Chapter 10 Quadratics and Non-linear Graphs PDF

Chapter 10 Quadratics and non-linear graphs Essential mathematics: why skills with quadratic equations are important Quadratic equation algebra skills are widely applied in business, science and e...

Chapter 10 Quadratics and non-linear graphs Essential mathematics: why skills with quadratic equations are important Quadratic equation algebra skills are widely applied in business, science and engineering. For example: Crash investigators substitute the length of tyre skid marks into a quadratic equation. Solving it gives the vehicle’s speed before braking. A graph of sales revenue versus selling price is an inverted parabola modelled by a quadratic equation. Engineers use quadratic equations to model the parabolic shape of support cables on suspension bridges. A parabolic telescope dish focusses incoming parallel radio waves above its vertex. Examples include large radio telescopes and caravan satellite dishes. Algorithms that manipulate a robot use circle equations. A ‘joint’ can move in a semicircle (or circle) and the ‘hand’ can follow a circle (or sphere), centred on the joint. Aerospace engineers use a quadratic equation of height versus time for the flight path of a rocket moving under the influence of gravity. Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  In this chapter 10A Expanding binomial products 10B Factorising a difference of perfect squares 10C Factorising trinomials of the form x2 + bx + c 10D Solving equations of the form ax2 = c 10E Solving equations using the null factor law 10F Applications of quadratics 10G Exploring parabolas 10H Graphs of circles and exponentials Australian Curriculum NUMBER AND ALGEBRA Patterns and algebra Expand binomial products and factorise monic quadratic expressions using a variety of strategies (ACMNA233) Linear and non-linear relationships Explore the connection between algebraic and graphical representations of relations such as simple quadratics, circles and exponentials using digital technology as appropriate (ACMNA239) © Australian Curriculum, Assessment and Reporting Authority (ACARA) Online resources A host of additional online resources are included as part of your Interactive Textbook, including HOTmaths content, video demonstrations of all worked examples, auto-marked quizzes and much more. Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 624 Chapter 10 Quadratics and non-linear graphs 1 Consider the expression 5 + 2ab − b. Warm-up quiz a How many terms are there? b What is the coefficient of the second term? c What is the value of the constant term? 2 Simplify each of the following by collecting like terms. a 7x + 2y − 3x b 3xy + 4x − xy − 5x c 4ab − 2ba 3 Simplify: a 4a b −24mn c 6a × 3a 2 12n d −2x × 3xy e x × (−3) ÷ (9x) f 4x2 ÷ (2x) 4 Expand and simplify by collecting like terms where possible. a 4(m + n) b −3(2x − 4) c 2x(3x + 1) d 4a(1 − 2a) e 5 + 3(x − 4) f 5 − 2(x + 3) + 2 5 Factorise each of the following by taking out a common factor. a 7x + 7 b −9x − 27x2 c a2 + ab 6 Solve: a 2x + 1 = 0 b 2(x − 3) = 0 c 3x + 1 = 4 4 7 a Complete this table for y = x2. x −3 −2 −1 0 1 2 3 y 9 b Plot a graph of y = x2 and state if its shape is: A linear (i.e. a straight line) B non-linear (i.e. not a straight line) y 9 x −3 O 3 −1 8 Answer true (T) or false (F) to the following. a x = 3 is a solution of x2 = 9. b x = −2 is a solution of x2 = −4. c x = 0 is a solution of x(x + 4) = 0. d x = −5 is a solution of x2 = 25. e x = 2 is a solution of x(x + 2) = 0. Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  10A Expanding binomial products 625 10A 10A Expanding binomial products Learning intentions To know the distributive law in order to expand binomial products To be able to expand and simplify expressions including brackets To be able to expand perfect squares and be able to form a difference of perfect squares Key vocabulary: expand, binomial, product, like terms, distributive law, perfect square, difference of perfect squares Expressions that include numerals and variables (or pronumerals) are central to the topic of algebra. Sound skills in algebra are essential for solving most mathematical problems and this includes the ability to expand expressions involving brackets, such as binomial products, perfect squares and the difference of perfect squares. Exploring how projectiles fly subject to Earth’s gravity, for example, can be modelled with expressions with and without brackets. Lesson starter: Why does (x + 1)(x + 2) = x2 + 3x + 2? x 2 Look at this rectangle with side lengths x + 1 and x + 2. What are the areas of the four regions? x Add up the areas to find an expression for the total area. Why does this explain that (x + 1)(x + 2) = x2 + 3x + 2? 