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UTS Mathematics Bridging Book

Dr Narelle Smith Worked solutions by Dr Van Ha Do © Department of Mathematical Sciences, Faculty of Science, University of Technology, Sydney Reprinted 2007

Introduction: How to use this booklet This booklet* has been designed for students wishing to practice the mathematical skills required for either 2 Unit or 3 Unit level Mathematics at the Higher School Certificate. It would be particularly suitable for students in maths bridging courses, as it concentrates on those topics in the HSC syllabus that are of most benefit to prospective tertiary students. The booklet is divided into 8 chapters, each representing a broad topic area; and each chapter further divided into “A”, “B”, and “C” sections. It is recommended that students who have a background of Year 12 Maths in Society concentrate on the “A’ material only. This will be sufficient to cover the basic 2 Unit at the HSC work. Students who are more confident, or who have already gained 2 Unit at the HSC, should only skim through the “A” material and should concentrate on the “B” material that covers 3 Unit only topics. A table of contents can be found on the following page. Each subsection includes a formula summary, worked examples and a series of exercises. Worked solutions to all of the exercises are included at the back. The booklet however has not been designed as a teach yourself textbook, but as a source of exercises to complement our recommended bridging course text “Introductory Mathematics” by Petocz, Petocz and Wood. In particular, the emphasis is on drill exercises to practice essential techniques. For more detailed explanations and a wealth of interesting real-life applications of these techniques, students are directed to the textbook.

*In writing this booklet, I drew freely on the material previously complied by lecturers in the School of Mathematical Sciences, including Brian Stephenson, Dr. Peter Petocz and Leigh Wood. The “C” Section for Business students was written by Judyth Hayne.

(b) 4 × 45 = 41+5 = 46 .

ANSWERS TO EXERCISES

(c) 716 ÷ 712 = 7 16−12 = 74 .

1

(d) a2 × a4 = a2+4 = a6 .

Algebra

(e) 54 ÷ 53 = 5 4−3 = 51 = 5.

1A.1.3. Index Notation (p. 2).

(f) t10 ÷ t7 = t10−7 = t3 .

1.

(g) 36 × 3−2 = 3 6+(−2) = 36−2 = 34 .

5

(a) 2 = 2 × 2 × 2 × 2 × 2 = 32. 3

(h) 48 ÷ 4−3 = 4 8−(−3) = 48+3 = 411 .

2

(b) 2 × 3 = (2 × 2 × 2).(3 × 3) = 72.

(i) 67 ×

(c) 22 × 33 = (2 × 2).(3 × 3 × 3) = 108. 0

67 ×

2

(d) 6 × 3 = 1 × 9 = 9. =

1 32

=

1 . 9

(f) 5−3 =

1 53

=

1 . 125

(e) 3

−2

1

1

(k) b9 ÷

1

1

1 1 92

(l) 9− 2 = 1

3 4

1 4

3 4

3

3

(p) 32 (q)

4−2 3

(r) 3 + (s)

= 32

=

1 42

3

1 2−4

1 (2 14 )2

=

4 10

=

(a)

4 3

= 1 13 .

=

=

1 x

1

= x−1 .

1 16×3

=

(d)

3 x2

(e)

5 x2 +6

(f)

1 (x+5)2

(g)

2 (3x−4)6

1 . 48

3 2

=1

= 3x−2 . = 5(x2 + 6) −1 . = (x + 5)−2 . = 2(3x − 4)−6 .

1. 2

1

(j)

√1 a

= a−2 .

(k)

4 √ y

= 4y− 2 .

= 3 + 24 = 19.

3 4

2

√ 1 4 x + 3 = (x + 3) 4 .  2 (i) 3 (x + 1) 2 = (x + 1) 3 .

= 32 = 4.

1 ( 94 ) 2

5

(h)

2 5

3

3

√ 1 a = a2. √ 3 4 (c) c3 = c 4 .

1 4

1 16

5

3.

16 = (16 ) = 4096 = 8.

