4ach04(more About Equations)

Certificate Mathematics in Action Full Solutions 4A

4 More about Equations

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

Follow-up Exercise

Checking: When x = –2,

p. 162 1.

2.

3.

( −2) 2 −4 ×( −2) −8 = 2

By substituting x2 = u into the equation x4 – 10x2 + 9 = 0, we have u2 – 10u + 9 = 0 (u – 1)(u – 9) = 0 u = 1 or u = 9 ∵ x2 = u ∴ x2 = 1 or x2 = 9 x = ± 1 or x = ± 3 ∴ The real roots of the equation are –3, –1, 1 and 3.

2

2.

4

∵

x2 = u

∴

x2 =

1 4

or

2.

Checking:

1.

∴ 3.

= 22

x2 – 4x – 8 = 4 x2 – 4x – 12 = 0 (x + 2)(x – 6) = 0 x = –2 or

83

x – 3 x −2 = 0 x = 3 x −2

1 1 , and 2 2

(

)

x2 = 3 x −2 2 x2 = 9x – 18 2 x – 9x + 18 = 0 (x – 3)(x – 6) = 0 x = 3 or x = 6 Checking: When x = 3, x – 3 x −2 = 3 – 3 3 −2 = 0 When x = 6, x – 3 x −2 = 6 – 3 6 −2 = 0 ∴ The real roots of the equation are 3 and 6. 4.

x+2

x −2 = 10 x −2 = 10 – x

)

2 = (10 – x)2 x −2 4x – 8 = 100 – 20x + x2 x2 – 24x + 108 = 0 (x – 6)(x – 18) = 0 x=6 or x = 18 Checking: When x = 6, x + 2 x −2 = 6 + 2 6 −2 = 10

x −4 x −8 = 2 2

x =1+3 1 =4

x = 1or x = –4 (rejected) x=1 The real root of the equation is 1.

(2

2

  2  x −4 x −8   

When x = 1, x + 3

Alternative Solution By substituting x = u into the equation x + 3 we have u2 + 3u = 4 u2 + 3u – 4 = 0 (u – 1)(u + 4) = 0 u = 1 or u = –4 ∵ = u x

2

p. 163

)

When x = 16, x + 3 x = 16 + 3 16 = 28 ≠ 4 ∴ The real root of the equation is 1.

∴

By substituting x3 = u into the equation x6 + 2x3 + 1 = 0, we have u2 + 2u + 1 = 0 (u + 1)2 = 0 u = –1 ∵ x3 = u ∴ x3 = –1 x = –1 ∴ The real root of the equation is –1.

(

x

(x – 4) = −3 x 2 x – 8x + 16 = 9x x2 – 17x + 16 = 0 (x – 1)(x – 16) = 0 x = 1 or x = 16 2

2

1 or x2 = 4 4 1 x=± or x = ± 2 2

x =4

2

u=4

The real roots of the equation are –2, −

x+3

x–4=–3

By substituting x = u into the equation 4x – 17x + 4 = 0, we have 4u2 – 17u + 4 = 0 (4u – 1)(u – 4) = 0 u=

4.

When x = 6, x 2 −4 x −8 = 6 2 −4 ×6 −8 = 2 ∴ The real roots of the equation are –2 and 6.

By substituting x2 = u into the equation x4 + 3x2 – 4 = 0, we have u2 + 3u – 4 = 0 (u – 1)(u + 4) = 0 u = 1 or u = – 4 ∵ x2 = u ∴ x2 = 1 or x2 = –4 (rejected) x=±1 ∴ The real roots of the equation are –1 and 1.

∴

x 2 −4 x −8 =

x=6

x = 4,

4 More about Equations y = 4x.

When x = 18, x + 2 x −2 = 18 + 2 18 −2 = 26 ≠ 10 ∴ The real root of the equation is 6.

(b) ∵ ∴

p.168 (c) ∵

For questions 1 to 4, refer to the graph below:

x2 + 3x – 2 = 0 x2 = –3x + 2 The equation of the required straight line is y = –3x + 2. 3x2 + 6x – 1 = 0 3x2 = –6x + 1 x2 = –2x +

∴

The equation of the required straight line is y = –2x +

2.

(a) ∵ ∴ ∴ (b) ∵

1.

x y ∵ ∴ 2.

1 –2

∴

3 4

–2 2

–1 0

0 –2

The two graphs do not intersect. The simultaneous equations have no real solutions.

y = x2 y=x–3 x2 = x – 3 x2 – x + 3 = 0 The quadratic equation that can be solved is x2 – x + 3 = 0. y = 3x2 y = 4x + 1 3x2 = 4x + 1 3x2 – 4x – 1 = 0 The quadratic equation that can be solved is 3x2 – 4x – 1 = 0.

3.

–x + y = 2 x y ∵ ∴

–1 1

0 2

1 3

The two graphs intersect at (–1, 1) and (3, 5). The solutions of the simultaneous equations are (–1, 1) and (3, 5).

x–y–4=0 x y ∵ ∴

1 –3

2 –2

3 –1

The two graphs do not intersect. The simultaneous equations have no real solutions.

(a) ∵ ∴

p. 175 1.

1 . 3

(c) y = –2x2 + x ……(1) 2x + y = 1 ……(2) By substituting (1) into (2), we have 2x + (–2x2 + x) = 1 2x2 – 3x + 1 = 0 ∴ The quadratic equation that can be solved is 2x2 – 3x + 1 = 0.

y = –2x – 2

∵ ∴

4.

2 1

The two graphs intersect at only one point (2, 1). The solution of the simultaneous equations is (2, 1).

x y

3.

∴

y = 3x – 5

1 3

(a) ∵ ∴

2

x – 4x = 0 x2 = 4x The equation of the required straight line is

x2 – 2x – 3 = 0 x2 – 2x = 3 The corresponding simultaneous equations are

y = x 2 − 2 x .  y = 3 Draw the straight line y = 3 on the graph of

84

Certificate Mathematics in Action Full Solutions 4A y = x2 – 2x. From the graphs, the roots of x2 – 2x – 3 = 0 are –1 and 3. (b) ∵ ∴

3.

x2 + 2x + 1 = 0 x2 – 2x + 4x +1 = 0 x2 – 2x = –4x –1 The corresponding simultaneous equations are

From (2), we have y = 4x + 7……(3) By substituting (3) into (1), we have 2x2 + 4x + 4x + 7 + 1 = 0 2x2 + 8x + 8 = 0 x2 + 4x + 4 = 0 (x + 2) 2 = 0 x = –2 By substituting x = –2 into (3), we have y = 4(–2) +7 = –1 ∴ The solution of the simultaneous equations is (–2, –1).

y = x 2 − 2 x .  y = −4 x −1 Draw the straight line y = –4x –1 on the graph of y = x2 – 2x. From the graphs, the root of x2 + 2x + 1 = 0 is –1. (c) ∵

2x2 + x – 6 = 0 2x + x – 5x – 6 = – 5x 2x2 – 4x – 6 = – 5x

4.

2

5 x 2 5 x2 – 2x = − x + 3 2

The corresponding simultaneous equations are

 y = x 2 − 2x  . 5  y = − x + 3  2  Draw the straight line y = −

5 x + 3 on the graph 2

5.

p.183

x 2 + y 2 = 20  (1)   ( 2) y = 6 − x By substituting (2) into (1), we have x2 + (6 – x)2 = 20 2 x + 36 – 12x + x2 = 20 2x2 – 12x + 16 = 0 x2 – 6x + 8 = 0 (x – 2)(x – 4) = 0 x–2=0 or x – 4 = 0 x=2 or x=4 By substituting x = 2 into (2), we have y = 6 – 2 = 4 By substituting x = 4 into (2), we have y = 6 – 4 = 2 ∴ The solutions of the simultaneous equations are (2, 4) and (4, 2).

2.

p.185 1.

x − 9 y + 6 = 0  (1)  2  ( 2) 3 y = x From (1), we have x = 9y – 6 ……(3) By substituting (3) into (2), we have 3y2 = 9y – 6 3y2 – 9y + 6 = 0 y2 – 3y + 2 = 0 (y – 1)(y – 2) = 0 y = 1 or y = 2 By substituting y = 1 into (3), we have x = 9(1) – 6 = 3 By substituting y = 2 into (3), we have x = 9(2) – 6 = 12 ∴ The solutions of the simultaneous equations are (3, 1) and (12, 2).

85

y = 2 x 2 + x +1  (1)  5 x + y −1 = 0  ( 2) From (2), we have y = –5x + 1……(3) By substituting (3) into (1), we have –5x + 1 = 2x2 + x + 1 2x2 + 6x = 0 Consider the discriminant of 2x2 + 6x = 0. ∆ = 62 – 4(2)(0) = 36 > 0 ∴ 2x2 + 6x = 0 has two distinct real roots. ∴ The simultaneous equations have two real solutions.

of y = x2 – 2x. From the graphs, the roots of 2x2 + x – 6 = 0 are –2 and 1.5.

1.

x 2 + y 2 =1  (1)  2 x − y + 7 = 0  ( 2) From (2), we have y = 2x + 7……(3) By substituting (3) into (1), we have x2 + (2x + 7) 2 = 1 2 x + 4x2 + 28x + 49 = 1 5x2 + 28x + 48 = 0 Consider the discriminant of 5x2 + 28x + 48 = 0. ∆ = 282 – 4(5)(48) = –176 < 0 ∴ 5x2 + 28x + 48 = 0 has no real roots. ∴ The simultaneous equations have no real solutions.

x2 – 2x – 3 = −

∴

2 x 2 + 4 x + y +1 = 0  (1)   ( 2) 4 x − y + 7 = 0

2.

Let x cm and y cm be the length and the width of the rectangle respectively. ∵ The perimeter of the rectangle is 46 cm. ∴ 2(x + y) = 46 x + y = 23 y = 23 – x ……(1) ∵ The area of the rectangle is 120 cm2. ∴ xy = 120 ……(2) By substituting (1) into (2), we have x(23 – x) = 120 23x – x2 = 120 x2 – 23x + 120 = 0 (x – 8)(x – 15) = 0 x=8 or x = 15 By substituting x = 8 into (1), we have y = 23 – 8 = 15 By substituting x = 15 into (1), we have y = 23 – 15 = 8 ∴ The dimensions of the rectangle are 8 cm × 15 cm. (a) ∵ The length of the string is 14 cm. ∴ PT + TQ = 14 cm x + y = 14 PT 2 + TQ 2 = PQ 2 (Pyth. theorem) x2 + y2 = 102 x2 + y2 = 100 ∴ The required simultaneous equations are

4 More about Equations ∴

(x – 2)(x + 1)(x + 5) = 0 x – 2 = 0 or x + 1 = 0 or x + 5 = 0 1 5 x = 2 or x= − or x = −

(a) ∵ ∴ ∵

f(1) = 14 + 13 – 14(1)2 – 8(1) + 48 = 28 ≠ 0 x – 1 is not a factor of f(x). f(–1) = (–1)4 + (–1)3 – 14(–1)2 – 8(–1) + 48 = 42 ≠ 0 x + 1 is not a factor of f(x). f(2) = 24 + 23 – 14(2)2 – 8(2) + 48 = 0 x – 2 is a factor of f(x). f(–2) = (–2)4 + (–2)3 – 14(–2)2 – 8(–2) + 48 = 16 ≠ 0 x + 2 is not a factor of f(x). f(3) = 34 + 33 – 14(3)2 – 8(3) + 48 = 6 ≠ 0 x – 3 is not a factor of f(x). f(–3) = (–3)4 + (–3)3 – 14(–3)2 – 8(–3) + 48 =0 x + 3 is a factor of f(x). The required linear factors of f(x) are x – 2 and x + 3.

 (1) x + y = 14 .  2 2  ( 2) x + y = 100  (b) From (1), we have x + y = 14 y = 14 – x ……(3) By substituting (3) into (2), we have x2 + (14 – x)2 = 100 x2 + 196 – 28x + x 2 = 100 2x2 – 28x + 96 = 0 x2 – 14x + 48 = 0 (x – 6)(x – 8) = 0 x = 6 or x = 8 By substituting x = 6 into (3), we have y = 14 – 6 = 8 By substituting x = 8 into (3), we have y = 14 – 8 = 6 ∴ 3.

2.

∴ ∵ ∴ ∵ ∴ ∵ ∴ ∵

x = 6 x = 8 The solutions are  or  . y = 8 y = 6

Let x be the tens digit and y be the units digit of the original number. ∴ The original number is 10x + y, and the reversed number is 10y + x. ∵ The number is increased by 36 when the digits are reversed. ∴ (10y + x) – (10x + y) = 36 –9x + 9y = 36 –x + y = 4 y = x + 4 ……(1) ∵ The product of the digits is 32. ∴ xy = 32 …..(2) By substituting (1) into (2), we have x(x + 4) = 32 x2 + 4 x = 32 2 x + 4 x – 32 = 0 (x – 4)(x + 8) = 0 x = 4 or x = –8 (rejected) By substituting x = 4 into (1), we have y = 4 + 4 = 8 ∴ The number is 48.

