r = r = 150 r = 50 r = 5 A pencil costs 0p and a rubber costs 5p. Substitute this value of p into equation 1 An alternative method of solving simultaneous equations is called substitution. This can be the easier method to use in cases where one equation gives one of the variables in terms of the other. This is shown in the next example. Example 3 Solve the simultaneous equations 3x y = 11 y = 5 x MEI, 03/03/05 /4

24 3x (5 x) = 11 3x 10+ 4x= 11 7x = 1 x = 3 y = 5 3 = 5 6 = 1 Core 1 Substitute the expression for y given in the second equation, into the first equation. Multiply out the brackets Substitute the value for x into the original second equation The solution is x = 3, y = -1 For some practice in examples like the ones above, try the interactive resources Solving linear simultaneous equations and Forming and solving linear simultaneous equations. One linear and one quadratic equation When you need to solve a pair of simultaneous equations, one of which is linear and one of which is quadratic, you need to substitute the linear equation into the quadratic equation. Example 4 Solve the simultaneous equations x + y = 6 x y = 1 Start by using the linear equation to write one variable in terms of the other. x = y + 1 ( 1) 6 y+ + y = y + y+ 1+ y = y + y = (3y+ 5)( y 1) = 0 y = 5 y = 3 or 1 Now substitute this expression for y into the first equation Multiply out, simplify and factorise Sometimes you will need to use the quadratic formula to solve the resulting quadratic equation. 5 5 y = 3 x= y+ 1= 3+ 1= 3 y = 1 x= y+ 1= 1+ 1= 5 The solutions are x=, y = and x =, y = Now substitute each value for y into the linear equation to find the corresponding values of x MEI, 03/03/05 3/4

25 When you try to solve a quadratic equation and a linear equation simultaneously, you usually get solutions. However, 0 or 1 solutions are also possible. Can you explain why? [Hint: think about the graphs] For some practice in examples like the ones above, involving a quadratic equation, try the interactive resource Simultaneous equations involving quadratics. MEI, 03/03/05 4/4

26 AS Core Basic Algebra Section 3: Simultaneous equations Crucial points 1. Be careful with signs when using the elimination method. Always check your solution Just substitute your solution into both of the original equations to make sure that it fits. 3. Remember that for non-linear simultaneous equations there may be more than one solution When you solve simultaneous equations where one is linear and one is quadratic, you should normally end up with two solutions unless: there is a repeated root (in which case the graph of the linear function would be a tangent to the graph of the quadratic) or there are no solutions (in which case the graph of the linear function would not cross or touch the graph of the quadratic). MEI, 03/03/05 1/1

27 Coordinate Geometry Glossary Asymptote A straight line which is approached by a curve, but the curve never reaches the line. Asymptotes are usually marked on graphs as dotted lines. Chord A straight line joining two points on the circumference of a circle Continuous graph A graph which has no breaks is said to be continuous. Coordinates A means of describing a position relative to some fixed point (the origin). In Cartesian coordinates position is given in terms of two perpendicular directions, x and y. Discontinuous graph A graph which has two or more separate branches which do not meet is said to be discontinuous. Equation of a circle The equation of a circle with centre (a, b) and radius r is given by ( ) ( ) x a + y b = r When the centre of the circle is at the origin, the equation simplifies to x + y = r Equation of a straight line The equation of a straight line with gradient m which intercepts the y-axis at the point (0, c) is given by y = mx + c The equation of a straight line with gradient m which passes through the point (, ) given by ( ) y y = m x x 1 1 x y is 1 1 MEI, 11/04/05 1/1

