7146992 4ACh01Quadratic Equations in One Unknown

7/31/2019 7146992 4ACh01Quadratic Equations in One Unknown

1/32

Certificate Mathematics in Action Full Solutions 4A

1 Quadratic Equations in One Unknown

Activity

Activity 1.1 (p. 28)

1.

2.

3. 3, 2

4.

0)2)(3(

062

=+=+

xx

xx

2or3

02or03

====+

xx

xx

5. The quadratic equation ax2 + bx + c = 0 can be solvedgraphically by reading thex-intercepts of the graph of

y = ax2 + bx + c.

Follow-up Exercise

p. 3

1. Let ,8.0 =x

(2)888888.810

(1)888888.0i.e.

==

x

x

9

880.

9

8

89,(1)(2)

=

=

=

x

x

2. Let ,61.0 =x

(2)666666.110

(1)666166.0i.e.

==

x

x

6

160.1

6

1

90

15

1590

5.19,(1)(2)

=

=

=

==

x

x

x

3. Let ,21.0 =x

(2)212121.12100

(1)212121.0i.e.

==

x

x

33

4210.

33

4

99

121299,(1)(2)

=

=

==

x

x

4. Let ,321.0 =x

(2)123123.1231000

(1)123123.0i.e.

==

x

x

333

413210.

333

41

999

123

123999,(1)(2)

=

=

=

=

x

x

p.6

Quadratic equation General formax2 + bx + c = 0

Value of

a b c

(a) 5x2 = 6 x 5x2 +x 6 = 0 5 1 6(b) x2 4 = 5x x2 5x 4 = 0 1 5 4

(c) 3x2 = 4 3x2 + 0x 4 = 0 3 0 4

(d) 8x=x2 x2 8x + 0 = 0 1 8 0

(e) 2x(3 x) = 0 x2 3x + 2 = 0 1 3 2

(f) (1+x)(1 x) + 3x = 2 x2 3x + 1 = 0 1 3 1

p.8

1. (a) x2 10x + 16 = 0(x 2)(x 8) = 0

8or2

08or02

====

xx

xx

(b) 2x2 + 13x + 15 = 0

1

x 4 3 2 1 0 1 2 3

x2 16 9 4 1 0 1 4 9

x 4 3 2 1 0 1 2 3

6 6 6 6 6 6 6 6 6

y 6 0 4 6 6 4 0 6


2/32


(2x + 3)(x + 5) = 0

5or2

3

05or032

==

=+=+

xx

xx

(c)

0)2)(13(

0253 2

=+

=

xx

xx

2or3

1

02or013

==

==+

xx

xx

2. (a)

0)23)(34(

0612 2

=+=+

xx

xx

3

2or

4

3

023or034

==

==+

xx

xx

(b)

0)52)(52(

0254254

2

2

=+==

xx

xx

2

5or

2

5

052or052

==

==+

xx

xx

(c)

0)3)(5(

0152

152

15)2(

2

2

=+=+

=+

=+

xx

xx

xx

xx

3or503or05

== ==+ xxxx

(d)

0)3)(1(

034

6232

)3(2)3)(1(

2

2

==+

=

=+

xx

xx

xxx

xxx

3or1

03or01

====

xx

xx

Alternative Solution

0)1)(3(

0]2)1)[(3(

0)3(2)3)(1(

)3(2)3)(1(

==+ =+

=+

xx

xx

xxx

xxx

1or3

01or03

====

xx

xx

p.10

1. (a)

.0134isequationrequiredThe

0134

0)14)(1(

014or01

04

1or01

4

1or1

2

2

=+

=+

=+==+

==+

==

xx

xx

xx

xx

xx

xx

(b) 2

1or

3

2 == xx

012or023

02

1or0

3

2

=+=

=+=

xx

xx

026

0)12)(2(3

2 =

=+

xx

xx

The required equation is 6x2 x 2 = 0.

