Mathematics In Action Ch1 Answer.doc 4g1541

1 Quadratic Equations in One Unknown (I)

1 Quadratic Equations in One Unknown (I)

12.

x 2  2ax  x  2a

 x ( x  2a )  ( x  2a )  ( x  2a )( x  1)

Review Exercise 1 (p. 1.4) 1.

13.

( 3 x  4 x)  ( x  3) 2

( 2 x  1) 2  x 2  ( 2 x  1  x )(2 x  1  x )  (3 x  1)( x  1)

 3 x 2  4 x  x  3   3x  3x  3 2

14. 2.

 ( x  6)( x  3)

( x 2  2 x  5)  ( x 2  x  2)  x2  2x  5  x 2  x  2  x 3

3.

15.

( x  2)  ( x 2  3 x  4)  ( x 2  1)

16.

 2( x  2)(3 x  1)

(  x 2  2 x  6)  ( x  6 x 2 )  ( 7  5 x  4 x 2 )

17.

x 2  5 x ( 2 x  3)  x 2  10 x 2  15 x   9 x 2  15 x

18.

 2x 2  6x  5

19.

9.

 4 x( x  1) 2

20.

 y 2 ( x  y )  x 2 ( y  x)

 2 x 2  8 x  7 x  28  ( 2 x 2  8 x  7 x  28)

 y 2 ( x  y)  x 2 ( x  y)

 2 x 2  x  28  2 x 2  15 x  28   14 x  56

 ( x  y )( y 2  x 2 )  ( x  y )( x  y )( y  x ) (or  ( x  y ) 2 ( x  y ))

4 x 2  20 x  25

Let’s Discuss Let’s Discuss (p. 1.23) Angel’s method:

 ( 2 x  5) 2

1 2 x  x 1 5

x 2  13 x  36  ( x  4)( x  9)

11.

xy 2  y 3  x 2 y  x 3

( x  4)(2 x  7)  ( x  4)(2 x  7)

 ( 2 x ) 2  2( 2 x )(5)  52

10.

4x3  8x 2  4x  4 x( x 2  2 x  1)

(3  x ) 2  x ( x  1)  9  6x  x 2  x 2  x   5x  9

8.

(3 x  4)(4 x  3)  ( x  3)( 4 x  3)  ( 4 x  3)[(3 x  4)  ( x  3)]  ( 4 x  3)( 2 x  7)

( 2 x  1)( x  2)  ( x  3)  2x 2  x  4x  2  x  3

7.

4( x  5) 2  (3 x  2) 2  [ 2( x  5)]2  (3 x  2) 2  [ 2( x  5)  (3 x  2)][2( x  5)  (3 x  2)]  (5 x  12)( x  8)   (5 x  12)( x  8)

 x 2  8 x  13

6.

6 x 2  14 x  4  2(3 x 2  7 x  2)

  x 2  2x  6  x  6x 2  7  5x  4 x 2

5.

2 x 2  7 x  15  ( x  5)( 2 x  3)

 x  2  x 2  3x  4  x 2  1  4x 1

4.

x 2  3 x  18

1 2 x  x 1  0 5

2 x 2  x ( x  3)  2 x 2  x 2  3x  3x 2  3x  3 x ( x  1)

1

NSS Mathematics in Action 4A Full Solutions L.H.S. = (3) 2  6(3)  9 = 0 R.H.S. = 0 Since L.H.S. = R.H.S., 3 is a root of the equation.

Using the quadratic formula,

 1 ( 1) 2  4  (1)  5  1 2   5

 ( 1)  x



1 5

1 2 5



5 5  2 2



5 5 2

or

5 5 2

Ken’s method:

1 2 x  x 1 5 x2  5x  5  0

Using the quadratic formula,

x

( 5) 2  4(1)(5) 2(1)

 ( 5) 



5 5 2



5 5 2

or

5 5 2

Let’s Discuss (p. 1.30) The solution obtained by using the factor method is the exact value of the root. However, the solution obtained by using the graphical method is an approximation only.

Classwork Classwork (p. 1.8) (a) Integer

24, 0, 100,  8

(b)

Natural number

24, 100

(c)

Negative integer

8

(d)

Terminating decimal

6.3

(e)

Recurring decimal

5   ,  5.2 1.2

(f)

Fraction

1 2 , 5 3

(g)

Irrational number

2,

31

Classwork (p. 1.11) 1. x2  6x  9  0 (a) When x = 3, 2

1 Quadratic Equations in One Unknown (I) (b) When x = 6, L.H.S. = (6) 2  6(6)  9 = 9 R.H.S. = 0 Since L.H.S.  R.H.S., 6 is not a root of the equation. (c) When x = –3 , L.H.S. = ( 3) 2  6( 3)  9 = 36 R.H.S. = 0 Since L.H.S.  R.H.S., –3 is not a root of the equation. 2.

∴ (c)

10 x  5.444 444 (6)  (5) : 9 x  4.9 x

(a) When x = –4, L.H.S. = 2( 4) 2  5( 4)  12 = 0 R.H.S. = 0 Since L.H.S. = R.H.S., –4 is a root of the equation.

∴

3 , 2



L.H.S. = 2 



3  2

2

(a)

3   5    12  15 2 

Since L.H.S.  R.H.S., 

3 is not a root of the 2

(b)

x 2  5x  3 ∴ a  1, b  5, c  3

2

 3  5   12  0  2

(c)

R.H.S. = 0 Since L.H.S. = R.H.S.,

( x  3) 2  4 x x 2  6 x  9  4 x x2  2x  9  0

3 is a root of the equation. 2

∴ a  1, b  2, c  9

( 2 x  1)(2 x  1)  3 x  1

Classwork (p. 1.26) (a) The x-intercepts of the graph are 2.5 and 1.

(d)

(b) The roots of the equation 2 x  3 x  5  0 are 2

(2 x) 2  1  3 x  1 4 x 2  3x

2.5 and 1.

4 x 2  3x  0 ∴ a  4, b  3, c  0

Quick Practice

Quick Practice 1.3 (p. 1.13) x 2  9 x  20  0 (a) ( x  4)( x  5)  0

Quick Practice 1.1 (p. 1.6)

, 0.7 0.777 777  7.777 777  7 7 x 9

x x 10 x (a) ( 2)  (1) : 9 x

   

 (1 )  ( 2)

x40

or

x5 0

x4

or

x5

2x 2  x  3  0 ( x  1)( 2 x  3)  0 (b) x 1  0

  7 0 .7 9

or 2 x  3  0

x   1 or

(b)

Let i.e.

x ( x  5)  3 x 2  5x  3  0

L.H.S. = 2

∴

6x  x2  2 ∴ a  1, b  6, c  2

3 , 2

Let i.e.

  49 0 .5 4 90

x2  6x  2  0

equation.

 3   2

49 90

Quick Practice 1.2 (p. 1.10)

R.H.S. = 0

(c) When x 

, x  0.54 x  0.544 444

Let i.e.

2 x 2  5 x  12  0

(b) When x  

7   37 0.3 99

7 ,  0.3  0.373 737   37.373 737   37 37 x 99

x x 100 x ( 4)  (3) : 99 x

x

3 2

3x 2  5  2 x 3x 2  2 x  5  0 (c) ( x  1)(3 x  5)  0 x 1  0

 ( 3 )  ( 4)

or 3x  5  0 5 x   1 or x 3

3

 ( 5)  (6)

NSS Mathematics in Action 4A Full Solutions Quick Practice 1.4 (p. 1.14) x(7 x  4)  3 x 2

7 x 2  4 x  3x 2 (a)

Quick Practice 1.7 (p. 1.21) (a) Using the quadratic formula,

4x2  4x  0 4 x( x  1)  0 x  0 or

x 1  0

x

x  1

5 9 4 53  4 2

x  3  0 or  2 x  4  0 x  3 or x2

x

36 x 2  12 x  1  0 (6 x)  2(6 x)(1)  1  0 (6 x  1)(6 x  1)  0 (a) 6x  1  0 2

2(36)

 12  72  12  72 1  6

1 6

0

(c) Using the quadratic formula,

( x  7)  6( x  7)  9  0 2

x

( x  7) 2  2( x  7)(3)  32  0 ( x  7)  3 ( x  7)  3  0 (b) ( x  4)( x  4)  0



3

32  4(5)(1) 2(5)

 3   11 10

∵  11 is not a real number. ∴ The equation has no real roots.

x  4  0 or x  4  0 x  4 or x  4 ∴ x = –4

Quick Practice 1.8 (p. 1.21) (a)

Quick Practice 1.6 (p. 1.16)

25 x 2  4  0 (5 x  2)(5 x  2)  0 5x  2  0 2 x 5

x x40 x

or or

5x  2  0 2 x 5



16( x  2) 2  49  0 ( x  2) 2 

x ( x  1)  4

2

Using the quadratic formula,

(5 x ) 2  2 2  0

(b)

 12  12 2  4(36)(1)



or 6 x  1  0 1 1 x   or x 6 6

(a)

1 2

or

(b) Using the quadratic formula,

Quick Practice 1.5 (p. 1.15)

∴ x

( 5) 2  4( 2)( 2) 2( 2)



5( x  3)  (2 x  1)( x  3) 5( x  3)  (2 x  1)( x  3)  0 ( x  3) 5  ( 2 x  1)  0 (b) ( x  3)(2 x  4)  0

2

 ( 5) 

49 16

(b)

49 49 or  16 16 7 7 x   2 or   2 4 4 15 1  or 4 4

x2

4

 ( 1)  1

17 2

( 1) 2  4(1)( 4) 2(1)

1 Quadratic Equations in One Unknown (I) Using the quadratic formula,

( x  2)( x  3)  4 x 2  2 x  3x  6  4

x

x 2  5x  2  0

Using the quadratic formula,

x 

5



5 2  4(1)( 2) 2(1) (b)

 5  17 2

( 6) 2  4(1)( 4) 2(1)

6

20

(or 3 

2

5)

( x  1) 2  (2 x  1)(2 x  1) x 2  2 x  1  (2 x ) 2  12 3x 2  2 x  2  0

Using the quadratic formula,

Quick Practice 1.9 (p. 1.22) (a)

 ( 6) 

x 2  2(3x  2)

x

x 2  6x  4 

x 2  6x  4  0

 ( 2) 

( 2) 2  4(3)(2) 2(3)

2

28 6



  or 1  7    3  

Quick Practice 1.10 (p. 1.27) (a) x –3 –2 y 21 5

–1 –3

0 –3

1 5

2 21

(b)

The xintercepts of the graph of y = 4x2 + 4x – 3 are –1.5 and 0.5. Therefore, the roots of 4x2 + 4x – 3 = 0 are –1.5 and 0.5. Quick Practice 1.11 (p. 1.28) (a) x –1 0 1 y –7 –1 3