0.4

5

1

(b)

1 4

1

= b9 ÷ b−6 = b9−(−6) = b15 .

(m) a− 4 × a 4 = a− 4 + 4 = a4 = a 2 .

(n) 16 = (16 )3 = 2 3 = 8, OR

)2 = (o) (1 79 )2 = ( 16 9

= a2 , OR

= a5+(−3) = a2 .

1

= 13 .

1

a5 = a5−3 a3 a5 × a−3

(l) 25 2 × 252 == 25 2 +1 = 25 2 (= 55 ).

= 13 .

1 1 27 3

(m) 27− 3 =

1 b6

= 6 7−2 = 65 , OR

= 6 × 6−2 = 67+(−2) = 6 5 .

a5 × a13 =

= 52 .

42

25 2

67 62 7

=

(j) a5 × a13 =

(g) (33 )2 = 33×2 = 36 = 729. √ 1 (h) 4 2 = 4 = 2. √ 1 (i) 81 4 = 4 81 = 3. √ 1 (j) 64 3 = 3 64 = 4. 4 )2 = (k) ( 25

1 62 1 62

1

1A.2.2. Collecting Like terms, Expanding Brackets, multiplication of Expressions (p.3).

1

(t) 810.75 = 81 = (81 4 )3 = 33 = 27. 2. (a) 53 × 52 = 53+2 = 55 .

1. 1

(e) 3b3 (2ab − 5) + a(4b4 − 1) = 6ab4 − 15b3 + 4ab4 − a = 10ab4 − 15b3 − a.

(a) 4a + 7 − 3a + 10 = a + 17. (b) 9x2 − 4x − 5x2 − 9x = 4x2 − 13x = x(4x − 13).

(f) 4c(2c + 8) − 3c(c − 1) = 8c2 + 32c − 3c2 + 3c = 5c2 + 35c.

(c) 2b + 5b2 + 8 − 7b2 − 10b = − 2b2 − 8b + 8 = −2(b2 + 4b − 4).

(g) 2x2 (3x2 + 2x + 5) − 5x(4x2 + 6x) = 6x4 + 4x3 + 10x2 − 20x3 − 30x2 = 6x4 − 16x3 − 20x2 .

(d) 3(x − 4) + 5(x + 4) = 3x − 12 + 5x + 20 = 8x + 8 = 8(x + 1). (e) 5(c + 3) − 2(c − 4) = 5c + 15 − 2c + 8 = 3 c + 23.

1A.3.3. Binomial Products (p. 4).

(f) 12 − 4(7 − 2m) = 12 − 28 + 8m = 8m − 16 = 8(m − 2).

1. (a) (a + 3)(a + 5) = a2 + 8a + 15.

(g) 8(a2 + 2ab + b2 ) − 4(2a2 − 4ab + 3b2 ) = 8a2 + 16ab + 8b2 − 8a2 + 16ab − 12b2 = 32ab − 4b2 = 4b(8a − b).

(b) (2c + 3)(c − 5) = 2c2 − 7c − 15. (c) (d − 4)2 = d 2 − 8d + 16.

2.

(d) (h − 7)(h − 4) = h2 − 11h + 28.

(a) 2a × 5 = 10a. 2

(e) (9 + 2k)(7 − 3k) = 63 − 13k − 6k 2 .

2

(b) 4 × 3x = 12x .

(f) (x + 4)(x − 4) = x2 − 16.

2

(c) 2x × 3x = 6x .

(g) (c + 2) 2 = c2 + 4c + 4.

2

2

(d) 5x × −4y = −20x y. 2

3

(h) (2c − 3)(2c + 3) = 4c2 − 9.

5

(e) 2c × 6c = 12c . 2

(f) (3x) =

32 2

(i) (x + 2y)(x − 2y) = x2 − 4y2 .

2

x = 9x .

(j) (4d + 1)(4d + 1) = 16d 2 + 8d + 1.

2

(g) −4a × −6a = 24a . 2

2

(k) (xy − 4)2 = (xy − 4)(xy − 4) = x2 y2 − 8xy + 16.