28 Gradient of a line The gradient of a line, often denoted by m, is a measure of its slope., x, y on a straight line, the gradient of the line is given For two points A ( x y ) and B ( ) 1 1 by gradient = = 1 increase in y from A to B increase in x from A to B y y1 x x Intersection of two lines or curves The point(s) where two lines or curves intersect can be found by solving their equations simultaneously. If two points of intersection are the same, then the line just touches the curve (i.e. it is a tangent to the curve). Midpoint of a line The point halfway between the two ends of a line., x, y, the midpoint M of the line AB is given by For any two points A ( x y ) and B ( ) 1 1 x1+ x y1+ y M =,. Parallel lines If two lines are parallel, they have the same gradient. Perpendicular bisector of a line For a line AB, the perpendicular bisector of AB is perpendicular to AB and passes through the midpoint of AB. Perpendicular lines Two perpendicular lines are at right angles to each other. If two lines with gradients m 1 and m are perpendicular, then m 1 m = -1. Plotting a graph This involves marking points on graph paper and joining them up as accurately as possible. MEI, 11/04/05 /

29 Pythagoras theorem In a right-angled triangle with side lengths a, b and c (c being the hypoteneuse) a c b Pythagoras theorem states that a + b = c Sketching a graph This means drawing the right general shape of a graph, usually marking the coordinates of important points such as intersections with the coordinates axes and turning points. Tangent A tangent to a curve is a straight line which just touches the curve. The tangent to a circle at any point is perpendicular to the radius of the circle through that point. MEI, 11/04/05 3/3

30 Coordinate Geometry Section 1: Points and straight lines Notes and Examples These notes contain sub-sections on: Gradients, distances and mid-points The equation of a straight line The intersection of two lines Gradients, distances and mid-points You will have met gradients before at GCSE. To revise finding the gradient of a line from a diagram, use the interactive resource The gradient of a line. Remember that lines which go downhill have negative gradients. To find the gradient of a straight line between two points ( x, y ) and (, ) use the formula y y1 gradient =. x x 1 If two lines are parallel, they have the same gradient. If two lines with gradients m 1 and m are perpendicular, then mm 1 = x y, Example 1 P is the point (-3, 7). Q is the point (5, 1). Calculate (i) the gradient of PQ (ii) the gradient of a line parallel to PQ (iii) the gradient of a line perpendicular to PQ. (i) Choose P as ( x 1, y 1) and Q as ( x, y ). or vice versa: it will still give the same answer (WHY?) Gradient of PQ = y x y1 x 1 = ( 3) = 6 8 = 3 4 Notes: (1) Draw a sketch and check that your answer is sensible (e.g. has negative gradient). () Check that you get the same result when you choose Q as ( x 1, y 1 ) and P as ( x, y ). MEI, 11/04/05 1/1

31 (ii) When two lines are parallel their gradients are equal. ( m 1 = m ) 3 So the gradient of the line parallel to PQ is also. 4 (iii) When two lines are perpendicular mm 1 = 1. (WHY?) 3 So m = m = 3 The gradient of a line perpendicular to PQ is 3 4. For further practice in examples like the one above, try the interactive resources The gradient of a line between two points and The gradient of a perpendicular. The midpoint of a line joining two points ( x, y ) and (, ) 1 1 Midpoint x + x, y + y = 1 1 x y is given by ( x, y ) The x-coordinate of M is halfway between x 1 and x. The y-coordinate of M is halfway between y 1 and y. M The length of a line joining two points ( x, y ) and (, ) Pythagoras Theorem. ( x, y ) x y can be found using Length = ( x x ) + ( y y ) 1 1 ( x, y ) y y 1 ( x, y ) 1 1 x x 1 MEI, 11/04/05 /

32 Example A is the point (, -6). B is the point (-3, 4). Calculate (i) the midpoint of AB (ii) the length of AB. or vice versa, it will still give the same answer (WHY?) Choose A as ( x 1, y 1) and B as ( x, y ). (i) Midpoint is x1 + x y1 + y, i.e. + ( 3) 6 + 4, 1 =, 1 (ii) The distance AB is given by d = ( x x ) + ( y y ) 1 1 = ( ( 3)) + (( 6) 4) = (5) + ( 10) = = 15 Note: The answer is often left like this if the square root is not exact. However since 15 = 5 5 then 15 = 5 5 = 5 5 is perhaps a simpler form. For further practice in examples like the one above, try the interactive resources The distance between two points and The midpoint between two points. The equation of a straight line The equation of a straight line is often written in the form y = mx + c, where m is the gradient and c is the intercept with the y-axis. Example 3 Find (i) the gradient and (ii) the y-intercept of the following straight-line equations. (a) 5y = 7x 3 (b) 3x+ 8y 7= 0 MEI, 11/04/05 3/3

