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1 More about Quadratic Equations

1

1 More about Quadratic

Equations

Review Exercise 1 (p. 1.5)

1.

10 or 2

010or 02

0)10)(2(

02082

=−=

=−=+

=−+

=−−

xx

xx

xx

xx

2.

2

5 or 3

052or 03

0)52)(3(

0152 2

=−=

=−=+

=−+

=−+

xx

xx

xx

xx

3.

2

3

0)32(

09124

02

962

2

2

2

=

=−

=+−

=+−

x

x

xx

xx

4. Consider the discriminant of the equation

0423

1 2=++ xx .

03

4

3

164

)4(3

1422

<−=

−=

−=∆

∴ The equation has no real roots.

5.

3

4 or

2

1

043or 012

0)43)(12(

0456

645

2

2

=−=

=−=+

=−+

=−−

=+

xx

xx

xx

xx

xx

6.

0153

351

2

2

=+−

−=−

xx

xx

Using the quadratic formula,

6

135

)3(2

)1)(3(4)5()5( 2

±=

−−±−−=x

7.

∴

4

55

6255

4

=

=

=

x

x

x

8.

∴

5

)3()3(

243)3(

5

=

−=−

−=−

x

x

x

9.

∴

6

3

2

3

2

3

2

3

2

729

64

3

2

6

6

6

=

=

=

=

x

x

x

x

10.

∴2

7

922

22

2)2(

512

14

922

912

1

−=

−=−

=

=

=

−−

−−

−

x

x

x

x

x

11.

fig.) sig. 3 to(cor. 18.4

23 log

11 log

3 log

11 log2

11 log3 2)log(

11 log3 log

113

2

2

=

+=

=−

=−

=

=

−

−

x

x

x

x

x

12.


19 log

99 log

2

1

9 log

99 log12

99 log 9log)12(

99 log9 log

999

12

12

=

−=

=+

=+

=

=

+

+

x

x

x

x

x

13.

∴ 3

55

1255

500)15(5

500555

50055

3

1

=

=

=

=−

=−⋅

=−+

x

x

x

x

xx

xx

NSS Mathematics in Action 5A Full Solutions

2

14.

17

1282

324

142

322222

3222

22

22

=

=

+

=⋅+⋅

=+

−

−+

x

x

xx

xx


2 log

17

128log

17

128log2 log

=

=

=

x

x

15.

4

5

4

11

21

2)1(log

02)1(log

2

2

2

=

=−

=−

−=−

=+−

−

x

x

x

x

x

16.

2

1

1084

12)4]log[(

14 log2)log(

=

=+

=+

=++

x

x

x

x

17.

9

20

2010

10

12

102

12

log

1 log)2log(

1

=

=−

=−

=−

−=

−

−=−−

−

x

xx

x

x

x

x

x

x

xx

18.

5

3

3050

275434

)12(2734

]3)12[(log)34(log

3log)12(log)34(log

3)12(log)34(log

333

3333

33

=

=

−=+

−=+

−=+

+−=+

+−=+

x

x

xx

xx

xx

xx

xx

Classwork

Classwork (p. 1.18)

1. (a) i5155 =−•=−

(b) i3199 =−•=−

(c) i411616 −=−•−=−−

(d) i4

1

8

14

8

4=

−•=

−

2. (a)

i

x

x

x

6

36

36

036

2

2

±=

−±=

−=

=+

(b)

i

x

x

x

3

2

9

4

27

12

01227

2

2

±=

−±=

−=

=+

Classwork (p. 1.21)

1. (a) Real part : 3 , imaginary part : 4−

(b) Real part : 5− , imaginary part : 7

(c) Real part : 8 , imaginary part : 0

(d) Real part : 0 , imaginary part : 2

3

(e) Real part : 2− , imaginary part : 2

2. (a) False (b) True (c) True (d) False

Classwork (p. 1.22)

(a)

i

i

53

259259

+=

+=−+

∴ They are equal.

(b)

i

i

i

316

316

9292 44

−≠

+=

+=−+

∴ They are not equal.

(c)

i

i

iii

74

74

)(7)1(474 32

+≠

−=

−+−−=+−


(d)

i

i

ii

iii

≠

=

+=

+=+−

5

32

3434


Quick Practice

Quick Practice 1.1 (p. 1.6) By substituting x2 = u into the equation x4 + 3x2 – 4 = 0, we have

4or 1

0)4)(1(

0432

−==

=+−

=−+

uu

uu

uu


3

Since x2 = u, we have

1

(rejected) 4or 1 22

±=

−==

x

xx ∴ The real roots of the equation are −1 and 1.

Quick Practice 1.2 (p. 1.7) By substituting x3 = u into the equation x6 + 9x3 +8 = 0, we have

1 or 8

0)1)(8(

0892

−=−=

=++

=++

uu

uu

uu


1or 2

1or 8 33

−=−=

−=−=

xx

xx ∴ The real roots of the equation are −2 and −1.

Quick Practice 1.3 (p. 1.8)

(a)

(1) 0910or 0

0)910(

0910

24

24

35

KK=+−=

=+−

=+−

xxx

xxx

xxx

By substituting x2 = u into equation (1), we have

9 or 1

0)9)(1(

09102

==

=−−

=+−

uu

uu

uu


3or 1

9or 1 22

±=±=

==

xx

xx ∴ The real roots of the equation are −3, −1, 0, 1 and 3.

(b) By substituting x2 + x = u into the equation

012)(8)( 222=++−+ xxxx , we have

06or 02

0)6)(2(

01282

=−=−

=−−

=+−

u u

uu

uu

Since x2 + x= u, we have

0)2)(3(or 0)1)(2(

06or 02 22

=−+=−+

=−+=−+

xxxx

xxxx

2or 3or 1or 2 =−==−= xxxx ∴ The real roots of the equation are −3, −2, 1 and 2.


1or 6

0)1)(6(

065

06

5

426

3

4)3(2

1

2

=−=

=−+

=−+

=−+

−=−+−

−=+

−

xx

xx

xx

xx

xx

xx


049

247

)2(5)2(2

1)2(

5)2(2

12

52

2

2

=+−

−=−

−=+−

=−

+−

=−

+

xx

xxx

xxxx

xx

xx

xx

∴ 2

659

)1(2

)4)(1(4)9()9( 2

±=

−−±−−=x


1682

)4()2(

42

42

2

22

+−=−

−=−

−=−

−=−−

xxx

xx

xx

xx

6or 3

0)6)(3(

01892

==

=−−

=+−

xx

xx

xx

Checking : When x = 3,

4

2

3232

−≠

−=

−−=−− xx

Hence, 3 is not a root of the equation

.42 −=−− xx

When x = 6,

4

6262

−=

−−=−− xx ∴ The real root of the equation is 6.


9or 4

1

0)9)(14(

09374

499124

)7()32(

732

0372

2

2

22

==

=−−

=+−

=++

=+

=+

=+−

xx

xx

xx

xxx

xx

xx

xx

Checking: When

0

34

17

4

12372,

4

1

=

+−

=+−= xxx

0

397)9(23729,When

=

+−=+−= xxx ∴ The real roots of the equation are

4

1 and 9.

Alternative Solution

By substituting ux = into the equation 0372 =+− xx ,

we have

3or 2

1

0)3)(12(

0372 2

==

=−−

=+−

uu

uu

uu


4

Since ux = , we have

9or 4

1

3or 2

1

==

==

xx

xx

∴ The real roots of the equation are 4

1 and 9.


)1(04)2(5)2(

04)2(52

2

2

KK=+−

=+−

xx

xx

By substituting 2x = u into (1), we have

4or 1

0)4)(1(

0452

==

=−−

=+−

uu

uu

uu

Since 2x = u, we have

2or 0

22or 22

42or 12

20

==

==

==

xx

xx

xx

∴ The real roots of the equation are 0 and 2.

Quick Practice 1.9 ( p. 1.12)

)1(027)5(26)5(

027)5(2625

2KK=−−

=−−

xx

xx


1or 27

0)1)(27(

027262

−==

=+−

=−−

uu

uu

uu


(rejected) 15or 275 −==xx


5 log

27 log

27 log5 log

27 log5 log

=

=

=

=

x

x

x

∴ The real root of the equation is 2.05.