2 3

(h) a b × −2b c = −2a b c. 2

3

(l) (y2 − 1)2 = (y2 − 1)(y2 − 1) = y4 − 2y2 + 1.

3 2 4

(i) pqr × p qr = p q r .

(m) (a2 + b2 )2 = a4 + 2a2 b2 + b4 .

(j) (−2x)2 = 4x2 . 3

(n) (1 − 16c2 )2 = 1 − 32c2 + 256c4

3

(k) (−2x) = −8x .

1A.4.3. Factorising (pp. 5-6).

(l) (7y)2 × (2y)3 = 49y2 × 8y3 = 392y5 .

1.

3.

(a) 6x + 12 = 6(x + 2).

(a) b(b + 6) − 2b(b + 1) = b2 + 6b − 2b2 − 2b = −b2 + 4b.

(b) a2 + 4a = a(a + 4).

(b) a2 (3a − 2) + 7a(a − 5) = 3a3 − 2a2 + 7a2 − 35a = 3a3 + 5a2 − 35a.

(c) 2c2 − 12c = 2c(c − 6). (d) x3 − 2xy = x(x2 − 2y).

(c) 4xy(2xy − 3) = 8x2 y2 − 12xy.

(e) x2 y3 − x3 y2 = x2 y2 (y − x).

(d) −b(2 − b) + 4(b + 16) = − 2b + b2 + 4b + 64 = b2 + 2 b + 64.

(f) 6a3 b + 9ab3 = 3ab(2a2 + 3b2 ). 2

(g) 8t2 + 46t − 12 = 2(4t2 + 23t − 6) = 2(4t − 1)(t + 6).

(g) a2 − 16 = (a + 4)(a − 4). (h) 4e2 − 9 = (2e)2 − 32 = (2e + 3)(2e − 3).

(h) 30y2 + 21y − 36 = 3(10y2 + 7y − 12) = 3(2y + 3)(5y − 4).

(i) 4x2 −9y2 = (2x)2 −(3y)2 = (2x+3y)(2x−3y). (j) 2m2 − 32 = 2(m2 − 42 ) = 2(m + 4)(m − 4).

4.

(k) 6d 2 − 54 = 6(d 2 − 32 ) = 6(d + 3)(d − 3).

(a) xy + xz + ay + az = x(y + z) + a(y + z) = (y + z)(x + a).

(l) 48p2 − 75 = 3(16p2 − 25) = 3(4p + 5)(4p − 5).

(b) x2 − y2 + 5x − 5y = (x + y)(x − y) + 5(x − y) = (x − y)(x + y + 5).

2. (a) a2 + 8a + 12 = (a + 2)(a + 6).

(c) x3 + 4x2 + 5x + 20 = x2 (x + 4) + 5(x + 4) = (x + 4)(x2 + 5).

(b) ab2 + 11b + 28 = (b + 4)(b + 7).

(d) a2 − b2 − 3a − 3b = (a + b)(a − b) − 3(a + b) = (a + b)(a − b − 3).

(c) g 2 + 5g − 14 = (g + 7)(g − 2). (d) k 2 + 18k − 63 = (k + 21)(k − 3).

(e) 4a2 − 9b2 + 4a + 6b = (2a + 3b)(2a − 3b) + 2(2a + 3b) = (2a + 3b)(2a − 3b + 2).

(e) l2 − 15l − 54 = (l + 3)(l − 18). (f) n2 − 10n + 21 = (n − 3)(n − 7).

1A.5.2. Simplifying Algebraic Fractions (pp. 6-7).

(g) p2 + 10p − 56 = (p + 14)(p − 4). (h) 4m2 + 4m − 24 = 4(m2 + m − 6) = 4(m + 3)(m − 2).

1. 6(x+4) 3x

2(x+4) . x

(a)

6x+24 3x

(b)

(j) a + 3a − 10a = a(a2 + 3a − 10) = a(a + 5)(a − 2).

a2 a2 +a

(c)

3a−3b 2b−2a

=

3(a−b) −2(a−b)

(k) 3k 2 + 9k − 30 = 3(k 2 + 3k − 10) = 3(k + 5)(k − 2).