∴ ∴

(b) ∵ x – 2 and x + 3 are factors of f(x). ∴ (x – 2)(x + 3) is also a factor of f(x). Divide f(x) by (x – 2)(x + 3), i.e. x2 + x – 6. x2 − 8 x 2 + x − 6 x4 + x3 − 1 4x2 − 8x + 4 8 x4 + x3 − 6 x 2 − 8 x2 − 8x + 4 8 − 8 x2 − 8x + 4 8

∴ ∵ ∴

±2

(a) The possible factors of f(x) are x ± 1, x ± 2, x ± 5 and x ± 10. ∵ f(1) = 13 + 4(1)2 – 7(1) – 10 = –12 ≠ 0 ∴ x – 1 is not a factor of f(x). ∵ f(–1) = (–1)3 + 4(–1)2 – 7(–1) – 10 = 0 ∴ x + 1 is a factor of f(x). ∵ f(2) = 23 + 4(2)2 – 7(2) – 10 = 0 ∴ x – 2 is a factor of f(x). ∵ f(–2) = (–2)3 + 4(–2)2 – 7(–2) – 10 = 12 ≠ 0 ∴ x + 2 is not a factor of f(x). ∵ f(5) = 53 + 4(5)2 – 7(5) – 10 = 180 ≠ 0 ∴ x – 5 is not a factor of f(x). ∵ f(–5) = (–5)3 + 4(–5)2 – 7(–5) – 10 = 0 ∴ x + 5 is a factor of f(x). ∴ The factors of f(x) are x – 2, x +1 and x + 5. ∴ A factor of f(x) is x – 2, x +1 or x + 5.(any one) (b) ∵ ∴ (c) ∵

The factors of f(x) are (x – 2), (x +1) and (x + 5). 1)( x + 5) f(x) = ( x −2)( x +

2

3.

2x4 + x3 – 3x2 – x + 1 = (2x4 – 3x2 + 1) + (x3 – x) = [2(x2)2 – 3x2 + 1] + x(x2 – 1) = (2x2 – 1)(x2 – 1) + x(x2 – 1) = (x2 – 1)(2x2 – 1 + x) = (x + 1)(x – 1)(2x2 + x – 1) = (x + 1)(x – 1)(2x – 1)(x + 1) = (x + 1)2(x – 1)(2x – 1) ∵ 2x4 + x3 – 3x2 – x + 1 = 0 ∴ (x + 1)2(x – 1)(2x – 1) = 0 (x + 1)2 = 0 or x – 1 = 0 or 2x – 1 = 0 1 1 or x= − x = 1 or x = 2

4.

8(x + 1)3 – (x + 2)3 = [2(x + 1)]3 – (x + 2)3 = [2(x + 1) – (x + 2)]{[2(x + 1)]2 + 2(x + 1)(x + 2) + (x + 2)2} = (2x + 2 – x – 2)(4x2 + 8x + 4 + 2x2 + 6x + 4 + x2 + 4x + 4) = x(7x2 + 18x + 12)

p.192 1.

x4 + x3 – 14x2 – 8x + 48 = (x – 2)(x + 3)(x2 – 8) f(x) = 0 (x – 2)(x + 3)(x2 – 8) = 0 x – 2 = 0 or x + 3 = 0 or x2 – 8 = 0 3 or x= 2 or x= − x=

∵ ∴ ∴

8(x + 1)3 – (x + 2)3 = 0 x(7x2 + 18x + 12) = 0 x = 0 or 7x2 + 18x + 12 = 0 x=0

or

x=

−18 ± 18 2 − 4(7)(12 ) 2(7)

f(x) = 0

86

Certificate Mathematics in Action Full Solutions 4A

= ∴

we have u2 + 9u + 8 = 0 (u + 1)(u + 8) = 0 u = –1 or u = –8 ∵ x3 = u ∴ x3 = –1 or x3 = –8 x = –1 or x = –2 ∴ The real roots of the equation are –2 and –1.

−18 ± −12 (rejected) 14

x=0

Exercise 8.

Exercise 4A (p.164)

2

  2 2  x +2 x +1  =3  

Level 1 1.

2.

3.

4.

5.

6.

7.

x2 + 2x + 1 = 9 x2 + 2x – 8 = 0 (x + 4)(x – 2) = 0 x = –4 or

By substituting x2 = u into the equation x4 – 17x2 + 16 = 0, we have u2 – 17u + 16 = 0 (u – 1)(u – 16) = 0 u = 1 or u = 16 ∵ x2 = u ∴ x2 = 1 or x2 = 16 x = ± 1 or x = ± 4 ∴ The real roots of the equation are –4, –1, 1 and 4. By substituting x2 = u into the equation x4 – 26x2 + 25 = 0, we have u2 – 26u + 25 = 0 (u – 1)(u – 25) = 0 u=1 or u = 25 ∵ x2 = u ∴ x2 = 1 or x2 = 25 x = ± 1 or x = ± 5 ∴ The real roots of the equation are –5, –1, 1 and 5.

Checking: When x = –4,

3

x3 – 8x2 + 7x = 0 x(x2 – 8x + 7) = 0 x(x – 1)(x – 7) = 0 x=0 or x = 1 or x = 7 ∴ The real roots of the equation are 0, 1 and 7. x5 – 6x4 + 5x3 = 0 x3(x2 – 6x + 5) = 0 x3(x – 1)(x – 5) = 0 x=0 or x = 1 or x = 5 ∴ The real roots of the equation are 0, 1 and 5. By substituting x3 = u into the equation x6 + 9x3 + 8 = 0,

87

x 2 +2 x +1 =

x 2 +2 x +1 = 2 2 +2( 2) +1 = 3 The real roots of the equation are –4 and 2.

When x = 2, ∴ 9.

x–

x – 12 = 0

x – 12 =

x

2

(x – 12) =

( x)

2

x2 – 24x + 144 = x x2 – 25x + 144 = 0 (x – 9)(x – 16) = 0 x=9 Checking: When x = 9, x –

or

x = 16

x – 12 = 9 –

9 – 12 = –6 ≠ 0

When x = 16, x – x – 12 = 16 – 16 – 12 = 0 ∴ The real root of the equation is 16. Alternative Solution By substituting x = u into the equation x – 0, we have u2 – u – 12 = 0 (u – 4)(u + 3) = 0 u=4 or u = –3 ∵ x =u

By substituting x2 = u into the equation x4 – 12x2 +27 = 0, we have u2 – 12u + 27 = 0 (u – 3)(u – 9) = 0 u = 3 or u = 9 ∵ x2 = u ∴ x2 = 3 or x2 = 9 x = ± 3 or x = ± 3 The real roots of the equation are –3, − 3 , and 3.

x=2

(− 4) 2 +2( − 4) +1 = 3

By substituting x2 = u into the equation x4 – 11x2 – 80 = 0, we have u2 – 11u – 80 = 0 (u – 16)(u + 5) = 0 u = 16 or u = –5 ∵ x2 = u ∴ x2 = 16 or x2 = –5 (rejected) x=±4 ∴ The real roots of the equation are –4 and 4.

∴

x 2 +2 x +1 =3

∴ ∴ 10.

x = 4or x = –3 (rejected) x = 16 The real root of the equation is 16.

x–5

x +6=0

x+6=5

(

x

)

(x + 6) = 5 x 2 x + 12x + 36 = 25x x2 – 13x + 36 = 0 (x – 4)(x – 9) = 0 x=4 or 2

2

Checking: When x = 4, x – 5

x=9

x +6=4–5

4 +6=0

When x = 9, x – 5 x + 6 = 9 – 5 9 + 6 = 0 ∴ The real roots of the equation are 4 and 9.

x – 12 =

4 More about Equations Alternative Solution By substituting x = u into the equation x – 5 = 0, we have u2 – 5u + 6 = 0 (u – 2)(u – 3) = 0 u=2 or u = 3 ∵ x =u ∴ ∴ 11.

∴ x +6

x = 2 or x =3 x = 4 or x=9 The real roots of the equation are 4 and 9.

x +2 = –x + 10

)

2

2

x +2 = (–x + 10) x + 2 = x2 – 20x +100 x2 – 21x + 98 = 0 (x – 7)(x – 14) = 0 x=7 or x = 14

Checking: When x = 7,

x +2 + x =

u=1

7 +2 + 7 = 10

When x = 14, x +2 + x = 14 +2 +14 = 18 ≠ 10 ∴ The real root of the equation is 7. 12.

x–

Level 2 15. By substituting x2 = u into the equation 4x4 + 5x2 – 9 = 0, we have 4u2 + 5u – 9 = 0 (u – 1)(4u + 9) = 0

x +2 + x = 10

(

∴

∆ <0 [– (2k + 4)]2 – 4(1)(k2 + 8) < 0 4k2 + 16k + 16 – 4k2 – 32 < 0 16k – 16 < 0 k<1 A possible value of k is –1 or –2.(or any other reasonable answers)

x +1 = 1

x–1=

x +1 = 0 –

∴

x2 = 1

∴

x=±1 The real roots of the equation are –1and 1.

1 is a root of the equation ax4 + bx2 + c = 0. a(1)4 + b(1)2 + c = 0 a+b+c=0 By substituting x2 = u into the equation ax4 + bx2 + c = 0, we have au2 + bu + c = 0 1 is also a root of the equation au2 + bu + c = 0. ∴ ∆ ≥ 0 b2 – 4ac ≥ 0 Let a = 1, b = –5, c = 4. a + b + c = 1 + (–5) + 4 = 0 b2 – 4ac = (–5)2 – 4(1)(4) = 25 – 16 = 9 ≥ 0 Let a = 3, b = 4, c = –7, a + b + c = 3 +4 + (–7) = 0 b2 – 4ac = 42 – 4(3)( –7) = 100 ≥ 0 ∴ A possible set of values of a, b and c is a = 1, b = –5, c = 4 or a = 3, b = 4, c = –7. (or any other reasonable answers) x+2 2

(2

∵

x −2 = k x −2 = –x + k

)

2

2

x −2 = (–x + k) 4x – 8 = x2 – 2kx + k2 x2 – (2k + 4)x + (k2 + 8) = 0 The equation has no real roots.

x2 = −

9 (rejected) 4

16. By substituting x2 = u into the equation 9x4 – 37x2 + 4 = 0, we have 9u2 – 37u + 4 = 0 (9u – 1)(u – 4) = 0

1 9

or

u=4

∵

x2 = u

∴

x2 =

∴

The real roots of the equation are –2, −

0 +1 = –1 ≠ 1

13. ∵ ∴

14.

or

u=

When x = 3, x – x +1 = 3 – 3 +1 = 1 ∴ The real root of the equation is 3.

9 4

x2 = u

2

Checking: When x = 0, x –

u=−

∵

x +1

(x – 1) = ( x +1 )2 2 x – 2x + 1 = x + 1 x2 – 3x = 0 x(x – 3) = 0 x = 0 or x = 3

or

1 or x2 = 4 9 1 x=± or x = ± 2 3 1 1 , and 3 3

2. 17. By substituting x2 = u into the equation 4x4 – 101x2 + 25 = 0, we have 4u2 – 101u + 25 = 0 (4u – 1)(u – 25) = 0 u=

1 or u = 25 4

∵

x2 = u

∴

x2 =

∴

The real roots of the equation are –5, −

1 or x2 = 25 4 1 x=± or x = ± 5 2 1 1 , and 2 2

5. 18. By substituting x2 = u into the equation 25x4 + 99x2 – 4 = 0, we have 25u2 + 99u – 4 = 0 (25u – 1)(u + 4) = 0 u= ∵

1 25

or

u = –4

x2 = u

88

Certificate Mathematics in Action Full Solutions 4A

1 25 1 x=± 5

(

x2 =

∴

1 1 The real roots of the equation are − and . 5 5

or

x2 = –4 (rejected)

∵

x2 = u

∴

x2 =

∴

or

x2 = 3

7 x=± 2

or

x=± 3

∴

7 , 2

3 .

3

23. By substituting x2 – 5x = u into the equation (x2 – 5x)2 + 8(x2 – 5x) + 16 = 0, we have u2 + 8u + 16 = 0 (u + 4)2 = 0 ∵ u = –4 ∴ x2 – 5x = u x2 – 5x = –4 x2 – 5x + 4 = 0 (x – 1)(x – 4) = 0 x = 1 or x = 4 ∴ The real roots of the equation are 1 and 4. 24. (a) 3x2 + 15x + 2

4 .

= 3(x2 + 5x + 1) + 2

)

2 2 3 x +3 = (3x – 3) 3x + 3 = 9x2 – 18x + 9 9x2 – 21x + 6 = 0 3x2 – 7x + 2 = 0 (3x – 1)(x – 2) = 0

or

3x2 + 15x + 2 2

x=2

3 x +3 + 2 =

1  3  + 3 + 2 = 4 3 

u=1

∴ 22.

3 x −3 + 5 = 2x 3 x −3 = 2x – 5

89

3( 2) +3 + 2 = 5

3x – 1 = 3(2) – 1 = 5 The real root of the equation is 2.

x 2 +5 x +1 – 3

(b) By substituting x 2 +5 x +1 = u into the equation 3x2 + 15x + 2 x 2 +5 x +1 = 2, we have 3u2 + 2u – 3 = 2 (By (a)) 3u2 + 2u – 5 = 0 (u – 1)(3u + 5) = 0

1  3x – 1 = 3   – 1 = 0 ≠ 4 3 3 x +3 + 2 =

x 2 +5 x +1 – 3

x 2 +5 x +1

= 3(x + 5x + 1) + 2 = 3u2 + 2u – 3

Checking:

1 , 3

x 2 +5 x +1

By substituting x 2 +5 x +1 = u into the expression 3x2 + 15x + 2 x 2 +5 x +1 , we have

3 x +3 = 3x – 3

1 3

x 2 +5 x +1

= 3x2 + 15x + 3 – 3 + 2

3 x +3 + 2 = 3x – 1

x=

3 x −3 + 5 = 3(4 ) −3 + 5 = 8 2x = 2(4) = 8 The real root of the equation is 4.