Quick Practice 1.10 (p. 1.12) By substituting log x = u into the equation

02 log 3) (log 2=+− xx , we have

2or 1

0)2)(1(

0232

==

=−−

=+−

uu

uu

uu

Since log x = u, we have

100or 10

10or 10

2 logor 1 log

21

==

==

==

xx

xx

xx



(a)

∴ (rejected) 3

5or 3

0)53)(3(

01543

16143

2)1)(13(

4)]1)(13[(log

4)1(log)13(log

2

2

4

2

22

−==

=+−

=−−

=+−

=−−

=−−

=−+−

xx

xx

xx

xx

xx

xx

xx

(b)

10

1

9

19

log

1)9log( log

2

2

2

=+

−=

+

−=+−

x

x

x

x

xx

∴ 9or 1

0)9)(1(

0910

910

2

2

==

=−−

=+−

+=

xx

xx

xx

xx

Quick Practice 1.12 (p. 1.15) Let x be the original number of children in the group.

(rejected) 30or 18

0)30)(18(

054012

42282160180

)4228()12(180

12

4228180

412

180180

2

2

−==

=+−

=−+

+=+

+=+

+

+=

++

=

xx

xx

xx

xxx

xxx

x

x

x

xx

∴ The original number of children in the group is 18.

Quick Practice 1.13 (p. 1.16) Let x be the number of days Mr Tung worked in the project.

(rejected) 12or 9

0)12)(9(

01083

3910836

)39()3(36

3

3936

3

)3(3636

1003

36003600

2

2

−==

=+−

=−+

+=+

+=+

+

+=

+

++=

++

=

xx

xx

xx

xxx

xxx

x

x

x

x

x

x

xx

∴ Mr Tung worked 9 days in the project.


(a)

i

ii

11

47

116149

1649

=

+=

−•+−•=

−+−


5

(b)

i

ii

5

7

25

3

14125

9

425

9

−=

−=

−•−−•=

−−−


(a)

i

i

ii

ii

−=

−•=

•=

=+•

)()1(

)(

4

344

34419

(b)

1

)1()1(

)(

22

2224

222490

−=

−•=

•=

=+•

ii

ii

Quick Practice 1.16 (p. 1.22) Using the quadratic formula,

i

i

x

21

2

42

2

162

)1(2

)5)(1(422 2

±−=

±−=

−±−=

−±−=


ibai 85215 +=− ∴ 3

515

=

=

b

b

and

4

82

−=

=−

a

a


(a)

i

i

iiii

92

)112()46(

11426)114()26(

+=

+−+−=

+−−=+−+−

(b)

i

i

iiii

1618

)97()513(

95713)95()713(

+−=

++−−=

+−+−=−−+−


(a)

i

ii

iii

iiiii

1113

)1(25615

25615

))(25()3)(25()3)(25(

2

+−=

−−++−=

−++−=

−+−−=+−−

(b)

i

i

ii

iii

125

)1(4129

4129

)2()2)(3(2)3()23(

2

222

−=

−+−=

+−=

+−+−=+−

(c)

34

)1(925

925

)3()5()35)(35(

2

22

=

−−=

−=

−−=−−+−

i

iii


(a)

22 )3(1

3010

31

31

31

10

31

10

i

i

i

i

ii

−

−=

−

−•

+=

+

i

i

i

31

10

3010

)9(1

3010

−=

−=

−−

−=

(b)

i

i

i

i

iii

i

i

i

i

i

i

32

2

64

)1(1

)1(65

1

55

1

1

1

5

1

5

22

2

+=

+=

−−

−++=

−

+++=

+

+•

−

+=

−

+

Quick Practice 1.21 (p. 1.26) ∵ 2 − i is a root of the equation x2 + px + q = 0. ∴

0)4()23(

02)1(44

0)2()44(

0)2()2(

2

2

=−−+++

=+−+−+−

=+−++−

=+−+−

ipqp

qpipi

qpipii

qipi

∴

=−−

=++

)2(04

)1(023

KK

KK

p

qp

From (2), p = −4

By substituting p = −4 into (1), we have

5

0)4(23

=

=+−+

q

q ∴

5 and 4 =−= qp

Further Practice

Further Practice (p. 1.11)

1. By substituting x3 = u into the equation x6 + 2x3 + 1 = 0, we have

1

0)1(

012

2

2

−=

=+

=++

u

u

uu


1

13

−=

−=

x

x ∴ The real root of the equation is −1.


6

2. By substituting x2 = u into the equation 2x4 − 7x2 − 4 = 0, we have

2

1or 4

0)12)(4(

0472 2

−==

=+−

=−−

uu

uu

uu


2

(rejected) 2

1or 4 22

±=

−==

x

xx ∴ The real roots of the equation are −2 and 2.

3.

2or 2

9

0)2)(92(

01852

036104

942710

)32)(32(211464

1)32)(32(

)32(7)32(2

132

7

32

2

2

2

2

=−=

=−+

=−+

=−+

−=+−

+−=+−+

=+−

−−+

=+

−−

xx

xx

xx

xx

xx

xxxx

xx

xx

xx

4.

6or 3

0)6)(3(

0189

)2(9

)23(

23

023

2

2

22

==

=−−

=+−

−=

−=

−=

=−−

xx

xx

xx

xx

xx

xx

xx

Checking: When x = 3,

0

233323

=

−−=−− xx

When x = 6,

0

263623

=

−−=−− xx


5.

4

9or

4

1

0)94)(14(

094016

6492416

)8()34(

834

83

4

2

2

22

==

=−−

=+−

=++

=+

=+

=+

xx

xx

xx

xxx

xx

xx

xx

Checking: When 4

1=x ,

8

62

4

1

3

4

14

34

=

+=

+=+

xx

When 4

9=x ,

8

26

4

9

3

4

94

34

=

+=

+=+

xx


1 and

4

9.

Alternative Solution

By substituting ux = into the equation

83

4 =+

xx , we have

2

3or

2

1

0)32)(12(

0384

834

83

4

2

2

==

=−−

=+−

=+

=+

uu

uu

uu

uu

uu


4

9or

4

1

2

3or

2

1

==

==

xx

xx


1 and

4

9.

6. By substituting uxx =−+ 242 into the equation

060)24(4)24( 222=−−+−−+ xxxx , we have

10or 6

0)10)(6(

06042

=−=

=−+

=−−

uu

uu

uu

Since uxx =−+ 242 , we have

2or 6or2

0)2)(6(or0)2(

0124or044

1024or624

2

22

22

=−=−=

=−+=+

=−+=++

=−+−=−+

xxx

xxx

xxxx

xxxx

∴ The real roots of the equation are −6, −2 and 2.


1.

......(1) 0433)3(

0433

2

12

=−•−

=−−+

xx

xx


1or 4

0)1)(4(

0432

−==

=+−

=−−

uu

uu

uu


3x = 4 or 3x = −1 (rejected)


3 log

4 log

4 log3 log

4 log3 log

=

=

=

=

x

x

x

∴ The real root of the equation is 1.26.


7

2.

......(1) 05)5(26)5(5

05)5(26)25(5

2=+−

=+−

xx

xx


5or 5

1

0)5)(15(

05265 2

==

=−−

=+−

uu

uu

uu


5

15 =

x or 5x = 5

155 −=

x or 5x = 51

x = −1 or x = 1

∴ The real roots of the equation are −1 and 1.

3. By substituting log(x + 1) = u into the equation

,01)1log(2)]1[log( 2=++++ xx we have

1

0)1(

012

2

2

−=

=+

=++

u

u

uu

Since log(x + 1) = u, we have

10

9

10

11

1)1log(

−=

=+

−=+

x

x

x

∴ The real root of the equation is 10

9− .

4.

∴ 2or 1

0)2)(1(

023

623

6)1)(2(

6log)1)(2(

log

6loglog)1(log)2(log

2

2

55

5555

==

=−−

=+−

=++

=++

=

++

=−+++

xx

xx

xx

xxx

x

xx

x

xx

xxx


1.

i

x

x

x

x

xx

7

149

49

49

049

0501

050)1)(1(

2

2

2

±=

−•±=

−±=

−=

=+

=+−

=+−+

2. (a)

i

iii

7

1098

1100181164

1008164

=

−+=

−•−−•+−•=

−−−+−

(b)

i

i

ii

3

1

6

2

6

5

6

7

136

251

36

49

36

25

36

49

=

=

−=

−•−−•=−−−

Further Practice (p. 1.23) 1. (a) Using the quadratic formula,

i

x

6

23

6

1

6

231

)3(2

)2)(3(4)1()1( 2

±=

−±=

−−±−−=

(b)

0342

324

3)2(2

2

2

=+−

=−

=−

xx

xx

xx


i

i

x

2

21

4

84

4

84

)2(2

)3)(2(4)4()4( 2

±=

±=

−±=

−−±−−=

2. (a) iyix )2(47)1( +−=+−

∴

5

41

=

=−

x

x

and

9

27

)2(7

−=

−−=

+−=

y

y

y

(b)

xiy

yix

yixii

2

)1()(2

21410 23

−=

−−−=

−=+

∴

7

214

−=

−=

x

x

and 10=y


1. (a)

i

i

iii

63

)758()423(

)74()52()83(

−−=

+−−+−−=

−−−−+−


8

(b)