33 (a) Rearrange the equation into the form y = mx + c y = 7x 3 becomes y = x so m = 7 and 3 5 c = Note the minus sign (i) The gradient is (ii) The y-intercept is 5. (b) Rearrange the equation into the form y = mx + c. 3x+ 8y 7= 0 becomes 8y = 3x+ 7 3 giving y = 8 x+ 7 8 Note the minus sign 3 so m = 8 and c = (i) The gradient is 8 (ii) The y-intercept is 7. 8 Sometimes you may need to sketch the graph of a line. A sketch is a simple diagram showing the line in relation to the origin. It should also show the coordinates of the points where it cuts one or both axes. Example 4 Sketch the lines (a) 5y = 7x 3 (b) 3x+ 8y 7= 0 (a) From Example 3 you know that 5y = 7x 3 has gradient 7 and y-intercept Sketch of 5y = 7x 3 The gradient is positive, so the line slopes upwards from left to right. It s also greater than 1 and so steeper than 45 degrees. y This means the line goes through (0, -0.6) which is below the origin. 3 5 x MEI, 11/04/05 4/4

34 3 (b) From Example 3 you know that 3x+ 8y 7= 0 has gradient and y-intercept The gradient is negative, so the line slopes downwards from left to right. It is also and less than 1 and so less steep than 45 degrees. This means the line goes through (0, 0.875) which is above the origin. Sketch of 3x+ 8y 7= 0 y 7 8 x Sometimes you may need to find the equation of a line given certain information about it. If you are given the gradient and intercept, this is easy: you can simply use the form y = mx + c. However, more often you will be given the information in a different form, such as the gradient of the line and the coordinates of one point on the line (as in Example 5) or just the coordinates of two points on the line (as in Example 6). In such cases you can use the alternative form of the equation of a straight line. For a line with gradient m passing through the point ( x1, y 1), the equation of the line is given by y y = m x x. ( ) 1 1 Example 5 (i) Find the equation of the line with gradient and passing through (3, -1). (ii) Find the equation of the line perpendicular to the line in (i) and passing through (3, -1). (i) The equation of the line is y y1 = m( x x1) y ( 1) = ( x 3) y+ 1= x 6 y = x 7 m = and ( x 1, y 1 ) is (3, -1) You should check that the point (3, -1) satisfies your line. If it doesn t, you must have made a mistake! MEI, 11/04/05 5/5

35 1 (ii) For two perpendicular lines mm 1 = 1, so the gradient of the new line is. The equation of the line is y y1 = m( x x1) y ( 1) = 1 ( x 3) y = x 3 y = x 1 y = x m = and x, y is (3, -1) ( ) 1 1 The final form of the equation can be written in various different ways: e.g. y = -x + 1 (This form has no fractions.) e.g. y + x = 1 (This has no fractions and avoids having a negative sign at the start of the right hand side.) In the next example, you are given the coordinates of two points on the line. Example 6 P is the point (3, 8). Q is the point (-1, 5). Find the equation of PQ. Choose P as ( x 1, y 1) and Q as ( x, y ). One method is to find the gradient and then use this value and one of the points in y y = m( x x ) 1 1 Gradient of PQ = y x y1 x = 1 3 = 3 4 = 4 3 Now use y y1 = m( x x1) 3 y 8= 4 ( x 3) 4( y 8) = 3( x 3) 4y 3= 3x 9 4y = 3x+ 3 You should check that P and Q satisfy your line. An alternative approach to the above examples is to put the formula for m into the straight line equation to obtain y y1 y y1 = ( x x1) x x 1 MEI, 11/04/05 6/6