2. (a)

0)8)(12(08152 2

=+=

xx

xx

8or2

1

08or012

==

==+

xx

xx

(b) The roots of the required equation are

2

1

1

and 81

i.e. 2 and8

1.

018or02

08

1or02

8

1or2

==+

==+

==

xx

xx

xx

02158

0)18)(2(

2 =+

=+

xx

xx

The required equation is 8x2 + 15x 2 = 0.

3. (a)

0)4)(34(

012134 2

=+=

xx

xx

4or

4

3

04or034

==

==+

xx

xx


24

3and

2

4, i.e.

8

3 and 2.

02or0

8

3

2or8

3

==+

==

xx

xx

8x + 3 = 0 or x 2 = 0

06138

0)2)(38(

2 =

=+

xx

xx

The required equation is 8x2 13x 6 0.

2


3/32


p.13

1. (a)

31

9)1( 2

=+=+

x

x

231

=+=x

oror

431

(b)

232

4)32( 2

==

x

x

or2

5

or232

=

+=

x

x

2

1

23

2. (a)

52

5)2( 2

=

=

x

x

52or52 +=x

(b)

64

1

64

12

=+

=

+

x

x

64

1or6

4

1 +=x

3. (a)

83

8)3( 2

=

=

x

x

d.p.)2to(cor.0.17ord.p.)2.to(cor.83.5

83or83

=+=x

(b)4

5)2( 2 =+x

)d.p.2tocor.(12.3or)d.p.2tocor.(88.0

2

52or

2

52

2

52

4

52

=

+=

=+

=+

x

x

x

p. 15

1. 22

2 )9(2

1818 +=

++ xxx

2. 22

2 )6(2

1212 =

+ xxx

3.

22

2

2

7

2

77

+=

++ xxx

4. 22

2 )4(2

88 +=

++ xxx

5.22

2

29

299 =+

xxx

6.22

2

4

1

4

1

2

1

=

+ xxx

7.22

2

6

1

6

1

3

1

=

+ xxx

8.22

2

4

5

4

5

2

5

+=

++ xxx

p. 16

1. (a)

1or9

54

5425)4(

2

89

2

88

98

098

2

22

2

2

2

==

==

+=

+

=

=

x

x

xx

xx

xx

xx

(b)

2

1or3

4

7

4

5

4

7

4

5

16

49

4

5

4

5

2

3

4

5

2

5

2

3

2

5

02

3

2

5

0352

2

22

2

2

2

2

=

=

=

=

+=

+

=

=

=

x

x

x

xx

xx

xx

xx

2. (a)

2

537or

2

537

2

53

2

7

2

53

2

7

4

53

2

7

2

71

2

77

17

017

2

222

2

2

+=

=

=+

=

+

+=

++

=+

=+

x

x

x

xx

xx

xx

3


4/32


(b)

3

51or

3

51

3

51

3

5)1(

2

2

3

2

2

2

2

3

22

03

22

0263

2

222

2

2

2

+=

=

=

+=+

=

=

=

x

x

x

xx

xx

xx

xx

p.20

1. Using the quadratic formula,

8or4

2

124

2

1444

)1(2

)32)(1(444

24

2

2

=

=

=

=

=a

acbbx


2

1or2

4

35

4

95

)2(2

)2)(2(4)5()5(

24

2

2

=

=

=

=

=a

acbbx

3.

08103

8103

2

2

=+

+=

xx

xx

Using the quadratic formula,

4or32

6

1410

6

19610

)3(2

)8)(3(41010

2

4

2

2

=

=

=

=

=a

acbbx


134or134

2

1328

2

528

)1(2

)3)(1(488

2

4

2

2

+=

=

=

=

=a

acbbx

5. 34 2 += xx0342 =+ xx


72or72

2

724

2

284

)1(2

)3)(1(444

2

4

2

2

+=

=

=

=

=a

acbbx

p. 25

1.