1 Key ideas Like terms have the same pronumeral part. They can be collected (added and subtracted) to form a single term. For example: 7x − 11x = −4x and 4a2 b − 7ba2 = −3a2 b. The distributive law is used to expand brackets. a(b + c) = ab + ac (a + b)(c + d) = ac + ad + bc + bd and a(b − c) = ab − ac a a b b ab c ac bc c ac d ad bd (a + b)(c + d) is called a binomial product because each expression in the brackets has two terms. Perfect squares (a + b)2 = (a + b)(a + b) = a2 + ab + ba + b 2 = a2 + 2ab + b2 (a − b)2 = (a − b)(a − b) = a2 − ab − ba + b 2 = a2 − 2ab + b2 Difference of perfect squares (DOPS) (a + b)(a − b) = a2 −  + ab  − b2 ba = a2 − b2 Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 626 Chapter 10 Quadratics and non-linear graphs Exercise 10A Understanding 1–3 3 1 Decide whether each of the following is a perfect square (PS) or a difference of perfect squares (DOPS). a (x + 1)2 b x2 − 16 c 4x2 − 25 d (2x − 3)2 2 Copy and complete. a a(b + c) = b a(b − c) = c (a + b)(c + d) = d (a + b)(a − b) = 3 Use each diagram to help expand the expressions. Hint: Simply add up all the areas a x(x + 2) b (x + 3)(x + 1) c (x + 4)2 inside the rectangular diagram. x x 3 x 4 x x2 x x2 3x x x2 4x 2 2x 1 x 3 4 4x 16 Fluency 4–7(½) 4–7(½) Example 1 Expanding simple expressions Expand and simplify. a −3(x − 5) b −2x(1 − x) Solution Explanation a −3(x − 5) = −3 × x − (−3) × 5 Use the distributive law a(b − c) = ab − ac. = −3x + 15 A negative times a negative is a positive. b −2x(1 − x) = −2x × 1 − (−2x) × x Recall: x × x = x2. = −2x + 2x2 Now you try Expand and simplify. a −2(x + 3) b −5x(3x − 1) 4 Expand and simplify where possible. a 2(x + 5) b 3(x − 4) c −5(x + 3) Hint: a(b + c) = ab + ac d −4(x − 2) e 3(2x − 1) f 4(3x + 1) g −2(5x − 3) h −5(4x + 3) i x(2x + 5) j x(3x − 1) k 2x(1 − x) l 3x(2 − x) m −2x(3x + 2) n −3x(6x − 2) o −5x(2 − 2x) Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  10A Expanding binomial products 627 Example 2 Expanding binomial products Expand the following. a (x + 5)(x + 4) b (2x − 1)(3x + 5) Solution Explanation a (x + 5)(x + 4) = x2 + 4x + 5x + 20 For binomial products use (a + b)(c + d) = 2 = x + 9x + 20 ac + ad + bc + bd. Simplify by collecting like terms: 4x + 5x = 9x b (2x − 1)(3x + 5) = 6x2 + 10x − 3x − 5 Expand using the distributive law and simplify. = 6x2 + 7x − 5 Note that 2x × 3x = 2 × 3 × x × x = 6x2 and −1 × 3x = −3x, −1 × 5 = −5. Now you try Expand the following. a (x − 1)(x + 7) b (3x − 2)(4x − 1) 5 Expand the following. a (x + 2)(x + 8) b (x + 3)(x + 4) Hint: (a + b)(c + d) = ac + ad + bc + bd c (x + 7)(x + 5) d (x + 8)(x − 3) e (x + 6)(x − 5) f (x − 2)(x + 3) g (x − 7)(x + 3) h (x − 4)(x − 6) i (x − 8)(x − 5) j (2x + 1)(3x + 5) k (4x + 5)(3x + 2) l (5x + 3)(2x + 7) m (3x + 2)(3x − 5) n (5x + 3)(4x − 2) o (2x + 5)(3x − 5) Example 3 Expanding perfect squares Expand these perfect squares. a (x + 2)2 b (x − 4)2 Solution Explanation a (x + 2)2 = (x + 2)(x + 2) First write in expanded form, then use the distributive law. = x2 + 2x + 2x + 4 = x2 + 4x + 4 or (x + 2)2 = x2 + 2(x)(2) + 22 (a + b)2 = a2 + 2ab + b2 with a = x and b = 2. = x2 + 4x + 4 b (x − 4)2 = (x − 4)(x − 4) Rewrite and expand, using the distributive law. 2 = x − 4x − 4x + 16 −4 × (−4) = 16 = x2 − 8x + 16 or (x − 4)2 = x2 − 2(x)(4) + 42 Alternatively for perfect squares = x2 − 8x + 16 (a − b)2 = a2 − 2ab + b2. Here a = x and b = 4. Now you try Expand these perfect squares. a (x + 3)2 b (2x − 1)2 Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 628 Chapter 10 Quadratics and non-linear graphs 10A 6 Expand these perfect squares. a (x + 5)2 b (x + 7)2 c (x + 6)2 Hint: Recall: d (x − 3)2 e (x − 8)2 f (x − 10)2 (x + 5)2 = (x + 5)(x + 5) g (2x + 5)2 h (5x + 6)2 i (7x − 1)2 =… Example 4 Expanding to form a difference of perfect squares Expand to form a difference of perfect squares. a (x − 3)(x + 3) b (2x + 1)(2x − 1) Solution Explanation a (x − 3)(x + 3) = x2 + 3x − 3x − 9 x × x = x2 , x × 3 = 3x, −3 × x = −3x, = x2 − 9 −3 × 3 = −9 Note that the two middle terms cancel. or (x − 3)(x + 3) = x2 − 32 (a − b)(a + b) = a2 − b2 = x2 − 9 b (2x + 1)(2x − 1) = 4x2 − 2x + 2x − 1 Expand, recalling that 2x × 2x = 4x2. Cancel the = 4x2 − 1 −2x and + 2x terms. or (2x + 1)(2x − 1) = (2x)2 − (1)2 Alternatively for difference of perfect squares 2 = 4x − 1 (a − b)(a + b) = a2 − b2. Here a = 2x and b = 1 and (2x)2 = 2x × 2x = 4x2. Now you try Expand to form a difference of perfect squares. a (x + 5)(x − 5) b (3x − 2)(3x + 2) 7 Expand to form a difference of perfect squares. Hint: The two middle terms a (x + 4)(x − 4) b (x + 9)(x − 9) will cancel to give c (x + 8)(x − 8) d (3x + 4)(3x − 4) (a + b)(a − b) = a2 − b2 e (2x − 3)(2x + 3) f (8x − 7)(8x + 7) g (4x − 5)(4x + 5) h (2x − 9)(2x + 9) i (5x − 7)(5x + 7) j (7x + 11)(7x − 11) Problem-solving and reasoning 8-9(½) 8-9(½), 10, 11 8 Write the missing number. a (x + 2)(x − 3) = x2 − x − Hint: Expand if you need to. b (x − 4)(x − 3) = x2 − x + 12 c (x − 4)(x + 4) = x2 − d (2x − 1)(2x + 1) = x2 − 1 e (x + 2)2 = x2 + x + 4 f (3x − 1)2 = 9x2 − x + 1 Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  10A Expanding binomial products 629 9 Write the missing number. a (x + )(x + 2) = x2 + 5x + 6 Hint: Notice how the two numerals in b (x + )(x + 5) = x2 + 8x + 15 the brackets multiply to give the constant term. c (x + 7)(x − ) = x2 + 4x − 21 d (x + 4)(x − ) = x2 − 4x − 32 e (x − 6)(x − ) = x2 − 7x + 6 f (x − )(x − 8) = x2 − 10x + 16 10 Find an expanded expression for the area of the pictures centred in these square and rectangular frames. Hint: For part a, the side length of the a b picture will be (x − 6) cm. 5 cm 3 cm Picture x cm Picture x cm 5 cm (x + 30) cm 11 Each problem below has an incorrect answer. Find the error and give the correct answer. a −x(x − 7) = −x2 − 7x b 3a − 7(4 − a) = −4a − 28 c (x − 9)(x + 9) = x2 − 18x − 81 d (2x + 3)2 = 4x2 + 9 Swimming pool algebra — 12 12 A pool company builds rectangular pools that are 5 m longer than they 2m are wide. The company then paves around the outside of the pool using a width of 2 m. Pool a Find an expanded expression for the: xm 2m i pool area (x + 5) m ii total area (including the pool and paving) Paving iii paved area b Find the area of the following when x = 4. i the pool ii the paved area Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  