25

)9(16

)3(4

)34)(34(

)34)](1(232[

)34)(242(

)34)(21)(2(

22

2

=

−−=

−=

−+=

−−−+=

−−+−=

−+−

i

ii

ii

iiii

iii

(c)

)1(4

)1(56

2

326

2

2

2

3

2

3

22

2

−−

−++=

−

+++=

+

+•

−

+=

−

+

i

i

iii

i

i

i

i

i

i

i

i

+=

+=

1

5

55

2. (a)

iyix

iyiixi

)1(102)2(

)6()4()3()52(

−+=++

+−−+=−++

∴

8

102

=

=+

x

x

and

3

12

=

−=

y

y

(b)

yiixx

yixix

yixiixi

yiixi

=−++

=−−−+

=−−+

=−+

)182()312(

)1(3)182(12

318212

)32)(6(

2

∴

=−

=+

)2( ......182

)1( ......0312

yx

x

From (1), x = −4

By substituting x = −4 into (2), we have

26

18)4(2

−=

=−−

y

y

∴ 4−=x and 26−=y

3. ∵ p + i is a root of the equation x2 + 4x + q = 0.

∴

0)42()14(

04412

0442

0)(4)(

2

2

22

2

=++−++

=+++−+

=+++++

=++++

ipqpp

qippip

qipipip

qipip

∴

=+

=−++

......(2)042

......(1)0142

p

qpp

From (2), p = −2

By substituting p = −2 into (1), we have

5

01)2(4)2( 2

=

=−+−+−

q

q

∴ 2−=p and 5=q

Exercise

Exercise 1A (p. 1.14)

Level 1

1. By substituting x2 = u into the equation x4 − 17x2 + 16 = 0, we have

16or 1

0)16)(1(

016172

==

=−−

=+−

uu

uu

uu

Since ux =2 , we have

4or1

16or1 22

±=±=

==

xx

xx ∴ The real roots of the equation are −4, −1, 1 and 4.

2. By substituting x2 = u into the equation 08011 24=−− xx ,

we have

5or 16

0)5)(16(

080112

−==

=+−

=−−

uu

uu

uu


x2 = 16 or x2 = −5 (rejected)

x = ±4


3.

......(1) 0352or 0

0)352(

0352

24

24

35

=−−=

=−−

=−−

xxx

xxx

xxx


5or 7

0)5)(7(

03522

−==

=+−

=−−

uu

uu

uu


x2 = 7 or x2 = −5 (rejected)

7±=x

∴ The real roots of the equation are 7− , 0 and 7 .

4. By substituting x3 = u into the equation x6 – 26x3 − 27 = 0, we have

27or 1

0)27)(1(

027262

=−=

=−+

=−−

uu

uu

uu


3or1

27or1 33

=−=

=−=

xx

xx


5. By substituting x3 = u into the equation x6 – 5x3 + 4 = 0, we have

4or 1

0)4)(1(

0452

==

=−−

=+−

uu

uu

uu


3

33

4or1

4or1

==

==

xx

xx

∴ The real roots of the equation are 1 and 3 4 .


9

6.

)1( ......02712or0

02712or0

0)2712(

02712

24

242

242

246

=+−=

=+−=

=+−

=+−

xxx

xxx

xxx

xxx


9or3

0)9)(3(

027122

==

=−−

=+−

uu

uu

uu


3or3

9or3 22

±=±=

==

xx

xx

∴ The real roots of the equation are −3, 3− , 0, 3

and 3.

7. By substituting x2 = u into equation 0954 24=−+ xx , we

have

4

9or1

0)94)(1(

0954 2

−==

=+−

=−+

uu

uu

uu


1

(rejected) 4

9or1 22

±=

−==

x

xx


8. By substituting x2 = u into equation 0251014 24=+− xx ,

we have

25or4

1

0)25)(14(

0251014 2

==

=−−

=+−

uu

uu

uu


5or2

1

25or4

1 22

±=±=

==

xx

xx

∴ The real roots of the equation are −5, 2

1− ,

2

1

and 5.

9. By substituting x3 = u into equation 0198 36=+− xx , we

have

1or8

1

0)1)(18(

0198 2

==

=−−

=+−

uu

uu

uu


1or2

1

1or8

1 33

==

==

xx

xx


1 and 1.

10.

4or 12

0)4)(12(

0488

848

24

12

2

2

=−=

=−+

=−+

=−

=−

xx

xx

xx

xx

x

x

11.

4or 1

0)4)(1(

043

622

)2)(3()1(2

1

3

2

2

2

2

=−=

=−+

=−−

−−=−

+−=−

−

−=

+

xx

xx

xx

xxx

xxx

x

x

x

12.

3or 2

0)3)(2(

065

06

5

126

16

12)1(6

1

2

==

=−−

=+−

=+−

=+++

=+

+

xx

xx

xx

xx

xx

xx

13.

4or 4

16

016

016

616

28

6)2(8

1

2

2

=−=

=

=−

=−

−=−+−

−=+

−

xx

x

x

xx

xx

xx

14.

2or 9

0)2)(9(

0187

31810

1)3(

)3(64

16

3

4

2

2

=−=

=−+

=−+

−=−

=−

−−

=−−

xx

xx

xx

xxx

xx

xx

xx

15.

1or 5

11

0)1)(115(

01165

314588

)15)(3(88

1)15)(3(

)3(315

115

3

3

1

2

2

=−=

=−+

=−+

−+=+

−+=+

=−+

++−

=−

++

xx

xx

xx

xxx

xxx

xx

xx

xx


10

16.

8or 2

0)8)(2(

0166

166

4)6(

46

2

2

222

2

=−=

=−+

=−−

=−

=−

=−

xx

xx

xx

xx

xx

xx

Checking: When x = −2,

4

16

)2(6)2(6 22

=

=

−−−=− xx

When x = 8,

4

16

)8(686 22

=

=

−=− xx

∴ The real roots of the equation are –2 and 8.

17.

14or 7

0)14)(7(

09821

201002

)10()2(

102

102

2

2

22

==

=−−

=+−

+−=+

−=+

−=+

=++

xx

xx

xx

xxx

xx

xx

xx


10

7272

=

++=++ xx

When x = 14,

10

18

142142

≠

=

++=++ xx

Hence, 14 is not a root of the equation 102 =++ xx .

∴ The real root of the equation is 7.

18.

4or 4

7

0)4)(74(

028234

2520433

)52()33(

5233

2533

2

2

22

==

=−−

=+−

+−=−

−=−

−=−

=+−

xx

xx

xx

xxx

xx

xx

xx

Checking:

When 4

7=x ,

2

13

52

3

54

9

534

73533

=

+=

+=

+−

=+−x

2

13

2

7

4

722

≠

=

=x

Hence, 4

7 is not a root of the equation xx 2533 =+− .

When x = 4,

8

53

53)4(3533

=

+=

+−=+−x

8

)4(22

=

=x


19.

9or 4

0)9)(4(

03613

253612

)5()6(

56

065

2

2

22

==

=−−

=+−

=++

=+

=+

=+−

xx

xx

xx

xxx

xx

xx

xx


0

645465

=

+−=+− xx

When x = 9,

0

695965

=

+−=+− xx ∴ The real roots of the equation are 4 and 9.

20.

16or 9

0)16)(9(

014425

14424

)()12(

12

112

112

2

2

22

==

=−−

=+−

=+−

=−

=−

=−

=−

xx

xx

xx

xxx

xx

xx

x

x

xx

Checking:

When x = 9,

1

1

43

9

129

12

≠

−=

−=

−=−

xx

Hence, 9 is not a root of the equation 113

=−

xx .

When x = 16,

1

34

16

1216

12

=

−=

−=−

xx



11

21.

)1( ......02)2(3)2(

02)2(32

2

2

=+−

=+−

xx

xx


2or 1

0)2)(1(

0232

==

=−−

=+−

uu

uu

uu


1or0

22or12

==

==

xx

xx


22.

)1( ......09)5(26)5(3

09)5(26)5(3

2

2

=−+

=−+

xx

xx


9or 3

1

0)9)(13(

09263 2

−==

=+−

=−+

uu

uu

uu



5 log

3

1log

3

1log5 log

3

1log5 log

(rejected) 95or 3

15

−=

=

=

=

−==

x

x

x

xx

∴ The real root of the equation is −0.683.