2. The equationx2 9x + k= 0 has a double real root.

4

81

0481

0))(1(4)9(

0

2

=

==

=

k

k

k

3. The equation (2k 1)x2 + 3x 6 = 0 has no realroots.

4

Quadratic equations Value ofx2 + 4x + 2 = 0 = 42 4(1)(2) = 8 > 02x2 3x 5 = 0 = (3)2 4(2)(5) = 49 > 0x2 + 8x + 16 = 0 = 82 4(1)(16) = 02x2 + 5x = 0 = 52 4(2)(0) = 25 > 04x2 + 25 = 0 = 02 4(4)(25) = 400 < 0

Nature of roots

2 distinctreal roots

1 double real root No real roots


5/32


16

548

15

01548

024489

0)6)(12(43

0

2

>

k

k

k

The range of possible values ofkis k< 89

.

10. The equation 5x2 + 3x + (k+ 1) = 0 has two distinct

real roots.

20

11

02011020209

0)1)(5(43

0

2

>

>+

>

k

kk

k

The range of possible values ofkis k< .2011

11. The quadratic equation 3x2 + 4x + k = 0 has real

roots.

0

i.e. 0))(3(442 k

3

4

1612

01216

k

k

k

3

4isofvaluestheofrangeThe kk

.

12. The quadratic equation (k+ 1)x2 2kx + (k 2) = 0

has real roots.

0

i.e.0)2)(1(4)2( 2 + kkk

2

84

084

08444 22

+++

k

k

k

kkk

The range of the values ofkis k2.

15


16/32


13. The equation 3x2 + 5xk= 0 has no real roots.

12

25

01225

0))(3(45

0

2


25/32


(b)

16. (a) two

(b) Thex-intercepts ofy = 4x2 4x 3 are 0.5 and1.5.

Therefore, the roots of the equation 4x2 4x 3 = 0are 0.5 and 1.5.

17. (a) one

(b) Thex-intercepts ofy =x2 6x + 9 is 3.Therefore, the roots of the equationx2 6x = 9 is3.

18. Letx be one of the number, then 27 x is the othernumber.

0)12)(15(

018027

18027

180)27(

2

2

==+

=

=

xx

xx

xx

xx

15015

== xx

or

or

12012

== xx

The two numbers are 12 and 15.


Letx be one of the number, thenx

180is the other

number.

0)12)(15(

018027

27180

27180

2

2

==+

=+

=+

xx

xx

xx

xx

15

015

==

x

xor

or

12

012

==

x

x

The two numbers are 12 and 15.

19. Letx cm be the length of the rectangle, then

cm)19(2

238x

x = is the width of the rectangle.

0)8)(11(

08819

8819

88)19(

2

2

==+

=

=

xx

xx

xx

xx

11

011

==

x

x

or

or

8

08

==

x

x

The length and width of the rectangle are 11 cm and8 cm respectively.


Letx cm be the length of the rectangle, thenx

88cm is

the width of the rectangle.

0)8)(11(

08819

1988

1988

3888

2

2

2

==+

=+

=+

=

+

xx

xx

xx

xx

xx

11

011

==

x

x

or

or

8

08

==

x

x

The length and width of the rectangle are 11 cm and 8 cmrespectively.

20.

0)32)(12(

0384

348

2

2

==+

=

xx

xx

xx

5.0

012

==

x

xor

or

5.1

032

==

x

x

After 0.5 seconds and 1.5 seconds, the ball is 3 mabove the ground.

21. The equationx2ax 40 = 0 has two distinct realroots.

160

0160

0)40)(1(4)(

0

2

2

2

>

>+

>

>

a

a

a

The square of any numbers is always positive.

The equationx2ax 40 = 0 has two distinct realroots for any real values ofa.

a = 1 or 2 or 3 (or any other reasonable answers)

22.