2 5

3 5

4 3

5 –1

6 –7

(b) The x-intercepts of the graph of y   x 2  5 x  1 are 0.2 and 4.8. Therefore, the roots of  x 2  5 x  1  0 are 0.2 and 4.8. Quick Practice 1.12 (p. 1.29) The x-intercept of the graph of y   x 2  x  Therefore, the root of  x 2  x 

1 is 0.5. 4

1  0 is 0.5. 4

Quick Practice 1.13 (p. 1.31) The graph of y   x 2  4 x  5 does not intersect the x-axis, i.e. there is no x-intercepts. Therefore, the equation  x 2  4 x  5  0 has no real roots. 5

NSS Mathematics in Action 4A Full Solutions Quick Practice 1.17 (p.1.39)

Quick Practice 1.14 (p. 1.32) (a) ∵ C(0, 3) is a point on the graph of y = –x2 + 2x + k. 2 ∴ 3  0  2(0)  k

3x 2  4x  4  0 ( x  2)(3x  2)  0 x20

k  3

(b) ∵ ∴

or 3 x  2  0

x  2 or

The graph of y   x 2  2 x  3 cuts the x-axis at A and B. The x-coordinates of A and B are the roots of the equation  x 2  2 x  3  0 .

x

2 3

1

1 ∴ The roots of the required equation are and 2 , 2

3

 x 2  2x  3  0  ( x  1)( x  3)  0 x  1  0 or x  3  0 x  1 or x3 ∴ The coordinates of A and B are (–1, 0) and (3, 0) respectively.  3  (1)  (3  0) sq. units  Area of ABC 2  6 sq. units

1 3 i.e.  and . 2 2 The required quadratic equation is



 1   x      2   1   x  2  

 x  

 x 

3  0 2 3  0 2

1 3   2 x    2  x    4(0) 2 2    (2 x  1)(2 x  3)  0

Quick Practice 1.15 (p. 1.38) The required quadratic equation is

 x  (4)  x  (5)  0

4x2  2x  6x  3  0

( x  4)( x  5)  0

4x2  4x  3  0

x  4 x  5 x  20  0 2

Quick Practice 1.18 (p. 1.41) Let x be the smaller integer, then x + 1 is the larger integer.

x 2  9 x  20  0

x 2  ( x  1) 2  145

Quick Practice 1.16 (p. 1.38) The required quadratic equation is

x 2  x 2  2 x  1  145



 1   x      ( x  3)  0  4   1   x   ( x  3)  0 4 

∴

2 x 2  2 x  144  0 x 2  x  72  0 ( x  8)( x  9)  0 x  8  0 or x  9  0 x  8 or x  9 (rejected)

1  4 x   ( x  3)  4(0) 4  (4 x  1)( x  3)  0

When x = 8, x + 1 = 9 ∴ The two consecutive positive integers are 8 and 9.

4 x 2  x  12 x  3  0

Quick Practice 1.19 (p. 1.42) (a)

4 x  11x  3  0 2

Area of the garden   (10)(8)  (10  x)(8  x ) m 2





 80  (80  8 x  10 x  x 2 ) m 2  ( x  18 x) m 2

2

 x  18 x  30 2

x  18 x  30  0 2

(b)

x

 (18)  (18) 2  4(1)(30) 2(1)

18  204 18  204 or (rejected) 2 2  1.86 (cor. to 3 sig. fig.) 

Quick Practice 1.20 (p. 1.43) Let x km/h be the speed of the faster ship, then (x – 4) km/h is 6

1 Quadratic Equations in One Unknown (I) the speed of the slower ship. Distance travelled by the faster ship in 2 hours = 2x km Distance travelled by the slower ship in 2 hours = 2(x – 4) km

(2 x) 2   2( x  4) 2  40 2

 40  x  ( x  4)     2 2

2

2

x 2  ( x 2  8 x  16)  400 2 x 2  8 x  384  0 x 2  4 x  192  0 ( x  16)( x  12)  0 x  16  0 or x  12  0 x  16 or x  12 (rejected) ∴ The speed of the faster ship is 16 km/h.

Further Practice Further Practice (p. 1.11) 1. (a) The equation is not in the general form.

x  2x2  5 2x2  x  5  0 Coefficient of x = –1 (b) (i) When x = 2, L.H.S. = 2 R.H.S. = 2(2)2 + 5 = 13 Since L.H.S.  R.H.S., 2 is not a root of the equation. (ii) When x = –6, L.H.S. = –6 R.H.S. = 2(–6)2 + 5 = 77 Since L.H.S.  R.H.S., –6 is not a root of the equation. 2.

(a) The equation is not in the general form.

( x  6) 2  0 x 2  2( x )(6)  6 2  0 x 2  12 x  36  0

Coefficient of x = 12 (b) (i) When x = 2, L.H.S. = (2 + 6)2 = 82 = 64 R.H.S. = 0 Since L.H.S.  R.H.S., 2 is not a root of the equation. (ii) When x = –6, L.H.S. = (–6 + 6)2 = 02 = 0 R.H.S. = 0 Since L.H.S. = R.H.S., –6 is a root of the equation. 3.

(a) The equation is in the general form. (b) (i) When x = 2, L.H.S. = –22 + 4(2) – 4 = – 4 + 8 – 4 = 0 R.H.S. = 0 Since L.H.S. = R.H.S., 2 is a root of the equation. (ii) When x = –6, L.H.S. = – (–6)2 + 4(–6) – 4 = – 36 – 24 – 4 = – 64 R.H.S. = 0 Since L.H.S.  R.H.S., –6 is not a root of the equation. 7

NSS Mathematics in Action 4A Full Solutions 4.

12  ( x  4)(3x  12)  0

(a) The equation is not in the general form.

( x  6)( x  2)  0

4  ( x  4) 2  0

x  6 x  2 x  12  0 2

5.

x  4 x  12  0 2

Coefficient of x = 4 (b) (i) When x = 2, L.H.S. = (2 + 6)(2 – 2) = 8 × 0 = 0 R.H.S. = 0 Since L.H.S. = R.H.S., 2 is a root of the equation. (ii) When x = –6, L.H.S. = (–6 + 6)(–6 – 2) = 0 × (–8) = 0 R.H.S. = 0 Since L.H.S. = R.H.S., –6 is a root of the equation.

[2  ( x  4)][2  ( x  4)]  0 ( x  2)(6  x)  0

x  3 or

2.

3.

1.

2x 1  0

or x  6  0 1 x   or x6 2

x  3

5 4 x 2  11x   0 2 5  2 4 x 2  11x    2  0 2  2 8 x  22 x  5  0 (4 x  1)(2 x  5)  0 4 x  1  0 or 2 x  5  0 1 5 x or x 4 2

Using the quadratic formula,

x  

 ( 7)  7

( 7) 2  4(3)( 16) 2(3)

241 6

7

241

or

25 0 4 5 5 (2 x)2  2(2 x)( )  ( )2  0 2 2 5 5 3. (2 x  )(2 x  )  0 2 2 5 2x   0 2 5 x 4 4 x 2  10 x 

x 2  6 x  27  0 ( x  9)( x  3)  0 or x  3  0 x3



x

or or

5 2x   0 2 5 x 4

5 4

3x( x  3)  x  8 12( x  33)  x 2  9

3x 2  9 x  x  8 4.

12 x  396  x 2  9

3x 2  10 x  8  0 (3x  4)( x  2)  0

4.

3x  4  0 or x  2  0 x

4 or 3

7

241

6 6   1.42 (cor. to 3 sig. fig.) or 3.75 (cor. to 3 sig. fig.

x2  1  2(5  x) 3 x2   1  3  2(5  x) 3  x 2  3  30  6 x

x   9 or

( x  4)( x  8)  0 x  4  0 or x  8  0 x   4 or x  8

2.

x9  0

x6

Further Practice (p. 1.23) x 2  12 x  32  0

x  3

 3 

x  2 or

(2 x  1)( x  6)  0

6.

2 x 2  12 x  18  0 1 2 1 (2 x  12 x  18)   0 2 2 2 x  6x  9  0 ( x  3)( x  3)  0 x  3  0 or x  3  0 ∴

6x 0

( x  5)(2 x  1)  (1  2 x)  0 (2 x  1) ( x  5)  1  0

Further Practice (p. 1.16)

1.

x  2  0 or

x2

8

x 2  12 x  405  0 ( x  15)( x  27)  0 x  15  0 or x  27  0 x   15 or x  27

1 Quadratic Equations in One Unknown (I)

(1  2 x) 2  25  0 (1  2 x) 2  5 2  0 5. [(1  2 x)  5][(1  2 x)  5]  0 (6  2 x)(4  2 x)  0 6  2x  0 x3

or  4  2 x  0 or x  2

( x  3)(4 x  5)  ( x  2)(2 x  6)  0 ( x  3)(4 x  5)  ( x  2)  2( x  3)  0 ( x  3) (4 x  5)  2( x  2)  0 ( x  3) 4 x  5  2 x  4  0 ( x  3)(6 x  9)  0 x3 0

6.

or 6 x  9  0 3 x   3 or x 2

Further Practice (p. 1.39) 1. The required quadratic equation is

2   1    x     0 5  2    2  1   x   x  0 5  2  2 1   5 x    2 x    10(0) 5 2   (5 x  2)(2 x  1)  0 

 x

10 x 2  4 x  5 x  2  0 10 x 2  x  2  0

2.

(a)

x 2  8 x  12  0 ( x  2)( x  6)  0 x  2  0 or

x60

x  2 or

x6

(b) The roots of the required equation are

2 6 and , 2 2

i.e. 1 and 3. The required quadratic equation is

( x  1)( x  3)  0 x  x  3x  3  0 2

x2  4 x  3  0

3.

(a)

8 x 2  10 x  3  0 (4 x  3)(2 x  1)  0 4x  3  0

or 2 x  1  0 3 1 x or x 4 2

9

NSS Mathematics in Action 4A Full Solutions (b) The roots of the required equation are 



3  1 and 4

y 2  ( y  3) 2  ( y  6) 2 y 2  ( y 2  6 y  9)  y 2  12 y  36

1 7 3  1 , i.e.  and  . 2 4 2

∴

The required quadratic equation is



y  9  0 or y  3  0 y  9 or y  3 (rejected)

7    3   x      x      0  4    2   7  3   x   x  0 4  2  7 3   4 x    2 x    8(0) 4 2    ( 4 x  7)(2 x  3)  0 

8 x 2  14 x  12 x  21  0 8 x 2  26 x  21  0 Further Practice (p. 1.43) 1. Let x cm be the length of the base of the parallelogram, then (x – 2) cm is its height.

x( x  2)  15 x 2  2 x  15

∴

x 2  2 x  15  0 ( x  5)( x  3)  0 x  5  0 or x  3  0 x  5 or x  3 (rejected)

∴ 2.

The length of the base is 5 cm.

The difference in age between the man and his daughter is (5x – 4) years.