(d)

x2 −y 2 5x+5y

=

(x+y )(x−y ) 5(x+y)

(e)

b2 −2b b2 +4b−12

b(b−2) = = (b+6)(b−2)

b . b+6

(f)

d2 +3d d2 +8d+15

d(d+3) = (d+5)(d+3) =

d . d+5

(i) 2a2 + 6a − 20 = 2(a2 + 3a − 10) = 2(a + 5)(a − 2). 3

2

(l) x3 − 10x2 + 24x = x(x2 − 10x + 24) = x(x − 4)(x − 6). 3.

(g)

2

(a) 2b + 5b + 3 = (2b + 3)(b + 1). (b) 3c2 + 7c + 4 = (3c + 4)(c + 1).

(h)

2

(c) 4b + 4b − 15 = (2b + 5)(2b − 3). (d) 6x2 − 19x + 10 = (3x − 2)(2x − 5).

=

=

2

= a(aa+1) =

a . a+1

= − 32 = −1 12 . =

x−y 5

.

2 +4)(x2 −4) x4 −16 = (x(x+2)(x−6) = x2 −4x−12 (x2 +4)(x+2)(x−2) (x2 +4)(x−2) . = x−6 (x+2)(x−6) 2 +2x−15) 2x2 +4x−30 = 2(x4(x 2 −9) 4x2 −36 2(x+5)(x−3) x+5 = 2(x+3) . 4(x+3)(x−3)

=

2.

(e) 15k 2 − 21k + 6 = 3(5k 2 − 7k + 2) = 3(5k − 2)(k − 1). (f) 3a2 + 13a − 30 = (3a − 5)(a + 6). 3

(a)

2 a

3 + 2a =

(b)

x 2

+

1−4x 3

4+3 2a

=

7 = 2a .

3x+2(1−4x) 6

=

2 −5 x . 6

(x−2)+(x+3) (x+3)(x−2)

(c)

1 x+3

(d)

5 x2 −4

(e)

2 7 2 − x+4 = (x+4)(x−2) x2 +2x−8 2−7(x−2) 16−7x = (x+4)(x−2) = (x+4)(x−2) .

(f)

3(x+y)+4 4 2 − 3 + = 2(x−y)x−2 −y 2 x+y x−y x2 −y 2 2x−2y−3x−3y+4 −x−5y+4 = = . x2 −y 2 x2 −y 2

(j) 14a + 3 = 2(5a − 6) ⇒ 14a + 3 = 10a − 12 ⇒ 14a − 10a = −12 − 3 ⇒ 4a = −15 ⇒ a = − 15 . 4

(g)

u(u+2)−3 3 u + = (u+2)(u−2) u−2 4−u2 2 +3)(u−1) u +2u−3 . = (u = (u+2)(u−2) (u+2)(u−2)

(k)

− 3a = 4 ⇒ 3a−62a = 4 ⇒ a6 = 4 ⇒ a = 24.

(l)

x+7 3

(h)

+

1 x−2

+

=

5+3(x−2) = (x+2)(x−2) =

3 x+2

2x+1 . (x+3)(x−2)

=

(g) −d = 7 ⇒ d = −7. (h)

3x−1 . (x+2)(x−2)

−

7 x+4

=

−

3a 7

×

14 9a

= 32 .

(b)

2x−4 3

×

6x x2 −4

=

12x(x−2) 3(x+2)(x−2)

=

2

4x . x+2

5y+15 y −9 = 5y+15 ÷ y2 +9 × y y+9y+14 2 −9 y +14 y+2 y+2 (y+7)(y +2) 5(y+7) 5(y+3) = (y+2) × (y+3)(y−3) = y−3 .

(d)

a3 +a2 a3 +a2 a2 −1 ÷ aa+1 2 −1 = a 2 −a × a+1 a2 −a 2 (a+1)(a−1) a (a+1) = a(a + = a(a−1) × a+1

(e)

2

(n)

2

(c)

1).