09

1

3

44

3

12

2

2

=

+

=

mxx

mx


8

m163

4

)4(2

9

1)4(4

3

4

3

4

2

4

2

2

=

=

=

m

a

acbbx

To have two rational roots of different signs, we need

9

1

9

16m16

3

4m16

>

>

>

m

25


26/32


m = 1 or 9 or 16 (or any other reasonable answers)

23. The graph ofy = ax2 + 4x + c intersects thex-axis atone point.

The equation ax2 + 4x + c = 0 has a double realroot.

4

0416

04)4(0

2

=== =

ac

ac

ac

a = 2, c = 2 ora = 1, c = 4 ora = 2, c = 2.(or any other reasonable answers)

24. The graphy = x2 + 2x + kdoes not intersect thex-axis.

The equation x2 + 2x + k= 0 has no real roots.

1

044

0))(1(42

0

2


27/32


9

821or

9

821

18

822

18

3282

)9(2

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

2

4

2

2

+=

=

=

=

=

a

acbbx

31.

0)3)(12(

0352

44816

)12)(1(4)13)(12(

2

22

=+=

=+

+=+

xx

xx

xxxx

xxxx

3or2

1

03or012

==

==+

xx

xx


0)3)(12(

0)3)(12(

0)]1(4)13)[(12(

0)12)(1(4)13)(12(

)12)(1(4)13)(12(

=+=++=+=++

+=+

xx

xx

xxx

xxxx

xxxx

3or2

1

03or012

==

==+

xx

xx

32.

1

01

0)1(

012

044412

04)1(4)1(

2

2

2

2

==+=+

=++

=+++

=++

x

x

x

xx

xxx

xx


[ ]

1

01

0)1(

02)1(

04)1(4)1(

2

2

2

==+=+=+

=++

x

x

x

x

xx

33. (a)

0)4)(32(

012112 2

==+

xx

xx

4or2

3

04or032

==

==

xx

xx


2

32 and

2(4), i.e. 3 and 8.

02411

0)8)(3(

08or03

8or3

2 =+

=====

xx

xx

xx

xx

The required equation is 024112 =+ xx .

34. (a)

0)3)(13(

0383 2

=+=+

xx

xx

3or3

1

03or013

==

=+=

xx

xx


3

11

and3

1

.

i.e. 3 and3

1 .

0383

0)13)(3(

013or03

03

1or03

3

1or3

2

=

==+=

=+=

==

xx

xx

xx

xx

xx

The required equation is 0383 2 = xx .

35. (a) The equation 016)5(22 =+ xkx has adouble real root.

1or945

45

16)5(

016)5(

064)5(4

0)16)(1(4)]5(2[

0

2

2

2

2

= =

==

=

=

=

=

k

k

k

k

k

k

(b) For 9=k ,

4

04

0)4(

0168

016)59(2

2

2

2

===

=+

=+

x

x

x

xx

xx

For 1=k ,

27


28/32


4

04

0)4(

0168

016)51(2

2

2

2

==+=+

=++

=+

x

x

x

xx

xx

36. (a) The equation 0)2(32 =+ kxx has realroots.

4

17

0174

0)2(49

0)]2()[1(4)3(

0

2

++++

k

k

k

k

The range of possible values ofkis4

17k .

(b)4

17k

Minimum value ofkis4

17 .

(c)

2

3

023

02

3

04

93

024

173

2

2

2

=

=

=

=+

=

+

x

x

x

xx

xx

37. (a)

0)2(44

244

2

2

=+

=+

kxx

kxx

The equation 0)2(44 2 =+ kxx has noreal roots.

3

048160)2(1616

0)2)(4(4)4(

0

2

>


29/32


(b) (i) 1 double real root

(ii) Thex-intercept of 1682 += xxy is 4.Therefore, the root of the equation

16)8( =xx is 4.

41. The graph of kxkxy ++= 52 touches thex-axis..

The equation 052 =++ kxkx has a double realroot.