23. By substituting log x = u into the equation

04 log 5) (log 2=+− xx , we have

4or 1

0)4)(1(

0452

==

=−−

=+−

uu

uu

uu


000 10

10or10

4 logor1 log4

=

==

==

xx

xx

24.

16or 1

0)16)(1(

01617

1716

1716

17 log16

log

17 log log)16(log

2

2

2

2

2

==

=−−

=+−

=+

=+

=

+

=−+

xx

xx

xx

xx

x

x

x

x

xx

25.

(rejected) 3

5or 6

0)53)(6(

030233

1040233

10)5)(83(

1)]5)(83[(log

1)5( log)83(log

2

2

==

=−−

=+−

=+−

=−−

=−−

=−+−

xx

xx

xx

xx

xx

xx

xx

26.

(rejected) 5or 5

25

169

4)3)(3(

2)]3)(3[(log

2)3( log)3(log

2

2

2

4

44

−==

=

=−

=−+

=−+

=−++

xx

x

x

xx

xx

xx

Level 2

27. By substituting uxx =− 42 into the equation

048)4(8)4( 222=−−−− xxxx , we have

012or 04

0)12)(4(

04882

=−=+

=−+

=−−

uu

uu

uu

Since uxx =− 42 , we have

6or 2or2

0)6)(2(or0)2(

0124or0442

22

=−==

=−+=−

=−−=+−

xxx

xxx

xxxx

∴ The real roots of the equation are −2, 2 and 6.

28. By substituting x2 + 2x = u into the equation

04)2(5)2( 222=++++ xxxx , we have

04or 01

0)4)(1(

0452

=+=+

=++

=++

uu

uu

uu

Since uxx =+ 22 , we have

042or 012 22=++=++ xxxx

Consider 0122=++ xx .

1

0)1(

012

2

2

−=

=+

=++

x

x

xx

Consider

.0422=++ xx

0

12

)4)(1(422

<

−=

−=∆

∴ x2 + 2x + 4 = 0 has no real roots.

∴ The real root of the equation is −1.

29. By substituting ux

=− 2

1 into the equation

062

1

)2(

12

=−−

+− xx

, we have

2or 3

0)2)(3(

062

=−=

=−+

=−+

uu

uu

uu


12

Since ux

=− 2

1, we have

2

5or

3

52

12or

3

12

22

1or3

2

1

==

=−−=−

=−

−=−

xx

xx

xx


5 and

2

5.


=−12

1 into the equation

0712

6

)12(

12

=−−

−− xx

, we have

7or 1

0)7)(1(

0762

=−=

=−+

=−−

uu

uu

uu

Since ux

=−12

1, we have

7

4or0

7

112or112

712

1or1

12

1

==

=−−=−

=−

−=−

xx

xx

xx

∴ The real roots of the equation are 0 and 7

4.


=

+1

1 into the equation

031

5

1

2=+

+

−+ xx

, we have

1or 2

3

0)1)(32(

0352 2

==

=−−

=+−

uu

uu

uu

Since ux

=

+1

1, we have

0or9

5

11or3

21

11or3

21

11

1or

2

3

1

1

2

2

=−=

=+

=+

=+=+

=

+

=

+

xx

xx

xx

xx


5− and 0.

32.

)1......(027)3(12)3(

027)3)(3(4)3(

027)3(43

2

2

12

=+−

=+−

=+−+

xx

xx

xx

By substituting ux=3 into (1), we have

9or 3

0)9)(3(

027122

==

=−−

=+−

uu

uu

uu

Since ux=3 , we have

∴ 2or1

93or33

==

==

xx

xx

33.

......(1) 025)5(24)5(

25)5(245

55

524

5

5

5)5(245

2

2

2

112

=−−

=−

=

−

=−−−

xx

xx

xx

xx


(rejected) 1or 25

0)1)(25(

025242

−==

=+−

=−−

uu

uu

uu


∴ 2

55

255

2

=

=

=

x

x

x

34.

)1( ......07)7(6)7(

07)7(67

07)7(649

2

2

=−−

=−−

=−−

xx

xx

xx


(rejected) 1or 7

0)1)(7(

0762

−==

=+−

=−−

uu

uu

uu


∴ 1

77

=

=

x

x

35. By substituting log(x + 2) = u into the equation

02)2log()]2[log( 2=−+−+ xx , we have

2or 1

0)2)(1(

022

=−=

=−+

=−−

uu

uu

uu

Since ux =+ )2log( , we have

98or10

19

1002or10

12

2)2log(or1)2log(

=−=

=+=+

=+−=+

xx

xx

xx

36. By substituting log2(2x − 1) = u into the equation

02)12(log 3)]12([log 22

2 =+−−− xx , we have

2or 1

0)2)(1(

0232

==

=−−

=+−

uu

uu

uu

Since ux =− )12(log2 , we have

2

5or

2

3412or212

2)12(logor1)12(log 22

==

=−=−

=−=−

xx

xx

xx


13

37.

(rejected) 2or 12

0)2)(12(

02410

8242

8)4)(6(

8)4)(6(

3)4)(6(

log

3log)4(log)6(log

2

2

2

222

−==

=+−

=−−

=−−

=+−

=+−

=

+−

=−++−

xx

xx

xx

xxx

xxx

x

xx

x

xx

xxx

38. (a)

323

.)15 Substitute(233

1523)15(3

15233153

152153

2

22

22

22

22

−+=

=+++−=

+++−++=

+++−++=

++++

uu

uxxuu

xxxx

xxxx

xxxx

(b) By substituting uxx =++ 152 into the equation

2152153 22=++++ xxxx , we have

(rejected) 3

5or 1

0)53)(1(

0523

(a)) (from2323

2

2

−==

=+−

=−+

=−+

uu

uu

uu

uu

Since uxx =++ 152 , we have

0or 5

0)5(

05

115

115

2

2

2

−=

=+

=+

=++

=++

x

xx

xx

xx

xx

39. (a) R.H.S.

L.H.S.

22

2)2(22

2)2(

2

2

2

=

+++=

−++++=

−++=

xxxx

xxxx

xx

∴ 2)2(22 22−++≡+++ xxxxxx

(b)

16)2(

142)2(

1422

2

2

2

=++

=−++

=+++

xx

xx

xxxx

(from (a))

22 )]4([2or)4(2

)4(2or42

42or42

xxxx

xxxx

xxxx

+−=+−=+

+−=+−=+

−=++=++

22 8162or8162 xxxxxx ++=++−=+

(2) ...... 0147or(1) ...... 0149 22=++=+− xxxx

From (1),

7or 2

0)7)(2(

==

=−−

xx

xx

When x = 2,

14

22)2(22222 22

=

+++=+++ xxxx

When x = 7,

14

98

27)7(27722 22

≠

=

+++=+++ xxxx

Hence, 7 is not a root of the equation. From (2), consider the discriminant of

01472=++ xx .

0

7

)14)(1(472

<

−=

−=∆

∴ (2) has no real roots.


Exercise 1B (p. 1.17)

Level 1 1. Let x be the number.

2

5or

5

2

0)52)(25(

0102910

291010

10

291

10

291

2

2

2

==

=−−

=+−

=+

=+

=+

xx

xx

xx

xx

xx

xx

∴ The possible values of the number are 5

2 and

2

5.

2. Let x and x + 1 be the two consecutive integers.

2or (rejected) 5

3

0)2)(35(

0675

55612

)1(5)12(6

6

5

)1(

1

6

5

1

11

2

2

=−=

=−+

=−−

+=+

+=+

=+

++

=+

+

xx

xx

xx

xxx

xxx

xx

xx

xx

∴ The two consecutive integers are 2 and 3.

3. Let x be the number.

2or 6

0)2)(6(

0124

12)4(

1124

2

=−=

=−+

=−+

=+

=+

xx

xx

xx

xx

xx

∴ The possible values of the number are −6 and 2.


14

4. Let x be the smaller positive even integer.

Then, the larger even integer is 2+x .

4or (rejected) 1

0)4)(1(

045

944

)3()2(

32

2

2

22

==

=−−

=+−

=++

=+

=+

xx

xx

xx

xxx

xx

xx

∴ The two consecutive positive even integers are

4 and 6. 5. From the question, we have

18or (rejected) 6

0)18)(6(

010812

6610854

)66()2(54

2

6654

2

26454

12

105454

2

2

=−=

=−+

=−−

−=−

−=−

−

−=

−

+−=

−−

+=

NN

NN

NN

NNN

NNN

N

N

N

N

N

N

NN

6. Let $x be the original price of an apple.

3or (rejected) 2

5

0)3)(52(

0152

05.375.25

55.775.3775

)55.77()5.0(75

5.0

5.257575

55.0

7575

2

2

2

=−=

=−+

=−−

=−−

−=−

−=−

−

+−=

−−

=

xx

xx

xx

xx

xxx

xxx

x

x

x

xx

∴ The original price of an apple is $3.