When x = 4,

The real roots of the equation are 1 and

When x = 2,

7  3  − 3 + 5 = 4 

3 x −3 + 5 =

13 7 7 ≠  = 2 4 2

The real roots of the equation are − 3 , −

When x =

x=4

13 2

7 or u = 3 4

7 4

(

7 , 4

When x =

20. By substituting x3 = u into the equation x6 – 5x3 + 4 = 0, we have u2 – 5u + 4 = 0 (u – 1)(u – 4) = 0 u = 1 or u = 4 ∵ x3 = u ∴ x3 = 1 or x3 = 4 x=1 or x = 3 4

21.

or

2x = 2 

7 and 2

∴

7 4

x= Checking:

19. By substituting x2 = u into the equation 4x4 – 19x2 + 21 = 0, we have 4u2 – 19u + 21 = 0 (4u – 7)(u – 3) = 0 u=

)

2 = (2x – 5)2 3 x −3 3x – 3 = 4x2 – 20x + 25 2 4x – 23x + 28 = 0 (4x – 7)(x – 4) = 0

∴

−

or

∵

x 2 +5 x +1 = u

∴

x 2 +5 x +1 = 1 or

5 (rejected) 3

u=−

5 3

x 2 +5 x +1 =

4 More about Equations Level 1

( x 2 + 5 x + 1 ) 2 = 12 x 2 + 5x + 1 = 1

For questions 1 to 3, refer to the graph below:

2

x + 5x = 0 x( x + 5) = 0 x = 0 or x = − 5 Checking: When x = 0, 3x2 + 15x + 2 x 2 +5 x +1 = 3(0)2 + 15(0) + 2 0 2 +5(0) +1 =2 When x = –5, 3x2 + 15x + 2 x 2 +5 x +1 = 3( − 5) 2 +15 ( − 5) +2

(− 5) 2 +5( − 5) +1

=2 ∴ The real roots of the equation are –5 and 0.

(

)

2

25. (a) RHS = x + x +2 = x2 + 2 x

= x + x + 2x = LHS x2 + x + 2 x

(

x +2 ≡ x +

(x +

(a))

x y

x +2

x2 + x + 2 x

(b)

(x +

x +2

)

2

∵ ∴

–2

x +2 = 14

x +2 )

2

x +2

)

2

2.

– 2 = 14

– 16 = 0

4)2

or

x + x +2 – 4 =

(

or

(

)

2

= (–x – 4)2

x + 2 = x2 + 8x + 16 x2 + 7x + 14 = 0 x=

or or

x +2

)

2

= (–x +

∵ ∴ 3.

or

2(1)

–1 2

0 3

1 4

The two graphs intersect at (0, 3) and (3, 6). The solutions of the simultaneous equations are (0, 3) and (3, 6).

x y ∵ ∴

(x – 2)(x –

3 6

y = 2x – 1

x + 2 = x2 – 8x + 16 x2 – 9x + 14 = 0

−7 ± 7 2 −4(1)(14 )

2 6

The two graphs intersect at (–1, 6) and (3, 6). The solutions of the simultaneous equations are (–1, 6) and (3, 6).

x y

x + x +2 + 4 = 0

x +2

1 6

y=x+3

(By

( x + x +2 + 4)( x + x +2 – 4) = 0 0

y=6

x +2 + x + 2 – 2

2

∴

1.

–2

0 –1

1 1

2 3

The two graphs intersect at only one point (2, 3). The solution of the simultaneous equations is (2, 3).

7) = 0 x=

−7 ± −7 (rejected) 2

or

x=2

or

x

For questions 4 to 6, refer to the graph below:

=7 Checking: When x = 2, x2 + x + 2 x 2( 2)

x +2 = 22 + 2 +

2 +2

= 14 When x = 7, x2 + x + 2 x 2(7)

∴

x +2 = 72 + 7 +

7 +2

= 98 ≠ 14 The real root of the equation is 2.

Exercise 4B (p.176)

4.

y=x–1 x y

–1 –2

0 –1

1 0

90

Certificate Mathematics in Action Full Solutions 4A ∵ ∴ 5.

The two graphs intersect at (–3.0, –4.0) and (1.0, 0). The solutions of the simultaneous equations are (–3.0, –4.0) and (1.0, 0).

y = 3x x y ∵

∵ ∴

0 3

0 1

y = 4x – 2 x 1 2 y 2 6

3 10

1 2

2 5

3 10

1 5

The two graphs intersect at (–3.2, –3.2) and (2.2, 7.2). The solutions of the simultaneous equations are approximately (–3.2, –3.2) and (2.2, 7.2). y = x2 y=9

∴

x2 = 9 x –9=0 The quadratic equation that can be solved is x2 – 9 = 0. 2

∵ ∴

∵

y = x2 y = –2x + 4

∴

x2 = –2x + 4 x2 + 2x – 4 = 0 The quadratic equation that can be solved is x2 + 2x – 4 = 0.

∴ 9.

–1 1

∵

∴ 8.

1 3

13. y = x2 + 1 x –2 –1 y 5 2

y = 2x + 3 x y

7.

0 0

The two graphs intersect at (–2.0, –6.0) and (2.0, 6.0). The solutions of the simultaneous equations are (–2.0, –6.0) and (2.0, 6.0).

∴ 6.

–1 –3

Level 2

y = x2 ……(1) 2y = 5x – 1 ……(2) By substituting (1) into (2), we have 2x2 = 5x – 1 2x2 – 5x + 1 = 0 ∴ The quadratic equation that can be solved is 2x2 – 5x + 1 = 0.

10. ∵ ∴ 11. ∵ ∴ 12. ∵

14. y = x2 – 1 x –2 –1 y 3 0

0 –1

y = 2x – 2 x 1 2 y 0 2

3 4

2 3

3 8

x2 + 5x + 1 = 0 x2 = –5x – 1 The equation of the required straight line is y = –5x – 1. 2x2 + 3x – 4 = 0 2x2 = –3x + 4

3 x+2 2

The equation of the required straight line is y=−

3 x + 2. 2

∵ ∴

The two graphs intersect at only one point (1.0, 0). The solution of the simultaneous equations is (1.0, 0).

15. y = x2

91

1 0

x2 – 4x – 4 = 0 x2 = 4x + 4 The equation of the required straight line is y = 4x + 4.

x2 = − ∴

The two graphs intersect at (1.0, 2.0) and (3.0, 10.0). The solutions of the simultaneous equations are (1.0, 2.0) and (3.0, 10.0).

4 More about Equations x y

–3 –2 –1 9 4 1

0 0

1 1

2 4

4x + y + 5 = 0 x –3 –2 –1 Y 7 3 –1

17. ∵ ∴ 18. ∵

∴ 19. ∵

2x2 – 3 = 0 2x – 3 + x – 1 = x – 1 2x2 + x – 4 = x – 1 The equation of the required straight line is y = x – 1. 2

– 2x2 + 5x + 1 = 0 2x2 – 5x – 1 = 0 2x2 – 5x – 1 + 6x – 3 = 6x – 3 2x2 + x – 4 = 6x – 3 The equation of the required straight line is y = 6x – 3. 4x2 + 3x – 9 = 0 4x2 + 2x + x – 8 – 1 = 0 4x2 + 2x – 8 = –x + 1 2x2 + x – 4 = −

∴

The equation of the required straight line is y= −

20. ∵ ∵ ∴

The two graphs do not intersect. The simultaneous equations have no real solutions.

16. y = x2 – x – 1 x –2 –1 0 1 y 5 1 –1 –1

∴ 2 1

3 5

2x + 3y = –1 x –2 –0.5 1 y 1 0 –1

∵ ∴

1 1 x+ 2 2

1 1 x+ . 2 2

x2 + 5x + 4 = 0 2x2 + 10x + 8 = 0 2x2 + x + 9x – 4 + 12 = 0 2x2 + x – 4 = –9x – 12 The equation of the required straight line is y = –9x – 12.

For questions 21 to 23, refer to the graph below: y = x2 – 3x – 3 x –2 –1 0 1 2 3 y 7 1 –3 –5 –5 –3

4 1

5 7

The two graphs intersect at (–0.7, 0.1) and (1.0, –1.0). The solutions of the simultaneous equations are approximately (–0.7, 0.1) and (1.0, –1.0).

92

Certificate Mathematics in Action Full Solutions 4A

21. ∵ ∴

x2 – 2x + 1 = 0 x – 2x – x – 4 + 1 = – x – 4 x2 – 3x – 3 = –x – 4 The corresponding simultaneous equations are 2

y = x 2 − 3 x − 3 .  y = −x − 4 Draw the straight line y = –x – 4 on the graph of y = x2 – 3x – 3. From the graphs, the root of x2 – 2x + 1 = 0 is 1.0. 22. ∵ ∴

x2 – 4x + 5 = 0 x2 – 3x – x + 8 – 3 = 0 x2 – 3x – 3 = x – 8 The corresponding simultaneous equations are

y = x 2 − 3 x − 3 .  y = x − 8 Draw the straight line y = x – 8 on the graph of y = x2 – 3x – 3. From the graphs, x2 – 4x + 5 = 0 has no real roots. 23. ∵

2x2 – 5x – 10 = 0 2x – 5x – x – 6 – 4 = – x 2x2 – 6x – 6 = – x + 4 2

x2 – 3x – 3 = − ∴

(c) (i) ∵

1 x+2 2

∴

Draw the straight line y = − 2

1 x + 2 on the graph of 2

y = x – 3x – 3. From the graphs, the roots of 2x2 – 5x – 10 = 0 are approximately –1.3 and 3.8. 24. (a) y = x2 – 4x x –1 0 y 5 0 (b)

y = x 2 − 4x . y = −1

The corresponding simultaneous equations are

 y = x 2 − 3x − 3  . 1  y = − x + 2  2 

1 2 3 –3 –4 –3

4 0

5 5

x2 – 4x + 1 = 0 x2 – 4x = –1 The corresponding simultaneous equations are 

Draw the straight line y = –1 on the graph of y = x2 – 4x. From the graphs, the roots of x2 – 4x + 1 = 0 are approximately 0.3 and 3.7. (ii) ∵ ∴

x2 – 3x – 3 = 0 x2 – 3x – x + x – 3 = 0 x2 – 4x = – x + 3 The corresponding simultaneous equations

y = x 2 − 4x . y = −x + 3

are 

Draw the straight line y = – x + 3 on the graph of y = x2 – 4x. From the graphs, the roots of x2 – 3x – 3 = 0 are approximately –0.8 and 3.8. (iii) ∵

2x2 – 7x – 2 = 0 2x2 – 7x – x – 2 = – x 2x2 – 8x = – x + 2 x2 – 4x = −

∴

1 x+1 2

The corresponding simultaneous equations are

 y = x 2 − 4x  . 1  y = − x +1  2 

Draw the straight line y = −

1 x + 1 on the 2

graph of y = x2 – 4x. From the graphs, the roots of 2x2 – 7x – 2 = 0 are approximately –0.3 and 3.8. 25. (a) y = 2x2 – x

93

4 More about Equations x y

–1 3

–0.5 1

0 0

0.5 0

1 1

1.5 3

are

(b)

 y = 2x 2 − x  3   y = 2

Draw the straight line y =

.

3 on the graph of 2

y = 2x2 – x. From the graphs, the roots of 4x2 – 2x – 3 = 0 are approximately –0.7 and 1.2. 26. (a) ∵

x2 + (b – m)x + (c – n) = 0 x2 + bx – mx + c – n = 0 x2 + bx + c = mx + n ∴ The corresponding simultaneous equations are

y = x 2 + bx + c .  y = mx + n From the graphs, the roots of x2 + (b – m)x + (c – n) = 0 are 1 and 4.

(c) (i) ∵ ∴

(b) (i) Read from the graph of y = x2 + bx + c. When x = 1, y = 6. ∴ 6 = 12 + b(1) + c b+c=5 c = –b + 5 ……(1) When x = 4, y = 3. ∴ 3 = 42 + b(4) + c 4b + c = –13 ……(2) By substituting (1) into (2), we have 4b + (–b + 5) = –13 3b = –18 6 b= −

2x2 – 3x + 1 = 0 2x2 – x – 2x + 1 = 0 2x2 – x = 2x – 1 The corresponding simultaneous equations

y = 2 x 2 − x . y = 2 x −1

are 

Draw the straight line y = 2x – 1 on the graph of y = 2x2 – x. From the graphs, the roots of 2x2 – 3x + 1 = 0 are 0.5 and 1.0. (ii) ∵

By substituting b = –6 into (1), we have c = 11 (ii) Read from the graphs of y = mx + n. When x = 1, y = 6. ∴ 6 = m(1) + n m+n=6 n = –m + 6 ……(1) When x = 4, y = 3. ∴ 3 = m(4) + n 4m + n = 3 ……(2) By substituting (1) into (2), we have 4m + (–m + 6) = 3 3m = –3 1 m= −

3x2 – 5x + 1 = 0

10 2 x+ =0 3 3 7 2 2x2 – x – x + = 0 3 3 7 2 x− 2x2 – x = 3 3 2x2 –

∴

The corresponding simultaneous equations are

 y = 2x2 − x  7 2  y = x−  3 3 

7 2 x − on the Draw the straight line y = 3 3 graph of y = 2x2 – x. From the graphs, the roots of 3x2 – 5x + 1 = 0 are approximately 0.2 and 1.4. (iii) ∵

4x2 – 2x – 3 = 0 4x2 – 2x = 3 2x2 – x =

∴

By substituting m = –1 into (1), we have n = 7

.

3 2

The corresponding simultaneous equations

Exercise 4C (p.186) Level 1 1.

y = x 2 + 1  (1)  y = 2 x + 4  ( 2) By substituting (2) into (1), we have 2x + 4 = x2 + 1 x2 – 2x – 3 = 0 (x + 1)(x – 3) = 0 x+1=0 or x–3=0 x = –1 or x=3 By substituting x = –1 into (2), we have y = 2(–1) + 4 = 2 By substituting x = 3 into (2), we have y = 2(3) + 4 = 10 ∴ The solutions of the simultaneous equations are

94

Certificate Mathematics in Action Full Solutions 4A (–1, 2) and (3, 10). 2.

6.

 (1)  y = 4 x +12  2  ( 2) y = x + 2 x + 13  By substituting (2) into (1), we have x2 + 2x + 13 = 4x +12 x2 – 2x + 1 = 0 (x – 1)2 = 0 x=1 By substituting x = 1 into (1), we have y = 4(1) + 12 = 16 ∴ The solution of the simultaneous equations is (1, 16).

3.

