1 AS-Level Maths: Core 1 for EdexcelC1.2 Algebra and functions 2 This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 50 © Boardworks Ltd 2005

2 Quadratic expressionsFactorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions Contents 2 of 50 © Boardworks Ltd 2005

3 Quadratic expressionsA quadratic expression is an expression in which the highest power of the variable is 2. For example: t2 2 x2 – 2 w2 + 3w + 1 4 – 5g2 ax2 + bx + c (where a ≠ 0) The general form of a quadratic expression in x is: x is a variable. As well as the highest power being 2, no power in a quadratic expression can be negative or fractional. (That would be an example of a polynomial.) Compare each of the quadratic expressions given with the general form. In x2 – 2, a = 1, b = 0 and c = –2. In w2 + 3w + 1, a = 1, b = 3 and c = 1. This is a quadratic in w. In 4 – 5g2, a = –5, b = 0 and c = 4. This is a quadratic in g. In t2/2, a = ½, b = 0 and c = 0. This is a quadratic in t. a is the coefficient of x2. b is the coefficient of x. c is a constant term.

4 Factorizing quadraticsQuadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions Contents 4 of 50 © Boardworks Ltd 2005

5 Factorizing quadratic expressionsFactorizing an expression is the inverse of expanding it. Expanding or multiplying out Factorizing (x + 1)(x + 2) x2 + 3x + 2 In this example, (x + 1) and (x + 2) are factors of x2 + 3x + 2. When we expand an expression we multiply out the brackets. When we factorize an expression we write it with brackets.

6 Factorizing quadratic expressionsNo constant term Quadratic expressions of the form ax2 + bx can always be factorized by taking out the common factor x. For example: 3x2 – 5x = x(3x – 5) The difference between two squares When a quadratic has no term in x and the other two terms can be written as the difference between two squares, we can use the identity Explain to the students that the approach we take to factorizing a quadratic depends on what form it takes. Later, when the discriminant is introduced, students will be able to check whether a quadratic expression factorizes by checking to see whether b2 – 4ac is a perfect square. a2 – b2 = (a + b)(a – b) to factorize it. For example: 9x2 – 49 = (3x + 7)(3x – 7)

7 Factorizing quadratic expressionsQuadratic expressions with a = 1 Quadratic expressions of the form x2 + bx + c can be factorized if they can be written using brackets as (x + d)(x + e) where d and e are integers. If we expand (x + d)(x + e), we have (x + d)(x + e) = x2 + dx + ex + de = x2 + (d + e)x + de This slide explains why, when we factorize an expression in the form x2 + bx + c to the form (x + d)(x + e), the values of d and e must be chosen so that d + e = b and de = c. To check whether quadratic expressions of the form x2 + bx + c factorize, check whether b2 – 4c is a perfect square. This is demonstrated later when the discriminant is introduced. Use the embedded flash activity to demonstrate factorizing quadratics written in this form. In each example the lower of the two hidden integers will be given first.

8 Factorizing quadratic expressionsThe general form Quadratic expressions of the general form ax2 + bx + c can be factorized if they can be written using brackets as (dx + e)(fx + g) where d, e, f and g are integers. If we expand (dx + e)(fx + g), we have (dx + e)(fx + g)= dfx2 + dgx + efx + eg = dfx2 + (dg + ef)x + eg To check whether quadratic expressions of the form ax2 + bx + c factorize, check whether b2 – 4ac is a perfect square. This is demonstrated later when the discriminant is introduced. Factorizing quadratic expressions where the coefficient of x2 is not 1 requires some practice. Students can start by using the identity shown and using trial and error to find values of d, e, f and g such that df = a, (dg + ef) = b and eg = c.

9 Contents Completing the square Quadratic expressionsFactorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions Contents 9 of 50 © Boardworks Ltd 2005

10 Perfect squares Some quadratic expressions can be written as perfect squares. For example: x2 + 2x + 1 = (x + 1)2 x2 – 2x + 1 = (x – 1)2 x2 + 4x + 4 = (x + 2)2 x2 – 4x + 4 = (x – 2)2 x2 + 6x + 9 = (x + 3)2 x2 – 6x + 9 = (x – 3)2 In general: x2 + 2ax + a2 = (x + a) or x2 – 2ax + a2 = (x – a)2 Encourage students to derive the general form of quadratics that are perfect squares before revealing the perfect square itself. This general form could also be written as x2 + bx + (b/2)2 = (x + b/2)2. How could the quadratic expression x2 + 8x be made into a perfect square? We could add 16 to it.