(b)

x 4

(c) |3x − 2| = 4 ⇒ 3x − 2 = 4 or 3x − 2 = −4 3x = 6 or 3x = −2 x = 2 or x = − 23 .

=

(d) |x + 4| = 7 ⇒ x + 4 = 7 or x + 4 = −7 x = 3 or x = −11. (e) |3x + 1| = |5x − 7| ⇒

= 5.

3x + 1 1+7 2x x

= 3 ⇒ x = 3 × 4 = 12.

(c) a + 11 = 20 ⇒ a = 20 − 11 = 9. (d) d − 5 = 6 ⇒ d = 11. 3x 7

= = = =

5x − 7 5x − 3x 8 84

1A.7.3. Surds (p. 9).

(e) 2x − 5 = 17 ⇒ 2x = 22 ⇒ x = 11. (f)

= 5 − x−1 3 ⇒ 3(x + 1) = 30 − 2(x − 1) ⇒ 3x + 3 = 30 − 2x + 2 . ⇒ 5x = 29 ⇒ x = 29 5

(b) |2x| = 12 ⇒ 2x = ±12 ⇒ x = ±6.

1. 10 2

x+1 2

(a) |x| = 9 ⇒ x = ±9.

1A.6.2. Solving Linear Equations and Absolute Value Equations (pp 7-8).

(a) 2x = 10 ⇒ x =

= x−5 ⇒ 2(x + 7) = 3(x − 5) 2 ⇒ 2x + 14 = 3x − 15 ⇒ 14 + 15 = 3x − 2x ⇒ x = 29.

2.

2

a −b −5a+5b ÷ aa+b−5 2 −7ab = a2 −8ab+7b2 a2 −b2 −5a+5b a2 −7ab × = a2 −8ab+7b2 a+b−5 a(a−7b) (a+b(a−b)−5(a−b) × a+b−5 (a−b)(a−7b) (a−b)(a+b−5) a = a. × a+b−5 a−b

a 2

(m) 20 − 3(y − 4) = 7(y − 1) ⇒ 20 − 3y + 12 = 7y − 7 ⇒ 32 + 7 = 7y + 3y ⇒ 10y = 39 ⇒ y = 3.9.

3. (a)

= 2 ⇒ 9a − 5 = 6 ⇒ 9a = 11 ⇒ a = 11 . 9

(i) 23x − 7 = 19x + 37 ⇒ 23x − 19x = 37 + 7 ⇒ 4x = 44 ⇒ x = 11.

−3(x2 −8x−9) 3 = 2x(2x−19) (x2 −8x−9)(2x−19) 2x−19 4x2 −38x−3x2 +24x+27 x2 −14x+27 . = (x+1)(x−9)(2x−19) (x+1)(x−9)(2x−19) 2x

x2 −8x−9

9a−5 3

= 9 ⇒ 3x = 63 ⇒ x = 21.

1. 4

or 3x + 1 or 3x + 5x or 8x or x

= −(5x − 7) = 7−1 = 6 = 34 .

(a) (b) (c) (d) (e) (f)

√ √ √ 32 = 42 × 2 = 4 2. √ √ √ 75 = 52 × 3 = 5 3. √ √ √ 28 = 22 × 7 = 2 7. √ √ √ 45 = 32 × 5 = 3 5. √ √ √ 98 = 72 × 2 = 7 2.    162 92 ×2 92 = = = 3 8 2 22

(g)



48 27

(h)



3 147

=



=

42 ×3 33



=

3 72 ×3



=

42 32



=

1 72

(f)

1A.8.3. 10).

(b) (c) (d) (e) (f)

(g) (h)

4 3

=

=

√ 2 5 . 5

√1 3

(b)

√2 5

2.