By substituting (1) into (2), we have x2 – x = 1 – 2x x2 + x – 1 = 0 Consider the discriminant of x2 + x – 1 = 0. ∆ = 12 – 4(1)(–1) = 5 > 0 ∴ x2 + x – 1 = 0 has two real roots. ∴ The simultaneous equations have two real solutions. 7.

x = 3 y − 7  (1)  2 2 x − y = 9  ( 2) By substituting (1) into (2), we have (3y – 7)2 – y2 = 9 9y2 – 42y + 49 – y2 = 9 8y2 – 42y + 40 = 0 4y2 – 21y + 20 = 0 (4y – 5)(y – 4) = 0 4y – 5 = 0 or

5 y= 4

y=4

8.

By substituting y = 4 into (1), we have x = 3(4) – 7 = 5 ∴ The solutions of the simultaneous equations are

4.

x 2 + y 2 = 4  (1)   ( 2) y = 5 − x By substituting (2) into (1), we have x2 + (5 – x)2 = 4 2 x + 25 – 10x + x2 = 4 2x2 – 10x + 21 = 0 Using the quadratic formula, we have

x=

= ∴ 5.

−( −10 ) ± ( −10 ) 2 −4( 2)( 21 ) 2( 2)

10 ± − 68 (rejected) 4

The simultaneous equations have no real solutions.

y = 2 x 2 +1  (1)   ( 2) y = 2 x By substituting (1) into (2), we have 2x2 + 1 = 2x 2x2 – 2x + 1 = 0 Consider the discriminant of 2x2 – 2x + 1 = 0. ∆ = (–2)2 – 4(2)(1) = –4 < 0 ∴ 2x2 – 2x + 1 = 0 has no real roots. ∴ The simultaneous equations have no real solutions.

y +1 = x 2 − x  (1)   ( 2) 2 x + y = 3 From (2), we have y = –2x + 3……(3) By substituting (3) into (1), we have –2x + 3 + 1 = x2 – x x2 + x – 4 = 0 Consider the discriminant of x2 + x – 4 = 0. ∆ = 12 – 4(1)(–4) = 17 > 0 ∴ x2 + x – 4 = 0 has two real roots. ∴ The simultaneous equations have two real solutions.

5 into (1), we have 4 13 5  x = 3  − 7 = − 4 4

By substituting y =

 13 5  ,  and (5, 4). −  4 4

3 x + y = 2 x 2 +10  (1)   ( 2) y = 9 x − 8 By substituting (2) into (1), we have 3x + (9x – 8) = 2x2 + 10 2x2 – 12x + 18 = 0 x2 – 6x + 9 = 0 Consider the discriminant of x2 – 6x + 9 = 0. ∆ = (–6) 2 – 4(1)(9) = 0 ∴ x2 – 6x + 9 = 0 has only one real root. ∴ The simultaneous equations have one real solution.

y–4=0

or

y = x 2 − x  (1)  y =1 − 2 x  ( 2)

9.

 (1) y = x  2  y = x − 3x + k  ( 2) By substituting (2) into (1), we have x2 – 3x + k = x x2 – 4x + k = 0……(3) ∵ The simultaneous equations have only one solution. ∴ (3) has only one real root. ∴ ∆ =0 (–4)2 – 4(1)(k) = 0 16 – 4k = 0 4k = 16 k= 4

10.  y = 4 x − 5

 (1)

 2  y = x − 6 x + k  ( 2)

By substituting (2) into (1), we have x2 – 6x + k = 4x – 5 2 x – 10x + (k + 5) = 0……(3) ∵ The simultaneous equations have only one solution. ∴ (3) has only one real root. ∴ ∆ =0 (–10)2 – 4(1)(k + 5) = 0 100 – 4k – 20 = 0 4k = 80 k = 20 11. 7 = 6 x − y

 (1)

 2  y = x − 4 x − k  ( 2)

95

4 More about Equations By substituting (2) into (1), we have 7 = 6x – (x2 – 4x – k) 2 x – 10x + (7 – k) = 0……(3) ∵ The simultaneous equations have only one solution. ∴ (3) has only one real root. ∴ ∆ =0 (–10)2 – 4(1)(7 – k) = 0 100 – 28 + 4k = 0 4k = –72 18 k=−

 2 12. x +16 = 5 x + y  (1)  ( 2) k + y = 5 x From (2), we have y = 5x – k ……(3) By substituting (3) into (1), we have x2 +16 = 5x + 5x – k x2 – 10x + (16 + k) = 0 ……(4) ∵ The simultaneous equations have only one solution. ∴ (4) has only one real root. ∴ ∆ =0 (–10)2 – 4(1)(16 + k) = 0 100 – 64 – 4k = 0 4k = 36 k=9 13. (a) ∵ ∴ ∵ ∴ ∴

The length of the rectangle is longer than the width by 10 cm. x – y = 10 The area of the rectangle is 200 cm2. xy = 200 The required simultaneous equations are x − y =10  (1) .  xy = 200  ( 2)

(b) From (1), we have x – y = 10 x = y + 10……(3) By substituting (3) into (2), we have (y + 10)y = 200 y2 + 10y – 200 = 0 (y – 10)(y + 20) = 0 y = 10 or y = –20 (rejected) By substituting y = 10 into (3), we have x = 10 + 10 = 20 ∴ The dimensions of the rectangle are 10 cm × 20 cm. 14. Let x cm and y cm be the length and the width of the rectangle respectively. Then the length of the equilateral triangle is also y cm. ∵ The perimeter of the figure is 40 cm. ∴ 2x + 4y = 40 x + 2y = 20 x = 20 – 2y ……(1) ∵ The area of the rectangle is 50 cm2. ∴ xy = 50 ……(2) By substituting (1) into (2), we have (20 – 2y)y = 50 20y – 2y2 = 50 2 2y – 20y + 50 = 0 y2 – 10y + 25 = 0 (y – 5)2 = 0 y=5 By substituting y = 5 into (1), we have x = 20 – 2(5) = 10 ∴ The dimensions of the rectangle are 5 cm × 10 cm.

15. Let x m and y m be the length and the width of the garden respectively. ∵ The area of the garden is 150 m2. ∴ xy = 150 ……(1) ∵ The area of the path is 186 m2. ∴ [x + 2(3)][y + 2(3)] – xy = 186 xy + 6x + 6y + 36 – xy = 186 6x + 6y = 150 x + y = 25 y = 25 – x ……(2) By substituting (2) into (1), we have x(25 – x) = 150 25x – x2 = 150 x2 – 25x + 150 = 0 (x – 10)(x – 15) = 0 x = 10 or x = 15 By substituting x = 10 into (2), we have y = 25 – 10 = 15 By substituting x = 15 into (2), we have y = 25 – 15 = 10 ∴ The dimensions of the garden are 10 m × 15 m. 2  16. y = x + k  (1)  ( 2) y = mx

By substituting (1) into (2), we have x2 + k = mx x2 – mx + k = 0 ……(3) ∵ The simultaneous equations have only one solution. ∴ (3) has only one real root. ∴ ∆ =0 (–m)2 – 4(1)k = 0 m2 = 4k From the graphs, y = mx has a positive slope. ∴ m>0 Let m = 2, then k = 1. Let m = 4, then k = 4. Let m = 8, then k = 16. Two pairs of possible values of m and k are: m = 2, k = 1 or m = 4, k = 4 or m = 8, k = 16. (or any other reasonable answers) 17. Let the constant term of the quadratic equation be k.

 (1) y = 1  2  y = x − 2 x + k  ( 2) By substituting (2) into (1), we have x2 – 2x + k = 1 2 x – 2x + (k – 1) = 0 ……(3) ∵ The simultaneous equations have two real solutions. ∴ (3) has two real roots. ∴ ∆ >0 (–2)2 – 4(1)(k – 1) > 0 4 – 4k + 4 > 0 8 > 4k k<2 ∴ A possible value of the constant term is –1 or 1. (or any other reasonable answers)

Level 2 18. x − y −1 = 0

 (1)

 2 2 x − xy + y = 7  ( 2)

From (1), we have x = 1 + y ……(3) By substituting (3) into (2), we have (1 + y)2 – (1 + y)y + y2 = 7 1 + 2y + y2 – y – y2 + y2 = 7

96

Certificate Mathematics in Action Full Solutions 4A y2 + y – 6 = 0 (y + 3)(y – 2) = 0 y = –3 or y = 2 By substituting y = –3 into (3), we have x = 1 + (–3) = –2 By substituting y = 2 into (3), we have x = 1 + 2 = 3 ∴ The solutions of the simultaneous equations are (–2, –3) and (3, 2). 2  2 19. x + y = 25

 (1)

3 x − 4 y + 25 = 0  ( 2) 3 x + 25 From (2), we have y = ……(3) 4 By substituting (3) into (1), we have 2

 3 x + 25  x 2 +  = 25 4   16x2 + 9x2 +150x + 625 = 400 25x2 + 150x + 225 = 0 x2 + 6x + 9 = 0 (x + 3)2 = 0 x = –3 By substituting x = –3 into (3), we have y=

3( −3) + 25 =4 4

∴

The solution of the simultaneous equations is (–3, 4).

 20. x + y + 3 x + 2 y = −2 2

2

2 x − 3 y = 2

From (2), we have x =

3 y +1 2

 (1)  ( 2) ……(3)

By substituting (3) into (1), we have 2

3  3  2  y +1 + y + 3 y +1 + 2 y = −2 2 2     9 2 9 y + 3 y +1 + y 2 + y + 3 + 2 y = −2 4 2 13 2 19 y + y +6 = 0 4 2 13y2 + 38y + 24 = 0 (y + 2)(13y + 12) = 0 y = –2 or y =

12 − 13 By substituting y = –2 into (3), we have

3 ( −2) +1 = –2 2 12 By substituting y = − into (3), we have 13 3  12  5 x = −  +1 = − 2  13  13 x=

∴

The solutions of the simultaneous equations are



5 12  ,−  .  13 13  2 2  21. 2 x − 3 xy − 2 y −12 = 0  (1)  ( 2) − 2 x + 3 y + 4 = 0 (–2,–2) and  −

97

From (2), we have x =

3 y +2 2

……(3)

By substituting (3) into (1), we have 2

3  3  2 y + 2  − 3 y + 2  y − 2 y 2 −12 = 0 2  2  9 2 9 y +12 y + 8 − y 2 − 6 y − 2 y 2 −12 = 0 2 2 2y2 – 6y + 4 = 0 y2 – 3y + 2 = 0 (y – 1)(y – 2) = 0 y = 1 or y = 2 By substituting y = 1 into (3), we have x =

3 7 (1) + 2 = 2 2 By substituting y = 2 into (3), we have x = 5 ∴

3 ( 2) + 2 = 2

The solutions of the simultaneous equations are

7   , 1 and (5, 2). 2  22. (a)  y = mx

 (1)

 2  y = x − 2 x + 4  ( 2)

By substituting (2) into (1), we have x2 – 2x + 4 = mx x2 – (2 + m)x + 4 = 0 ……(3) ∵ The simultaneous equations have only one solution. ∴ (3) has only one real root. ∴ ∆ =0 [– (2 + m)]2 – 4(1)(4) = 0 4 + 4m + m2 – 16 = 0 m2 + 4m – 12 = 0 (m + 6)(m – 2) = 0 6 or m = 2 m= − (b) When m = –6, (3) becomes x2 – [2 + (–6)] x + 4 = 0 x2 + 4x + 4 = 0 (x + 2)2 = 0 x = –2 By substituting m = –6 and x = –2 into (1), we have y = –6(–2) = 12 ∴ The coordinates of P are (–2, 12). When m = 2, (3) becomes x2 – (2 + 2)x + 4 = 0 x2 – 4x + 4 = 0 (x – 2)2 = 0 x=2 By substituting m = 2 and x = 2 into (1), we have y = 2(2) = 4 ∴ The coordinates of P are (2, 4). 23.  y + k = 2 x

 (1)

 2  y = x − 8 x + 9  ( 2)

By substituting (2) into (1), we have x2 – 8x + 9 + k = 2x x2 – 10x + (9 + k) = 0……(3) ∵ The simultaneous equations have real solutions.

4 More about Equations ∴ ∴

∴

27. (a) ∵

(3) has real roots.

∆ ≥ 0 (–10)2 – 4(1)(9 + k) ≥ 0 100 – 36 – 4k ≥ 0 4k ≤ 64 k ≤ 16 The maximum value of k is 16.

24. (a)  y = x + 4

The area of the rhombus is 15 cm2.

1 xy = 15 2

∴

xy = 30 The sum of the lengths of its diagonals is 11 cm. x + y = 11 The required simultaneous equations are  (1) xy = 30 .  x + y =11  ( 2)

∵ ∴ ∴

 (1)

 2  y = kx + 3 x + 2  ( 2)

By substituting (2) into (1), we have kx2 + 3x + 2 = x + 4 kx2 + 2x – 2 = 0 ……(3) ∵ The simultaneous equations have real solutions. ∴ (3) has real roots. ∴ ∆ ≥ 0 22 – 4(k)(–2) ≥ 0 4 + 8k ≥ 0 8k ≥ –4 k≥ −

(b) From (2), we have x + y = 11 y = 11 – x……(3) By substituting (3) into (1), we have x(11 – x) = 30 11x – x2 = 30 x2 – 11x + 30 = 0 (x – 5)(x – 6) = 0 x = 5 or x = 6 By substituting x = 5 into (3), we have y = 11 – 5 = 6 (rejected ∵ AC > BD) By substituting x = 6 into (3), we have y = 11 – 6 = 5 ∴ x=6 ,y=5

1 2

(b) By substituting the minimum value of k, i.e. −

1 , 2

into (3), we have

−

1 2 x + 2x – 2 = 0 2

Exercise 4D (p.193)

2

– x + 4x – 4 = 0 x2 – 4x + 4 = 0 (x – 2)2 = 0 x=2 By substituting x = 2 into (1), we have y = 2 + 4 = 6 ∴ The solution of the simultaneous equations is (2, 6).

Level 1 1.