( x  1) 2  (2 x  11)  5 x  4 ( x 2  2 x  1)  (2 x  11)  5 x  4 ∴

x2  5x  6  0 ( x  6)( x  1)  0 x  6  0 or x  1  0 x  6 or 2

x  1 (rejected)

2

When x = 6, (x + 1) = (6 + 1) = 49 ∴ 3.

y 2  6 y  27  0 ( y  9)( y  3)  0

The man is 49 years old.

Draw a line perpendicular to the base as shown.

10

1 Quadratic Equations in One Unknown (I)

Exercise 1A (p. 1.8) Level 1 1. (a) Integer (b)

Fraction

(c)

Rational





(a)

number

3 2 (d) ,2 4 3 3 2  , 0, 13.245, 65,Let  ,  25 , 1.37 2 i.e.3 4

Irrational

2 ,

17



1.13

x



 (6)

6.

62 495

  0.322 222 0.32    0.323 232 0.32

Any rational number between these two numbers is acceptable, for example, 0.323 or 0.3232. Level 2 7.

0.71   0.717 171 1 0.7   0.711 111 0.71  is the largest. 1 ∴ 0 .7

, 1  0.7  0.717 171...  (1 )  71.717 171...  ( 2)  71 71 x 99

Let i.e.

99 x  12.4 124 x 990 62  495 5   0.12

∴

(a)

x x 100 x ( 2)  (1) : 99 x

23   1 , which is rational. 5 25 (b) 3 is irrational while 100 is rational, so 100 3 is irrational. (c) 2 is irrational while 1 is rational, so 2  1 is irrational. (d) ( 2  1)( 2  1)  ( 2 ) 2  12  1 , which is rational. (a)

(e)

4 9

(5 

5 ) 2  52  2  5 

5  ( 5 )2

 25  10 5  5  30  10 5

30 and 10 are rational while (5 

  71 1 0.7 99

∴

9 x

:

100 x  12.525 252

4,  3

1 3 1 (  0.025), (  0.12), (  0.125) 40 25 8 1  ), 1 (  0.0416 ) (  0.16 (b) 6 24 4.

(3)

 ( 5 )

 1.37

(b) 4, 16, 300 (c) 62.9 (d)  26 ,  2 5 3.

1.25

5 , x  0.12 x  0.125 252

(6)  (5) :

Recurring

4





decimal 2.



  113 0.125 900

∴

25 , 0, 65

number (e)



x

i.e. ( 4)

(d)

x 10 x

Let

Exercise

(f)

5 is irrational, so

5 ) is irrational. 2

5  and Both 2.3

2 are rational, so their difference is 3

also rational. 5.

(a)

0.125 

125 1  1000 8

8.

(b)

 25 ,  0.1  0.125 125  125.125 125  125 125 x 999

Let i.e.

x x 1000 x ( 2)  (1) : 999 x

∴

, 0.2 0.222 222 2.222 222 2 2   9   2 0.2 9

  10 ( 2)  (1) : 9

Let i.e.  (1 )  ( 2)

∴

 25   125 0.1 999

(c) 11

   

 (1 )  ( 2 )

0 .1

0 .1

113 900

NSS Mathematics in Action 4A Full Solutions

 i.e.  100  ( 4)  (3) : 99  Let

∴



6 , 0.1

 0.161 616  16.161 616

( a  1) 2  4a

 (3)  ( 4)

10.

 16

16   99 6   16 0.1 99



( a 2  2a  1)  4a



a 2  2a  1



( a  1) 2

, which is also a rational

 a 1

number. Exercise 1B (p. 1.17) Level 1 1. When x = 4, L.H.S. = 3(4)2 – 11(4) – 4 = 0 R.H.S. = 0 Since L.H.S. = R.H.S., 4 is a root of the equation.

The equation becomes

2 16 x 9 99 16 2 x  99 9 8  11

2.

When x = –3, L.H.S. = (–3)2 + 13(–3) + 36 = 6 R.H.S. = 0 Since L.H.S.  R.H.S., –3 is not a root of the equation. x( x  5)  0

3.

x0

or

9.

x 5  0 x5

  1000 ( 2)  (1) : 999

Let i.e.

∴

Let i.e.

 04 ,  0.1  0.104 104  104.104 104

 104 104   999  04   104 0.1 999 ,   0 .5 7

  0.577 777  10   5.777 777  ( 4)  (3) : 9   5.2

∴

 (1 ) 4.  ( 2)

5.

 ( 3)  ( 4) 6.

52   90 26  45 26   0.57 45

(2 x  3)( x  8)  0 2 x  3  0 or x  8  0 3 x or x  8 2 (9  t )(3t  12)  0 9  t  0 or 3t  12  0 t  9 or t4

3(2 p  5) 2  0 1 1  3(2 p  5) 2   0 3 3 2 (2 p  5)  0 2 p  5  0 or 5 p   or 2 ∴

p

2p 5  0 p

5 2

The equation becomes

104 26  x0 999 45

7.

104 26 x  999 45 20  111

x 2  3x  0 x( x  3)  0 x  0 or

x3 0 x  3

k 2  16k 8.

k 2  16k  0 k ( k  16)  0 k  0 or

k  16  0 k  16

12

5 2

1 Quadratic Equations in One Unknown (I)

9.

x 2  7 x  12  0 ( x  4)( x  3)  0

 3  16.

x  4  0 or x  3  0 x   4 or x  3 y  8 y  12  0 2

10. ( y  2)( y  6)  0 y  2  0 or

y 6  0

y  2 or

y6

17.

( h  6) 2  0 h  6  0 or h  6  0 h  6 or h  6

5x  1  0

2t 2  5t  3  0 (2t  3)(t  1)  0 18. 2t  3  0

d 2  2(d )(9)  9 2  0

t

( d  9) 2  0 d  9  0 or d  9  0 d  9 or d 9

or x  2  0 x2

or t  1  0 3 or 2

t  1

4 y 2  12  13 y

d=9 x 2  49  0

4 y 2  13 y  12  0 19. ( y  4)(4 y  3)  0 y40

x2  72  0 ( x  7)( x  7)  0

or 4 y  3  0 3 y   4 or y 4

x  7  0 or x  7  0 x   7 or x7

17 m  12m 2  6

9 y 2  16  0

14.

(5 x  1)( x  2)  0 1 x or 5

d 2  18d  81  0

13.

x6

5x 2  9x  2  0

h 2  2(h)(6)  6 2  0

∴

or x  6  0

x   3 or

∴ h = –6

12.

x 2  3x  18  0 ( x  3)( x  6)  0 x30

h 2  12h  36  0 11.

1 2 x  x6  0 3 1 2  x  x  6  3 0 3 

12m 2  17 m  6  0 20. (3m  2)(4m  3)  0 3m  2  0 or 4m  3  0 2 3 m or m 3 4

9 y 2  16 16 y2  9 16 16 or  9 9 4 4  or  3 3

y

21. (a) Let  and  be the roots of the equation. From the equation, we have  = –28 Thus any pair of integers  and  satisfying the above relation is acceptable. ∴ The two integers can be –4 and 7, –2 and 14. (or any other reasonable answers)

2 x 2  8 x  42  0 1 2 1 (2 x  8 x  42)   0 2 2 2 15. x  4 x  21  0

(b) For the equation whose roots are –4 and 7, the corresponding value of k is k  ( 4  7)

( x  7)( x  3)  0 x  7  0 or x  3  0 x   7 or x3

 3 For the equation whose roots are –2 and 14, the corresponding value of k is k  ( 2  14)

  12 (or reasonable answer corresponding to (a))

13

NSS Mathematics in Action 4A Full Solutions 22.

25 p 12 12(1  p 2 )  25 p 1 p2 

x

–p

x

–q

–px

12  12 p 2  25 p 25.

–qx = –(p + q)x

(4 p  3)(3 p  4)  0 4 p  3  0 or 3 p  4  0

( p  q )  3 pq 3 ∴ The pairs of values for p and q can be p  4, q  1 or p  3, q  0 or p  2, q  1 . (or any other reasonable answers)

p

25 4 25  5x   4

26.

2

a

(2 x  5)  0 2 x  5  0 or 2 x  5  0 5 5 x  or x 2 2 2

27.

25  10m 3 25   3 3m 2    3  10m 3  2 9m  25  30m 3m 2 

24.

or a  8  0 1 or 2

a 8

5 x( x  2)  ( x  2) 2

5 2

x

2a 2  8  15a

2a  1  0

(2 x)  2(2 x)(5)  5  0

∴

5 x( x  2)  ( x  2) 2  0 ( x  2) 5 x  ( x  2)  0 ( x  2)(4 x  2)  0 x  2  0 or 4 x  2  0 x  2 or

x

(2 x  5) 2  3( x  3)(2 x  5)

9m  30m  25  0

(2 x  5) 2  3( x  3)(2 x  5)  0

(3m)  2(3m)(5)  5  0

28. (2 x  5) (2 x  5)  3( x  3)  0

(3m  5)  0

(2 x  5)( x  4)  0

2 2

2 x  5  0 or  x  4  0

3m  5  0 or 3m  5  0 5 5 m  or m 3 3 ∴

m

1 2

(2 x  5) 2  ( x  3)(6 x  15)

2

2

4 3

2a 2  15a  8  0 ( 2a  1)(a  8)  0

4 x 2  20 x  25  0 2

p

15 a 2 15 a 2  22  a 2 2 2(a  4)  15a

x 2  5x 

 4  x 2  4  2 4 x  20 x  25

3 or 4

(a  2)(a  2) 

Level 2

23.

12 p 2  25 p  12  0

x

5 3

5 or 2

( x  1) 2  3( x  1)  4  0  ( x  1)  4 ( x  1)  1  0 29. ( x  5) x  0 x  5  0 or x  0 x  5

14

x4

1 Quadratic Equations in One Unknown (I)

(2 x  3) 2  x(2 x  23)

(4  x) 2  5(4  x)  14  0 (4  x)  2 (4  x)  7  0 30. (6  x)(3  x)  0

4 x 2  12 x  9  2 x 2  23x 36.

6  x  0 or  3  x  0 x  6 or x  3

2 x 2  11x  9  0 ( x  1)(2 x  9)  0 x  1  0 or 2 x  9  0

31.

4( x  3) 2  100

( x  2) 2  ( 4  5 x ) 2

( x  3) 2  25 x  3  5 or  5

( x  2) 2  (4  5 x) 2  0

x  8 or  2

37.

 ( x  2)  (4  5 x) ( x  2)  (4  5 x)  0 (4 x  2)(6 x  6)  0

( x  2) 2  81x 2  0

 4 x  2  0 or 6 x  6  0

( x  2) 2  (9 x) 2  0

x

[( x  2)  9 x] [( x  2)  9 x]  0 32. (10 x  2)( 8 x  2)  0 10 x  2  0 x

or  8 x  2  0 1 or 5

x

2x2  6x  x  3  3  0

x  2 or 2 x  3  0

or

 1 2 3 4

6x 2  x  5  0 (6 x  5)( x  1)  0 6x  5  0 x

or x  1  0 5 or 6

 24 12 8 6

k –25 –14 –11 –10

∴ k = 25, 14, 11, 10

x 1

Exercise 1C (p. 1.24) Level 1 ( x  3) 2  16

( x  4)(6 x  3)  3(3  2 x) ( x  4)  3(2 x  1)  3(3  2 x)

1.