(a) x2 + 6x + 4 = 0 ⇒ x2 + 6x = −4 x2 + 6x + 32 = −4 + 9 (x + 3)2 = 5 √ x+3 = ± 5 √ x = −3 ± 5. Checking with the quadratic formula: (a = 1, b = 6, c = 4) x

(c)

√1 5−2

=

√ 5+2 √ √ ( 5−2)( 5+2)

=

√ 5+2 5−4

(d)

√ 2 3−1

=

√ √ 2( 3+1) √ ( 3−1)( 3+1)

=

√ 2( 3+1) 3−1

=

√ −b± b2 −4ac √2a −6± 62 −4×1×4 √ √ 2×1 −6± 20 = −6±2 5 2 √ 2 √ 2(−3± 5) = −3 ± 5. 2

(b) x2 + 5x = −5 ⇒ x2 + 5x + ( 25 )2 = −5 + ( 52 )2 (x + 25 )2 = 45  √ x + 25 = ± 54 = ± 25 √

x = − 52 ± 25 . Checking with the quadratic formula: (a = 1, b = 5, c = 5)

√ 5 + 2. =

= = = =

x

√ √ 2( 3− 2) √ 2√ = √ √ √ √ 3+ 2 ( 3+ 2)( 3− 2) √ √ √ √ 2( 3− 2) = 2( 3 − 2). 3−2

Solving Quadratic Equations (p.

(c) 2x2 + 11x − 21 = 0 ⇒ (x + 7)(2x − 3) = 0 ⇒ x = −7 or x = 23.

= 17 .

(a)

(e)

√ 5).

(b) a2 + 2a − 35 = 0 ⇒ (a + 7)(a − 5) = 0 ⇒ a = −7 or a = 5.

= 1 13 .

3. √ 3 . 3

−6(2 +

(a) x2 + 3x + 2 = 0 ⇒ (x + 1)(x + 2) = 0 ⇒ x = −1 or x = −2.

4 12 .

√ √ √ √ 8 3 − 3 3 = (8 − 3) 3 = 5 3. √ √ √ √ 2 5 − 22√× 5 = √80 − √20 = 4 × √ 4 5 − 2 5 = (4 − 2) 5 = 2 5. √ √ √ √ 3 × 2 = 3 × 2 = 6. √ √ √ √ 4 3 × 5 2 = (4 × 5) 3 × 2 = 20 6. √ √ √ √ 12 + √ 75 = √ 22 × 3 + 52 × 3 = √ 2 3 + 5 3 = 7 3. √ √ √ √24 − 96√+ 54 = √ 22 × 6 √ − 42 √ × 6 + 32 × 6 = √ √ √ 2 6 − 4 6 + 3 6 = (2 − 4 + 3) 6 = 6. √ √ √ √ √ 27 × 50 = 3 3 × 5 2 = 15 6. √ √ √ √ √45 + √28 − √20 + √63 = √ √ 3 5 + 2 7 − 2 5 + 3 7 = 5 + 5 7. =

=

1. 9 2

2. (a)

√ 6(2+ 5) 6√ = √ √ 2− 5 (2− 5)(2+ 5) √ √ 6(2+ 5) 6(2+ 5) = = 4−5 −1

= = = =

√ 3 + 1.

√ −b± b2 −4ac √2a −5± 52 −4×1×5 √ 2×1 −5± 5 2 √ − 52 ± 25 .

(c) x2 − 6x = −1 ⇒ x2 − 6x + 32 = −1 + 3 2 (x − 3)2 = 8 √ √ x − 3 = ± 8√ = ±2 2 x = 3 ± 2 2.

=

5

Checking with the quadratic formula: (a = 1, b = −6, c = 1) x

= = = = =

2. (a) (5 + 3a)3 = 53 + 3.52 .(3a) + 3.5.(3a)2 + (3a)3 = 125 + 225a + 135a2 + 27a.

√ −b± b2 −4ac √ 2a 2 6± (−6) −4×1×1 √ 2×1 6± 32 2√ √ 2) 6±4 2 = 2(3±2 2 2 √

(b) (2x − 1)4 = (2x)4 + 4.(2x)3 .(−1) + 6.(2x)2 .(−1)2 + 4.(2x).(−1)3 + (−1)4 = 16x4 − 32x3 + 24x2 − 8x + 1.