7. Let x km/h be the original speed of Peter.

(rejected) 4or 5

0)4)(5(

020

212020

)21()1(20

1

12020

11

2020

2

2

−==

=+−

=−−

−=−

−=−

−

+−=

−−

=

xx

xx

xx

xxx

xxx

x

x

x

xx

∴ The original speed of Peter is 5 km/h.

8. Let x km/h be the usual walking speed of David.

(rejected) 3or 4

0)3)(4(

012

131212

)13()1(12

1

11212

11

1212

5.01

66

2

2

−==

=+−

=−−

−=−

−=−

−

+−=

−−

=

−−

=

xx

xx

xx

xxx

xxx

x

x

x

xx

xx

∴ The usual walking speed of David is 4 km/h.

Level 2 9. From the question, we have

(rejected) 4or 5

0)4)(5(

020

04022

2424040

)242()1(40

1

224040

21

4040

2

2

2

−==

=+−

=−−

=−−

−=−

−=−

−

+−=

−−

=

NN

NN

NN

NN

NNN

NNN

N

N

N

NN

10. From the question, we have

(rejected) 20or 30

0)20)(30(

060010

720720060012

720)10)(72012(

720)10(720

12

2

2

−==

=+−

=−−

=−+

=−+

=−

+

NN

NN

NN

NNN

NNN

NN

11. Let x be the number of rabbits that the man bought.

(rejected) 10or 20

0)10)(20(

020010

0800040040

)5)(401600(1800

)5(401600

2001600

2

2

−==

=+−

=−−

=−−

−+=

−

+=+

xx

xx

xx

xx

xxx

xx

∴ The man bought 20 rabbits.


15

12. Let x km/h be the man’s rowing speed in still water.

(rejected) 2or 8

0)2)(8(

0166

166

)4)(4()4(3)4(3

14

3

4

3

44

12

4

12

2

2

−==

=+−

=−−

−=

−+=−++

=+

+−

=+

+−

xx

xx

xx

xx

xxxx

xx

xx

∴ The man’s rowing speed in still water is 8 km/h.


)2(6)6)(22(

6

6

)2(

2

)6(6

1

1

2

11

+=++

+=

+

++

+

=+

+

NNNN

NNN

NN

NNN

2or (rejected) 2

3

0)2)(32(

062

6367

)2(3)6)(1(

2

22

=−=

=−+

=−−

+=++

+=++

NN

NN

NN

NNNN

NNNN

Exercise 1C (p. 1.27) Level 1

1. (a)

i

ii

5

3294

=

+=−+−

(b)

i

ii

2

2

1

2

5

4

1

4

25

=

−=−−−

2. (a)

i

i

ii

−=

=

=+×

3

3254103

(b)

1

)1(

2

216466

=

−−=

−=

−=−+×

i

ii

3. (a) Using the quadratic formula,

i

i

x

±=

±=

−±=

−−±−−=

1

2

22

2

42

)1(2

)2)(1(4)2()2( 2

(b)

0102

1332

13)3)(1(

2

2

=+−

−=−−

−=−+

xx

xx

xx


i

i

x

31

2

62

2

362

)1(2

)10)(1(4)2()2( 2

±=

±=

−±=

−−±−−=

4. (a) iyxi 6714 +=−

∴

2

714

=

=

y

y

and

6

6

−=

=−

x

x

(b)

yiix

yix

−−=+−

+−=−+−

832

892 3

∴

4

82

=

−=−

x

x

and

3

3

−=

−=

y

y

5.

i

iii

35

)58()32()53()82(

+=

−++=−++

6.

i

iii

61

)82()54()85()24(

−=

−++−=−++−

7.

i

iii

68

)42()35()43()25(

−=

−−++=+−−−

8.

i

iii

136

)67()42()64()72(

+−=

++−−=−−+−

9. (a)

7)3()5(

7)3()5(

−=−++

−=−++

ixy

iyxi

∴

12

75

−=

−=+

y

y

and

3

03

=

=−

x

x

(b)

0)105()63(

0)106()53(

=−+−

=+−+

iyx

yiix

∴

2

063

=

=−

x

x

and

2

1

0105

=

=−

y

y

10.

i

i

iiii

35

)1(53

53)53( 2

+=

−−=

−=−


16

11.

i

i

iiiii

514

)1(2512

28312)23)(4( 2

+=

−−+=

−+−=+−

12.

i

i

iii

125

)1(4129

)2()2)(3(23)23( 222

−=

−+−=

+−=−

13.

34

25)1(9

5)3()53)(53( 22

−=

−−=

−=−+ iii

14.

i

i

i

i

ii

ii

i

i

i

5

4

5

3

5

43

)1(4

144

2

))(2(22

)2)(2(

)2(

2

2

22

22

2

+=

+=

−−

−+=

−

++=

+−

+=

−

+

15.

i

i

ii

ii

ii

i

i

−=

−=

−−

−−−+=

−+

−+=

+

+

3

29

2987

)1(254

)1(65552622

)52)(52(

)52)(1311(

52

1311

16.

i

i

ii

ii

ii

i

i

−=

−=

−−

−−+−=

−−+−

−−+−=

+−

+−

4

5

520

)1(4

)1(671214

)2)(2(

)2)(67(

2

67

17.

i

i

i

ii

i

i

i

5

62

5

1

5

621

)1(23

)1(2623

)23)(23(

)23(

23

23 2

−=

−=

−−

−+−=

−+

−=

+

−

Level 2

18.

i

ii

ii

iiiiii

−=

−−−+=

−+=

−−+−−=−+−

157

)1(32221154

)7)(322(

)7](520)1(28[)7)(52)(4(

19.

i

ii

ii

ii

i

i

ii

i

ii

i

25

1

25

7

)1(916

)1(3344

)34)(34(

)34)(1(

34

1

)1(242

1

)21)(2(

1

+=

−−

−−−+=

−+

−+=

+

+=

−−+−

+=

+−

+

20.

i

ii

iii

iii

3449

)2442(107

)]1(18123624[107

)36)(64()107(

+−=

−−+−=

−−+−−+−=

+−−+−

21.

i

ii

ii

ii

i

i

i

i

i

i

25

66

25

112

)1(169

)1(40963072

)43)(43(

)43)(1024(

43

1024

)1(44

)1(1025

2

52

+=

−−

−−+−=

+−

+−=

−

−=

−+−

−+−=

−

−

22.

i

i

iiiiii

80

)10(8

)]54()54)][(54()54[()54()54( 22

=

=

−−+−++=−−+

23.

2

11

2

2

2

2

)1(21

)1(21

)1(21

)1(21

1

1

1

122

−=

−−=

−+

−=

−+−

−+++

−++

−+−=

−

+−

+

−

i

i

i

i

i

i

i

i

i

i

i

i

24.

xixy

xyixi

xyi

xi

xyi

xi

=−++

=++−

=

−

−−−

=

−−−

)26(7

627

)1(

6)27(

6)27(

∴

3

026

=

=−

x

x

and

4

37

7

−=

=+

=+

y

y

xy


17

25.

iixy

iyixi

iyiixi

32)1()1(

321

32)1()1(

+=++−

+=−++

+=+++

∴

1

31

−=

=+

x

x

and

1

21

−=

=−

y

y

26.

iyixx

iyixix

iyixi

5)2()2(

5)1(22

5)2)(1(

+=−++

+=−−−+

+=−+

∴

7

52

=

=−

x

x

and

9

27

2

=

=+

=+

y

y

yx

27.

xiiy

xiyi

xiyii

yixi

i

22

22)1(

)1(2)1(

1

2

1

+=−

+=−−−

+=−−

−−=

+

∴ 2=y

and

2

1

21

−=

=−

x

x

28. ∵ 4 − i is a root of the equation 02=++ nmxx .

∴

0)8(415

04)1(816

0)4()4( 2

=−−+++

=+−+−+−

=+−+−

imnm

nmimi

nimi

∴

=−−

=++

)2( ......08

)1( ......0415

m

nm

From (2), m = −8

By substituting m = −8 into (1), we have

17

0)8(415

=

=+−+

n

n

∴ 8−=m and 17=n

29. ∵ a + 2i is a root of the equation 062 2=++ bxx .