36 and then make the substitutions. This is equivalent to the first method, but does not involve calculating m separately first. For further practice in examples like the one above, try the interactive resource The equation of a line between two points. The intersection of two lines The point of intersection of two lines is found by solving the equations of the lines simultaneously. This can be done in a variety of ways. When both equations are given in the form y = then equating the right hand sides is a good approach (see below). If both equations are not in this form, you can rearrange them into this form first, then apply the same method. Alternatively, you can use the elimination method (see Basic Algebra section 3) if the equations are in an appropriate form. Example 7 Find the point of intersection of the lines y = 3x and y = 5x 8. 3x = 5x 8 = x 8 6= x x = 3 Substitute into one of the equations to find y Substituting x = 3 into y = 3x gives y = 3 3 = 7 The point of intersection is (3, 7) Check that (3, 7) satisfies the second equation. MEI, 11/04/05 7/7

37 Coordinate geometry Section 1: Points and straight lines Crucial points 1. Ensure you can calculate the gradient of the line correctly. The gradient of a line, m, is given by change in y m = change in x The gradient, m, of the line joining two points, ( x 1, y 1 ) and ( x, y ) is given by y y1 m = x x 1 Don t get the gradient calculation upside-down! The gradient tells you by how much y changes when x increases by 1.. Make sure you can calculate the y-intercept of a straight-line graph. The y-intercept of a line is where it crosses the y-axis. It is the value of y when x = Make sure you understand how the standard straight-line equation works. An equation which can be written in the form y = mx+ c represents a straight line. m is the gradient and c is the y-intercept. 4. Make sure you understand the conditions on the gradients of lines for the lines to be parallel or perpendicular. If two lines have gradients m 1 and m then: the lines are parallel if m1 = m. 1 the lines are perpendicular if mm 1 = 1 (i.e. if = m = m ). 5. Make sure you understand and can remember how to calculate the distance between two points, x, y is given by The distance, d, between two points, ( x y ) and ( ) d = ( x x ) + ( y y ) 1 1 This is just from applying Pythagoras s theorem MEI, 11/04/05 1/1

38 6. Make sure you understand and can remember how to calculate the midpoint of the line between two points. x, y and The coordinates of the midpoint, M, of the line joining ( 1 1) ( x, y ) are given by 1 1 M = x + x, y + y 7. Make sure you can calculate the equation of a straight line. from the coordinates of two points on it. from its gradient and the coordinates of a point on it. MEI, 11/04/05 /

39 Notes and Examples Core 1 Coordinate Geometry Section : Curves and circles These notes and examples contain subsections on The equation of a circle Finding the equation of a circle The intersection of a line and a curve The intersection of two curves The equation of a circle Start this section by looking at the Circles dynamic spreadsheet. Select the Circle Equations sheet. First, set the centre of the circle to be the origin and vary the radius. Look at how the equation of the circle changes. Now vary the coordinates of the centre of the circle, and look at how the equation of the circle changes. The general equation of a circle, centre (0, 0) and radius r is x² + y² = r². The general equation of a circle, centre (a, b) and radius r is (x a)² + (y b)² = r². Make sure you understand why these equations describe circles. See page 6 in the textbook for help. Example 1 For each of the following circles find (i) the coordinates of the centre and (ii) the radius. (a) x² + y² = 49 (b) (x + )² + (y 6)² = 9 (a) x² + y² = 49 can be written as x² + y² = 7². (i) The coordinates of the centre are (0, 0) (ii) The radius is 7. This is a particular case of the general form x² + y² = r² which has centre (0, 0) and radius r. This is a particular case of the general form (x a)² + (y b)² = r² which has centre (a, b) and radius r. (b) (x + )² + (y 6)² = 9 can be written as (x ( ))² + (y 6)² = 3². (i) The coordinates of the centre are (, 6) (ii) The radius is 3. MEI, 11/04/05 1/1