630 Chapter 10 Quadratics and non-linear graphs 0B 10B Factorising a difference of perfect squares Learning intentions To be able to identify a highest common factor and factorise simple expressions To be able to factorise a difference of perfect squares Key vocabulary: factorise, highest common factor, difference of perfect squares A common and key step in the simplification and solution of equations involves factorisation. Factorisation is the process of writing a number or expression as a product of its factors. For example: 6 = 2 × 3, 2x + 6 = 2(x + 3), x2 − x = x(x − 1) and x2 − 9 = (x + 3)(x − 3). In this section we look at expressions in which each term has a common factor and expressions that are a difference of perfect squares. Lesson starter: It’s just a DOPS expansion in reverse Complete each column to see the connection when expanding or factorising a DOPS. Expand Factorise (x + 2)(x − 2) = x2 − 4 x2 − 4 = (x )(x ) (x − 3)(x + 3) = x2 − 9 x2 − 9 = (x )( ) (2x + 3)(2x − 3) = 4x2 − 9 4x2 − 9 = (2x )( ) (7x − 6)(7x + 6) = =( )( ) Key ideas Factorise expressions with common factors by ‘taking out’ the highest common factor. For example: −5x − 20 = −5(x + 4) and 4x2 − 8x = 4x(x − 2). Factorise a difference of perfect squares (DOPS) using a2 − b2 = (a + b)(a − b) For example: x2 − 16 = x2 − 42 and 9x2 − 25 = (3x)2 − 52 = (x + 4)(x − 4) = (3x + 5)(3x − 5) Exercise 10B Understanding 1, 2 2 1 Complete these statements. a 2(x + 3) = 2x + 6, so 2x + 6 = 2( ) b −4(x − 1) = −4x + 4, so −4x + 4 = −4( ) c (x + 2)(x − 2) = x2 − 4, so x2 − 4 = ( )( ) d (3x + 2)(3x − 2) = 9x2 − 4, so 9x2 − 4 = ( )( ) Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  10B Factorising a difference of perfect squares 631 2 Determine the highest common factor of these pairs of terms. a 7x and 14 b 12x and 30 c −8y and 40 Hint: Include a common negative. d −5y and −25 2 e 4a and 2a f 12a2 and 9a g −5a2 and −50a h −3x2 y and −6xy i −2ab and −6a2 b Fluency 3–4(½) 3–5(½) Example 5 Taking out common factors Factorise by taking out the highest common factor. a −3x − 12 b 20a2 + 30a Solution Explanation a −3x − 12 = −3(x + 4) −3 is common to both −3x and −12. b 20a2 + 30a = 10a(2a + 3) The HCF of 20a2 and 30a is 10a. Place 10a out the front and divide each term by 10a. Now you try Factorise by taking out the highest common factor. a −4x − 36 b 14b2 − 21b 3 Factorise by taking out the highest common factor. a 3x − 18 b 4x + 20 c 7a + 7b Hint: Find the highest common factor, then take it out. d 9a − 15 e −5x − 30 f −4y − 2 g −12a − 3 h −2ab − bc i 4x2 + x j 5x2 − 2x k 6b2 − 18b l 14a2 − 21a m 10a − 5a2 n 12x − 30x2 o −2x − x2 Example 6 Factorising a difference of perfect squares Factorise the following differences of perfect squares. a x2 − 16 b 9a2 − 4b2 Solution Explanation a x2 − 16 = (x)2 − (4)2 Use a2 − b2 = (a − b)(a + b) with a = x and = (x − 4)(x + 4) b = 4. b 9a2 − 4b2 = (3a)2 − (2b)2 9a2 = (3a)2 since 3a × 3a = 9a2 and 4b2 = (2b)2. = (3a − 2b)(3a + 2b) Now you try Factorise the following differences of perfect squares. a x2 − 100 b 16a2 − 81b2 Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 632 Chapter 10 Quadratics and non-linear graphs 10B 4 Factorise the following differences of perfect squares. a x2 − 9 b x2 − 25 c y2 − 49 Hint: a2 − b2 = (a + b)(a − b) d y2 − 1 e a2 − 16 f b2 − 36 g y2 − 144 h z2 − 400 i 4x2 − 9 2 j 36a − 25 k 1 − 81y 2 l 100 − 9x2 m 25x2 − 4y2 n 64x2 − 25y2 o 9a2 − 49b2 5 Factorise these differences of perfect squares. a 4 − x2 b 9 − y2 Hint: 4 − x2 = 22 − x2 now use c 36 − a 2 d 100 − 9x2 a = 2 and b = x. e b2 − a2 f 400 − 25a2 g 4a2 − 9b2 h 16y2 − 121x2 Problem-solving and reasoning 6(½), 7 6(½), 7, 8(½) Example 7 Factorising by first taking out a common factor Factorise 12y2 − 1200 by first taking out a common factor. Solution Explanation 12y2 − 1200 = 12(y2 − 100) First take out the common factor of 12. = 12(y − 10)(y + 10) 100 = (10)2 , use a2 − b2 = (a − b)(a + b). Now you try Factorise 4a2 − 36 by first taking out a common factor. 6 Factorise the following by first taking out a common factor. Hint: As a first step, take out a a 2x2 − 32 b 5x2 − 45 c 6y2 − 24 common factor and then factorise d 3y2 − 48 e 3x2 − 75y2 f 3a2 − 300b2 the DOPS. g 12x2 − 27y2 h 63a2 − 112b2 i 108x2 − 147y2 7 The height (in metres) of a falling object above ground level is given by 100 − t2 , where t is in seconds. a Find the height of the object: i initially (i.e. at t = 0) ii after 2 seconds iii after 8 seconds b Factorise the expression 100 − t2. c Use your factorised expression from part b to find the height of the object: i initially (i.e. at t = 0) ii after 2 seconds iii after 8 seconds d How long does it take for the object to hit the ground? Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  10B Factorising a difference of perfect squares 633 8 We can work out problems such as 192 − 172 without a calculator, like this: Hint: Factorise first, using 192 − 172 = (19 + 17)(19 − 17) a2 − b2 = (a + b)(a − b), = 36 × 2 then evaluate. = 72 Use this idea to evaluate the following by first factorising, without the use of a calculator. a 162 − 142 b 182 − 172 c 132 − 102 d 152 − 112 e 172 − 152 f 112 − 92 g 272 − 242 h 522 − 382 Flexible framing — 9 9 A special square picture frame can hold a square picture of any size up Frame to 40 cm. a Using a picture side length of x cm, write expressions for the area of: i the picture Picture ii the frame (not including the picture) x cm b Factorise your expression for the frame area. c Find the frame area when: 40 cm i x = 20 ii x = 10 d Using trial and error, what value of x is required if the frame area is to be 700 cm2 ? Using technology 10B: Expanding and factorising This activity is available on the companion website as a printable PDF. Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  634 Chapter 10 Quadratics and non-linear graphs 0C 10C Factorising trinomials of the form x2 + bx + c Learning intentions To know the form of a monic quadratic trinomial To be able to factorise a monic quadratic trinomial Key vocabulary: monic quadratic trinomial, factor, constant, coefficient, integer A quadratic trinomial of the form x2 + bx + c is called a monic quadratic because the coefficient of x2 is 1. (‘Monic’ comes from the word ‘mono’, which means ‘one’.) Now consider: (x + m)(x + n) = x2 + xn + mx + mn = x2 + (m + n)x + mn We can see from this expansion that mn gives the constant term (c) and m + n is the coefficient of x. This tells us that to factorise a monic quadratic, we should look for factors of the constant term (c) that add to give the coefficient of the middle term (b). Lesson starter: So many choices of factors? We know that to factorise x2 − 5x − 24 we must choose a pair of numbers that multiply to give −24. Look at the following equations and discuss which of them are true. x2 − 5x − 24 = (x + 8)(x + 3) x2 − 5x − 24 = (x + 6)(x − 4) x2 − 5x − 24 = (x − 12)(x − 2) x2 − 5x − 24 = (x − 6)(x + 4) x2 − 5x − 24 = (x − 12)(x + 2) x2 − 5x − 24 = (x − 8)(x + 3) x2 − 5x − 24 = (x − 8)(x − 3) x2 − 5x − 24 = (x − 24)(x + 1) Key ideas Monic quadratics have a coefficient of x2 equal to 1. Monic quadratics of the form x2 + bx + c can be factorised by finding the two numbers that multiply to give the constant term (c) and add to give the coefficient of x (b). x2 + 5x + 6 = (x + 3)(x + 2) x2 − x − 6 = (x − 3)(x + 2) 2+3 2×3 −3 + 2 −3 × 2 x2 − 5x + 6 = (x − 3)(x − 2) x2 + x − 6 = (x + 3)(x −2) −2 + (−3) −2 × (−3) 3 + (−2) 3 × (−2) Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  10C Factorising trinomials of the form x2 + bx + c 635 Exercise 10C Understanding 1–3 3 1 Give the missing words or letters. a A quadratic has the coefficient of x2 equal to 1. b To factorise a monic quadratic, we look for factors of the term (c) that add to the of x (b). 2 Find two integers that multiply to give the first number and add to give the second number. a 18, 11 b 20, 12 c −15, 2 Hint: The integers include d −12, 1 e −24, −5 f −30, −7 …, −3, −2, −1, 0, 1, 2, 3, … g 10, −7 h 36, −15 i −8, 7 3 a i Which two numbers multiply to give 15 and add to give 8? ii Complete x2 + 8x + 15 = ( )( ) b i Which two numbers multiply to give −10 and add to give 3? ii Complete x2 + 3x − 10 = ( )( ) c i Which two numbers multiply to give 8 and add to give −6? ii Complete x2 − 6x + 8 = ( )( ) Fluency 4–5(½) 4–5(½) Example 8 Factorising trinomials of the form x2 + bx + c Factorise: a x2 + 7x + 12 b x2 + x − 6 c x2 − 5x + 6 Solution Explanation a x2 + 7x + 12 = (x + 4)(x + 3) 3 × 4 = 12 and 3 + 4 = 7. Check: (x + 4)(x + 3) = x2 + 3x + 4x + 12 = x2 + 7x + 12. b x2 + x − 6 = (x − 2)(x + 3) Since the numbers must multiply to −6, one must be positive and one negative. −2 × 3 = −6 and −2 + 3 = 1. Check: (x − 2)(x + 3) = x2 + 3x − 2x − 6 = x2 + x − 6. c x2 − 5x + 6 = (x − 3)(x − 2) −3 × (−2) = 6 and −3 + (−2) = −5. Check: (x − 3)(x − 2) = x2 − 2x − 3x + 6 = x2 − 5x + 6. Now you try Factorise: a x2 + 13x + 36 b x2 + 7x − 8 c x2 − 8x − 20 Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 636 Chapter 10 Quadratics and non-linear graphs 10C 4 Factorise these quadratic trinomials. a x2 + 7x + 6 b x2 + 5x + 6 c x2 + 6x + 9 Hint: For x2 + bx + c, look for factors of c that add to give b. d x2 + 7x + 10 e x2 + 7x + 12 f x2 + 11x + 18 g x2 + 5x − 6 h x2 + x − 6 i x2 + 2x − 8 j x2 + 3x − 4 k x2 + 7x − 30 l x2 + 9x − 22 m x2 − 7x + 10 n x2 − 6x + 8 o x2 − 7x + 12 p x2 − 2x + 1 q x2 − 9x + 18 r x2 − 11x + 18 s x2 − 4x − 12 t x2 − x − 20 u x2 − 5x − 14 v x2 − x − 12 w x2 + 4x − 32 x x2 − 3x − 10 Example 9 Factorising perfect squares Factorise x2 − 8x + 16 to form a perfect square. Solution Explanation x2 − 8x + 16 = (x − 4)(x − 4) −4 × (−4) = 16 and −4 + (−4) = −8. = (x − 4) 2 (x − 4)(x − 4) = (x − 4)2 is a perfect square. Now you try Factorise x2 − 10x + 25 to form a perfect square. 5 Factorise these perfect squares. a x2 − 4x + 4 b x2 + 6x + 9 Hint: Factorise perfect squares just like any trinomial but finish by c x2 + 12x + 36 d x2 − 14x + 49 writing them in the form (x + a)2. e x2 − 18x + 81 f x2 − 20x + 100 g x2 + 8x + 16 h x2 + 20x + 100 Problem-solving and reasoning 6, 7(½) 6–7(½), 8 Example 10 Factorising by first taking out a common factor Factorise 2x2 − 10x − 28 by first taking out a common factor. Solution Explanation 2x2 − 10x − 28 = 2(x2 − 5x − 14) First take out the common factor of 2. = 2(x − 7)(x + 2) −7 × 2 = −14 and −7 + 2 = −5. Now you try Factorise 3x2 − 18x + 15 by first taking out a common factor. Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  10C Factorising trinomials of the form x2 + bx + c 637 6 Factorise by first taking out the common factor. a 2x2 + 14x + 20 b 3x2 + 21x + 36 Hint: Factor out the c 2x2 + 22x + 36 d 5x2 − 5x − 10 coefficient of x2. e 4x2 − 16x − 20 f 3x2 − 9x − 30 g −2x2 − 14x − 24 h −3x2 + 9x − 6 i −2x2 + 10x + 28 j −4x2 + 4x + 8 k −5x2 − 20x − 15 l −7x2 + 49x − 42 7 Factorise these as perfect squares after first taking out the common factor. a 2x2 + 44x + 242 b 3x2 − 24x + 48 c 5x2 − 50x + 125 d −3x2 + 36x − 108 e −2x2 + 28x − 98 f −4x2 − 72x − 324 8 A rectangular garden has length 3 m more than its width, x m. There is a pond of area 10 m2 in the centre. Garden Pond xm 10 m2 (x + 3) m a Find an expression for: i the entire area (Expand your answer.) ii the garden area, excluding the pond b Factorise your answer from part a ii. c What is the area of the garden, excluding the pond, when: i x = 5? ii x = 7? Algebraic fractions — 9(½) 9 Some algebraic fractions can be simplified using factorisation. Here is an example: x2 − x − 12 =  − (x  + 3) 4)(x Hint: First factorise the numerator or  =x+3 denominator, then cancel. x−4 x−  4 Use this idea to simplify these fractions. 2 2 2 a x − 3x − 54 b x + x − 12 c x − 6x + 9 x−9 x+4 x−3 x+2 x−3 x+1 d e f x2 + 9x + 14 x2 − 8x + 15 x2 − 5x − 6 2 2 x2 − 16x + 64 g x − 4x + 4 h x + 2x + 1 i x−2 x+1 x−8 Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  638 Chapter 10 Quadratics and non-linear graphs 0D 10D Solving equations of the form ax2 = c Learning intentions To know that a quadratic equation can have 0, 1 or 2 solutions To be able to solve a quadratic equation of the form ax2 = c Key vocabulary: quadratic equation, solve, exact, surd, square root A quadratic equation can be expressed in the general form ax2 + bx + c = 0, where a, b and c are real numbers with a ¢ 0. In a quadratic equation the highest power is 2. The simplest quadratic equation can be expressed in the form x2 = c and these will be considered in this section. We will see that since 52 = 5 × 5 = 25 and (−5)2 = −5 × (−5) = 25, there are two possible solutions to the equation x2 = 25; i.e. x = 5 or x = −5. However, note that x2 = −25 has no solutions. This simplest form of a quadratic equation arises in many circumstances, including in Pythagoras’ theorem and in area and volume formulas. Formulas involving motion, including the velocity of a constantly accelerating object and the time taken for an object to fall due to gravity, also give rise to quadratic equations. Lesson starter: How many possible solutions? List all the pairs of numbers that multiply to 16. List all the pairs of numbers that multiply to −16. What do you notice about the signs of your pairs of numbers that multiply to 16? What about the signs of the numbers that multiply to −16? How many of your listed pairs satisfy x × x = 16? Are there any pairs that satisfy x × x = −16? From what you have found above, how many values of x would make the following true? a x2 = 9 b x2 = −9 Key ideas A quadratic equation is an equation in which the highest power is 2. For example: x2 = 9, 2x2 − 6 = 0 and x2 − 2x = 8. A quadratic equation may have 0, 1 or 2 possible solutions. The square of a number is always positive or 0; i.e. x2 Å 0 for all values of x. For example: (2)2 = 2 × 2 = 4 and (−3)2 = −3 × (−3) = 9. The inverse operation of squaring is the square root, Ë. Ë4 = 2 since 2 × 2 = 4. Ë7 × Ë7 = 7 When x2 = c, then x = ± Ëc provided that c Å 0. ± represents two possible solutions: +Ëc and −Ëc. c = 0 gives just one solution since Ë0 = 0. When x2 = c and c < 0, there are no real solutions. Square roots that do not reduce to whole numbers are called surds and can be left in this exact form; i.e. Ë2, Ë10, … etc. For example: Ë25 is not a surd as 25 is a square number and Ë25 = 5. Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  10D Solving equations of the form ax2 = c 639 To solve equations of the form ax2 = c, make x2 the subject and then take the square root of both sides to solve for x. In practical problems such as those involving measurement, it may make sense to reject the negative solution obtained. Exercise 10D Understanding 1–4 4 1 List the first 12 square numbers. Hint: Start at 12 = 1, 22 = … 2 Evaluate the following. a 42 b 92 c 72 Hint: 42 = 4 × 4 = … d (−3)2 e (−8)2 f (−1)2 (−3)2 = −3 × (−3) = … 3 Evaluate the following. a Ë36 b −Ë49 c −Ë100 Hint: Ë4 = 2 since 22 = 4. d Ë0 e Ë400 f Ë1600 4 Complete the following. The equation x2 = c has: a two solutions when b one solution when c no solutions when Fluency 5–7(½) 5–7(½) Example 11 Solving equations of the form ax2 = c Find all possible solutions to the following equations. a x2 = 16 b 3x2 = 12 c x2 = −25 Solution Explanation a x2 = 16 Take the square root of both sides to solve for x. If x2 = c, Â x = ± Ë16 where c > 0, then x = ± Ëc. x=±4 Since 16 is a square number, Ë16 is a whole number. ± represents two solutions: + 4 since (+ 4)2 = 16 and −4 since (−4)2 = 16. b 3x2 = 12 First make x2 the subject by dividing both sides by 3. x2 = 4 Take the square root of both sides to solve for x. This gives two possible values for x : + 2 and −2. Â x = ± Ë4 x=±2 c x2 = −25 Since x2 Å 0 for all values of x, there are no values of x that There are no solutions. will make x2 = −25. Continued on next page Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 640 Chapter 10 Quadratics and non-linear graphs 10D Now you try Find all possible solutions to the following equations. a x2 = 64 b 2x2 = 18 c x2 = −36 5 Find the possible solutions for x in the following equations. a x2 = 25 b x2 = 81 c x2 = 36 Hint: When taking the square root, don’t forget ±. d x2 = 49 e x2 = −16 f x2 = −100 2 g x = 400 2 h x = 144 i 2x2 = 50 j 3x2 = 48 k −5x2 = −5 l −5x2 = 20 Hint: 1 x2 is the same as x. 2 2 2 2 2 m x = 32 n x = 27 o 1 x2 = 18 2 3 2 Example 12 Solving equations with answers that are surds Find all possible solutions to the following equations. a x2 = 10 (Give your answer in exact surd form.) b 3x2 = 9 (Give your answer to one decimal place.) Solution Explanation a x2 = 10 Take the square root of both sides. Â x = ± Ë10 Since 10 is not a square number, leave your answer in exact surd form ± Ë10, as required. b 3x2 = 9 Divide both sides by 3. 