( x  4)(2 x  1)  3  2 x 2x 2  8x  x  4  3  2 x

x  3   16 x  3  4 or  3  4 1 or  7

2x 2  5x  7  0 ( x  1)( 2 x  7)  0 x 1  0

( x  4) 2  49

or 2 x  7  0

x   1 or

x

5 2

39. Let  and  be the roots of the equation. From the equation, we have  = 24 and k = –( + ) We now list the possible values of  and , and the corresponding value of k.

3 x 2

3 x  1  6 x 2  2 x  4

35.

or 4 x  10  0

( x  2)(2 x  3)  0

(3 x  1)(1  2 x)  4

34.

(3 x  3  7  x)(3x  3  7  x)  0 (2 x  4)(4 x  10)  0 2x  4  0

2x 2  7 x  6  0

x   2 or

x 1

3( x  1) 2  (7  x) 2  0

1 4

( x  3)(2 x  1)  3  0

x20

1 or 2

9( x  1) 2  (7  x) 2

38.

33.

9 2

x  1 or x 

4( x  3) 2  100  0

x

2.

7 2

15

x  4   49 x  4  7 or 4  7  11 or  3

 –1 –2 –3 –4

 –24 –12 –8 –6

k 25 14 11 10

NSS Mathematics in Action 4A Full Solutions 2



1 25  2x    2 4  

3.

1 25 2x    2 4 1 5 1 5 2x   or  2 2 2 2 2x  3 or  2 3 x or  1 2

8.

Using the quadratic formula,

x

7

2(5)

 7  169 10  7  13  10 3  or  2 5 

2

2  25 x    9 5   2

4.

2  52   x    9 5  (5 x  2) 2  9

3x 2  5  3 x

9.

3x 2  3 x  5  0 Using the quadratic formula,

5x  2   9

x

5 x  2  3 or  2  3 5x  1 1 x 5 5.





or  1

 ( 2)  2

10.

2 x 2  5x  6  0 x 

3

( 7) 2  4(1)(11) 2(1)

x 

5 2  4( 2)( 6) 2( 2)

 5  73 4

 ( 9 ) 

( 9) 2  4( 2)( 15) 2( 2)

9

201 4

9

201 4  5.79 (cor. to 2 d.p.)

Using the quadratic formula,



 14  14  4( 1)(32) 2( 1) 2

x

5

11. Using the quadratic formula,

7 5  2 7.

 3  69 6

2 x 2  4  10  5 x

64

 ( 7) 

3 2  4(3)(5) 2(3)

Using the quadratic formula,

Using the quadratic formula,

x

3

2( x 2  2)  5( 2  x)

( 2) 2  4(1)( 15) 2(1)

2 28  2  5 or 6.

or  5

Using the quadratic formula,

x

7 2  4(5)(6)

 14  324 2  14  18  2   2 or 16

or or

9

201 4  1.29 (cor. to 2 d.p.)

12. Using the quadratic formula,



x 

 ( 23) 

( 23) 2  4( 17)(5) 2( 17)

23  869  34

23  869  34   1.54 (cor. to 2 d.p.) 

16

or or

23  869  34 0.19 (cor. to 2 d.p.)

1 Quadratic Equations in One Unknown (I)

6 x 2  11x  2  0

1 0 3  2x 2  x   0 18. 3   3 2 x 2  x    3  0 3  2 6 x  3x  1  0 x( 2 x  1) 

( x  2)(6 x  1)  0

13.

x20

or 6 x  1  0 1 x   2 or x 6

3 x( x  2)  8  6 x

14.

3x 2  6 x  8  6 x 3 x 2  12 x  8  0

Using the quadratic formula,

x 

 ( 12) 

( 12) 2  4(3)(8) 2(3)

12  48 6

12  48 6  3.15 (cor. to 2 d.p.) 

(x  d )2  15.

or or

12  48 6 0.85 (cor. to 2 d.p.)

1 4

xd  

1 4

1 1 x   d or   d 2 2 To ensure both roots are integers, the value of d can be

1

1 1 1 .(or any other reasonable answers) , or  2 2 2

Level 2

5 1 x 0 4 2 16. 5 1  4 x 2  x    4  0 4 2  2 4x  5x  2  0 x2 

Using the quadratic formula,

x 

17.

 ( 5)  5

( 5) 2  4( 4)(2) 2( 4)

57 8

0.1x 2  0.5 x  0.2  0 10(0.1x 2  0.5 x  0.2)  10  0 x 2  5x  2  0

Using the quadratic formula,

x 

 ( 5)  5

( 5) 2  4(1)(2) 2(1)

17 2

17

NSS Mathematics in Action 4A Full Solutions Using the quadratic formula,

x 

3

Using the quadratic formula,

3 2  4(6)(1) 2( 6)

3

x

 3  33 12

2(1)  3  53 2





1 3  ( x  2)  x  4 2 2  19. ( 2 x  1)( x  2)  3 x  8 2 x 2  5 x  2  3x  8

 3  53 2  2.14 (cor. to 2 d.p.)

 x



24.

x 2  4x  3  0



4

28 2

2 p  5 p  8 p  20  p 2  5 2 p2  3p  5  0 Using the quadratic formula,

( 4) 2  4(1)(3) 2(1) (or 2 

p

7)



(2 x  5)  ( x  2)(2 x  5) (2 x  5)  ( x  2)(2 x  5)  0 (2 x  5)1  ( x  2)  0 20. (2 x  5)(3  x)  0 2 x  5  0 or 3  x  0 5 x or x3 2

( 3) 2  4(1)(5) 2(1)

3

 11 2

( m  3)(5m  4)  m2 2 ( m  3)(5m  4)  2m 2

25.

5m 2  15m  4m  12  2m 2 3m 2  11m  12  0

Using the quadratic formula,

3( x  2)  2( x  2)( x  1)  0 ( x  2) 3  2( x  1)  0 21. ( x  2)(5  2 x)  0

m 

x20

or 5  2 x  0 5 x   2 or x 2

 11  112  4(3)( 12) 2(3)  11  265 6

 11  265 or 6  0.88 (cor. to 2 d.p.) or 

x(3x  4)  ( x  1)( x  2)  0

 11  265 6  4.55 (cor. to 2 d.p.)

y ( y  1)  6  ( y  5)( y  3)

(3x  4 x)  ( x 2  x  2 x  2)  0 2

22.

 ( 3) 

∵  11 is not a real number. ∴ The equation has no real roots.

3( x  2)  (2 x  4)( x  1) 3( x  2)  2( x  2)( x  1)

2x 2  5x  2  0 ( x  2)(2 x  1)  0 x20

or

 3  53 2  5.14 (cor. to 2 d.p.)

2

Using the quadratic formula,

 ( 4) 

or

( 2 p  5)( p  4)  ( p  5)( p  5)

2x 2  8x  6  0

x

3 2  4(1)(11)

26.

y 2  y  6  ( y 2  5 y  3 y  15) 2y2  7 y  9  0 Using the quadratic formula,

or 2 x  1  0

x   2 or

1 x 2

y 

( x  3) 2  ( 2 x  1)(2  x )  0

 ( 7 )  7

( 7) 2  4( 2)(9) 2( 2)

 23 4

∵  23 is not a real number. ∴ The equation has no real roots.

2 2 23. ( x  6 x  9)  ( 4 x  2  2 x  x )  0

 x 2  3 x  11  0 x 2  3 x  11  0

18

1 Quadratic Equations in One Unknown (I) ∵ The graph of y  3 x 2  8 x  7 does not

1  ( k  3) k    1  2( k  1) 5  27. ( k  3)(5k  1)  5  10( k  1)

intersect the x-axis. The equation  3 x 2  8 x  7  0 has no real roots. ∴ The equation 3 x 2  8 x  7  0 also has no real ∴

5k 2  14k  3  5  10k  10 5k 2  24k  2  0

roots.

Using the quadratic formula,

k

 ( 24) 

( 24) 2  4(5)(2) 2(5)

7.

24  536  10

(a) x y

24  536 10  4.72 (cor. to 2 d.p.) 

or or

–2 5

–1 1

0 –1

1 –1

2 1

24  536 10 0.08 (cor. to 2 d.p.)

3 5 (b)

Exercise 1D (p. 1.33) Level 1 1. The x-intercepts of the graph of y  2 x 2  6 x are –3 and 0. Therefore, the roots of 2 x 2  6 x  0 are –3 and 0. 2.

The x-intercepts of the graph of y  2 x 2  x  3 are The x-intercepts of the graph of y  x 2  x  1 are

–1.5 and 1. Therefore, the roots of  2 x 2  x  3  0 are –1.5 and

–0.6 and 1.6. Therefore, the roots of x 2  x  1  0 are –0.6 and 1.6.

1. 3.

The x-intercepts of the graph of y  12 x 2  29 x  14 are 0.7 and 1.8.

8.

(a)

Therefore, the roots of  12 x 2  29 x  14  0 are 0.7 and 1.8. 4.

x y

0 25

1 9

2 1

3 1

4 9 (b)

The x-intercepts of the graph of y  3 x 2  7 x  5 are – 0.6 and 2.9. Therefore, the roots of 3 x 2  7 x  5  0 are –0.6 and

2.9.

4 x 2  12 x  9  0 5.

1 1 ( 4 x 2  12 x  9)   0 4 4 9 x 2  3x   0 4 ∵ The x-intercept of the graph of y  x 2  3 x 

9 is 4

The x-intercept of the graph of y  4 x 2  20 x  25 is 2.5.

1.5.

9  0 is 1.5. 4 ∴ The root of 4 x 2  12 x  9  0 is also 1.5. ∴

Therefore, the root of 4 x 2  20 x  25  0 is 2.5.

The root of x  3 x  2

9.

(a) x y

3x  8 x  7  0 2

6.

(b)

 (3 x 2  8 x  7 )  0  3x 2  8 x  7  0 19

0 –2

1 1

2 2

3 1

4 –2

NSS Mathematics in Action 4A Full Solutions

x 2  3x 1  0 9 4 (b) 2  x  3x 1   36    36  0 9 4  2  4 x  12 x  9  0 ∵ The x-intercept of the graph of y  4 x 2  12 x  9 is 1.5. ∴ The root of  4 x 2  12 x  9  0 is 1.5.

The x-intercepts of the graph of y   x 2  4 x  2 are 0.6 and 3.4. Therefore, the roots of  x 2  4 x  2  0 are 0.6 and 3.4.