26. Let x cm and y cm be the lengths of the sides of the squares ABCD and DEFG respectively. ∵ EC = 4 cm ∴ x–y=4 y = x – 4……(1) ∵ The sum of their areas is 400 cm2. ∴ x2 + y2 = 400……(2) By substituting (1) into (2), we have x2 + (x – 4)2 = 400 x2 + x2 – 8x + 16 = 400 2x2 – 8x – 384 = 0 x2 – 4x – 192 = 0 (x – 16)(x + 12) = 0 x = 16 or x = –12 (rejected) By substituting x = 16 into (1), we have y = 16 – 4 = 12 ∴ The length of the sides of the square ABCD is 16 cm. The length of the sides of the square DEFG is 12 cm.

f(1) = 13 + 2(1)2 – 1 – 2 =1+2–1–2 =0 x – 1 is a factor of f(x).

∴

(b) By long division,

2  25. y = −x + kx + 4  (1)  ( 2) 4 x + y − 3 = 0

By substituting (1) into (2), we have 4x + (–x2 + kx + 4) – 3 = 0 –x2 + (4 + k)x + 1 = 0 Consider the discriminant of –x2 + (4 + k)x + 1 = 0. ∆ = (4 + k)2 – 4(–1)(1)= (4 + k)2 + 4 ∵ (4 + k)2 ≥ 0 for all real values of k. ∴ ∆ >0 ∴ The simultaneous equations have two real solutions for all real values of k.

(a) ∵

x 2 + 3x + 2 x −1 x 3 + 2 x 2 − x − 2 x3 − x2 3x 2 − x 3x 2 − 3 x 2x − 2 2x − 2

∴

x3 + 2x2 – x – 2 = (x – 1)(x2 + 3x + 2) =( x − 1)( x + 1)( x +2)

(c) ∵ ∴

f(x) = 0 (x – 1)(x + 1)(x + 2) = 0 x – 1 = 0 or x + 1 = 0 or 1 x = 1 or x= −

x+2=0 or x =

− 2

2.

(a) ∵

f(–1) = (–1)3 – (–1)2 – 10(–1) – 8 = –1 – 1 + 10 – 8 =0 x + 1 is a factor of f(x).

∴

(b) By long division, x 2 − 2x − 8 x + 1 x 3 − x 2 − 10 x − 8 x 3 + x2 − 2 x 2 − 10 x − 2x 2 − 2x − 8x − 8 − 8x − 8

∴

x3 – x2 – 10x – 8 = (x + 1)(x2 – 2x – 8) =( x + 1)( x +2)( x −4)

(c) ∵ ∴

f(x) = 0 (x + 1)(x + 2)(x – 4) = 0 x + 1 = 0 or x + 2 = 0 1 x= − or

or x – 4 = 0 2 x= − or

=4

98

x

Certificate Mathematics in Action Full Solutions 4A

3.

(a) ∵

f(2) = 23 – 22 – 4(2) + 4 =8–4–8+4 =0 x – 2 is a factor of f(x).

∴

∴

(b) ∵ x – 1 and x + 2 are factors of f(x). ∴ (x – 1)(x + 2) is also a factor of f(x). Divide f(x) by (x – 1)(x + 2), i.e. x2 + x – 2.

(b) By long division, x2 + x − 2 x − 2 x 3 − x 2 − 4x + 4 3

x − 2x

= 16 + 24 – 8 – 24 – 8 =0 x + 2 is a factor of f(x).

x 2 − 4x + 4 x 2 + x − 2 x 4 − 3 x3 − 2 x 2 + 1 2x − 8

2

x 4 + x 3 − 2x 2

2

x −4x

− 4x

x2 − 2x

+ 12x 2

4x 2 + 4x − 8

− 2x + 4

∴

3 3

− 4x − 4x + 8x

− 2x + 4

4x 2 + 4x − 8

x3 – x2 – 4x + 4 = (x – 2)(x2 + x – 2)

∴

=( x − 2)( x − 1)( x +2)

x4 – 3x3 – 2x2 + 12x – 8 = (x – 1)(x + 2)(x2 – 4x + 4) =( x −1)( x +2)( x −2) 2

(c) ∵ ∴

4.

f(x) = 0 (x – 2)(x – 1)(x + 2) = 0 x – 2 = 0 or x – 1 = 0 or x = 2 or x = 1 or

(a) ∵

x+2=0 2 x= −

f(–3) = (–3)3 + 2(–3)2 – 5(–3) – 6 = –27 + 18 + 15 – 6 =0 x + 3 is a factor of f(x).

∴

(c) ∵ ∴

f (x) = 0 (x – 1)(x + 2)(x – 2)2 = 0 x – 1 = 0 or x + 2 = 0 or 2 x = 1 or x= −

(a) ∵

f (1) = 14 + 5(1)3 + 5(1)2 – 5(1) – 6 =1+5+5–5–6 =0 x – 1 is a factor of f(x). f (–2) = (–2)4 + 5(–2)3 + 5(–2)2 – 5(–2) – 6 = 16 – 40 + 20 + 10 – 6 =0 x + 2 is a factor of f(x).

2

(b) By long division,

7.

x2 − x − 2 x + 3 x 3 + 2x 2 − 5x − 6 x 3 + 3x 2 − x 2 − 5x − x 2 − 3x −2x − 6 −2x − 6

∴

x3 + 2x2 – 5x – 6 = (x + 3)(x2 – x – 2)

∴ ∵

=( x +3)( x + 1)( x −2)

(c) ∵ ∴

f (x) = 0 (x + 3)(x + 1)(x – 2) = 0 x + 3 = 0 or x + 1 = 0 or 3 or 1 x= − x= −

(x – 2)2 = 0 or x =

∴

(b) ∵ x – 1 and x + 2 are factors of f(x). ∴ (x – 1)(x + 2) is also a factor of f(x). Divide f(x) by (x – 1)(x + 2), i.e. x2 + x – 2.

x–2=0 or x =

2

x 2 + 4x + 3 x 2 + x − 2 x 4 + 5 x3 + 5x 2 − 5x − 6 x 4 + x3 − 2 x 2 4x 3 + 7x 2 − 5x 4x 3 + 4x 2 − 8x 3x 2 + 3x − 6 3x 2 + 3x − 6

5.

(a) ∵

3

2

f (1) = 3(1) + 4(1) – 5(1) – 2 =3+4–5–2 =0 x – 1 is a factor of f(x). f (–2) = 3(–2)3 + 4(–2)2 – 5(–2) – 2 = –24 + 16 + 10 – 2 =0 x + 2 is a factor of f(x).

∴ ∵ ∴

∴

x4 + 5x3 + 5x2 – 5x – 6 = (x – 1)(x + 2)(x2 + 4x + 3)

=( x − 1)( x + 2)( x + 1)( x + 3)

(c) ∵ f(x) = 0 ∴ (x – 1)(x + 2)(x + 1)(x + 3) = 0 x – 1 = 0 or x + 2 = 0 or x + 1 = 0 or x + 3 = 0 2 or x = − 1 or x = x = 1 or x= −

(b) ∵ x – 1 and x + 2 are factors of f(x). ∴ (x – 1)(x + 2) is also a factor of f(x). Divide f(x) by (x – 1)(x + 2), i.e. x2 + x – 2.

− 3

8.

3x + 1 x2 + x − 2 3x 3 + 4 x 2 − 5 x − 2 3x 3 + 3 x2 − 6 x

(a) ∵

f(1) = 2(1)4 + 13 – 2(1)2 + 5(1) – 6 =2+1–2+5–6 =0 x – 1 is a factor of f(x). f(–2) = 2(–2)4 + (–2)3 – 2(–2)2 + 5(–2) – 6 = 32 – 8 – 8 – 10 – 6 =0 x + 2 is a factor of f(x).

x2 + x − 2 x2 + x − 2

∴

3x3 + 4x2 – 5x – 2

∴ ∵

=(3 x + 1)( x − 1)( x +2)

(c) ∵ ∴

f (x) = 0 (3x + 1)(x – 1)(x + 2) = 0 3x + 1 = 0 or x–1=0 1 x=− or 3

∴ or

x+2=0

x =1

or

2 =−

x

(b) ∵ x – 1 and x + 2 are factors of f(x). ∴ (x – 1)(x + 2) is also a factor of f(x). Divide f(x) by (x – 1)(x + 2), i.e. x2 + x – 2. 2 x2 − x + 3 x 2 + x − 2 2 x 4 + x3 − 2 x2 + 5x − 6 2 x 4 + 2 x3 − 4 x2 − x3 + 2x 2 + 5x − x3 − x2 + 2 x 3 x2 + 3x − 6 3 x2 + 3x − 6

6.

(a) ∵ ∴ ∵

99

f(1) = 14 – 3(1)3 – 2(1)2 + 12(1) – 8 = 1 – 3 – 2 + 12 – 8 =0 x – 1 is a factor of f(x). f(–2) = (–2)4 – 3(–2)3 – 2(–2)2 + 12(–2) – 8

∴

2x4 + x3 – 2x2 + 5x – 6 =( x − 1)( x +2)( 2 x 2 −x +3)

(c) ∵ ∴

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

4 More about Equations x – 1 = 0 or x + 2 = 0 x = 1 or x = –2 or x =

or

2x2 – x + 3 = 0

∴

3x + 1 = 0 x=−

−( −1) ± ( −1) 2 −4( 2)( 3) 2( 2) = ∴ 9.

1 ± − 23 (rejected) 4

∴ 10.

−5 ± −3 2

1 x=− 3

f(2) = 23 – 3(2)2 – 4(2) + 12 = 8 – 12 – 8 + 12 =0 x – 2 is a factor of f(x).

∴

By long division, x2 − x − 6 x − 2 x 3 − 3 x 2 − 4 x + 12 x3 − 2 x2 − x2 − 4 x − x2 + 2 x − 6 x + 12 − 6 x + 12

∴

8(x – 3) – 27 = 0 [2(x – 3)]3 – 33 = 0 [2(x – 3) – 3][4(x – 3)2 + 2(x – 3)(3) + 32] = 0 (2x – 9)(4x2 – 24x + 36 + 6x – 18 + 9) = 0 (2x – 9)(4x2 – 18x + 27) = 0 ∴ 2x – 9 = 0 or 4x2 – 18x + 27 = 0 or

x3 – 3x2 – 4x + 12 = (x – 2)(x2 – x – 6)

= x=

f(3) = 33 – 32 – 41(3) + 105 = 27 – 9 – 123 + 105 =0 x – 3 is a factor of f(x).

By long division,

9 2

x 2 + 2x − 35 x − 3 x 3 − x 2 − 41x + 105 x 3 − 3x 2 2 x 2 − 41x 2 x 2 − 6x − 35 x + 105 − 35 x + 105

2 or 5

∴

x3 – x2 – 41x + 105 = (x – 3)(x2 + 2x – 35)

=( x − 3)( x − 5)( x +7 )

(b) ∵ ∴

f(x) = 0 (x – 3)(x – 5)(x + 7) = 0 x – 3 = 0 or x – 5 = 0 or x = 3 or x = 5 or

x=

= x=

x+2=0 2 x= −

(rejected)

−( −110 ) ± ( −110 ) 2 −4(100 )( 31 ) 2(100 )

∴

f(x) = 0 (x – 2)(x – 3)(x + 2) = 0 x – 2 = 0 or x – 3 = 0 or x = 2 or x = 3 or

∴

18 ± −108 8

125(2x – 1)3 + 1 = 0 [5(2x – 1)]3 + (1)3 = 0 [5(2x – 1) + 1][25(2x – 1)2 – 5(2x – 1) + 1] = 0 (10x – 4)(100x2 – 100x + 25 – 10x + 5 + 1) = 0 (10x – 4)(100x2 – 110x + 31) = 0 ∴ 10x – 4 = 0 or 100x2 – 110x + 31 = 0 x=

(b) ∵ ∴

14. (a) ∵

2( 4)

∴

=( x −2)( x − 3)( x +2)

x=

−( −18 ) ± ( −18 ) 2 −4( 4)( 27 )

12.

− 3 ± −3 (rejected) 6

3

9 x= 2

11.

=

13. (a) ∵

(rejected)

1 x= −

x=

Level 2

−5 ± 5 2 −4(1)( 7) 2(1) =

or

2(3)

∴

(x + 2)3 – 1 = 0 [(x + 2) – 1][ (x + 2)2 + (x + 2) + 1] = 0 (x + 1)(x2 + 4x + 4 + x + 2 + 1) = 0 (x + 1)(x2 + 5x + 7) = 0 ∴ x+1=0 or x2 + 5x + 7 = 0 x = –1 or x=

1 3

3x2 + 3x + 1 = 0

−3 ± 3 2 −4(3)(1)

2 x= −

x = 1 or

or

110 ± −300 200

(rejected)

2 5

(2x + 1)3 + x3 = 0 [(2x + 1) + x][(2x + 1)2 – (2x + 1)x + x2] = 0 (3x + 1)(4x2 + 4x + 1 – 2x2 – x + x2) = 0 (3x + 1)(3x2 + 3x + 1) = 0

15. (a) ∵

x+7=0 7 x= −

f(4) = 44 + 43 – 26(4)2 – 6(4) + 120 = 256 + 64 – 416 – 24 + 120 =0 x – 4 is a factor of f(x). f(–5) = (–5)4 + (–5)3 – 26(–5)2 – 6(–5) + 120 = 625 – 125 – 650 + 30 + 120 =0 x + 5 is a factor of f(x). x – 4 and x + 5 are factors of f(x). (x – 4)(x + 5) is also a factor of f(x).