2 x =3 Take the square root of both sides. Â x = ± Ë3 Use a calculator to evaluate ± Ë3 = 1.73205… and round your x = ± 1.7 (to 1 d.p.) answer to one decimal place, as required. Now you try Find all possible solutions to the following equations. a x2 = 29 (Give your answer in exact surd form.) b 5x2 = 55 (Give your answer to one decimal place.) 6 a Solve the following equations, where possible. Leave your answer in exact surd form. i x2 = 14 ii x2 = 22 iii x2 = 17 Hint: Ë14 is exact surd form. iv x2 = −7 v 3x2 = 15 vi 4x2 = 24 2 2 vii x = 7 viii x = 3 ix 1 x2 = 3 3 2 5 b Solve these equations, giving your answers to one decimal place. i x2 = 12 ii x2 = 35 iii 3x2 = 90 2 iv x = 7 9 Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  10D Solving equations of the form ax2 = c 641 Example 13 Solving equations of the form ax2 + b = 0 Find the exact solution(s), where possible, to the following equations. a x2 − 9 = 0 b x2 + 12 = 0 2 c x − 3 = 4, x > 0 d 7 − x2 = 6, x < 0 2 Solution Explanation a x2 − 9 = 0 Make x2 the subject by adding 9 to both sides. x2 = 9 Solve for x by taking the square root of both sides. Â x = ± Ë9 x=±3 b x2 + 12 = 0 Make x2 the subject by subtracting 12 from both sides. x2 = −12 Since x2 is positive for all values of x, x2 = c, where c is negative There is no real solution. has no real solution. c x2 − 3 = 4 Solve for x2 by adding 3 to both sides and then multiplying both 2 sides by 2. x2 = 7 2 x2 = 14 Take the square root of both sides to solve for x. Ë14 is the exact form of the answer. Â x = ± Ë14 x = Ë14 since x > 0. Note the restriction that x must be a positive value (i.e. > 0), so reject the solution x = −Ë14. d 7 − x2 = 6 Subtract 7 from both sides and then divide both sides by −1. −x2 = −1 Alternatively, add x2 to both sides to give 7 = x2 + 6 and then solve. x2 = 1 Â x = ± Ë1 x=±1 Ë1 = 1 since 12 = 1. x = −1 since x < 0. Reject the solution x = 1 because x < 0. Now you try Find the exact solution(s), where possible, to the following equations. a x2 − 25 = 0 b x2 + 7 = 0 2 c x − 1 = 4, x > 0 d 3 − x2 = −6, x < 0 3 Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 642 Chapter 10 Quadratics and non-linear graphs 10D 7 Find the exact solution(s), where possible, of these equations. a x2 − 4 = 0 b x2 − 81 = 0 c x2 + 16 = 0 2 d x + 10 = 0 2 e x + 6 = 15 f x2 − 5 = 10 Hint: First get your equation in the 2 form x2 = c. 2 g 2x − 6 = 0, x > 0 2 h 3x + 10 = 13 i x + 4 = 5, x > 0 2 j 10 − x2 = 3, x < 0 k 5 − x2 = 7 l 4 − x2 = −1, x > 0 Problem-solving and reasoning 8–11 8(½), 11, 13–16 8 Solve these equations by first collecting like terms. Hint: Rewrite the equation in the form a 3x2 = 2x2 + 16 b 4x2 = 2x2 + 18 x2 = c by collecting x2 terms on one 2 2 c 4x − 11 = x + 16 d x2 = 50 − x2 side and the number on the other. e 2x2 + 5 = 17 − x2 f 4x2 + 8 = 2x2 + 5 3x2 = 2x2 + 16 −2x2 x2 = 16 −2x2 9 The area of a square is 36 m2. What is its perimeter? 10 The distance, d metres, that a bird dives in t seconds is given by d = 5t2. How long does the bird take to dive 80 m? 11 Recall Pythagoras’ theorem: c2 = a2 + b2. c a b Use this to find the unknown side length in the following triangles. Give your answers in exact form. a b x Hint: Set up the Pythagorean 2 10 x equation first. 5 6 c d 2 x 8 x 9 Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  10D Solving equations of the form ax2 = c 643 12 The volume of a cylinder is V = pr2 h, where r is the radius and h is the height of the cylinder. If the volume of a cylindrical water Hint: pr2 (15) can be written as 15pr2. tank is 285 m3 and its height is 15 m, determine its radius, to one decimal place. 13 a Solve the equation pr2 = 24, to find two possible values for r, rounded to one decimal place. b Given that A = pr2 is the formula for the area of a circle of radius r, what restriction does this place on your answer to part a if the area of the circle is 24 units2 ? 14 Determine for which values of b the equation x2 − b = 0 has: Hint: For Questions 14 & 15, a two solutions b one solution c no solutions first solve for x2. 15 Determine for which values of b the equation x2 + b = 0 has: a two solutions b one solution c no solutions 16 a Solve the equation x2 − 25 = 0. b Hence, determine the range of values of x for which: Hint: In part b, make use of i x2 − 25 < 0 ii x2 − 25 > 0 your answer to part a. Square powers and brackets — 17(½) 17 Consider the following example. Solve (x + 1)2 = 16 x + 1 = ± Ë16 (Take the square root of both sides.) x+1=±4 x+1=4 or x + 1 = −4 (Write each equation separately.) x=3 or x = −5 (Subtract 1 from both sides.) Use this method to solve these equations. a (x + 1)2 = 9 b (x + 3)2 = 49 c (x − 2)2 = 4 d (x − 5)2 = 16 e (2x + 1)2 = 25 f (2x − 1)2 = 1 g (3x − 2)2 = 4 h (4x − 3)2 = 81 Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  644 Chapter 10 Quadratics and non-linear graphs 0E 10E Solving equations using the null factor law Learning intentions To understand how the null factor law gives solutions to an equation To be able to factorise and solve a quadratic equation using the null factor law Key vocabulary: null factor law, product, standard form, factor In previous chapters you would have solved linear equations such as 3x = 9 and 2x − 1 = 5, and you may have used ‘back tracking’ or inverse operations to solve them. For quadratic equations such as x2 − x = 0 or x2 − x − 20 = 