∴ The root of

Level 2 10. (a) x y

–3 8

–2 2

–1 –2

0 –4

1 –4

2 –2

3 2

4 8

x2  x  4

(b)

x2  x  4  0 ∵ The x-intercepts of the graph of y  x 2  x  4 are –1.6 and 2.6. ∴ The roots of x 2  x  4  0 are –1.6 and 2.6. ∴ The roots of x 2  x  4 are also –1.6 and 2.6. 11. (a) x y

–1 –25

0 –9

1 –1

2 –1

3 –9

4 –25

20

x 2  3x 1   0 is also 1.5. 9 4

1 Quadratic Equations in One Unknown (I) 12. (a) x y

–2 19

–1 9

0 3

1 1

2 3

3 9

4 19

15. ∵ P. ∴

The graph of y  x 2  6 x  9 touches the x-axis at The x-coordinate of P is the root of the equation x2  6x  9  0 .

x2  6x  9  0 ( x  3) 2  0 x  3  0 or x  3  0 x  3 or ∴ ∵ ∴

3  2 x(2  x)

(b)

3  4x  2x

x3

x-coordinate of P = 3 The graph of y  x 2  6 x  9 cuts the y-axis at Q.

y-coordinate of Q  0 2  6(0)  9 9

OP = (3 – 0) units = 3 units OQ = (9 – 0) units = 9 units ∴ OP : OQ = 3 : 9

2

2x  4x  3  0 2

 1: 3

∵ The graph of y  2 x  4 x  3 does not 2

16. ∵

intersect the x-axis. ∴ The equation 2 x 2  4 x  3  0 has no real roots. ∴ The equation 3 = 2x(2 – x) also has no real roots.

∴

and R. The x-coordinates of P and R are the roots of the equation x 2  3 x  10  0 .

x 2  3x  10  0 ( x  2)( x  5)  0 x  2  0 or x  5  0 x  2 or x5

13. (a) ∵ C(–1, 9) is a point on the graph of y  x 2  6x  k . ∴

The graph of y  x 2  3 x  10 cuts the x-axis at P

9  ( 1) 2  6( 1)  k 9  1  6  k k  16

∴ ∵

The x-coordinates of P and R are –2 and 5 respectively. The graph of y  x 2  3 x  10 cuts the y-axis at

∴

Q. y-coordinate of Q  0 2  3(0)  10

(b) ∵ The graph of y   x 2  6 x  16 cuts the xaxis at A and B. ∴ The x-coordinates of A and B are the roots of the equation  x 2  6 x  16  0 .

 10

Area of PQR

 x 2  6 x  16  0  ( x  2)( x  8)  0 x  2  0 or x  8  0 x  2 or x8



 5  (2)   0  (10)

2  35 sq. units

∴ The coordinates of A and B are (–2, 0) and (8, 0) respectively. 14. (a) ∵ The graph of y  2 x 2  6 x  20 cuts the x-axis at A and B. ∴ The x-coordinates of A and B are the roots of the equation 2 x 2  6 x  20  0 .

17. (a) ∵ R(–2, 0) and T(–1, q) are points on the graph of y   px 2  4 x  12 .

0   p ( 2) 2  4( 2)  12 0  4 p  8  12 4p  4 p1

∴

2 x 2  6 x  20  0

sq. units

and

x 2  3x  10  0 ( x  2)( x  5)  0

q  ( 1) 2  4( 1)  12  1  4  12  7

x  2  0 or x  5  0 x  2 or x5

(b) ∵ The graph of y   x 2  4 x  12 cuts the x-

∴ The coordinates of A and B are (–2, 0) and (5, 0) respectively. (b) AB = [5 – (–2)] units  7 units

axis at R and S. ∴ The x-coordinates of R and S are the roots of the equation  x 2  4 x  12  0 . 21

NSS Mathematics in Action 4A Full Solutions

 x 2  4 x  12  0  ( x  2)( x  6)  0 x  2  0 or x  6  0 x  2 or x6



 x 

1  4 x   ( x  3)  4(0) 4  ( 4 x  1)( x  3)  0

∴ The x-coordinate of S is 6. Length of RS = [6 – (–2)] units = 8 units Length of TS



 6  ( 1)

2

4 x 2  x  12 x  3  0 4 x 2  13 x  3  0

 (0  7) 2 units

 9.90 units (cor. to 3 sig. fig.) ∴ RS is shorter. Exercise 1E (p. 1.40) Level 1 1. The required quadratic equation is

( x  2)( x  5)  0 x  2 x  5 x  10  0 2

x 2  7 x  10  0 2.

The required quadratic equation is

 x  (5) x  (3)  0 ( x  5)( x  3)  0 x  5 x  3 x  15  0 2

x 2  8 x  15  0 3.

The required quadratic equation is

( x  2) x  ( 6)   0 ( x  2)( x  6)  0

x 2  2 x  6 x  12  0 x 2  4 x  12  0 4.

The required quadratic equation is

 x  (4) ( x  1)  0 ( x  4)( x  1)  0 x  4x  x  4  0 2

x 2  3x  4  0 5.

The required quadratic equation is

( x  0)( x  4)  0 x2  4x  0

6.

The required quadratic equation is

 x  (3) x  (3)  0 ( x  3)( x  3)  0 x2  6x  9  0

7.

1  ( x  3)  0 4

The required quadratic equation is

22

1 Quadratic Equations in One Unknown (I) 8.

The required quadratic equation is

Level 2



13.



 1  x    2 

 [ x  (3)]  0 



 x 

1  ( x  3)  0 2

( x  9)( x  6)  0 x  9  0 or x  6  0 x  9 or x  6

∴

The roots of the required equation are 9 + 2 and –6 + 2, i.e. 11 and –4. The required quadratic equation is

1  2 x   ( x  3)  2(0) 2  (2 x  1)( x  3)  0

( x  11) x  ( 4)  0 ( x  11)( x  4)  0

x 2  11x  4 x  44  0

2x 2  x  6x  3  0

x 2  7 x  44  0

2x 2  7 x  3  0 9.

(5 x  2)( x  4)  0 5x  2  0 14.

The required quadratic equation is



1   1    x     0 4   3  1 1  x    3 x    0 4 3  (4 x  1)(3 x  1)  0

 x 

 4 

x ∴

12 x 2  3 x  4 x  1  0 12 x  x  1  0

2 or 5

x4

The roots of the required equation are

, i.e. 

2

or x  4  0

1 1 2 and  4 5

5 1 and . 2 4

The required quadratic equation is



1  5    x      x    0 4  2    5 1   2 x    4 x    0 2 4   ( 2 x  5)(4 x  1)  0

10. The required quadratic equation is

2  3   x   x  0 3  2  2 3   3 x    2 x    0 3 2   (3 x  2)(2 x  3)  0

8 x 2  20 x  2 x  5  0

6x  4x  9x  6  0 2

8 x 2  18 x  5  0

6 x  13 x  6  0 2

4 x 2  20 x  25  0

11. The required quadratic equation is



(2 x) 2  2(2 x)(5)  5 2  0

5   1  x      x    0 2   5   5 1   2 x    5 x    0 2 5   (2 x  5)(5 x  1)  0 

(2 x  5) 2  0 2 x  5  0 or 2 x  5  0 5 5 x  or x 2 2

15. (a)

∴

10 x  25 x  2 x  5  0 2

10 x 2  23 x  5  0

x

5 2

(b) The roots of the required equation are 2 

12. The required quadratic equation is

2



1  1  x   x  0 3  3  1 1   3 x    3 x    0 3 3   (3 x  1)(3x  1)  0

5 , i.e. 5 and 5. 2

The required quadratic equation is

( x  5)( x  5)  0 x 2  10 x  25  0

9x 2  6x  1  0

23

5 and 2

NSS Mathematics in Action 4A Full Solutions

3x 2  8 x  4  0 16. (a)

2.

(3 x  2)( x  2)  0 3 x  2  0 or x  2  0 2 x or x2 3

(b) The roots of the required equation are

5t 2  12t  4  0 (5t  2)(t  2)  0 5t  2  0 or t  2  0 t  0.4 or t2

1 2  and 2 3

∴

1 1  2 , i.e. and 1. 2 3 The required quadratic equation is



 x 

1  ( x  1)  0 3

1  3 x   ( x  1)  0 3  (3 x  1)( x  1)  0 3x 2  x  3x  1  0 3x 2  4 x  1  0 7 x2  x  2  0 2 2 2x  7x  4  0 17. (a) ( x  4)(2 x  1)  0 x40

or 2 x  1  0

x   4 or

x

1 2

(b) The roots of the required equation are  4 

 4

1 and 2

7 1 , i.e.  and –2. 2 2

The required quadratic equation is



7    x       x  ( 2)  0 2    7   x   ( x  2)  0 2  7  2 x   ( x  2)  0 2  (2 x  7)( x  2)  0 2 x 2  7 x  4 x  14  0 2 x 2  11x  14  0 Exercise 1F (p. 1.44) Level 1 1. Let x be the positive integer. x 2  x  56 ∴

x 2  x  56  0 ( x  7)( x  8)  0 x  7  0 or x  8  0 x  7 or

∴

When the ball is 5 m above the ground, 5  1  12t  5t 2

x  8 (rejected)

The positive integer is 7. 24

The ball is 5 m above the ground when t = 0.4 s or t = 2 s.

1 Quadratic Equations in One Unknown (I) 3.

Distance sailed by the ship in the first t hours = t(t + 6) km Distance sailed by the ship in the remaining (t + 2) hours = (t + 2)(t – 2) km

7.

 60  2 x   cm 2    (30  x ) cm

The width of the rectangle  

t (t  6)  (t  2)(t  2)  356

∵

(t 2  6t )  (t 2  2 2 )  356 ∴

x(30  x)  216

2t 2  6t  360  0

30 x  x 2  216

t  3t  180  0 (t  12)(t  15)  0 2

2 ∴ x  30 x  216  0

( x  18)( x  12)  0 x  18  0 or x  12  0 x  18 or x  12 (rejected)

t  12  0 or t  15  0 t  12 or t  15 (rejected) ( x  14) 2  x 2  ( x  2) 2

8.

( x 2  28 x  196)  x 2  x 2  4 x  4

x 2  ( x  2) 2  10 2 x 2  ( x 2  4 x  4)  100

x  24  0 x  24 5.

Let x m be the length of the carpet, then (x – 2) m is its width.

x 2  32 x  192  0 ( x  24)( x  8)  0

4.

x8  0 x  8 (rejected)

or or

∴

Let x be the smallest integer, then (x + 1) and (x + 2) are the other two integers. ∴

(3 x  3) 2  x 2  ( x  1) 2  ( x  2) 2  94 9 x 2  18 x  9  x 2  ( x 2  2 x  1) 

9.