∴ ∵ ∴ ∵ ∴

Divide f(x) by (x – 4)(x + 5), i.e. x2 + x – 20. x2 − 6 x2 + x − 2 0x4 + x 3 − 2 6x2 − 6 x + 1 2 0 x4 + x 3 − 2 0x2 − 6 x2 − 6 x + 1 2 0 − 6 x2 − 6 x + 1 2 0

100

Certificate Mathematics in Action Full Solutions 4A ∴

∴

x4 + x3 – 26x2 – 6x + 120

–5x + 12 = 0 or

=( x −4)( x +5)( x 2 −6)

(b) ∵ ∴

12 5

x=

f(x) = 0 (x – 4)(x + 5)(x2 – 6) = 0 x – 4 = 0 or x + 5 = 0 or 5 x = 4 or x= −

or

x –6=0 or x =

2( 43 ) =

16. (a) ∵

f(3) = 34 + 2(3)3 – 30(3)2 + 38(3) + 21 = 81 + 54 – 270 + 114 + 21 =0 x – 3 is a factor of f(x). f(–7) = (–7)4 + 2(–7)3 – 30(–7)2 + 38(–7) + 21 = 2401 – 686 – 1470 – 266 + 21 =0 x + 7 is a factor of f(x). x – 3 and x + 7 are factors of f(x). (x – 3)(x + 7) is also a factor of f(x).

∴ ∵ ∴ ∵ ∴

Divide f(x) by (x – 3)(x + 7), i.e. x2 + 4x – 21.

∴

x=

− 2x 3 − 9x 2 + 3 8x − 2x 3 − 8 x 2 + 4 2x − x 2 − 4x + 2 1 − x 2 − 4x + 2 1

=( x −3)( x +7)( x

(b) ∵ ∴

−2 x −1)

f(x) = 0 (x – 3)(x + 7)(x2 – 2x – 1) = 0 x – 3 = 0 or x + 7 = 0 or 7 or x = 3 or x = − x=

x2 – 2x – 1 = 0

2(1)

x=−

∴ 20.

3 4

4(x – 3)3 – 256(3x + 1)3 = 0 (x – 3)3 – 64(3x + 1)3 = 0 (x – 3)3 – [4(3x + 1)]3 = 0 [(x – 3) – 4(3x + 1)][(x – 3)2 + 4(x – 3)(3x + 1) + 16(3x + 1)2] = 0 (–11x – 7)(x2 – 6x + 9 + 12x2 – 32x – 12 + 144x2 + 96x + 16) = 0 (–11x – 7)(157x2 + 58x + 13) = 0 ∴ –11x – 7 = 0 or 157x2 + 58x + 13 = 0 x=−

7 11

or

x=

−58 ± 58 2 −4(157 )(13 )

or

2(157 )

x=

−( −66 ) ± ( −66 ) 2 −4( 21 )( 61 ) 2( 21 )

=

1 x=− 3

− 58 ± − 4800 314

(rejected)

66 ± − 768 (rejected) 42

(x + 3)3 – 27(2x – 3)3 = 0 (x + 3)3 – [3(2x – 3)]3 = 0 2 [(x + 3) – 3(2x – 3)][(x + 3) + 3(x + 3)(2x – 3) + 9(2x – 3)2] = 0 (–5x + 12)(x2 + 6x + 9 + 6x2 + 9x – 27 + 36x2 – 108x + 81) =0 (–5x + 12)(43x2 – 93x + 63) = 0

101

− 6 ± − 5292 8

(rejected)

17. 64(x – 1)3 – (x – 5)3 = 0 [4(x – 1)]3 – (x – 5)3 = 0 [4(x – 1) – (x – 5)][16(x – 1)2 + 4(x – 1)(x – 5) + (x – 5)2] =0 (3x + 1)(16x2 – 32x + 16 + 4x2 – 24x + 20 + x2 – 10x + 25) =0 (3x + 1)(21x2 – 66x + 61) = 0 ∴ 3x + 1 = 0 or 21x2 – 66x + 61 = 0

=

x=

2( 4)

=1± 2

1 3

or

=

−( −2) ± ( −2) 2 −4(1)( −1)

x=−

3 4

− 6 ± 6 2 − 4( 4)( 333 )

x4 + 2x3 – 30x2 + 38x + 21 2

(rejected)

12 5

x=−

x 4 + 4x 3 − 2 1x 2

∴

93 ± − 2187 86

19. 8(x + 6)3 + (2x – 9)3 = 0 [2(x + 6)]3 + (2x – 9)3 = 0 [2(x + 6) + (2x – 9)][4(x + 6)2 – 2(x + 6)(2x – 9) + (2x – 9)2] = 0 (4x + 3)(4x2 + 48x + 144 – 4x2 – 6x + 108 + 4x2 – 36x + 81) = 0 (4x + 3)(4x2 + 6x + 333) = 0 ∴ 4x + 3 = 0 or 4x2 + 6x + 333 = 0

x 2 − 2x − 1 x 2 + 4x − 2 1x 4 + 2x 3 − 3 0x 2 + 3 8x + 2 1

18.

x=

−( −93 ) ± ( −93 ) 2 −4( 43 )( 63 ) 2

± 6

∴

43x2 – 93x + 63 = 0

7 x=− 11

∴ 21.

x4 + x3 – 6x2 – 4x + 8 = 0 (x – 6x2 + 8) + (x3 – 4x) = 0 [(x2)2 – 6x2 + 8] + x(x2 – 4) = 0 (x2 – 2)(x2 – 4) + x(x2 – 4) = 0 (x2 – 4)[(x2 – 2) + x] = 0 (x + 2)(x – 2)(x2 + x – 2) = 0 (x + 2)(x – 2)(x + 2)(x – 1) = 0 (x + 2)2(x – 1)(x – 2) = 0 (x + 2)2 = 0 or x – 1 = 0 or x – 2 = 0 2 x= − or x = 1 or x = 2 4

4 More about Equations

22.

x4 – 3x3 – 2x2 + 12x – 8 = 0 4 (x – 2x2 – 8) – (3x3 – 12x) = 0 [(x2)2 – 2x2 – 8] – 3x(x2 – 4) = 0 (x2 + 2)(x2 – 4) – 3x(x2 – 4) = 0 (x2 – 4)[(x2 + 2) – 3x] = 0 (x + 2)(x – 2)(x2 – 3x + 2) = 0 (x + 2)(x – 2)(x – 2)(x – 1) = 0 (x + 2)(x – 1)(x – 2)2 = 0 x+2=0 or x – 1 = 0 or 2 or x= − x =1 or

(x – 2)2 = 0 x= 2

u=

−1 ± 12 −4(1)( −3) 2(1)

u=

−1 + 13 2

∵

x =u

∴

x =

−1 + 13 2

or

or

u=

−1 − 13 2

x =

−1 − 13 2

(rejected) 2

23.

 −1 + 13   x=   2  

x4 – 3x3 – 3x2 + 3x + 2 = 0 (x4 – 3x2 + 2) – (3x3 – 3x) = 0 [(x2)2 – 3x2 + 2] – 3x(x2 – 1) = 0 (x2 – 1)(x2 – 2) – 3x(x2 – 1) = 0 (x2 – 1)[(x2 – 2) – 3x] = 0 (x2 – 1)(x2 – 3x – 2) = 0 (x + 1)(x – 1)(x2 – 3x – 2) = 0 x + 1 = 0 or x – 1 = 0 or x2 – 3x – 2 = 0 1 x= − or x = 1 or x=

∴

−( −3) ± ( −3) 2 −4(1)( −2) 2(1)

4.

3 ± 17 = 2

=

7 − 13 2

The real root of the equation is

(

By substituting x2 = u into the equation x4 – 9x2 + 20 = 0, we have u2 – 9u + 20 = 0 (u – 4)(u – 5) = 0 u = 4 or u = 5 ∵ x2 = u ∴ x2 = 4 or x2 = 5 x = ± 2 or x = ± 5

7 − 13 . 2

3 x +4 + x = 8

(x – 8)2 = − 3 x +4 x2 – 16x + 64 = 3x + 4 x2 – 19x + 60 = 0 (x – 4)(x – 15) = 0 x = 4 or x = 15

Level 1

∴

1 −2 13 +13 4

x – 8 = − 3 x +4

Revision Exercise 4 (p.196)

1.

=

Checking: When x = 4,

3 x +4 + x =

)

2

3( 4) +4 + 4= 8

When x = 15, 3 x +4 + x = 3(15 ) +4 + 15 = 22 ≠ 8 ∴ The real root of the equation is 4. For questions 5 to 6, refer to the graph below:

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

2.

By substituting x3 = u into the equation 27x6 – 1720x3 – 512 = 0, we have 27u2 – 1720u – 512 = 0 (27u + 8)(u – 64) = 0 u=− ∵

x3 = u

∴

x3 = −

∴ 3.

8 27 2 x=− 3

8 27

or

or

u = 64

x3 = 64 or

x=4

The real roots of the equation are −

2 and 4. 3

By substituting x = u into the equation x + 0, we have u2 + u – 3 = 0

5.

y = –x – 1 x y

x –3=

∵ ∴

–1 0

0 –1

1 –2

The two graphs intersect at (–1, 0) and (4, –5). The solutions of the simultaneous equations are (–1, 0) and (4, –5).

102

Certificate Mathematics in Action Full Solutions 4A

6.

x y

7.

2  11. y = 2 x − x +1  (1)  ( 2) 2 x − y = 3

y = –3x + 1 –1 4

0 1

1 –2

∵ ∴

The two graphs intersect at (–2, 7) and (3, –8). The solutions of the simultaneous equations are (–2, 7) and (3, –8).

∵

x2 + 5x + 2 = 0 x + 3x + 2x – 1 + 3 = 0 x2 + 3x – 1 = –2x – 3 The equation of the required straight line is y = –2x – 3. 2

∴ 8.

∵

∴

 (1) x − y − k = 0  ( 2)

From (2), we have y = x – k……(3) By substituting (3) into (1), we have x2 + (x – k)2 = 4 x2 + x2 – 2kx + k2 = 4 2x2 – 2kx + (k2 – 4) = 0……(4) ∵ The simultaneous equations have only one solution. ∴ (4) has only one real root. ∴ ∆ =0 (–2k)2 – 4(2)(k2 – 4) = 0 4k2 – 8k2 + 32 = 0 4k2 = 32 k = ±2 2

13 x −4 2

The equation of the required straight line is y=

9.

2  2 12. x + y = 4

2x2 – 7x + 6 = 0 2x2 + 6x – 13x – 2 + 8 = 0 2x2 + 6x – 2 = 13x– 8 x2 + 3x – 1 =

13 x −4 . 2

y = x 2 − 2 x + 3  (1)   ( 2) y = 3 x −1 By substituting (1) into (2), we have x2 – 2x + 3 = 3x – 1 x2 – 5x + 4 = 0 (x – 1)(x – 4) = 0 x–1=0 or x – 4 = 0 x=1 or x=4 By substituting x = 1 into (2), we have y = 3(1) – 1 = 2 By substituting x = 4 into (2), we have y = 3(4) – 1 = 11 ∴ The solutions of the simultaneous equations are (1, 2) and (4, 11).

2  2 10. x + y =1  (1) y = 2 x +1  ( 2)

By substituting (2) into (1), we have x2 + (2x + 1)2 = 1 x2 + 4x2 + 4x + 1 = 1 5x2 + 4x = 0 x(5x + 4) = 0 x=0 or 5x + 4 = 0 x=−

4 5

By substituting x = 0 into (2), we have y = 2(0) + 1 = 1 By substituting x = −

4 into (2), we have y = 5

 4 2 −  +1  5 =− ∴

 

103

3 5

The solutions of the simultaneous equations are (0, 1) and  −

4 3 ,−  . 5 5

From (2), we have y = 2x – 3 ……(3) By substituting (1) into (3), we have 2x2 – x + 1 = 2x – 3 2x2 – 3x + 4 = 0 Consider the discriminant of 2x2 – 3x + 4 = 0. ∆ = (–3)2 – 4(2)(4) = –23 < 0 ∴ 2x2 – 3x + 4 = 0 has no real roots. ∴ The simultaneous equations have no real solutions.

2  2 13. x − 4 y + 20 = 0  (1)  ( 2) x + 3 y − k = 0

From (2), we have x = k – 3y……(3) By substituting (3) into (1), we have (k – 3y) 2 – 4y2 + 20 = 0 k2 – 6ky + 9y2 – 4y2 + 20 = 0 5y2 – 6ky + (20 + k2) = 0……(4) ∵ The simultaneous equations have only one solution. ∴ (4) has only one real root. ∴ ∆ =0 (–6k)2 – 4(5)(20 + k2) = 0 36k2 – 400 – 20k2 = 0 16k2 = 400 5 k= ± 14. Let x be the tens digit and y be the units digit of the original number. ∴ The original number is 10x + y, and the reversed number is 10y + x. ∵ The number is increased by 18 when the digits are reversed. ∴ (10y + x) – (10x + y) = 18 –9x + 9y = 18 –x + y = 2 y=x+2 ……(1) ∵ The product of the two numbers is 1855. ∴ (10x + y)(10y + x) = 1855 10x2 + 101xy + 10y2 = 1855 .…..(2) By substituting (1) into (2), we have 10x2 + 101x(x + 2) + 10(x + 2)2 = 1855 2 10x + 101x2 + 202x + 10x2 + 40x + 40 = 1855 121x2 + 242 x – 1815 = 0 x2 + 2 x – 15 = 0 (x – 3)(x + 5) = 0 x = 3 or x = –5 (rejected) By substituting x = 3 into (1), we have y=3+2=5 ∴ The original number is 35.