∴

0)128()862(

0126)44(2

0)2(6)2(2

2

2

2

=++−++

=+++−+

=++++

iabaa

biaaia

biaia

∴

=+

=−++

)2( ......0128

)1( ......0862 2

a

baa

From (2), we have

2

3−=a

By substituting 2

3−=a into (1), we have

2

25

082

36

2

32

2

=

=−+

−+

−

b

b

∴ 2

3−=a and

2

25=b

30. 02510 24=++ xx …… (1)

By substituting x2 = u into (1), we have

5

0)5(

02510

2

2

−=

=+

=++

u

u

uu


5

52

ix

x

±=

−=

31. 06)2(5)2( 222=+−+− xxxx …… (1)

By substituting x2 − 2x = u into (1), we have

03or 02

0)3)(2(

0652

=+=+

=++

=++

uu

uu

uu


032or 022 22=+−=+− xxxx

Consider 0222=+− xx .


i

x

±=

−±=

−−±−−=

1

2

42

)1(2

)2)(1(4)2()2( 2

Consider 0322=+− xx .


i

x

21

2

82

)1(2

)3)(1(4)2()2( 2

±=

−±=

−−±−−=

Revision Exercise 1 (p. 1.31)

Level 1

1. 0209 24=+− xx …… (1)


5or 4

0)5)(4(

02092

==

=−−

=+−

uu

uu

uu


5or2

5or4 22

±=±=

==

xx

xx

∴ The real roots of the equation are 5− , −2, 2 and

5 .


18

2. 04359 24=−+ xx …… (1)


4or 9

1

0)4)(19(

04359 2

−==

=+−

=−+

uu

uu

uu


3

1

(rejected) 4or9

1 22

±=

−==

x

xx


1− and

3

1.

3.

(1) ...... 021194or 0

0)21194(

021194

24

24

35

=+−=

=+−

=+−

xxx

xxx

xxx


3or 4

7

0)3)(74(

021194 2

==

=−−

=+−

uu

uu

uu


3or2

7

3or4

7 22

±=±=

==

xx

xx

∴ The real roots of the equation are 3− , 2

7− , 0,

2

7 and 3 .

4. (1) ...... 012627 36=−− xx


1or 27

1

0)1)(127(

012627 2

=−=

=−+

=−−

uu

uu

uu


1or3

1

1or27

1 33

=−=

=−=

xx

xx


1− and 1.

5.

2or 3

20

0)2)(203(

040263

840343

8)43)(10(

84

3)10(

2

2

−=−=

=++

=++

=++

=++

=

++

xx

xx

xx

xxx

xxx

xx


20− and −2.

6.

1or 4

5

0)1)(54(

054

24431

)2(2)41)(1(

41

2

2

21

41

21

2

1

2

2

=−=

=−+

=−+

+=++−

+=−−−

−=

+

−−

−=−

+

xx

xx

xx

xxx

xxx

xx

x

xx


5− and 1.

7.

7or 11

0)7)(11(

0774

3480

40

1

34

2

40

1

)1)(3(

)3()1)(2(

40

1

13

2

2

2

2

=−=

=−+

=−+

++=

=++

=++

+−++

=+

−+

+

xx

xx

xx

xx

xx

xx

xxxx

x

x

x

x


8.

15or 4

0)15)(4(

06019

166443

)8()43(

843

843

2

2

22

==

=−−

=+−

+−=+

−=+

−=+

=++

xx

xx

xx

xxx

xx

xx

xx


8

44)4(343

=

++=++ xx

When x = 15,

8

22

154)15(343

≠

=

++=++ xx

Hence, 15 is not a root of the equation.


9.

2or 5

3

0)2)(35(

06135

0306525

65366025

)65()65(

6565

6655

2

2

2

22

==

=−−

=+−

=+−

+=+−

+=−

+=−

=+−

xx

xx

xx

xx

xxx

xx

xx

xx


19

Checking: When 5

3=x ,

6

0

65

35

5

35655

≠

=

+

−

=+− xx

Hence, 5

3 is not a root of the equation.

When x = 2,

6

6)2(5)2(5655

=

+−=+− xx


10. 03 =−+ xx …… (1)

By substituting ux = into (1), we have

032=−+ uu


2

131or

2

131

)1(2

)3)(1(411 2

−−+−=

−−±−=u


2

137

4

13214

4

131321

2

131

(rejected) 2

131or

2

131

2

−=

−=

+−=

+−=

−−=

+−=

x

xx

∴ The real root of the equation is 2

137 −.

11. 016)4(1042=+−

xx …… (1)


8or 2

0)8)(2(

016102

==

=−−

=+−

uu

uu

uu


2

3or

2

132or12

22or22

84or24322

==

==

==

==

xx

xx

xx

xx

12. 064)2(202 24=+−

xx …… (1)


16or 4

0)16)(4(

064202

==

=−−

=+−

uu

uu

uu


2or1

42or22

22or22

162or424222

22

==

==

==

==

xx

xx

xx

xx

13.

1or (rejected) 9

0)1)(9(

098

24158

24 log)158log(

)32log()]5)(3log[(

3 log2 log 3)5log()3log(

2

2

2

3

=−=

=−+

=−+

=++

=++

×=++

+=+++

xx

xx

xx

xx

xx

xx

xx

14.

3or (rejected) 2

3

0)3)(32(

0932

)1(3)3(2

2

3

1

3

2

3log

1

3log

2 log3 log)1log()3log(

2

2

2

2

2

=−=

=−+

=−−

+=−

=+

−

=

+

−

−=+−−

xx

xx

xx

xx

x

x

x

x

xx

15.

2or (rejected) 12

0)2)(12(

02414

164014

4)10)(4(

2)]10)(4[(log

2)10(log)4(log

2

2

2

4

44

−=−=

=++

=++

=++

=++

=++

=+++

xx

xx

xx

xx

xx

xx

xx

16. 4) log3)( log2( =−+ xx …… (1)

By substituting log x = u into (1), we have

1or 2

0)1)(2(

02

46

4)3)(2(

2

2

−==

=+−

=−−

=−+

=−+

uu

uu

uu

uu

uu


10

1or100

1logor2log

==

−==

xx

xx


20


(rejected) 32or 40

0)32)(40(

012808

06400405

58406400800

)5840()8(800

8

)8(5800800

58

800800

2

2

2

−==

=+−

=−−

=−−

−=−

−=−

−

−−=

−−

=

NN

NN

NN

NN

NNN

NNN

N

N

N

NN

18. Let x km/h be the original speed of the car.

(rejected) 75or 60

0)75)(60(

0450015

3154500300

)315()755(60

755

1530060

5

1

15

6060

5

4

15

240240

2

2

−==

=+−

=−+

+=+

+=+

+

++=

++

=

++

=

xx

xx

xx

xxx

xxx

x

x

x

xx

xx

∴ The original speed of the car is 60 km/h.


(rejected) 16or 5

0)16)(5(

08011

287410108639

)410)(7()4)(3(9

)4(9

410

7

3

)4(9

49

7

3

9

1

434

3

2

22

−==

=+−

=−+

++=++

++=++

+

+=

+

+

+

++=

+

+

++

=++

+

nn

nn

nn

nnnn

nnnn

n

n

n

n

n

nn

n

n

n

n

n

n

∴ The original fraction is 45

5

+, i.e.

9

5.

20. (a)

i

ii

=

=+× 14313

(b)

1

4520

=

=×ii

(c)

1

2

241042

−=

=

=+×

i

ii

(d)

i

i

i

ii

−=

=

=

=

+×

3

343

1553 )(

21. (a)

ix

x

x

x

32

92

9)2(

09)2(

2

2

±=

−±=−

−=−

=+−

(b)

0166

01642

0)4(4)2(

2

2

=++

=+++

=+++

xx

xxx

xxx


i

x

73

2

286

)1(2

)16)(1(466 2

±−=

−±−=

−±−=

22. (a) Using the quadratic formula,

i

x

65

2

2410

)1(2

)31)(1(41010 2

±−=

−±−=

−±−=

∴ The real part of the roots of the equation is −5.

(b) Using the quadratic formula,

i

x

22

1

8

1284

)4(2

)9)(4(4)4()4( 2

±=

−±=

−−±−−=

∴ The real part of the roots of the equation is 2

1.