( x 2  4 x  4)  94 6 x 2  12 x  90  0

x 2  2 x  48  0 ( x  8)( x  6)  0 x  8  0 or x  6  0 x  8 or x  6 (rejected)

Let x cm be the length of the longer wire, then (80 – x) cm is the length of the shorter wire.

x cm 4 80  x Length of the side of the smaller square = cm 4 Length of the side of the larger square =

x 2  2 x  15  0 or or

x5 0 x  5 (rejected)

2

 x  80  x       4  4 

When x = 3, x + 1 = 4 and x + 2 = 5 ∴ The consecutive positive integers are 3, 4 and 5. Let r cm be the radius of the smaller circle, then the radius of the larger circle is (2r – 3) cm. Area of smaller circle = r 2 cm2

x 2  80 x  1024  0 ( x  64)( x  16)  0 x  64  0 or x  16  0 x  64 or x  16 (rejected)

 ( 2r  3) 2  r 2  72

∴

(2r  3) 2  r 2  72 4r 2  12r  9  r 2  72

The length of the longer wire is 64 cm.

10. AE = [(x + 1) – 1] cm = x cm HB = (x – 1) cm

3r  12r  63  0 2

( x  1)( x  1)  3( x  1)

r 2  4r  21  0

x 2  12  3 x

(r  7)(r  3)  0 r  7  0 or r  3  0 r  7 or

 272

2 x 2  160 x  2048  0

Area of larger circle =  ( 2r  3) 2 cm2

∴

2

x 2  (6400  160 x  x 2 )  272  16 ∴

6.

2 x 2  4 x  96  0

When x = 8, x – 2 = 6 ∴ The length and the width of the carpet are 8 m and 6 m respectively.

 x  ( x  1)  ( x  2) 2  x 2  ( x  1) 2  ( x  2) 2  94

( x  3)( x  5)  0 x 3  0 x3

The area of the rectangle is 216 cm2.

x 2  3x  1  0 ∴

r  3 (rejected)

x

When r = 7, 2r – 3 = 2(7) – 3 = 11 ∴ The radii of the two circles are 7 cm and 11 cm respectively.



25

 (3)  (3) 2  4(1)(1) 2(1) 3  13 2

or

3  13 (rejected) 2

NSS Mathematics in Action 4A Full Solutions ∴ The 14th diagram has 105 black dots.

11. AG = (12 – x) cm, AE = (8 – x) cm

n( n  1)  254 2 n( n  1)  508

2 Area of ABCD  (12)(8  sin 30) cm  96 sin 30 cm 2

Area of AEFG

n 2  n  508

 (12  x ) (8  x )  sin 30 cm 2

(b)

 (12  x )(8  x ) sin 30 cm 2

n 2  n  508  0

∴

n

96 sin 30  2(12  x)(8  x ) sin 30 48  (12  x)(8  x)



48  96  8 x  12 x  x 2  ( 20)  (20) 2  4(1)( 48) 2(1)

20  208 2  2.8 (cor. to 1 d.p.) 

20  208 15. (a) Base area of the box 2  ( 26  2 x )(16  2 x ) cm 2 or 17.2 (rejected)  ( 416  32 x  52 x  4 x 2 ) cm 2

 ( 4 x 2  84 x  416) cm 2 (b) ∵ The base area of the box is 200 cm2. ∴

x 2  (92  x) 2  682

∴

 1  2033 2

or

Level 2 12. Let x m be the distance between P and Q, then (92 – x) m is the distance between Q and R.

∴

2(1)

Since 2033 is not a perfect square, n cannot be an integer. ∴ It is not possible to form a diagram consisting of 254 black dots.

x 2  20 x  48  0 x

 1  12  4(1)(508)

x 2  (8464  184 x  x 2 )  4624

4 x 2  84 x  416  200

2 x 2  184 x  3840  0

x 2  21x  104  50 x 2  21x  54  0

x 2  92 x  1920  0 ( x  32)( x  60)  0 x  32  0 or x  60  0 x  32 or x  60

( x  3)( x  18)  0 x  3  0 or x 3

Volume of the box  base area  height

20 2  12 2 cm  16 cm

 200 x cm 3

Consider A'B'O.

 200(3) cm 3

(16  x) 2  (12  x )2  20 2

 600 cm 3

(256  32 x  x 2 )  (144  24 x  x 2 )  400 2x2  8x  0

16. (a) Area of ABCD = 10  10 cm 2 = 100 cm2

x2  4 x  0

10(10  x ) cm 2 2  5(10  x ) cm 2

Area of ADF 

( x  4) x  0 x  4  0 or x  0 (rejected) x4

10(10  2 x ) cm 2 2  5(10  2 x ) cm 2 1 2 Area of CEF  2 ( x )(2 x ) cm  x 2 cm 2 Area of ABE

14. (a)

n(n  1)  105 2 n( n  1)  210 n 2  n  210 n 2  n  210  0 ( n  14)(n  15)  0 n  14  0 or n  14 or

x  18 (rejected)

∴

The distance between P and Q is 32 m or 60 m.

13. In ABO, AO =

or

x  18  0

n  15  0 n  15 (rejected) 26



1 Quadratic Equations in One Unknown (I)

Area of  AEF  area of ABCD  area of  ADF



(b) ∵ The area of AEF is 26 cm2. ∴

 area of  ABE  area of  CEF



 100  5(10  x )  5(10  2 x )  x 2 cm 2  (100  50  5 x  50  10 x  x 2 ) cm 2

15 x  x 2  26 x 2  15 x  26  0 ( x  2)( x  13)  0

 (15 x  x 2 ) cm 2

x2 0

or

x2

or

x  13  0 x  13 (rejected)

17. (a) Area of trapezium ABCD



 ( x  2)  (3x  1) (2 x  1) cm 2

(b) (i)

∵ The area of the trapezium is 76 cm2.

2 ( 4 x  1)(2 x  1)  cm 2 2 2 8x  2 x  4 x  1  cm 2 2 8x 2  2 x  1  cm 2 2 1    4 x 2  x   cm 2 2 

∴

1  76 2 8 x 2  2 x  1  152 4x2  x 

8 x 2  2 x  153  0 8 x 2  2 x  153  0 (2 x  9)(4 x  17)  0 (ii) 2 x  9  0 or 4 x  17  0 9 17 x  or x   (rejected) 2 4 Construct DE perpendicular to BC as shown.

13  9   2  cm  cm 2 2  

AD = 

9   AB   2   1 cm  8 cm 2   9 25   BC   3   1 cm  cm 2 2   EC  BC  BE

 25 13    cm 2  2  6 cm 

27

NSS Mathematics in Action 4A Full Solutions DC 

∴ Perimeter of the trapezium ABCD 25  13   8  10  cm 2 2  

DE 2  EC 2

 8 2  6 2 cm  10 cm

 37 cm 18. (a)

AB = ( x  6) cm BC = ( 4 x  3) cm

(b) Consider △ABP. BP  

AB 2  AP 2 cm ( x  6) 2  x 2 cm

 12 x  36 cm ∵ AB and BP are folded to AB' and B'P respectively. ∴ AB' = AB = (x + 6) cm and B'P = BP = 12 x  36 cm

∵ AB'CD' is a rhombus. AB  BC ∴

AB  BC  BP  B ' P

x  6  (4 x  3)  12 x  36  12 x  36 2 12 x  36  3x  3

2 12x  36 

2

 (3x  3) 2

4(12 x  36)  9 x 2  18 x  9 9 x 2  66 x  135  0 3x 2  22 x  45  0 ( x  9)(3x  5)  0 x  9  0 or 3x  5  0 x  9 or

x

5 (rejected) 3

When x = 9, B'C = (x + 6) cm = 15 cm 2 ∴ Area of the rhombus  (9  15) cm

 135 cm 2

Revision Exercise 1 (p. 1.49) Level 1 1. (a) 1, 8 (b) –6,  9 , 0, 1, 8 (c) –6,  (d)

2.

28

(a)

 , 9 , 0, 1, 2.4

9 , 4.71, 8 2

 , 10 3 3   0.131 313 0.1   0.133 333 0.13 0.133 3   0.133  0.13  ∴ 0.1

1 Quadratic Equations in One Unknown (I) (b) (d)

3 , 0.1

x

Let

x  0.131 313 100 x  13.131 313

i.e. ( 2)  (1) :

 ( 3)

10 x  1.333 333

 ( 4)

2 15

133 1000

7.

2

8 

28 

16  4 , which is rational.

(b)

2

3 

23 

6 , which is irrational since

8.

 4.2

9.

( x  7)(5 x  4)  0 x70 (a)

or 5 x  4  0 4 x  7 or x 5

( x  9 )( x  9 )  0 x 9 0

x 9

or

x 9

x  3

or

x 3

(7 x  1)( 2 x  2 )  0 7 x  1  0 or 1 or 7 The roots are not irrational. x

2m(m  1)  (m  2)(m  1)  0 (m  1) 2m  (m  2)  0 (m  1)(3m  2)  0 m 1  0

p2

or 3m  2  0 m

2 3

(a) ∵ 3 is a root of x2 – 8x + p = 0. 2 ∴ 3  8(3)  p  0 p  15

10. (a) ∵ –2 is a root of 8x2 + kx + 6 = 0. 8( 2) 2  k ( 2)  6  0 ∴  2k  38  0 k  19

The roots are not irrational. (c)

4 or 3

x 2  8 x  15  0 (b) ( x  3)( x  5)  0 x  3  0 or x  5  0 x  3 or x5 ∴ The other root is 5.

The roots are not irrational. or

or p  2  0

m   1 or

rational.  is rational, so its square is also rational. 3.7

x 9 0

3 2 p  p40 2 3p2  2 p  8  0 (3 p  4)( p  2)  0 3p  4  0 p

  are rational, so 3.7 is also  and 4.2 (e) Both 3.7

(b)

3y2  7 y  6  0 ( y  3)(3 y  2)  0 y 3 0

or 3 y  2  0 2 y   3 or y 3

6 is not a perfect square. (c) Both 2 and 3 are irrational, so their sum is also irrational.  are rational, so their sum is  and 4.2 (d) Both 3.7 also rational.

4.

or x  2  0 5 x or x 2 2

6  3y2  7 y 6.

(a)

(f)

(2 x  5)( x  2)  0 2x  5  0

3(2  y 2 )  7 y

9 x  1. 2 12 x 90 2  15

0.133 

3.

5.

, x  0.13 x  0.133 333

  0.13

∴

x 5

or

2 x 2  9 x  10  0

3   13 0.1 99

( 4)  (3) :

x 5 0

or

The roots are irrational and equal to

99 x  13

Let i.e.

x 5 0 x 5

 (1 )  ( 2)

13 x 99 ∴

(x  5 )2  0

2x  2  0 x  1

29

5 .

NSS Mathematics in Action 4A Full Solutions

8 x 2  19 x  6  0 ( x  2)(8 x  3)  0 (b) x20

11. The required quadratic equation is

 x  (3) ( x  2)  0 ( x  3)( x  2)  0

or 8 x  3  0

x  2 or ∴ The other root is 

x

x 2  3x  2 x  6  0

3 8

x2  x  6  0

3 . 8

12. The required quadratic equation is



5  2   x  0 2  5  5 2   2 x    5 x    0 2 5   ( 2 x  5)(5 x  2)  0  x

10 x 2  25 x  4 x  10  0 10 x 2  29 x  10  0 x 2  10 x  264  0 x 13.