4 More about Equations ∴ ∵

15. Let x be the present age of the younger brother and y be the present age of the older brother. ∵ The older brother is twice as old as the younger brother. ∴ 2x = y……(1) ∵ After four years, the sum of the squares of their ages is 277. ∴ (x + 4)2 + (y + 4)2 = 277 x2 + 8x + 16 + y2 + 8y + 16 = 277 x2 + y2 + 8x + 8y – 245 = 0 …..(2) By substituting (1) into (2), we have x2 + (2x)2 + 8x + 8(2x) – 245 = 0 5x2 + 24x – 245 = 0 (x – 5)(5x + 49) = 0 x=5

or

x=−

16. (a) Let y cm be the length of a side of the square. ∵ The perimeter of the rectangle is 18 cm greater than that of the square. ∴ 2(4 + x) – 4y = 18 8 + 2x – 4y = 18 x – 2y = 5 x = 5 + 2y ……(1) ∵ The rectangle and the square have equal areas. ∴ 4x = y2 ……(2) By substituting (1) into (2), we have 4(5 + 2y) = y2 2 y – 8y – 20 = 0 (y – 10)(y + 2) = 0 y = 10 or y = –2(rejected) ∴ The length of a side of the square is 10 cm. (b) By substituting y = 10 into (1), we have x = 5 + 2(10) = 25 17. (a) Let f(x) = x3 – 14x + 15. ∵ f(3) = 33 – 14(3) + 15 = 27 – 42 + 15 =0 ∴ x – 3 is a factor of x3 – 14x + 15. (b) By long division, x 2 + 3x − 5 x − 3 x 3 + 0 x 2 − 14 x + 15 x 3 − 3x 2 3x 2 − 14 x 3x 2 − 9x − 5 x + 15 − 5 x + 15

∴ (c) ∵ ∴

2 x3 – 14x + 15 =( x −3)( x +3 x −5)

x3 – 14x + 15 = 0 (x – 3)(x2 + 3x – 5) = 0 x–3=0 or x2 + 3x – 5 = 0 3 x= or x =

−3 ± 3 2 −4(1)( −5) 2(1) =

∴ (b) ∵ ∴

−3 ± 29 2

x2 − 2 x − 1 2 4 3 2 x − x − 2 x − 3x − x + 5x + 2 4

3

2

18. (a) Let f(x) = x – 3x – x + 5x + 2. ∵ f(–1) = (–1)4 – 3(–1)3 – (–1)2 + 5(–1) + 2 =1+3–1–5+2 =0

3

x − x − 2x

2

− 2 x3 + x 2 + 5 x − 2 x3 + 2 x 2 + 4 x − x2 + x + 2 − x2 + x + 2

∴ x4 – 3x3 – x2 + 5x + 2 = (x + 1)(x – 2)(x2 – 2x – 1) ∵ x4 – 3x3 – x2 + 5x + 2 = 0 ∴ (x + 1)(x – 2)(x2 – 2x – 1) = 0 x + 1 = 0 or x – 2 = 0 or x2 – 2x – 1 = 0 1 or x= − x = 2 or x=

−( −2) ± ( −2) 2 −4(1)( −1) 2(1) =1± 2 19.

1 – 8(x – 1)3 = 0 13 – [2(x – 1)]3 = 0 [1 –2(x – 1)][12 + 2(x – 1) + 4(x – 1)2] = 0 (3 – 2x)( 1 + 2x – 2 + 4x2 – 8x + 4) = 0 (3 – 2x)(4x2 – 6x + 3) = 0 ∴ 3 – 2x = 0 or 4x2 – 6x + 3 = 0 x=

3 2

or

x=

−( −6) ± ( −6) 2 −4( 4)( 3) 2( 4) =

6± − 12 8

(rejected) ∴

x=

3 2

20. Let a = 0, b = –1. The equation x4 + ax2 + b = 0 becomes x4 – 1 = 0. x4 – 1 = 0 (x2 – 1)(x2 + 1) = 0 (x + 1)(x – 1)(x2 + 1) = 0 ∴ x + 1 = 0 or x – 1 = 0 or x2 + 1 = 0 x = –1 or x = 1 or x2 = –1(rejected) ∴ The equation has only two real roots. Let a = 1, b = –2. The equation x4 + ax2 + b = 0 becomes x4 + x2 –2 = 0. x4 + x2 –2 = 0 2 (x – 1)(x2 + 2) = 0 x2 = 1 or x2 = –2 (rejected) x = –1 or x = 1 ∴ The equation has only two real roots. ∴

4

x + 1 and x – 2 are factors of f(x). (x + 1)(x – 2) is also a factor of f(x).

Divide f(x) by (x + 1)(x – 2), i.e. x2 – x – 2.

49 5

(rejected) By substituting x = 5 into (1), we have y = 2(5) = 10 ∴ The present ages of the brothers are 5 and 10.

x + 1 is a factor of x4 – 3x3 – x2 + 5x + 2. f(2) = 24 – 3(2)3 – 22 + 5(2) + 2 = 16 – 24 – 4 + 10 + 2 =0 x – 2 is a factor of x4 – 3x3 – x2 + 5x + 2.

a = 0, b = –1 or a = 1, b = –2 (or any other reasonable answers)

104

Certificate Mathematics in Action Full Solutions 4A 21.  y = c

 (1)

 2  y = x + 4 x + 2  ( 2)

By substituting (2) into (1), we have x2 + 4x + 2 = c x2 + 4x + (2 – c) = 0……(3) Let the two solutions of the simultaneous equations be h and k. Then x – h and x – k are the factors of (3). ∴ x2 + 4x + (2 – c) = (x – h)(x – k) = x2 – (h + k)x + hk By comparing constant terms, we have 2 – c = hk ∵ Both solutions are negative. ∴ hk > 0 ∴ 2–c>0 c<2 ∵ The simultaneous equations have two solutions. ∴ (3) has two real roots. ∴ ∆ >0 42 – 4(1)(2 – c) > 0 16 – 8 + 4c > 0 c > –2 ∴

A possible value of c is –1 or reasonable answers)

∵ ∴

−( −3) ± ( −3) 2 −4(1)( −1) 2(1) = ∴

3 ± 13 2

The real roots of the equation are

25.

3x –

2 x +1 = 9

3x – 9 =

2 x +1

(

(3x – 9)2 = 2 x +1 9x2 – 54x + 81 = 2x + 1 9x2 – 56x + 80 = 0 (9x – 20)(x – 4) = 0 x=

20 9

23. By substituting x2 + 2x = u into the equation (x2 + 2x)2 + 5(x2 + 2x) + 4 = 0, we have u2 + 5u + 4 = 0 (u + 1)(u + 4) = 0 u = –1 or u = –4 ∵ x2 + 2x = u ∴ x2 + 2x = –1 or x2 + 2x = –4 2 2 x + 2x + 1 = 0 or x + 2x + 4 = 0 (x + 1)2 = 0 or x =

−2 ± 2 2 −4(1)( 4) 2(1) ∴

x = –1 or x = −1 ± −3 (rejected) The real root of the equation is –1.

24. (x2 – 3x)2 – 2x2 + 6x + 1 = 0 (x2 – 3x)2 – 2(x2 – 3x) + 1 = 0 By substituting x2 – 3x = u into the equation (x2 – 3x)2 – 2(x2 – 3x) + 1 = 0, we have u2 – 2u + 1 = 0

105

)

2

or

x=4

Checking: When x =

 20  –  9 

20 , 3x – 9

2 x +1 = 3 

13  20  2  +1 = 3  9  ≠ 9 When x = 4, 3x – 2 x +1 = 3(4) – 2( 4) +1 = 9 ∴ The real root of the equation is 4. 26.

5x –

5 x +6 = 6

5x – 6 =

Level 2

3 − 13 and 2

3 + 13 . 2

1 . (or any other 4

22. By substituting x = u into the equation x + 2 x = k, we have u2 + 2u = k u2 + 2u – k = 0 …..(1) ∵ The equation x + 2 x = k has no real roots. ∴ The equation u2 + 2u – k = 0 also has no real roots. ∴ The discriminant of u2 + 2u – k = 0 is negative. ∆ <0 22 – 4(1)( –k) < 0 4 + 4k < 0 4k < –4 k < –1 ∴ A possible value of k is –2 or –3. (or any other reasonable answers)

(u – 1)2 = 0 u=1 x2 – 3x = u x2 – 3x = 1 2 x – 3x – 1 = 0 x=

5 x +6

(

(5x – 6)2 = 5 x +6 25x2 – 60x + 36 = 5x + 6 25x2 – 65x + 30 = 0 5x2 – 13x + 6 = 0 (5x – 3)(x – 2) = 0 x=

3 5

or

)

2

x=2

Checking: When x =

3 , 5x – 5

3 – 5

5 x +6 = 5 

3 5  + 6 = 0 ≠ 6 5 When x = 2, 5x – 5 x +6 = 5(2) – 5( 2) +6 = 6 ∴ The real root of the equation is 2. 27. y = x2 – x x –2 –1 y 6 2

0 0

1 0

2 2

3 6

4 More about Equations 2x – 3y = 1 x –1 0.5 y –1 0

∵ ∴

2 1

The two graphs intersect at (0.2, –0.2) and (1.4, 0.6). The solutions of the simultaneous equations are approximately (0.2, –0.2) and (1.4, 0.6).

106

Certificate Mathematics in Action Full Solutions 4A 28. y = x2 – 4x + 3 x 0 1 2 y 3 0 –1

3 0

4 3

x y + =1 2 3 x y

0 3

2 0

4 –3

29. ∵

2x2 – 5x + 2 = 0 2x2 – 4x + 2 = x x2 – 2x + 1 =

∴

∵ ∴

The two graphs intersect at (0, 3.0) and (2.5, –0.8). The solutions of the simultaneous equations are approximately (0, 3.0) and (2.5, –0.8).

For questions 29 to 30, refer to the graph below: y = x2 – 2x + 1 x y

–2 –1 9 4

0 1

1 0

2 1

3 4

4 9

x 2

The corresponding simultaneous equations are

 y = x2 − 2x +1  . x  y=  2  Draw the straight line y =

x on the graph of 2

y = x2 – 2x + 1. From the graphs, the roots of 2x2 – 5x + 2 = 0 are approximately 0.5 and 2.0. 30. ∵

2x2 + 4x + 1 = 0 2x2 + 4x – 8x + 1 + 1 = 1 – 8x 2x2 – 4x + 2 = 1 – 8x x2 – 2x + 1 =

∴

1 −4x 2

The corresponding simultaneous equations are

 y = x 2 − 2 x +1  . 1  y = − 4x  2  Draw the straight line y =

1 − 4 x on the graph of 2

y = x2 – 2x + 1. From the graphs, the roots of 2x2 + 4x + 1 = 0 are approximately –1.7 and –0.3.

107

4 More about Equations

31. (a) y = x2 + 2x x –2 –1 y 0 –1

∴ 0 0

1 3

2 8

x2 + 2x – 3x – 1 = 0 x2 + 2x = 3x + 1 The corresponding simultaneous equations are

y = x 2 + 2 x .  y = 3 x +1

3 4 5 15 24 35

Draw the straight line y = 3x + 1 on the graph of y = x2 + 2x. From the graphs, the roots of x2 – x – 1 = 0 are approximately –0.6 and 1.6.

(b)

(iii) ∵

2x2 + 3x – 2 = 0 2x + 4x – x – 2 = 0 2x2 + 4x = x + 2 2

x2 + 2x = ∴

x +1 2

The corresponding simultaneous equations are

 y = x 2 + 2x  . x   y = 2 + 1 Draw the straight line y =

x +1 on the graph of 2

y = x2 + 2x. From the graphs, the roots of 2x2 + 3x – 2 = 0 are approximately –2.0 and 0.5.

32. (a) From the graph, the y-intercept of y = ax2 + bx + c is 4. ∴ 4 = a(0)2 + b(0) + c c= 4 From the graph, the x-intercepts of y = ax2 + bx + c are – 3 and 1. ∴ 0 = a(–3)2 + b(–3) + c 0 = 9a –3b + c ……(1) and 0 = a(1)2 + b(1) + c 0=a+b+c ……(2) By substituting c = 4 into (1) and (2), we have 0 = 9a –3b + 4 ……(3) and 0 = a + b + 4 ……(4) By solving (3) and (4), we have 4 8 a=− ,b=− 3 3 (b) By substituting a = −

4 8 ,b=− and c = 4 3 3

into y = ax2 + bx + c, we have y=−

(c) (i) ∵ ∴

2

x – 3x + 1 = 0 x2 + 2x – 5x + 1 = 0 x2 + 2x = 5x – 1 The corresponding simultaneous equations are

y = x 2 + 2 x .  y = 5 x −1 Draw the straight line y = 5x – 1 on the graph of y = x2 + 2x. From the graphs, the roots of x2 – 3x + 1 = 0 are approximately 0.4 and 2.6. (ii) ∵

x2 – x – 1 = 0

4 2 8 x − x +4 3 3

We are going to solve the following simultaneous equations:

4 8   y = − x2 − x + 4  3 3  x − 2 y + 8 = 0

 (5)  (6)

By substituting (5) into (6), we have

8 4  x − 2 − x2 − x + 4 +8 = 0 3 3  8 2 19 x + x =0 3 3 8x2 + 19x = 0

108

Certificate Mathematics in Action Full Solutions 4A x(8x + 19) = 0 x = 0 (rejected)

or

19 8

x=−

19 into (6), we have 8

By substituting x = −

19 − – 2y + 8 = 0 8 45 y= 16 ∴

By substituting x = y=

d.p.)

y=

19 45 , ). 8 16

 33. x + y − x =1  (1) 2 x + 3 y =1  ( 2)

− 4 + 2 10 14 + 2 10 +6= = 6.77(cor. to 2 3 3

∴

The solutions of the simultaneous equations are (–3.44, 2.56) and (0.77, 6.77).

35. (a) ∵ ∴

1 −2x ……(3) 3

By substituting (3) into (1), we have x2 +

1 −2 x –x=1 3

∴

1 3

By substituting x = −

or

x=2

1 into (3), we have 3

 1 1 − 2 −   3 = 5 y= 3 9 By substituting x = 2 into (3), we have y = ∴

1 −2( 2) = –1 3

The solutions of the simultaneous equations are

∴

x − y + 6 = 0   3 x 2 + 3x − y + 4 = 0  4

 (1)  ( 2)

From (1), we have y = x + 6 ……(3) By substituting (3) into (2), we have

3 2 x + 3x – (x + 6) + 4 = 0 4

x= x=

−8 ± 8 2 −4(3)( −8) 2(3)

−4 − 2 10 3

x = –3.44(cor. to 2 d.p.)