11 Completing the square Adding 16 to the expression x2 + 8x to make it into a perfect square is called completing the square. x2 + 8x = x2 + 8x + 16 – 16 We can write If we add 16 we then have to subtract 16 so that both sides are still equal. By writing x2 + 8x + 16 we have completed the square and so we can write this as x2 + 8x = (x + 4)2 – 16 In general:

12 Completing the square Complete the square for x2 – 10x.Compare this expression to (x – 5)2 = x2 – 10x + 25 x2 – 10x = x2 – 10x + 25 – 25 = (x – 5)2 – 25 Complete the square for x2 + 3x. Compare this expression to

13 How can we complete the square forCompleting the square How can we complete the square for x2 – 8x + 7? Look at the coefficient of x. This is –8 so compare the expression to (x – 4)2 = x2 – 8x + 16. x2 – 8x + 7 = x2 – 8x + 16 – = (x – 4)2 – 9 In general: Students should not need to remember the generalization for the completed square form. With practice they should be able to recall that they need to put half the coefficient of x at the end of the bracket and then subtract the square of this from the constant.

14 Completing the square Complete the square for x2 + 12x – 5.Compare this expression to (x + 6)2 = x2 + 12x + 36 x2 + 12x – 5 = x2 + 12x + 36 – 36 – 5 = (x + 6)2 – 41 Complete the square for x2 – 5x + 7. The last example may need some revision of adding and subtracting fractions. Point out that when algebraic expressions involve fractions, whether they are coefficients or constants, it is preferable to leave them as improper fractions rather than decimals or mixed numbers. Compare this expression to

15 Complete the square for 2x2 + 8x + 3.Completing the square When the coefficient of x2 is not 1, quadratic equations in the form ax2 + bx + c can be rewritten in the form a(x + p)2 + q by completing the square. Complete the square for 2x2 + 8x + 3. Take out the coefficient of x2 as a factor from the terms in x: 2x2 + 8x + 3 = 2(x2 + 4x) + 3 By completing the square, x2 + 4x = (x + 2)2 – 4 so Explain that when the coefficient of x2 is not one we start by taking out the coefficient of x2 from the terms in x. We then complete the square for the expression in the brackets (shown in orange). 2x2 + 8x + 3 = 2((x + 2)2 – 4) + 3 = 2(x + 2)2 – 8 + 3 = 2(x + 2)2 – 5

16 Complete the square for 5 + 6x – 3x2.Completing the square Complete the square for 5 + 6x – 3x2. Take out the coefficient of x2 as a factor from the terms in x: 5 + 6x – 3x2 = 5 – 3(–2x + x2) = 5 – 3(x2 – 2x) By completing the square, x2 – 2x = (x – 1)2 – 1 so Explain that we must take –3 out as a factor. If we take 3 out as a factor then the coefficient of x will not be 1 but –1. Point out to students that dividing + 6x by –3 gives us –2x. Warn students to be very careful when multiplying by negatives throughout the problem. 5 + 6x – 3x2 = 5 – 3((x – 1)2 – 1) = 5 – 3(x – 1)2 + 3 = 8 – 3(x – 1)2

17 Complete the square

18 Solving quadratic equationsQuadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions Contents 18 of 50 © Boardworks Ltd 2005

19 The general form of a quadratic equation in x is:Quadratic equations ax2 + bx + c = (where a ≠ 0) The general form of a quadratic equation in x is: Quadratic equations can be solved by: factorization completing the square, or using the quadratic formula. The solutions to a quadratic equation are called the roots of the equation. A quadratic equation may have: two real distinct roots one repeated root, or no real roots.

20 The roots of a quadratic equationIf we sketch the graph of a quadratic function y = ax2 + bx + c the roots of the equation coincide with the points where the function cuts the x-axis. As can be seen here, this can happen twice, once or not at all. Drag the general parabola into different positions to demonstrate how many times it can cut the x-axis. This is explored in more detail when the discriminant is introduced.