23. (a)

iyxi

iyixi

iyixi

53

53

5)3(

2

−=+−

−=+

−=+

∴ 3−=y

and 5−=x

(b)

ixyi

iyxii

iiyixii

)4()6(32

6432

6432 432

−+−=+

+−−=+

+++=+

∴

4

26

=

=−

y

y

and

7

34

=

=−

x

x

24. (a)

i

i

iii

10

)415()235(

)42()3()55(

=

+++−−=

+−+−−+


21

(b)

i

ii

ii

ii

ii

ii

ii

24

13

2

223

)1(1

223

)1)(1(

)1(23

1

2)3(

2

+=

+++=

+++=

−−

+++=

+−

+++=

−++

(c)

i

i

i

i

ii

ii

i

i

i

ii

3

7

3

4

)74(3

1

2

148

3

1

)1(1

)1(31411

3

1

)1)(1(

)1)(311(

3

1

1

)1(310

3

1

33

)2)(5(

2

+=

+=

+•=

−−

−++•=

+−

++•=

−

−−+•=

−

+−

25. Let a = −2, b = 1.

The equation 024=++ baxx becomes 012 24

=+− xx .

1or 1

1

0)1(

012

2

22

24

=−=

=

=−

=+−

xx

x

x

xx

∴ The equation has only two real roots, −1 and 1.

Let a = −4, b = 4.

The equation 024=++ baxx becomes 044 24

=+− xx .

2or 2

2

0)2(

044

2

22

24

=−=

=

=−

=+−

xx

x

x

xx

∴ The equation has only two real roots, .2 and 2−

∴ a = −2, b = 1 or a = −4, b = 4

(or any other reasonable answers)

26. By substituting ux = into the equation kxx =+ 2 ,

we have

(1) ...... 02

2

2

2

=−+

=+

kuu

kuu

∵ The equation kxx =+ 2 has no real roots.

∴ The equation 022=−+ kuu also has no real roots.

∴ The discriminant of 022=−+ kuu is negative.

1

044

0))(1(42

0

2

−<

<+

<−−

<∆

k

k

k

∴ Possible values of k = −2 or −3.

(or any other reasonable answers)

Level 2

27. 01

1312 =

−

+−−

xx …… (1)

By substituting ux =−1 into (1), we have

0132

01

32

01

32

2=+−

=

+−

=+−

uu

uuu

uu

1or

2

1

0)1)(12(

==

=−−

uu

uu

Since ux =−1 , we have

2or4

5

11or4

11

11or2

11

==

=−=−

=−=−

xx

xx

xx

Checking: When 4

5=x ,

0

2

1

13

2

12

14

5

131

4

52

1

1312

=

+−

=

−

+−−=

−

+−−

xx

When x = 2,

0

1

13)1(2

12

13122

1

1312

=

+−=

−

+−−=

−

+−−

xx


5 and 2.

28. 0134 3

1

3

2

=−− xx …… (1)

By substituting ux =3

1

into (1), we have

1or 4

1

0)1)(14(

0134 2

=−=

=−+

=−−

uu

uu

uu

Since ux =3

1

, we have

1or64

1

1or4

1

1or4

1

3

3

3

1

3

1

=−=

=

−=

=−=

xx

xx

xx


1− and 1.


22

29.

(1) ...... 01)3(2)3(

0162)3(

222

222

=+−−−

=++−−

xxxx

xxxx

By substituting uxx =− 32 into (1), we have

1

0)1(

012

2

2

=

=−

=+−

u

u

uu


013

13

2

2

=−−

=−

xx

xx

Using the quadratic formula,,

2

133

)1(2

)1)(1(4)3()3( 2

±=

−−−±−−=x


133 − and

2

133 +.

30. (1) ...... 1523

1

3

1=

−

−− xx

By substituting ux

=− 3

1 into (1), we have

5or 3

0)5)(3(

0152

15)2(

2

=−=

=−+

=−−

=−

uu

uu

uu

uu

Since ux

=− 3

1, we have

5

16or

3

85

13or

3

13

53

1or3

3

1

==

=−−=−

=−

−=−

xx

xx

xx


8 and

5

16.

31. 056

46

2

=−

−+

−

xx

xx …… (1)

By substituting ux

x =−6

into (1), we have

1or 5

0)1)(5(

0542

=−=

=−+

=−+

uu

uu

uu

Since ux

x =−6

, we have

0)3)(2(or0)1)(6(

06or065

6or56

16

or56

22

22

=−+=−+

=−−=−+

=−−=−

=−−=−

xxxx

xxxx

xxxxx

xx

x

1or 6 =−= xx or 3or 2 =−= xx

∴ The real roots of the equation are −6, −2, 1 and 3.

32.

3or 1

0)3)(1(

032

32

33

27)3(

2

2

32

2

2

=−=

=−+

=−−

=−

=

=

−

−

xx

xx

xx

xx

xx

xx


33.

(1) ...... 0827)2(

278)2(

72

82

7282

722

2

2

3

=−⋅−

⋅=−

=−

=⋅−

=−

−

−

xx

xx

x

x

xx

xx


1or 8

0)1)(8(

0872

−==

=+−

=−−

uu

uu

uu


3

22

(rejected) 12or 82

3

=

=

−==

x

x

xx


34.

(1) ...... 02)3(7)3(9

02)3(73

02)3(79

2

)1(2

1

=−+

=−+

=−+

+

+

xx

xx

xx


1or 9

2

0)1)(29(

0279 2

−==

=+−

=−+

uu

uu

uu



3 log

9

2log

9

2log3 log

9

2log3 log

(rejected) 13or 9

23

−=

=

=

=

−==

x

x

x

xx

∴ The real root of the equation is −1.37.

35.

(1) ......08)1log( 2)]1[log(

08)1log()]1[log(

2

22

xx

xx

=−−−−

=−−−−

By substituting )1log( −x = u into (1), we have

4or 2

0)4)(2(

0822

=−=

=−+

=−−

uu

uu

uu


23

Since log(x − 1) = u, we have

001 10or100

101

000 101or100

11

4)1log(or2)1log(

==

=−=−

=−−=−

xx

xx

xx

∴ The real root of the equation are 100

101 and 10 001.

36.

1or (rejected) 3

4

0)1)(43(

043

43

23

2)13(

1)]13([log

113loglog

2

2

2

2

22

=−=

=−+

=−+

=+

=+

=+

=+

=++

xx

xx

xx

xx

xx

xx

xx

xx


1

2log1log

13log1log13loglog

22

12222

=

+=

++=++xx


37. (a) ∵ 3 is a root of the equation 115

222

=−

+− x

k

x.

∴ 12

2

3

8

184

2

11)3()3(5

222

=

=

=+−

=−

+−

k

k

k

k

(b)

(1) ...... 06316

565810

1)1)(5(

)5(12)1(2

11

12

5

2

24

242

22

22

22

=+−

−+−=+−

=−−

−+−

=−

+−

xx

xxx

xx

xx

xx


9or 7

0)9)(7(

063162

==

=−−

=+−

uu

uu

uu


3or7

9or7 22

±=±=

==

xx

xx

∴ The other roots of the equation are −3, 7−

and 7 .

38. Let x be the original number of fish that Mabel bought.

10or (rejected) 20

0)10)(20(

020010

0600303

)3630()1(600

1

33627600

31

27600600

2

2

=−=

=−+

=−+

=−+

+=+

+

++=

++

+=

xx

xx

xx

xx

xxx

x

x

x

xx

∴ Mabel bought 10 fish originally.


3or (rejected) 2

0)3)(2(

06

)3()32(2

2

1

)3(

3

2

1

3

11

2

=−=

=−+

=−−

+=+

=+

++

=+

+

HH

HH

HH

HHH

HH

HH

HH

40. Let x hours be the time taken by the smaller pipe to fill up the swimming pool.

12or (rejected) 2

0)12)(2(

02414

)6()62(4

4

1

)6(

6

4

1

6

11

2

==

=−−

=+−

−=−

=−

+−

=−

+

xx

xx

xx

xxx

xx

xx

xx

∴ The smaller pipe takes 12 hours to fill up the

swimming pool. 41. (a)

i

iii

iiii

518

)1(4)1(15652

)4()31)(52(

+=

−−+−−+−=

−++−

(b)

i

ii

iii

iii

43

4386

)]1(44[)1(393

)2()31)(3( 2

+=

−−+=

−++−−−+−=

+−+−

(c)

1

5

3

5

2

)1(2

)1(2

)1(21

)1(2

)2)(2(

)2)(1(

)21)(21(

)21(

2

1

21

222

=

−+

+=

−−

−−−+

−−

−−=

+−

+−+

−+

−=

−

−+

+

ii

ii

ii

ii

ii

ii

i

i

i

i


24

42. (a)

yiixx

yixix

yiixi

+=−++

+=−−−+

+=−+

4)3()3(

4)1()3(3

4)1)(3(

∴

1

43

=

=+

x

x

and

2

31

3

−=

−=

=−

y

yx

(b)

iyxyxi

yiyxxi

iyixi

iyix

i

)()(42

)1()(42

)1)((42

142

++−=−

−+++=−

++=−

+=+

−

−=+

=−

(2) ......4

(1) ......2

yx

yx

(1) + (2):

1

422

−=

−=

x

x

By substituting x = −1 into (1), we have

3

21

−=

=−−

y

y

∴ 1−=x and 3−=y

43. (a)

(1) ......0)3(6

586

5)4)(2(

2

2

=+++

−=++

−=++

mxx

mxx

mxx

∵ (1) has no real roots.