 10  10 2  4(1)(264) 2(1)

 10  1156 2  10  34  10  34  or 2 2  12 or  22 

1 x 5 5x 2  1  5x x2 

14.

5x 2  5x  1  0 x 

 (5)  (5) 2  4(5)(1) 2(5) 5  45 10



  or 5  3 5    10  

15.

x  2  4 x ( x  4) x  2  4 x 2  16 x 4 x 2  15 x  2  0 x 

 15  15 2  4( 4)(2) 2( 4)  15  257 8

 15  257  15  257 or 8 8  0.13 (cor. to 2 d.p.) or  3.88 (cor. to 2 d.p.) 

16.

30

1 Quadratic Equations in One Unknown (I)

9( x  1)  4  2( x  1) 2

x2 x20 8 x 2  8x  16  0



2( x  1) 2  9( x  1)  4  0 x 1 

 9  9 2  4( 2)(4)

x 2  2( x)(4)  4 2  0

2(2)

( x  4) 2  0 x  4  0 or x  4  0 x  4 or x4

 5  113 x 4  5  113  5  113 or 4 4  1.41 (cor.to 2d.p.) or  3.91 (cor. to 2 d.p.)  17. (a) x y

–2 5

–1 –2

0 –5

1 –4

∴

The x-coordinate of P is 4.

∵

The graph of y  

∴

y-coordinate of Q  

x2  x  2 cuts the y-axis at 8

Q.

2 1

02 02 8  2

(b) PQ  

( 4  0) 2   0  ( 2) 

2

units

20 units (or 2 5 units)

20. (a) ∵ A(4, 0) and C(14, c) are points on the graph of y  kx 2  14 x  40 .

∴

0  k ( 4) 2  14( 4)  40 0  16k  16 k 1 and c  k (14) 2  14(14)  40  14 2  14 2  40

The x-intercepts of the graph of y  2 x 2  x  5 are –1.4 and 1.9. Therefore, the roots of 2 x 2  x  5  0 are –1.4 and 1.9.

 40 (b) ∵ The graph of y  x 2  14 x  40 cuts the xaxis at A and B. ∴ The x-coordinates of A and B are the roots of the equation x 2  14 x  40  0 .

18. (a) x y

–2 –4

–1 1

0 4

1 5

2 4

3 1

4 –4

x 2  14 x  40  0 ( x  4)( x  10)  0 x  4  0 or x  10  0 x  4 or x  10 ∴ The coordinates of B are (10, 0).

(b)

(c) AB = (10 – 4) units = 6 units Area of ABC =

6  40 sq. units  120 sq. units 2 21. Let x be the smaller integer, then (x + 5) is the larger integer.

The x-intercepts of the graph of y = 4 – x(x – 2) are – 1.2 and 3.2. Therefore, the roots of 4 – x(x – 2) = 0 are –1.2 and 3.2.

2 x 2  ( x  5) 2  71 2 x 2  x 2  10 x  25  71 ∴

x2 19. ∵ The graph of y    x  2 touches the x-axis at 8

x  16  0 or x  6  0 x  16 or x  6 (rejected)

P. ∴

The x-coordinate of P is the root of the equation



x 2  10 x  96  0 ( x  16)( x  6)  0

x2  x20. 8

When x = 16, x + 5 = 21 ∴ The two integers are 16 and 21.

31

NSS Mathematics in Action 4A Full Solutions

x( x  2)  180  (3x  2)

(b) Let x cm be the length of AB, then BC = (3x – 5) cm. CE = CD = x cm BE = BC – CE = [(3x – 5) – x] cm = (2x – 5) cm ∵ Area of BEDF = 42 cm2

x  2 x  3 x  182 2

22.

23. (a)

x 2  x  182  0 ( x  14)( x  13)  0 x  14  0 or x  13  0 x  14 or x  13 (rejected)

( 2 x  5) x  42

∴

2 x 2  5 x  42 2 x 2  5 x  42  0

 7   ( x  6)  x       0 2    7  ( x  6) x    0 2  7  ( x  6)  2 x    0 2  ( x  6)(2 x  7)  0

From (a),

x6

or

x

7 (rejected) 2

When x = 6, BC = [3(6) – 5] cm = 13 cm ∴ The length of ABCD is 13 cm and its width is 6 cm.

1 x( x  1)  ( x  3) 2 2 x( x  1)  2( x  3) 2

2 x 2  12 x  7 x  42  0

x 2  x  2 x 2  12 x  18

24.

x 2  11x  18  0

2 x 2  5 x  42  0

( x  9)( x  2)  0 x  9  0 or x  2  0 x  9 or

x  2 (rejected)

Perimeter of the L-shaped figure  perimeter of the rectangle  2[ x  ( x  1)]cm  2[9  (9  1)]cm  34 cm 25. (a) The x-intercepts of the graph of y  x 2  20 x  40 are 2 and 18. Therefore, the roots of x 2  20 x  40  0 are 2 and 18. (b) (i) When T = 20, we have

20  0.25 x 2  5 x  30 0.25 x 2  5 x  10  0 4(0.25 x 2  5 x  10)  0 x 2  20 x  40  0 From (a), x  2 or x  18 After 2 hours, that is, at 8:00 p.m., the temperature in the greenhouse first falls to 20°C. (ii) The algebraic method gives the exact answer. 26. Let  be the root of ax 2  bx  c  0 . If  is positive, then a 2  b  c  0 . ∴ The equation does not have a positive root. 27. Let  and  be the roots of x 2  bx  c  0 .

c  (  )(  )    0

∴

32

It is not possible that c is positive.

1 Quadratic Equations in One Unknown (I) Level 2

(2 x  1)(3x  1)  4( x  1)(2 x  1)  0 (2 x  1) (3x  1)  4( x  1)  0

x(6 x  5)  4 6 x 2  5x  4

28.

(2 x  1)( x  3)  0 2x 1  0

33.

6x2  5x  4  0 (2 x  1)(3x  4)  0 2x  1  0 x

x

or 3x  4  0 1 or 2

x

4 3

( x  2) (2 x  1)  3( x  4)  0 ( x  2)( x  13)  0

34.

 (12)  (12)  4( 4)(7) 2



x  2  0 or  x  13  0

2( 4)

x  2 or

12  32  3  2  or  8 2  

2( x 2  12 )  4( x  1)  x 2x 2  2  4x  4  x

x 2  2 x  1  3x  6  0 x 2  5x  7  0

2x 2  5x  6  0

35.

 5  5 2  4(1)(7) x 2(1)  ∵ ∴

x

5 3 2



 3 is not a real number.

36. (a)

2 x 2  3x  8 x  12  3x  2

2(2) 5  73 4

(2 y  3)(2 y  1)  0 2y  3  0

or 2 y  1  0 3 1 y or y 2 2

2 x 2  8 x  10  0 x2  4x  5  0

(b) From (a), we have

1 3  2 2 5 x  1 1 x 5

( x  1)( x  5)  0

5x 

x  1  0 or x  5  0 x   1 or x5

( x  1)( x  2) x  3 2 3x  ( x  1)( x  2) 2

32.

 (5)  (5) 2  4(2)(6)

4 y2  4 y  3  0

The equation has no real roots.

( 2 x  3)( x  4)  3x  2

31.

x  13

2( x  1)( x  1)  4( x  1)  x

( x  1) 2  3( x  2)  0

30.

x3

( x  2)(2 x  1)  3( x  2)( x  4) ( x  2)(2 x  1)  3( x  2)( x  4)  0

4 x 2  12 x  7  0 x

1 or 2

( x  2)(2 x  1)  (3x  6)( x  4)

4 x( x  3)  7  0

29.

or  x  3  0

3x 2  x 2  x  2 x  2

37. (a)

2 x 2  3x  2  0 ( x  2)(2 x  1)  0 x20

or 2 x  1  0 1 x   2 or x 2

or

1 1  2 2 5x  1 1 x 5

5x 

or or

12 x 2  17 x  6  0 (3x  2)(4 x  3)  0 3x  2  0 or 4 x  3  0 2 3 x or x 3 4

17 1 ( w  1)  12 2 12( w  1) 2  17( w  1)  6 ( w  1) 2 

(b)

12( w  1) 2  17( w  1)  6  0 From (a), we have

w 1 

2 3

w 33

or 1 3

or

w 1 

3 4

w

1 4

NSS Mathematics in Action 4A Full Solutions

3x 2  8 x  3

(b)

3x  8 x  3  0 2

3x 2  8 x  3  0

( x  3)(3 x  1)  0 38. (a) x30

or 3 x  1  0 1 x   3 or x 3

From (a), the roots are –3 and

1 . 3

∴ The roots of the required equation are –(–3) and



1 1 , i.e. 3 and  . 3 3

The required quadratic equation is

  1  ( x  3)  x       0 3    1  ( x  3) x    0 3  1  ( x  3)  3 x    0 3  ( x  3)(3 x  1)  0 3x 2  9 x  x  3  0 3x 2  8 x  3  0 1  x  12 x 2  0 12 x 2  x  1  0 39. (a) (3 x  1)(4 x  1)  0 3x  1  0

or 4 x  1  0 1 1 x or x 3 4

(b)

x (12 x  1)  1 12 x 2  x  1 1  x  12 x 2  0 From (a), the roots are 

1 1 and . 4 3

3 ∴ The roots of the required equation are 1 and  3 3 1 , i.e. –9 and 12. 4 The required quadratic equation is

 x  (9) ( x  12)  0 ( x  9)( x  12)  0

x  9 x  12 x  108  0 2

x 2  3 x  108  0 40.

( (

p 

q )(

p 

p) ( q) 2

q)

2

 pq (

Since p and q are positive integers and p > q, q )( p  q ) is also a positive integer.

p 

41. (a)

34

1 Quadratic Equations in One Unknown (I)

    

L et i. e. ( 2)



( 1 )

10 9

:

0.4 16, 44. (a) 0.4 16 66 6  4.16 6 x –1 0 1 66 2 36 3.75 37 5  y 12 –8 –6 4 9 002 

  0.416

∴



   

5 12

5 12

 is a root of 0.416

∵

60 x  x  k  0 . 2

 5  60   12 

∴

2

 5   k 0  12  k   10



60 x 2  x  (10)  0 60 x 2  x  10  0 (b) (12 x  5)(5 x  2)  0

2 x ( x  3)  1  x

12 x  5  0

5 x or 12 ∴ The other root is 

2x 2  6x  1  x

(b)

or 5 x  2  0

2x 2  5x  1  0

2 x 5

4 x 2  10 x  2  0 ∵ The x-intercepts of the graph of y  4 x 2  10 x  2 are –0.2 and 2.7.

2 . 5

∴ The roots of 4 x 2  10 x  2  0 are –0.2 and 2.7. ∴ The roots of 2x(x – 3) = 1 – x are also –0.2 and 2.7.