21 Solving quadratic equations by factorizationSolve the equation 5x2 = 3x Start by rearranging the equation so that the terms are on the left-hand side: Don’t divide through by x! 5x2 – 3x = 0 Factorizing the left-hand side gives us x(5x – 3) = 0 Students might be tempted to start by dividing through by x. In a quadratic of this form (where c = 0 and there is always the solution x = 0), this leads to the non-zero root, but it makes it easy to overlook x = 0. If students suggest dividing through, encourage them to factorize instead. Tell students that, in general, when they are solving equations they should never divide through by a variable whose value may be 0. x = 0 So or 5x – 3 = 0 5x = 3

22 Solving quadratic equations by factorizationSolve the equation x2 – 5x = –4 by factorization. Start by rearranging the equation so that the terms are on the left-hand side. x2 – 5x + 4 = 0 We need to find two integers that add together to make –5 and multiply together to make 4. Because 4 is positive and –5 is negative, both the integers must be negative. These are –1 and –4. Factorizing the left-hand side gives us (x – 1)(x – 4) = 0 x – 1 = 0 or x – 4 = 0 x = 1 x = 4

23 Solving quadratics by completing the squareQuadratic equations that cannot be solved by factorization can be solved by completing the square. For example, the quadratic equation x2 – 4x – 3 = 0 can be solved by completing the square as follows: (x – 2)2 – 7 = 0 (x – 2)2 = 7 x – 2 = x = 2 + or x = 2 – x = 4.65 x = –0.646 (to 3 s.f.)

24 Solving quadratics by completing the squareSolve the equation 2x2 – 4x + 1 = 0 by completing the square. Write the answer to 3 significant figures. Start by completing the square for 2x2 – 4x + 1: 2x2 – 4x + 1 = 2(x2 – 2x) + 1 = 2((x – 1)2 – 1) + 1 With practice the square can be completed in fewer steps. This quadratic equation could also be solved using the quadratic equation formula. = 2(x – 1)2 – 2 + 1 = 2(x – 1)2 – 1

25 Solving quadratics by completing the squareNow solving the equation 2x2 – 4x + 1 = 0: 2(x – 1)2 – 1 = 0 2(x – 1)2 = 1 It is often sufficient and indeed more desirable to leave the solutions in surd form. Students should be encouraged to check questions carefully to see which form is required. or x = 1.71 x = (to 3 s.f.)

26 Using the quadratic equation formulaAny quadratic equation of the form ax2 + bx + c = 0 can be solved by substituting the values of a, b and c into the formula Tell students that they are not required to learn this formula since it is given in the examination. It is a complex formula, however, and so explain that some practice is needed to use it correctly. Ask students if they can see how this formula will lead to two solutions. Derivation of this formula by completing the square is left as a student exercise. This formula can be derived by completing the square on the general form of the quadratic equation.

27 Using the quadratic formulaUse the quadratic formula to solve 2x2 + 5x – 1 = 0. 2x2 + 5x – 1 = 0 As before point out that it is often sufficient to leave the solutions in surd form. Questions should be checked carefully to see which form is required. or x = 0.186 x = –2.69 (to 3 s.f.)

28 Using the quadratic formulaUse the quadratic formula to solve 9x2 – 12x + 4 = 0. 9x2 – 12x + 4 = 0 Point out that whenever b2 – 4ac is 0, there will be one repeated root. There is one repeated root:

29 Equations that reduce to a quadratic formSome equations, although not quadratic, can be written in quadratic form by using a substitution. For example: Solve the equation t4 – 5t2 + 6 = 0. This is an example of a quartic equation in t. Let’s substitute x for t2: x2 – 5x + 6 = 0 This gives us a quadratic equation that can be solved by factorization: Point out that a quartic equation will have, at most, four solutions as shown by this example. (x – 2)(x – 3) = 0 x = 2 or x = 3 So t2 = 2 t2 = 3 t = t = or

30 Contents The discriminant Quadratic expressions Factorizing quadraticsCompleting the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions Contents 30 of 50 © Boardworks Ltd 2005