40 For practice in examples like the one above, try the interactive resource Finding the radius and centre of a circle (circle equation in its simplest form). Sometimes the circle equation needs to be rearranged into its standard form before you can find the centre and radius. Example Show that the equation and radius. x y x y = 0 represents a circle, and find its centre The general equation of a circle is ( x a) + ( y b) = r Multiplying out: x ax+ a + y by+ b = r x + y ax by + a + b r = 0 Comparing with the original equation: a = 4 a = b= 6 b= 3 a + b r = r = 3 r = 16 The equation can be written as ( x+ ) + ( y 3) = 4 This is the equation of a circle, centre (-, 3), radius 4. For practice in examples like the one above, try the interactive resource Finding the radius and centre of a circle (circle equation in its expanded form). In the example above, you are using the technique of completing the square, which is covered briefly in Chapter 1 (pages 19 0), and in more depth in Chapter 3 (pages 98 99). Finding the equation of a circle In Section 1 you looked at different ways of finding the equation of a line. You can find the equation of a line from the gradient and the intercept, or from the gradient and one point on the line, or from two points on the line. In the same way, there are several ways of finding the equation of a circle, depending on the information available. Finding the equation of a circle from the radius and centre Example 3 Find the equation of each of the following. (a) a circle, centre (0, 0) and radius 4. (b) a circle, centre (3, 4) and radius 6. MEI, 11/04/05 /

41 (a) The equation of a circle centre the origin is x² + y² = r² r = 4 so the equation is x² + y² = 4² i.e. x² + y² = 16 (b) The equation of a circle centre (a, b) and radius r is (x a)² + (y b)² = r² a = 3, b = 4 and r = 6 so the equation is (x 3)² + (y ( 4))² = 6² i.e. ( x 3)² + ( y+ 4)² = 36 Finding the equation of a circle from its centre and one point on its circumference If you know the centre of the circle and one point on its circumference, you can find the radius by calculating the distance between these two points. You can then find the equation of the circle. Example 4 Find the equation of the circle, centre (1, -), which passes through the point (-, -3). The distance r between (1, -) and (-, -3) is given by: = = 10 ( 1 ( ) ) ( ( 3) ) r = + The radius of the circle is therefore 10. The equation of the circle is ( x 1) + ( y+ ) = 10 For practice in examples like the one above, try the interactive resource Find the equation of a circle. Finding the equation of a circle from three points on its circumference To find the equation of a line, you need the coordinates of two points on the line. To find the equation of a circle, you need the coordinates of three points on the circumference of the circle. One method is illustrated by the Circles dynamic spreadsheet. Select the sheet Circumcentre and follow the instructions on the sheet. This demonstration shows that the centre of the circle is the intersection of the perpendicular bisector of each pair of points. To find the centre of a circle through three points A, B and C, it is sufficient to find two of the perpendicular bisectors. For example, you can find the equations of the perpendicular bisectors of AB and BC, and then solve these MEI, 11/04/05 3/3

42 equations simultaneously to find the point of intersection, i.e. the centre of the circle. You can then use the coordinates of the centre and one of the three points A, B and C to find the radius of the circle (as in Example 4), and hence find the equation of the circle. Example 5 Find the equation of the circle passing through A (1, 3), B (9, 1) and C (8, 4). A sketch is often helpful You want to find the equation of the perpendicular bisector of AB. This is perpendicular to AB and passes through the midpoint M of AB. The gradient of AB is found by using m y y 1 = x x1 1 m = ( 3) 4 = = 1 Note: Looking at the sketch we expect the gradient of AB to be positive. Using mm 1 = 1, the gradient of the perpendicular bisector is. 1 1 The midpoint M of AB is found by using M = x + x, y + y You are given A(1, -3) and B(9, 1) so M is, = (5, 1) MEI, 11/04/05 4/4