42. Since r is a root of x 2  3 x  5  0, we have r 2  3r  5  0 .

r 5  3r 4  5r 3  2r 2  6r

45. The length of the base = (2x – 1) cm The length of the hypotenuse = [40 – x – (2x – 1)] cm = (41 – 3x) cm By Pythagoras’ theorem,

 r 3 ( r 2  3r  5)  2( r 2  3r  5)  10  r 3 (0)  2(0)  10  10

x 2  (2 x  1) 2  (41  3 x) 2 x 2  (4 x 2  4 x  1)  1681  246 x  9 x 2

43. (a) x y

–2 10

–1 3

0 –2

1 –5

2 –6

3 –5

4 –2

4 x 2  242 x  1680  0

5 3

2 x 2  121x  840  0 ( x  8)(2 x  105)  0 x  8  0 or 2 x  105  0 x  8 or

46. (a)

x2  2x  1 2 x 2  4x  2

(b)

105 (rejected) 2

25 x 2  25 x  6  0 (5 x  1)(5 x  6)  0 5x  1  0

or 5 x  6  0 1 6 x or x 5 5

(b)

120 000(1  r %) 2  120 000(1  r %)  28 800

x  4x  2  0 2

25(1  r %) 2  25(1  r %)  6  0

∵ The x-intercepts of the graph of y  x2  4x  2 are –0.4 and 4.4. ∴ The roots of x 2  4 x  2  0 are –0.4 and 4.4. ∴ The roots of

x

From (a), we have

6 1 or 1  r %   (rejected) 5 5 1 r%  5 r  20

1  r% 

x2  2 x  1 are also –0.4 and 2

4.4.

35

 

NSS Mathematics in Action 4A Full Solutions 47. (a) Consider PST and PQR. PTS = PRQ (corr. s, ST // QR) PST = PQR (corr. s, ST // QR) SPT = QPR (common ) ∴ PST  PQR (AAA) PS = PQ – SQ  (16  x ) cm

PS ST = QR PQ ST =

1  AP  AD 2 (b) Area of ADP 1  ( x  8) cm 2 2  4 x cm 2 1   BP  BQ 2 1 Area of BPQ  (8  x )(8  x ) cm 2 2 1  (8  x ) 2 cm 2 2 

(corr. sides,  s)

16  x 3   12 cm   12  x  cm 16 4  

(b) ∵ The area of trapezium TSQR is 72 cm2.

 1  3    12  x   12 x  72 2  4    (48  3x)  48 x  576 (96  3x) x  576

2 Area of △ ADP   area of △ BPQ 3 2 1 4 x   (8  x) 2 3 2 12 x  (8  x) 2

(32  x) x  192

∴

32 x  x 2  192 x 2  32 x  192  0 ( x  8)( x  24)  0

12 x  64  16 x  x 2

x  8  0 or x  24  0 x  8 or

x 2  28 x  64  0

x  24 (rejected)

x

Area of the original paper 48. (a)

 (3 x  1)(5 x  2) cm 2



 (15 x 2  5 x  6 x  2) cm 2  (15 x 2  x  2) cm 2

28  528 28  528 or (rejected) 2 2  2.51 (cor. to 3 sig. fig.)

3 4 cm 2  6 cm 2 2 Area of the shaded region



50. Let x m/s be the speed of Tom, then the speed of Mary is (x – 2) m/s.



 (15 x 2  x  2)  8  6 cm 2  (15 x  x  50) cm (b) (i)

(6 x  27) 2  [6( x  2)  27]2  60 2

2

(6 x  27) 2  (6 x  15) 2  3600

∵ The area of the shaded region is 186 cm2.

(2 x  9) 2  (2 x  5) 2  400

15 x 2  x  50  186

∴

4 x 2  36 x  81  4 x 2  20 x  25  400

15 x 2  x  236  0

8 x 2  56 x  294  0

15 x  x  236  0 ( x  4)(15 x  59)  0 (ii) x  4  0 or 15 x  59  0 2

x  4 or

28  528 2



Area of a triangle =

2

 ( 28)  ( 28) 2  4(1)(64) 2(1)

x

4 x 2  28 x  147  0 (2 x  7)(2 x  21)  0 2 x  7  0 or 2 x  21  0 x  3.5 or x  10.5 (rejected)

59 15 (rejected)

When x = 3.5, x – 2 = 1.5 ∴ The speed of Tom is 3.5 m/s and the speed of Mary is 1.5 m/s.

When x = 4, BC = (3x + 1) cm = 13 cm ∴ JK = (13 – 4 – 4) cm = 5 cm AB = (5x – 2) cm = 18 cm ∴ GH = (18 – 3 – 3) cm = 12 cm

51. (a) (i)

49. (a)

Each person sends 3 Christmas cards. There are 4 persons sending Christmas cards. ∴ The total number of Christmas cards that are sent  3  4  12

(ii) Each person sends (n – 1) Christmas cards. There are n persons sending Christmas cards. ∴ The total number of Christmas cards sent 36

1 Quadratic Equations in One Unknown (I)  n ( n  1)

(b) Let n be the number of students in the class.

n(n  1)  1260

n  n  1260  0 (n  36)(n  35)  0 n  36  0 2

or n  35  0

n  36 or

n  35 (rejected)

∴ The number of students in the class is 36. 52. (a) R(p, q) is a point on the graph of y = x2 – 6x + 10. ∴ q = p2 – 6p + 10

2( p  q)  12 pq 6

p  ( p  6 p  10)  6 2

(b)

p2  5 p  4  0 ( p  1)( p  4)  0 p  1  0 or p  4  0 p  1 or

p4

53. (a) ∵ k is a root of x 2  x  c  0. 2 ∴ k k c 0

c  k 2  k x2  x  k 2  k  0 (b)

x 2  x  k (k  1)  0 ( x  k )( x  k  1)  0 x  k  0 or x  k  1  0 x  k or x  k  1 ∴ The other root of the equation in (a) is –k – 1.  2 x 2  16 x  24  0 x 2  8 x  12  0 ( x  2)( x  6)  0

54. (a)

x  2  0 or x  6  0 x  2 or x6 ∴ The coordinates of M and N are (2, 0) and (6, 0) respectively. (b) Let (x1, y1) be the coordinates of P and (x2, y2) be the coordinates of Q. ∵ Area of PMN = area of QMN = 12 sq. units ∴

1 1 (6  2) y1  (6  2) y 2  12 2 2 y1  y 2  6

When y  6 , 6  2 x 2  16 x  24

2 x 2  16 x  30  0 x 2  8 x  15  0 ( x  3)( x  5)  0 ∴

x  3  0 or x  3 or

x 5  0 x5

x1  3 and

x2  5

∴ The coordinates of P and Q are (3, 6) and (5, 6) respectively.

37

NSS Mathematics in Action 4A Full Solutions

[ x  ( 1)][ x  ( 7)]  0 ( x  1)( x  7)  0

(c) We can find a point R on the graph with y-coordinate – 6. ∴ Area of RMN =

x 2  8x  7  0

1 (6  2)[0  (6)] sq. units 2 = 12 sq. units Multiple Choice Questions (p. 1.55) 1. Answer: C   0.344 444  0 .3 4 4   0.343 434  0.3   0.655 555  0 .6 5 5   0.656 565  0 .6

4   0.34   0.65 5   0 .6  . ∴ 0.3

2.

Answer: B I.

II.

7 can be converted into a recurring decimal which 6  . is 1.16  and 3.7  . Consider two recurring decimals 0.37  The difference of 3.7 and

  3 .7   0 .3 7  0 .3 7  3 .4

∴ The difference of two recurring decimals may not be a recurring decimal.

2 and 3 3 2 The product of 3 and 2  

III. Consider two fractions

3 . 2 2 3  3 2 1

∴ The product of two fractions may be an integer. ∴ III is true only. 3.

Answer: D I.  x 2  5 x  9  0 is in the general form. II.

( x  1) 2  2( x  1)  3  0 is not in the general form.

III.

2 2 x  x  0 is in the general form. 5

∴ I and III are true only. 4.

Answer: C

2x 2  x  6  0 (2 x  3)( x  2)  0 2x  3  0

or x  2  0 3 x   or x2 2

∴ The roots are  5.

3 and 2 . 2

Answer: A The required quadratic equation is

38

1 Quadratic Equations in One Unknown (I) 6.

Answer: D

Investigation Corner (p. 1.58)

[ x  ( 5)]( x  2)  0 ( x  5)( x  2)  0

1.

x  5 x  2 x  10  0 2

(a) Checking: When x 

3 , 4

x 2  3 x  10  0

7.

0

Answer: B ∵  is a root of 2 x 2  3 x  5  0 .

When x  

2  2  3  5  0

5 , 3

4  2  6   5  2(2  2  3  5)  5

∴

 2( 0)  5 5 8.

9 3  11   15 16 4

12 x 2  11x  15  12 

∴ y  x 2  3 x  10 may represent the graph.

12 x 2  11x  15  12 

25 5   11      15 9 3 

0

Answer: A ∵ 3 is a root of 5( x  k ) 2  20  0 .

∴ The roots of the equation are

5(3  k ) 2  20  0

3 5 and  . 4 3

(b)

(9  6k  k )  4  0 2

Step 1 :

k 2  6k  5  0 (k  1)(k  5)  0

∴

x 2  23 x  288  0 Step 2 : ( x  9)( x  32)  0 Step 3 : ( 24 x  9)(24 x  32)  0

k  1  0 or k  5  0 k  1 or 9.

1  x 2  23 x  12  24  0

k 5

Step4 : ( 24 x  9)( 24 x  32)  0 24 x  9  0

Answer: D

( x  3)( x  4)  (a  3)(a  4)

x

x 2  7 x  12  a 2  7 a  12 x 2  7 x  a 2  7a  0

9 or 24

∴

( x 2  a 2 )  7( x  a )  0 ( x  a)( x  a )  7( x  a)  0

3 x 8

( x  a )( x  a  7)  0 x  a  0 or x  a  7  0 x  a or x 7a

2.

HKMO (p. 1.57) 1.

or 24 x  32  0

AB BC  BC EB W 1  1 W 1

or

x

4 3

Consider ax 2  bx  c  0 , where a  0 . Step 1: Multiply the constant term by the leading coefficient a and change the leading coefficient a to 1.

1  x 2  bx  a  c  0 x 2  bx  ac  0......(*) Step 3:

Replacing x by ax in (*), we have

(ax ) 2  b(ax )  ac  0

W (W  1)  1

a 2 x 2  abx  ac  0

W  W 1  0 2

ax 2  bx  c  0



1 5 2



1 5 1 5 or (rejected) 2 2

(a  0)

Therefore, solving the quadratic equation (*) is equivalent to solve the quadratic equation ax 2  bx  c  0 . Hence, Angel’s approach can be used to solve any quadratic equation ax 2  bx  c  0 .

 (1)  ( 1)  4(1)(1) W 2(1) 2

39

x

32 24

NSS Mathematics in Action 4A Full Solutions

40