31 The discriminant By solving quadratic equations using the formulawe can see that we can use the expression under the square root sign, b2 – 4ac, to decide how many roots there are. When b2 – 4ac > 0, there are two real distinct roots. When b2 – 4ac = 0, there is one repeated root: When b2 – 4ac < 0, there are no real roots. Explain that when b2 – 4ac is greater than 0 and therefore positive √(b2 – 4ac) can be evaluated and so the quadratic equation will have two distinct (that is, different) roots. When b2 – 4ac is equal to 0, x = –b/2a. This is the only root. When b2 – 4ac is less than 0 and therefore negative √(b2 – 4ac) has no real value and so the quadratic equation has no real roots. Also, when b2 – 4ac is a perfect square, the roots of the equation will be rational and the quadratic will factorize. b2 – 4ac is called the discriminant of ax2 + bx + c

32 The discriminant We can demonstrate each of these possibilities graphically. By changing the values of a, b and c, demonstrate examples of functions that have two real roots, one repeated root and no real roots. Ask students to give their own examples and demonstrate them on the board.

33 Graphs of quadratic functionsQuadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions Contents 33 of 50 © Boardworks Ltd 2005

34 Plotting graphs of quadratic functionsy = ax2 + bx + c (where a ≠ 0) A quadratic function in x can be written in the form: We can plot the graph of a quadratic function using a table of values. For example: Plot the graph of y = x2 – 4x + 2 for –1 < x < 5. x x2 – 4x + 2 y = x2 – 4x + 2 –1 1 2 3 4 5 1 1 4 9 16 25 Talk through the substitution for each value of x to give the corresponding value of y. Ask students if they can tell you between which values of x the roots will be. + 4 + 0 – 4 – 8 – 12 – 16 – 20 + 2 + 2 + 2 + 2 + 2 + 2 + 2 7 2 –1 –2 –1 2 7

35 Plotting graphs of quadratic functionsx –1 1 2 3 4 5 7 –2 y = x2 – 4x + 2 The points given in the table are plotted … y 6 … and the points are then joined together with a smooth curve. 5 4 3 The shape of this curve is called a parabola. Point out that, at this level, graphs are rarely plotted in this way but are usually sketched to show their shape relative to the x- and y-axes and their general features. When a sketch is required we only find the coordinates of the points where the function crosses the axes and the coordinates of any turning points. 2 1 It is characteristic of a quadratic function. –1 1 2 3 4 5 x –1

36 Parabolas Parabolas have a vertical axis of symmetry ……and a turning point called the vertex. When the coefficient of x2 is positive the vertex is a minimum point and the graph is -shaped. When the coefficient of x2 is negative the vertex is a maximum point and the graph is -shaped.

37 Exploring graphs of the form y = ax2 + bx + cChange the values of a, b and c to observe how each one affects the shape and position of the parabola. In particular, draw students’ attention to the fact that when a is positive the parabola is -shaped and when a is negative the parabola is -shaped. Changing the value of a stretches or squashes the graph. Note, too, that when a = 0 the function is no longer quadratic but linear.

38 Sketching graphs of quadratic functionsWhen a quadratic function factorizes we can use its factorized form to find where it crosses the x-axis. For example: Sketch the graph of the function y = x2 – 2x – 3. The function crosses the x-axis when y = 0. x2 – 2x – 3 = 0 (x + 1)(x – 3) = 0 Explain that the equation of the x-axis is y = 0. If we therefore put y = 0 in the equation y = x2 – 2x – 3, the solutions will tell us where the function crosses the x-axis. Point out that, before sketching a graph of a quadratic function, it can also be helpful to find the discriminant to check whether it crosses the x-axis and to look at the coefficient of x2 to check whether the graph will be -shaped or -shaped. x + 1 = 0 or x – 3 = 0 x = –1 x = 3 The function crosses the x-axis at the points (–1, 0) and (3, 0).

39 Sketching graphs of quadratic functionsBy putting x = 0 in y = 2x2 – 5x – 3 we can also find where the function crosses the y-axis. y = 2(0)2 – 5(0) – 3 y = – 3 So the function crosses the y-axis at the point (0, –3). In general: The quadratic function y = ax2 + bx + c will cross the y-axis at the point (0, c). We now know that the function y = x2 – 2x – 3 passes through the points (–1, 0), (3, 0) and (0, –3) and so we can place these points on our sketch.