0, we need a new method, because there are different powers of x involved and ‘back tracking’ isn’t useful. The result of multiplying a number by zero is zero. Therefore, if an expression equals zero then at least one of its factors must be zero. This is called the null factor law and it provides us with an important method that can be utilised to solve a range of mathematical problems involving quadratic equations. Such equations relate to the engineering of bridges, for example, which use parabolic arches. Lesson starter: How does the null factor law work? Start this investigation by completing this table. x −5 −4 −3 −2 −1 0 1 2 (x − 1)(x + 4) 6 Which values of x make (x − 1)(x + 4) = 0? Why? Could you work out what values of x make (x − 1)(x + 4) = 0 without doing a table? Explain. What values of x make (x − 2)(x + 3) = 0 or (x + 5)(x − 7) = 0? Key ideas The null factor law states that if the product of two numbers is zero, then either or both of the two numbers is zero. If a × b = 0, then a = 0 and/or b = 0. To solve a quadratic equation, write it in standard form (i.e. ax2 + bx + c = 0) and factorise. Then use the null factor law. For example: x2 − 2x − 8 = 0 (x − 4)(x + 2) = 0 x − 4 = 0 or x + 2 = 0 x = 4 or x = −2 If the coefficients of all the terms have a common factor, then first divide by that common factor. Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  10E Solving equations using the null factor law 645 Exercise 10E Understanding 1–3 3 1 Complete the following. a If a × b = 0, the null factor law states that or. b If the coefficients of all terms in an equation have a then divide by that. 2 a Complete this table for the given values of x. x −3 −2 −1 0 1 2 (x + 2)(x − 1) b What values of x make (x + 2)(x − 1) = 0? c What values of x would make (x + 3)(x − 2) = 0? 3 Copy and complete: a x(x − 2) = 0 b (x − 1)(x + 4) = 0 x = 0 or =0 x − 1 = 0 or =0 x = 0 or x = x= or x = c (x + 6)(2x − 7) = 0 = 0 or =0 x= or 2x = x = −6 or x = Fluency 4–7(½) 4–7(½) Example 14 Using the null factor law Use the null factor law to solve these equations. a x(x − 1) = 0 b (x − 1)(2x + 5) = 0 Solution Explanation a x(x − 1) = 0 x = 0 or x − 1 = 0 Set each factor equal to zero. x = 0 or x = 1 For x − 1 = 0, add 1 to both sides to finish. b (x − 1)(2x + 5) = 0 Set each factor equal to zero and then solve x − 1 = 0 or 2x + 5 = 0 each linear equation. x = 1 or 2x = −5 x = 1 or x = − 5 2 Now you try Use the null factor law to solve these equations. a x(x + 7) = 0 b (x + 6)(3x − 2) = 0 Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 646 Chapter 10 Quadratics and non-linear graphs 10E 4 Use the null factor law to solve these equations. a x(x + 1) = 0 b x(x − 5) = 0 Hint: Null factor law: if a × b = 0, c 2x(x − 4) = 0 d (x − 3)(x + 2) = 0 then either a = 0 or b = 0. e (x + 5)(x − 4) = 0 f (x + 1)(x − 1) = 0 g (2x − 4)(x + 1) = 0 h (3x − 2)(x − 7) = 0 i 3x(4x + 5) = 0 j (2x − 1)(3x + 7) = 0 k (4x − 5)(5x + 2) = 0 l (8x + 3)(4x + 3) = 0 Example 15 Solving quadratic equations with a common factor Solve x2 − 2x = 0. Solution Explanation x2 − 2x = 0 Factorise by taking out the common factor x. Apply the null x(x − 2) = 0 factor law: if a × b = 0, then a = 0 or b = 0. Solve for x. Â x = 0 or x − 2 = 0 Â x = 0 or x = 2 Now you try Solve 5x2 + 30x = 0. 5 Solve the following quadratic equations. a x2 − 4x = 0 b x2 − 3x = 0 Hint: First take out the common 2 factor, then use the null factor law. c x + 2x = 0 d 3x2 − 12x = 0 e 2x2 − 10x = 0 f 4x2 + 8x = 0 Example 16 Solving with DOPS Solve x2 − 16 = 0 by factorising the DOPS. Solution Explanation x2 − 16 = 0 Note that x2 − 16 is a difference of perfect squares. (x + 4)(x − 4) = 0 x2 − 16 = x2 − 42 = (x + 4)(x − 4) x + 4 = 0 or x − 4 = 0 Solve each linear factor equal to zero to finish. x = −4 or x = 4 Now you try Solve x2 − 81 = 0 by factorising the DOPS. 6 Solve the following by factorising the DOPS. a x2 − 25 = 0 b x2 − 36 = 0 Hint: Recall 4x2 = (2x)2 2 c x − 100 = 0 d 4x2 − 9 = 0 e 9x2 − 16 = 0 f 49x2 − 81 = 0 Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  10E Solving equations using the null factor law 647 Example 17 Solving quadratic equations Solve the following quadratic equations. a x2 − 5x + 6 = 0 b x2 + 2x + 1 = 0 Solution Explanation a x2 − 5x + 6 = 0 Factorise by finding two numbers that multiply to 6 and (x − 3)(x − 2) = 0 add to −5: −3 × (−2) = 6 and −3 + (−2) = −5. Â x − 3 = 0 or x − 2 = 0 Apply the null factor law and solve for x. Â x = 3 or x = 2 b x2 + 2x + 1 = 0 1 × 1 = 1 and 1 + 1 = 2. (x + 1)(x + 1) = 0 (x + 1)2 = 0 (x + 1)(x + 1) = (x + 1)2 is a perfect square. Âx+1=0 This gives one solution for x. Â x = −1 Now you try Solve the following quadratic equations. a x2 + 11x + 30 = 0 b x2 − 8x + 16 = 0 7 Solve the following quadratic equations. a x2 + 3x + 2 = 0 b x2 + 5x + 6 = 0 Hint: Parts j to n are perfect squares, so you will find only one solution. c x2 − 6x + 8 = 0 d x2 − 7x + 10 = 0 2 e x + 4x − 12 = 0 f x2 + 2x − 15 = 0 g x2 − x − 20 = 0 h x2 − 5x − 24 = 0 i x2 − 12x + 32 = 0 j x2 + 4x + 4 = 0 2 k x + 10x + 25 = 0 l x2 − 8x + 16 = 0 m x2 − 14x + 49 = 0 n x2 − 24x + 144 = 0 Problem-solving and reasoning 8(½), 9 8(½), 9, 10 8 How many different solutions for x will these equations have? a (x − 2)(x − 1) = 0 b (x + 7)(x + 3) = 0 c (x + 1)(x + 1) = 0 d (x − 3)(x − 3) = 0 e (x + Ë2)(x − Ë2) = 0 f (x + 8)(x − Ë5) = 0 g (x + 2)2 = 0 h (x + 3)2 = 0 i 3(2x + 1)2 = 0 9 The height of a paper plane above floor level, in metres, is given by − 1 t(t − 10), where t is in seconds. 