3 ± 2 2.

(c) (4a − 2)5 = (4a)5 + 5.(4a)4 .(−2) + 10.(4a)3 .(−2)2 + 10.(4a)2 .(−2)3 + 5.(4a).(−2)4 + (−2)5 = 1024a5 − 2560a4 +2560a3 − 1280a2 +320a− 32.

(d) 4x2 − 12x − 27 = 0 ⇒ x2 − 12 x = 274 4 3 2 2 x − 3x + ( 2 ) = 274 + ( 23 )2 (x − 23 )2 = 9 x − 32 = ±3 x = 32 ± 3 x = 92 or − 32 .

(d) (b + b3 )5 = b5 + 5b4 .( 3b ) + 10b3 .( b3 )2 + 10b2 .( 3b )3 + 5b.( 3b )4 + ( 3b )5 = b5 + 15b3 + 90b +

Checking with the quadratic formula: (a = 4, b = −12, c = −27) x

= = = = =

= = = =

405 b3

+

243 . b5

(f) (2x + 3)6 = (2x)6 + 6.(2x)5 .3 + 15.(2x)4 .32 + 20.(2x)3 .33 + 15.(2x)2 .34 + 6.(2x).35 + 3 6 = 64x6 + 576x5 + 2160x4 + 4320x3 + 4860x2 + 2916x + 729. 3. (a) (4x − 2)4 = (4x)4 + 4.(4x)3 .(−2) + 6.(4x)2 .(−2)2 + 4.(4x).(−2)3 + (−2)4 . ⇒ Coefficient of x3 term is 4.43 .(−2) = −512. (b) (6x2+x3 )5 = (6x2 )5+5 .(6x2 )4 . x3 +10.(6x2 )3 .( 3x )2

Checking with the quadratic formula: (a = 2, b = 5, c = −12) =

+

(e) (a2 + 4) 4 = (a2 )4 + 4.(a2 )3 .4 + 6.(a2 )2 .42 + 4.a2 .43 + 4 4 = a8 + 16a4 + 96a4 + 256a2 + 256.

√ −b± b2 −4ac √2a 12± (−12)2 −4×4×(−27) 2×4 √ 12± 576 8 12±24 8 9 or − 32 . 2

(e) 2x2 + 5x − 12 = 0 x2 + 52 x = 12 2 5 2 x + 2 x + ( 54 )2 = 6 + ( 54 )2 (x + 54 )2 = 121 16 x + 54 = ± 11 4 x = − 45 ± 11 4 x = 32 or − 4.

x

270 b

+10.(6x2 )3 .( x3 )3 + 5.(6x2 ).( 3x )4 + ( x3 )5 . Coefficient of x3 term is 0 (there is no x3 term).

√ −b± b2 −4ac √2a −5± 52 −4×2×(−12) √ 2×2 −5± 121 4 −5±11 4 3 or − 4. 2

(c) (5 − x)3 (2 + x) =

[53 + 3.52 .(−x) + 3.5.(−x)2 + (−x)3 ](2 + x).

The term x3 is (−x)3 .2 + 3.5.(−x)2 .x = 13x3 . The coefficient of x3 term is 13.

1B.1.3. Binomial Expansion and Pascal’s 1B.2.2. Sigma Notation (pp. 12-13). Triangle (pp. 11-12). 1. 1. 6 2 2 2 2 2 2 2 (a) i=1 i = 1 + 2 + 3 + 4 + 5 + 6 . 2 1 5 10 10 5 1 (b) r=0(5r − 1) = (5.0 − 1) + (5.1 − 1) 1 6 15 20 15 6 1 +(5.2 − 1) = −1 + 4 + 9. 1 7 21 35 35 21 7 1 6

(c)

10

k=4 (8

• 7 C6 =

− k) = (8 − 4) + (8 − 5) + (8 − 6)

+(8 − 7) + (8 − 8) + (8 − 9) + (8 − 10)