43 The perpendicular bisector is found using y y1 = m( x x1) with ( x1, y1 ) = (5, 1) and m =. so y ( 1) = ( x 5 ) y + 1 = x + 10 y = x + 9 (equation I) Next, use the same method to find the perpendicular bisector of BC. The gradient of BC is = Note: Looking at the sketch, we expect the gradient of BC to be negative. Therefore the gradient of the perpendicular bisector of BC is The midpoint N of BC is, so N is (8.5,.5). The equation of the perpendicular bisector is y.5 = 1 3 (x 8.5) 3(y.5) = x 8.5 3y 7.5 = x 8.5 3y = x + 1 (equation II) y = x + 9 (equation I) 3y = x + 1 (equation II) Next, find the coordinates of the centre of the circle by solving equations (I) and (II) simultaneously. Substituting (I) into (II) 3( x + 9) = x + 1 6x + 7 = x = 7x x = 4 Substituting x = 4 into equation (I) gives y = (4) + 9 = 1 So the coordinates of the centre are (4, 1). Note: Looking at the sketch this appears to be a plausible result. The radius is the distance between the centre (4, 1) and a point on the circumference such as (9, 1). This can be found by using d = ( x x ) + ( y y ) 1 1 radius = ( 9 4) + ( 1 1) = 5 = 5. MEI, 11/04/05 5/5

44 Finally, using the general form (x a)² + (y b)² = r² with a = 4, b = 1 and r = 5 the equation of the circle is (x 4)² + (y 1)² = 5. Note: You should check that each of the points A, B and C satisfy this equation. The intersection of a line and a curve Just as the point of intersection of two straight lines can be found by solving the equations of the two lines simultaneously, the point(s) of intersection of a line and a curve can be found by solving their equations simultaneously. In many cases, the equations of both the line and the curve are given as an expression for y in terms of x. When this is the case, a sensible first step is to equate the expressions for y, as this leads to an equation in x only. Example 6 Find the coordinates of the points where the line y = x + meets the curve y = x² 3x + 5. x² 3x + 5 = x + x² 4x + 3 = 0 (x 3)(x 1) = 0 x = 3 or x = 1 When x = 3 then y = 3 + = 5 When x = 1 then y = 1 + = 3. The points where the line meets the curve are (3, 5) and (1, 3). Equate the expressions for y to give an equation in x only Substitute the x values into the equation of the line You should check that each of these points satisfies the equation of the curve. (You have already used the equation of the line to find the y-values). Notice that this problem involved solving a quadratic equation, which in this case had two solutions, showing that the line crossed the curve twice. However, the quadratic equation could have had no solutions, which would indicate that the line did not meet the curve at all, or one repeated solution, which would indicate that the line touches the curve. For practice in examples like the one above, try the interactive resource Quadratic and line intersection. MEI, 11/04/05 6/6

45 The next example looks at the intersection of a line and a circle. Before reading this example, look at the Circles dynamic spreadsheet. Select the sheet Circle and a line. Try varying the equation of the line and/or the circle, and make sure that you can see that there may be two intersections, no intersections or one intersection (in which case the line touches the circle). Example 7 Find the coordinates of the point(s) where the circle (i) the line y = 5 (ii) the line x = 1 (iii) the line y = x ( x ) ( y 1) = meets (i) Substituting y = 5 into the equation of the circle: ( x + ) + (5 1) = 9 ( x + ) + 16= 9 ( x + ) = 7 There are no solutions. The line does not meet the circle. The expression (x + )² cannot be negative (ii) (iii) Substituting x = 1 into the equation of the circle: (1+ ) + ( y 1) = ( y 1) = 9 The point is on the line x = 1, so its x-coordinate must be 1. ( y 1) = 0 y = 1 The line touches the circle at (1, 1). Substituting y = x into the equation of the circle: ( x+ ) + ( x 1) = 9 ( x+ ) + (1 x) = 9 x + 4x x+ x = 9 x + x 4= 0 x + x = 0 ( x 1)( x+ ) = 0 x = 1 or x = Substitute the x values into the equation of the line to find the y-coordinates. When x = 1, y = 1 = 1 When x =, y = ( ) = 4 The line crosses the circle at (1, 1) and (, 4). For practice in examples like the one above, try the interactive resource Circle and line intersection. MEI, 11/04/05 7/7

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