∴

6

244

012436

0)3)(1(46

0

2

>

>

<−−

<+−

<∆

m

m

m

m

∴ The range of possible values of m is m > 6.

(b) The smallest integral value of m in (a) is 7.

∴

0106

0)37(6

2

2

=++

=+++

xx

xx (from (1))


i

x

±−=

−±−=

−±−=

3

2

46

)1(2

)10)(1(466 2

44. (a) ∵ −1 + 4i is a root of the equation x2 + px + q = 0.

∴

0)84()15(

04)1(1681

0)41()41( 2

=−++−−

=++−−+−

=++−++−

ipqp

qpipi

qipi

∴

=−

=+−−

(2) ......084

(1) ......015

p

qp

From (2),

2

84

=

=

p

p

By substituting p = 2 into (1), we have

17

0215

=

=+−−

q

q

∴ 2=p and 17=q

(b)

0134

0)417()22(

2

2

=++

=−+++

xx

xx


i

x

32

2

364

)1(2

)13)(1(444 2

±−=

−±−=

−±−=

45. (a)

ia

aiiaaiaai

)1(

)1())(1(

2

2

+=

−+++=++

(b)

i

iiii

iiii

iiiiiii

34

171052

)14()13()12()11(

)4)(41()3)(31()2)(21()1(

2222

2

=

+++=

+++++++=

++++++++++

46. (a) Let x

xy2

+= .

4

42

44

44

2

2

2

2

2

2

−=

−

+=

−

++=+

y

xx

xx

xx

(b) Let x

xy2

+= .

3or 0

0)3(

03

0434

042

34

046

43

2

2

2

2

2

2

−==

=+

=+

=++−

=+

++

+

=++++

yy

yy

yy

yy

xx

xx

xxxx

Since x

xy2

+= , we have

2or 1

0)2)(1(or2

023or2

32

or02

22

−=−=

=++±=

=++−=

−=+=+

xx

xxix

xxxx

xx

x


25

Multiple Choice Questions (p. 1.33)

1. Answer: B

x4 − 6x2 − 27 = 0 …… (1)


9or 3

0)9)(3(

02762

=−=

=−+

=−−

uu

uu

uu


3or 3

9or (rejected) 3 22

=−=

=−=

xx

xx

2. Answer: A

3or 2

0)3)(2(

065

66

1)6(

26

16

21

2

2

==

=−−

=+−

−=−−

=−

−−

=−

−

xx

xx

xx

xxx

xx

xx

xx

3. Answer: C

(1) ...... 01)3(10)3(9

01)3(10)3(9

2

2

=+−

=+−

xx

xx


1or 9

1

0)1)(19(

01109 2

==

=−−

=+−

uu

uu

uu


9

13 =

x or 3x = 1

∴ 2−=x or 0=x

4. Answer: D

(1) ...... 03 log 4) (log

03 log 2) (log

2

22

=+−

=+−

xx

xx

By substituting log x = u into (1), we have

3or 1

0)3)(1(

0342

==

=−−

=+−

uu

uu

uu


1000or10

3 logor1 log

==

==

xx

xx

5. Answer: A

For I,

equation. quadratic anot is which ,013

3312

3)1(

21

31

21

23

32

2

2

2

=−+−

+=++

=

+

++

=

+

+

xxx

xxxx

xx

xx

xx

For II,

(1) ...... 1)3(2)3(

162)3(

222

222

−−=−

−−=−

xxxx

xxxx

By substituting uxx =− 32 into (1), we have

equation. quadratic a is which ,012

12

2

2

=+−

−=

uu

uu

For III,

9612

)3()12(

312

312

2

22

+−=−

−=−

−=−

=+−

xxx

xx

xx

xx

01082=+− xx , which is a quadratic equation.

∴ The answer is A.

6. Answer: A

(1) ...... 0233

233

2

2

=−

−+

−

+=+

−

xx

xx

xx

xx

By substituting ux

x =−3

into (1), we have

01or 02

0)1)(2(

022

=−=+

=−+

=−+

uu

uu

uu

Since ux

x =−3

, we have

023

=+−x

x or 013

=−−x

x

0322=−+ xx or 032

=−− xx

0)1)(3( =−+ xx or )1(2

)3)(1(4)1()1( 2−−−±−−

=x

x = −3 or x = 1 or 2

131±=x

∴ The equation has 4 distinct real roots.

7. Answer: D

Original number of eggs that can be boughtx

60=

Number of eggs that can be bought after price

reduction2.0

60

−=

x

∵ 10 more eggs can be bought when the price is

lowered.

∴ The following equation gives the value of x:

1060

2.0

60=−

− xx

8. Answer: B

i

ii

2

81064100

=

−=−−−


26

9. Answer: B

i

i

i

ii

ii

i

i

21

25

5025

)1(43

)1(85033

)43)(43(

)43)(211(

43

211

22

+=

+=

−−

−++=

+−

++=

−

+

10. Answer: A

yiixx

yixiix

yiiix

+=−++

+=−−+−

+=+−

31)12()43(

31)1(4123

31)3)(4(

=−

=+

(2) ...... 12

(1) ...... 3143

yx

x

From (1), x = 9

By substituting x = 9 into (2), we have

3

129

−=

=−

y

y

∴ 9=x , 3−=y

HKMO (p. 1.35)

1.

=−

=

(2) ...... 3

(1) ...... 4

xyzw

wxyz

From (1), w

xyz4

= …… (3)

By substituting (3) into (2), we have

4or (rejected) 1

0)4)(1(

043

34

34

2

2

=−=

=−+

=−−

=−

=−

ww

ww

ww

ww

ww

∵ The solution of w is P.

∴ 4=P

2.

)1)(1(

2

)3)(1(

22

1

2

)3)(1(

13

1

2

3

1

1

1

2

2

−+=

+−

+

−=

+−

−++

−=

++

−

xxxx

x

xxx

xx

xxx

Since x ≠ 1,

(rejected) 1or 2

0)1)(2(

02

312

3)1(

1

2

3

)1(2

2

2

2

=−=

=−+

=−+

+=++

+=+

+=

+

+

xx

xx

xx

xxx

xx

xx

x

∵ a is the real root of the equation.

∴ 2−=a

3.

(1) ...... 04232

232242

)322(242

)]32(2[log)42(log

)32log()2(log)42(log

)32(log)44(log

2

22

2

12

22

12

22

122

=−⋅−

⋅−⋅=+

−⋅=+

−=+

−+=+

−+=+

+

+

+

xx

xxx

xxx

xxx

xxx

xxx


4or 1

0)4)(1(

0432

=−=

=−+

=−−

uu

uu

uu


2

42or (rejected) 12

=

=−=

x

xx

∵ a satisfies the equation.

∴ 2=a

Investigation Corner (p. 1.36)

1. Consider a quadratic equation 02=++ CBxAx , where A,

B and C are real numbers.

∵ a + bi is one of its roots.

∴

0)2()(

0)()2(

0)()(

22

22

2

=++++−

=++++−

=++++

ibBabACBabaA

CbiaBabibaA

CbiaBbiaA

=+

=++−

02

0)( 22

bBabA

CBabaA

Consider

0

)2()(

)()2(

)()(

22

22

2

=

+−++−=

+−+−−=

+−+−

ibBabACBabaA

CbiaBabibaA

CbiaBbiaA

∴ a − bi is the other root of the quadratic equation.

2. For the two non-real roots a + bi and a − bi in question 1,

(1) ......0)(2

0))(()]()[(

0)]()][([

222

2

=++−

=−++−++−

=−−+−

baaxx

biabiaxbiabiax

biaxbiax

. ∴ The coefficients of the quadratic equation are

real numbers. 3. Consider the following pair of complex numbers:

1 + 2i, 1 − 2i

The equation formed is: (1)) from(0)21()1(2

0)]21()][21([

222=++−

=−−+−

xx

ixix

i.e. 0522=+− xx

Consider the following pair of complex numbers:

−1 + 2i, −1 − 2i The equation formed is:

(1)) from(0]2)1[()1(2

0)]21()][21([

222=+−+−−

=−−−+−−

xx

ixix

i.e. 0522=++ xx

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