5 a Find the height of the plane after: i 2 seconds ii 6 seconds b Solve − 1 t (t − 10) = 0 for t. 5 c How long does it take for the plane to hit the ground after it is launched? Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 648 Chapter 10 Quadratics and non-linear graphs 10E 10 Solve by first taking out a common factor. a 2x2 + 16x + 24 = 0 Hint: First take out the common b 2x2 − 20x − 22 = 0 factor, then factorise before using c 3x2 − 18x + 27 = 0 the null factor law. d 5x2 − 20x + 20 = 0 Photo albums — 11 11 A printer produces rectangular photo albums. Each page has a length Border 5 cm more than the width, and includes a spot in the middle for a standard 10 cm by 15 cm photo. a Find the area of the photo. 15 (x + 5) cm Photo cm b Find an expression for: i the total area of a page ii the border area of a page (i.e. excluding the photo) 10 cm c Factorise your expression for the border area. x cm d For what value of x is the border area equal to zero? e For what value of x is the border area equal to 350 cm2 ? Using technology 10E: Solving equations using the null factor law. This activity is available on the companion website as a printable PDF. Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  10F Applications of quadratics 649 10F 10F Applications of quadratics Learning intentions To be able to set up a quadratic equation from a simple worded problem To be able to solve a problem in a real context using a quadratic equation and the null factor law Key vocabulary: define, pronumeral, solve Defining variables, setting up equations, solving equations and interpreting solutions are all important elements of applying quadratics in problem solving. For example, the area of a rectangular paddock that can be fenced off using a limited length of fencing can be found by setting up a quadratic equation, solving it and then interpreting the solutions. Lesson starter: Rectangular quadratics The length of a rectangular room is 4 m longer than its width, which is x metres. Its area is 21 m2. Write an expression for the area using the variable x. (x + 4) m Use the 21 m2 area fact to set up an equation equal to zero. Factorise and solve the equation to find x and then find the dimensions of the rectangle. 21 m2 xm Key ideas When applying quadratic equations: Define a pronumeral; i.e. ‘Let x be …’. Write an equation. Solve the equation. Choose the solution(s) that solves the equation and answers the question. Exercise 10F Understanding 1–3 3 1 Write expressions for each of the following. a The length of a rectangle if it is 4 more than its width, x. Hint: An example of an expression is x + 3. b The length of a rectangle if it is 10 more than its width, x. c The length of a rectangle if it is 7 less than its width, x. d The height of a triangle if it is 2 less than its base, x. e The height of a triangle if it is 6 more than its base, x. 2 Rearrange these equations so that there is a zero on the right-hand side. Do not try to solve the equation. a x2 + 2x = 3 Hint: Subtract from both sides to 2 give a zero on the right-hand side. b x − 3x = 5 c x2 + 7x = 4 Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party. 650 Chapter 10 Quadratics and non-linear graphs 10F 3 The given steps for solving a mathematical problem are in the wrong order. Give the correct order. A Solve the equation B Define a pronumeral C Choose the solution to answer the question D Write an equation Fluency 4–6 4–7 Example 18 Finding dimensions The area of a rectangle is fixed at 28 m2 and its length is 3 m more than its width. Find the dimensions of the rectangle. Solution Explanation Let x m be the width of the (x + 3) m rectangle. Length = (x + 3) m xm 28 m2 x(x + 3) = 28 Write an equation for area using the given information. 2 x + 3x − 28 = 0 Then write with a zero on the right and solve for x. (x + 7)(x − 4) = 0 x = −7 or x = 4 Disregard x = −7 because x must be greater than zero. Choose x = 4. Rectangle has width 4 m Answer the question in full. and length 7 m. Now you try The area of a rectangle is fixed at 72 m2 and its length is 6 m more than its width. Find the dimensions of the rectangle. 4 A rectangle has an area of 24 m2. Its length is 5 m longer than its width. (x + 5) m xm 24 m2 a Copy this sentence: ‘Let x m be the width of the rectangle.’ b Write an expression for the rectangle’s length. c Write an equation using the rectangle’s area. d Write your equation from part c with a zero on the right-hand side, and solve for x. e Find the dimensions of the rectangle. Essential Mathematics for the Australian Curriculum ISBN 978-1-108-87886-9 © Greenwood et al. 2021 Cambridge University Press CORE Year 10 Photocopying is restricted under law and this material must not be transferred to another party.  10F Applications of quadratics 651 5 Repeat all the steps in Question 4 to find the dimensions of a rectangle with the following properties. Hint: Carefully set out each step, as a Its area is 60 m2 and its length is 4 m more than its width. in Example 18 and Question 4. b Its area is 63 m2 and its length is 2 m less than its width. c Its area is 154 mm2 and its length is 3 mm less than its width. 6 A triangle’s area is 4 cm2 and its height is 2 cm more than its base. a Write an expression for the area of the triangle, using A = 1 bh. 2 (x + 2) cm 2 b Write an equation using the 4 cm area fact. c Multiply both sides by 2 and write your equation with a zero on the right side. d Solve your equation to find the base and height dimensions of x cm the triangle. 7 Find the height and base l