40 Sketching graphs of quadratic functionsWe can also use the fact that a parabola is symmetrical to find the coordinates of the vertex. y x (–1, 0) (3, 0) The x coordinate of the vertex is half-way between –1 and 3. (0, –3) (1, –4) Ask students to tell you the equation for the axis of symmetry of the parabola. This is x = 1. When x = 1, y = (1)2 – 2(1) – 3 y = –4 So the coordinates of the vertex are (1, –4). We can now sketch the graph.

41 Sketching graphs of quadratic functionsIn general: When a quadratic function is written in the form y = a(x – α)(x – β), it will cut the x-axis at the points (α, 0) and (β, 0). α and β are the roots of the quadratic function. For example, write the quadratic function y = 3x2 + 4x – 4 in the form y = a(x – α)(x – β) and hence find the roots of the function. This function can be factorized as follows, In practice we would probably solve the equation (3x – 2)(x + 2) = 0 to find the roots. This example is intended to illustrate how the equation can be written in the form y = a(x – α)(x – β) where α and β are the roots of the equation. y = (3x – 2)(x + 2) It can be written in the form y = a(x – p)(x – q) as Therefore, the roots are

42 Sketching graphs by completing the squareWhen a function does not factorize we can write it in completed square form to find the coordinates of the vertex. For example: Sketch the graph of y = x2 + 4x – 1 by writing it in completed square form. x2 + 4x – 1 = (x + 2)2 – 5 The least value that (x + 2)2 can have is 0 because the square of a number cannot be negative. (x + 2)2 ≥ 0 Establish that the vertex of this function will be a minimum point since the coefficient of x is positive. (x + 2)2 – 5 ≥ – 5 Therefore The minimum value of the function y = x2 + 4x – 1 is therefore y = –5.

43 Sketching graphs by completing the squareWhen y = –5, we have, (x + 2)2 – 5 = –5 (x + 2)2 = 0 x = –2 The coordinates of the vertex are therefore (–2, –5). The equation of the axis of symmetry is x = –2. y x x = –2 Also, when x = 0 we have y = x2 + 4x – 1 y = –1 We can see from this sketch that the roots of the quadratic y = x2 + 4x – 1 lie between –4 and –5 and between 0 and 1. If the exact values of these roots are required they can be found by solving (x + 2)2 – 5 = 0 to give the roots as x = ±√5 – 2. So the curve cuts the y-axis at the point (–1, 0). (–1, 0) Using symmetry we can now sketch the graph. (–2, –5)

44 Sketching graphs by completing the squareIn general, when the quadratic function y = ax2 + bx + c is written in completed square form as a(x + p)2 + q The coordinates of the vertex will be (–p, q). The axis of symmetry will have the equation x = –p. Also: If a > 0 (–p, q) will be the minimum point. As an extension activity ask students to complete the square for y = ax2 + bx + c to show that p = b/2a and q = c – b2/4. Also, the roots can be written as x = ±√(–q/a) – p. If a < 0 (–p, q) will be the maximum point. Plotting the y-intercept, (0, c) will allow the curve to be sketched using symmetry.

45 Examination-style questionsQuadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions Contents 45 of 50 © Boardworks Ltd 2005

46 Examination-style questiona) Write 2x2 – 8x + 7 in the form a(x + b)2 + c. b) Write down the minimum value of f(x) = 2x2 – 8x + 7 and state the minimum value of x where this occurs. c) Solve the equation 2x2 – 8x + 7 = 0 leaving your answer in surd form. d) Sketch the graph of y = 2x2 – 8x + 7. a) 2x2 – 8x + 7 = 2(x2 – 4x) + 7 = 2((x – 2)2 – 4) + 7 = 2(x – 2)2 – 8 + 7 = 2(x – 2)2 – 1

47 Examination-style questionb) f(x) can be written as f(x) = 2(x – 2)2 – 1 From this we can see that the minimum value of f(x) is –1. This occurs when x = 2. c) 2x2 – 8x + 7 = 0 2(x – 2)2 – 1 = 0 2(x – 2)2 = 1 (x – 2)2 = x – 2 = x = 2 x = 2 or

48 Examination-style questiond) When y = 0, x = 2 – or x = 2 + When x = 0, y = 7 So the graph cuts the coordinate axes at ( , 0), (2 – , 0) and (0, 7). The parabola has a minimum at the point (2, –1). y x 7 –1 2 + 2 –