0)25)(25(

0425

0))((45

0

2

2

=+=

=

=

kk

k

kk

2

5or

2

5

025or025

==

==+

kk

kk

42. The graph of 1322

+= kxxy has two distinctx-intercepts.

The equation 0132 2 =+ kxx has two distinctreal roots.

8

1

081

0189

01243

0

2

>+>

>

k

k

)(k

)k)(()(

The range of possible values ofkis8

1


30/32


Area ofABC

2

2

2

cm180

cm2

409

cm2

)267()36(

2

=

=

+=

= BCAC

Perimeter ofABC

cm90

cm)94041(

cm)]36()267()566[(

=++=

++++=++= ACBCAB

46. (a) Area of shaded region

22

22

2

cm)5015(

cm)48215(

cm82

34)13)(25(

=

=

+=

xx

xx

xx

(b) (i) 1865015 2 =xx 023615 2 = xx

(ii)

0)4)(5915(

023615 2

=+=

xx

xx

4or(rejected)15

59

04or05915

==

==+

xx

xx

cm5

cm]44)143[(

=+=

= KCBJBCJK

cm12

cm]33)245[(

==

= HBAGABGH

Multiple Choice Questions (p. 48)

1. Answer: C

0)2)(32(062

2

=+=

xxxx

2or2

3

02or032

==

==+

xx

xx

2.

Answer: B

2

2

)14(

1816

+=

++

x

xx

3. Answer: D

Thex-intercepts of the graph are 5 and 2. The roots of the graph are 5 and 2.

02or05

2or5

==+==

xx

xx

0103

0)2)(5(

2 =+

=+

xx

xx

The required equation is 1032 += xxy .

4. Answer: D

For 0322 =++ xx ,

8

)3)(1(42 2

==

0 < The equation 0322 =++ xx has no real roots.

The graph 322 ++= xxy has no x-intercepts.For 0=x

3

3)0(202

=++=y

The graph 322 ++= xxy has positivey-intercept.

5. Answer: B

The graph cxxy += 42 touches thex-axis.

The equation 042 =+ cxx has a double realroot.

4

0416

0))(1(4)4(

0

2

===

=

c

c

c

6. Answer: AThe equation 08)8(

2 =+++ kxkx has a doubleroot.

8

08

0)8(

06416

0326416

0)8)(1(4)8(

0

2

2

2

2

===

=+

=++

=+

=

k

k

k

kk

kkk

kk

7. Answer: A

The equation 082 =+ pxx has no real roots.

16

0464

0))(1(4)8(

0

2

>


31/32


5or2

05or02

====

xx

xx

9. Answer: B

is a root of 0532 2 =+ xx .

5

5)0(2

5)532(2564

0532

22

=+=

++=+

=+

.

10. Answer: A

The graph cxxy += 82 has twox-intercepts.

The equation 082 =+ cxx has two distinct realroots.

16

0464

0))(1(4)8(

0

2

>

>

c

c

c

The range of possible values ofc is c < 16.

HKMO (p. 49)

04)4()241(

0)2(4441

0)()(

2

22

=+

=+++

=+

xkxk

xxkxx

xkgxf

The equation 0)()( =+ xkgxf has a single root.

0)40)(16(

064024

032656168

0)4)(241(4)]4([

0

2

2

2

=+=+

=++

==

kk

kk

kkk

kk

40or16

040or016

===+=

kk

kk

40=d

Lets Discuss

p. 20

Angels method:

04

1

3

2

2

5

4

1

3

2

2

5

2

2

=

=

xx

xx


30

1064or

30

1064

30

106

15

2

5

18

53

3

2

2

52

4

1

2

54

3

2

3

22

+=

=

=

=x

Kens method:

03830

3830

4

1

3

2

2

5

2

2

2

=

=

=

xx

xx

xx


30

1064or

30

1064

30

1064

60

10628

60

4248

)30(2

)3)(30(4)8()8( 2

+=

=

=

=

=x

31


32/32


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7146992 4ACh01Quadratic Equations in One Unknown