109

or or

x=

− 4 + 2 10 3

x = 0.77(cor. to 2 d.p.)

x = 20 y = 21

The solutions are 

or

x = 21 .  y = 20

36. Let x cm and y cm be the length and the width of the rectangle respectively. ∵ The perimeter of the rectangle is 28 cm. ∴ 2(x + y) = 28 y = 14 – x ……(1) The length of the diagonal = x 2 + y 2 cm ∵ The product of the lengths of the diagonals is 100 cm2. ∴

3x2 + 12x – 4x – 24 + 16 = 0 3x2 + 8x – 8 = 0 Using the quadratic formula, we have

 (1) .  ( 2)

(b) From (1), we have x + y = 41 y = 41 – x ……(3) By substituting (3) into (2), we have x2 + (41 – x)2 = 841 x2 + 1681 – 82x + x2 = 841 2x2 – 82x + 840 = 0 x2 – 41x + 420 = 0 (x – 20)(x – 21) = 0 x = 20 or x = 21 By substituting x = 20 into (3), we have y = 41 – 20 = 21 By substituting x = 21 into (3), we have y = 41 – 21 = 20

 1 5  − ,  and (2, –1).  3 9 34.

The length of the wire is 70 cm. x + y + 29 = 70 x + y = 41 AB 2 + BC 2 = AC 2 (Pyth. theorem) x2 + y2 = 292 x2 + y2 = 841 The required simultaneous equations are

x + y = 41  2 2 x + y = 841

3x2 + 1 – 2x – 3x = 3 3x2 – 5x – 2 = 0 (3x + 1)(x – 2) = 0 x=−

− 4 + 2 10 into (3), we have 3

d.p.)

2

From (2), we have y =

−4 − 2 10 14 −2 10 +6= = 2.56(cor. to 2 3 3

By substituting x =

The coordinates of P are ( −

−4 − 2 10 into (3), we have 3

  2 2  2 2   x +y  x +y =100   

x2 + y2 = 100 ……(2) By substituting (1) into (2), we have x2 + (14 – x)2 = 100 2 x + 196 – 28x + x2 = 100 2x2 – 28x + 96 = 0 x2 – 14x + 48 = 0 (x – 6)(x – 8) = 0 x = 6 or x = 8 By substituting x = 6 into (1), we have y = 14 – 6 = 8 By substituting x = 8 into (1), we have y = 14 – 8 = 6 ∴ The length and the width of the rectangle is 6 cm and

4 More about Equations 8 cm or the length and the width of the rectangle is 8 cm and 6 cm respectively. 37. ∵ ∴

The length of the wire is 36 cm. 8x + 4y = 36 y = 9 – 2x ……(1) ∵ The total surface area enclosed by the framework is 48 cm2. ∴ 2x2 + 4xy = 48 ……(2) By substituting (1) into (2), we have 2x2 + 4x(9 – 2x) = 48 2x2 + 36x – 8x2 = 48 6x2 – 36x + 48 = 0 x2 – 6x + 8 = 0 (x – 2)(x – 4) = 0 x = 2 or x = 4 By substituting x = 2 into (1), we have y = 9 – 2(2) = 5 By substituting x = 4 into (1), we have y = 9 – 2(4) = 1 ∴

x = 2 or y = 5

The solutions are 

x = 4 .  y =1

38. Let x cm and y cm be the length and width of each rectangle respectively. ∵ The length of the wire is 78 cm. ∴ 10x + 12y = 78 y=

39 −5 x 6

……(1)

∵ ∴

The area enclosed by the framework is 96 cm2. 8xy = 96 xy = 12 ……(2) By substituting (1) into (2), we have

 39 − 5 x   = 12 6  

x

39x – 5x2 = 72 5x2 – 39x + 72 = 0 (x – 3)(5x – 24) = 0 x=3 or x = 4.8 By substituting x = 3 into (1), we have y=

2

2

2

By substituting (2) into (1), we have x2 – 6x + 6 = 2x + k – 2 2 x – 8x + (8 – k) = 0 ……(3) ∵ The simultaneous equations have only one solution. ∴ (3) has only one real root. ∴ ∆ =0 (–8)2 – 4(1)(8 – k) = 0 64 – 32 + 4k = 0 32 + 4k = 0 8 k= − (b) By substituting k = –8 into (3), we have x2 – 8x + [8 – (–8)] = 0 x2 – 8x + 16 = 0 (x – 4)2 = 0 x=4 By substituting x = 4 and k = –8 into (1), we have y = 2(4) + (–8) – 2 = –2 ∴ The coordinates of P are (4, –2). f(–1) = (–1)3 + 7(–1)2 + 15(–1) + 9 = – 1 + 7 – 15 + 9 =0 x + 1 is a factor of f(x).

x3 + x 2 6x 2 + 15 x 6x 2 + 6 x 9x + 9 9x + 9

∴

x 2 + y 2 cm.

x3 + 7x2 + 15x + 9 = (x + 1)(x2 + 6x + 9) =( x +1)( x +3) 2

(b) ∵ ∴

f(x) = 0 (x + 1)(x + 3)2 = 0 x + 1 = 0 or (x + 3)2 = 0 1 3 x= − or x = −

42. (a) ∵

f(2) = 23 – 3(2)2 –10(2) + 24 = 8 – 12 – 20 + 24 =0 x – 2 is a factor of f(x).

∴

By long division,

= 18

x + y = 18 ∵ The perimeter of ABCD is 24 cm. ∴ 4(x + y) = 24 y=6–x By substituting (3) into (2), we have x2 + (6 – x) 2 = 18 x2 + 36 – 12x + x 2 = 18

 (1)

 2  y = x − 6 x + 6  ( 2)

x 2 + 6x + 9

39. (a) Consider △AHE. Let AH = x cm and AE = y cm. ∴ The length of the side of ABCD is (x + y) cm. ∵ ∠ A is a right angle. ∴ AH 2 + AE 2 = EH 2 (Pyth. theorem) (x cm)2 + (y cm)2 = EH 2 EH = x 2 + y 2 cm ……(1)

 2 2   x +y   

40. (a)  y = 2 x + k − 2

x + 1 x 3 + 7 x 2 + 15 x + 9

The dimensions of each rectangle are 3 cm × 4 cm or 4.8 cm × 2.5 cm.

∴

∴

By long division,

39 −5( 4.8) y= = 2.5 6

The length of the side of EFGH is The area of EFGH is 18 cm2.

The triangle has two equal sides and one right angle. It is an isosceles right-angled triangle.

∴

39 −5(3) =4 6

∴ ∵

(b) ∵

41. (a) ∵

By substituting x = 4.8 into (1), we have

∴

2x2 – 12x + 18 = 0 x2 – 6x + 9 = 0 (x – 3)2 = 0 x=3 By substituting x = 3 into (3), we have y = 6 – 3 = 3 By substituting x = 3 and y = 3 into (1), we have EH = x 2 + y 2 cm = 32 +32 cm =3 2 cm ∴ The sides of each triangle are 3 cm, 3 cm and 3 2 cm.

x 2 − x − 12 x − 2 x 3 − 3x 2 − 10 x + 24 x3 − 2 x2 − x 2 − 10x − x 2 + 2x − 12x + 24 − 12x + 24

……(2) ∴ ……(3) (b) ∵ ∴

x3 – 3x2 – 10x + 24 = (x – 2)(x2 – x – 12) =( x −2)( x −4)( x + 3)

f(x) = 0 (x – 2)(x – 4)(x + 3) = 0

110

Certificate Mathematics in Action Full Solutions 4A x – 2 = 0 or x = 2 or 43. (a) ∵

x – 4 = 0 or x = 4 or

x+3=0 3 x= −

−1 ± 12 −4(1)( −4) 2(1)

f(3) = 33 – 2(3)2 –23(3) + 60 = 27 – 18 – 69 + 60 =0 x – 3 is a factor of f(x).

∴

=

−1 ± 17 2

Multiple Choice Questions (p.200)

By long division, x 2 + x − 20 x − 3 x 3 − 2x 2 − 2 3x + 6 0 3

x − 3x

2

2

x − 2 3x x 2 − 3x − 20 x + 60 − 20 x + 60

∴

1.

Answer: A By substituting x2 = u into the equation x4 – 8x2 – 9 = 0, we have u2 – 8u – 9 = 0 (u – 9)(u + 1) = 0 u = 9 or u = –1 ∵ x2 = u ∴ x2 = 9 or x2 = –1 (rejected) 1 x= ±

2.

Answer: C By substituting x3 = u into the equation x6 + 9x3 +8 = 0, we have u2 + 9u + 8 = 0 (u + 8)(u + 1) = 0 u = –8 or u = –1 ∵ x3 = u ∴ x3 = –8 or x3 = –1 2 1 x= − or x = −

3.

Answer: C By substituting x = u into the equation x + we have u2 + u – 2 = 0 (u – 1)(u + 2) = 0 u = 1 or u = –2 ∵ x =u

x3 – 2x2 – 23x + 60 = (x – 3)(x2 + x – 20) =( x − 3)( x −4)( x +5)

(b) ∵ ∴

f(x) = 0 (x – 3)(x – 4)(x + 5) = 0 x – 3 = 0 or x – 4 = 0 or x = 3 or x = 4 or

44. (a) ∵

x+5=0 5 x= −

f(–1) = (–1)4 – 2(–1)3 – 7(–1)2 + 8(–1) + 12 = 1 + 2 – 7 – 8 + 12 =0 x + 1 is a factor of f(x). f(2) = 24 – 2(2)3 – 7(2)2 + 8(2) + 12 = 16 – 16 – 28 + 16 + 12 =0 x – 2 is a factor of f(x). x + 1 and x – 2 are factors of f(x). (x + 1)(x – 2) is also a factor of f(x).

∴ ∵ ∴ ∵ ∴

Divide f(x) by (x + 1)(x – 2), i.e. x2 – x – 2. x2 − x − 6 x 2 − x − 2 x4 − 2 x3 − 7 x 2 + 8 x + 1 2 x4 − x 3 − 2 x2 − x3 − 5 x2 + 8 x − x3 + x2 + 2 x − 6 x2 + 6 x + 12 − 6 x2 + 6 x + 12

∴

x4 – 2x3 – 7x2 + 8x + 12 = (x + 1)(x – 2)(x2 – x – 6)

=( x + 1)( x −2)( x +2)( x − 3)

(b) ∵ ∴

f(x) = 0 (x + 1)(x – 2)(x + 2)(x – 3) = 0 x + 1 = 0 or x – 2 = 0 or x + 2 = 0 or x – 3 = 0 1 2 x= − or x = 2 or x = − or

∴

x4 + 6x3 – 7x2 – 36x + 6 = 0 (x – 7x2 + 6) + (6x3 – 36x) = 0 [(x2)2 – 7x2 + 6] + 6x(x2 – 6) = 0 (x2 – 6)(x2 – 1) + 6x(x2 – 6) = 0 (x2 – 6)[(x2 – 1) + 6x] = 0 (x2 – 6)(x2 + 6x – 1) = 0 ∴ x2 – 6 = 0 or x2 + 6x – 1= 0 x=± 6 or x=

4.

4

Answer: C ∵ ax2 + (b – m)x + (c – d) = 0 ax2 + bx – mx + c – d = 0 ax2 + bx + c = mx + d ∴ The corresponding simultaneous equations are

y = ax 2 + bx + c .  y = mx + d From the graphs, the roots of ax2 + (b – m)x + (c – d) = 0 are 0 and 2. 5.

−6 ± 6 −4(1)( −1) 2

Answer: B ∵ y = x2

2(1)

y = 3x – = −3 ± 10

46.

4

3

111

∴

2

x + x – 5x – x + 4 = 0 (x4 – 5x2 + 4) + (x3 – x) = 0 [(x2)2 – 5x2 + 4] + x(x2 – 1) = 0 (x2 – 4)(x2 – 1) + x(x2 – 1) = 0 (x2 – 1)[(x2 – 4) + x] = 0 (x2 – 1)(x2 + x – 4) = 0 ∴ x2 – 1 = 0 or x2 + x – 4= 0 1 or x= ± x=

x = –2 (rejected)

x =1

x=3 45.

x = 1or

x – 2 = 0,

∴ 6.

3 2 x2 = 3x –

3 2

2x2 = 6x – 3 2x2 – 6x + 3 = 0 The quadratic equation can be solved is 2x2 – 6x + 3 = 0.

Answer: C ∵ The equation x2 + ax + b = 0 has no x-intercepts.

4 More about Equations ∴ It has no real roots. ∴ I is wrong.

∴

10. Answer: A

∵

x2 + (a – 1)x + b = 0 x2 + ax – x + b = 0 x2 + ax + b = x ∴ The corresponding simultaneous equations are

y = x 2 − 5 x + 6  (1)   ( 2) y = 2 x By substituting (1) into (2), we have x2 – 5x + 6 = 2x x2 – 7x + 6 = 0 (x – 1)(x – 6) = 0 x = 1 or x=6 By substituting x = 1 into (2), we have y = 2(1) = 2 By substituting x = 6 into (2), we have y = 2(6) = 12 ∴ The solutions of the simultaneous equations are 2) and (6, 12).

y = x 2 + ax + b .  y = x From the graphs, the simultaneous equations have one real solution. ∴ The equation x2 + (a – 1)x + b = 0 has a double real root. ∴ II is correct. The corresponding simultaneous equations of

2

x + ax

y = x 2 + ax + b + b = c are  . y = c

 (1) y = 2 x + k + 2  2 y = −x + 6 x − 6  ( 2) By substituting (2) into (1), we have –x2 + 6x – 6 = 2x + k + 2 x2 – 4x + (k + 8) = 0 ……(3) ∵ The simultaneous equations have only one solution. ∴ (3) has only one real root. ∴ ∆ =0 (–4)2 – 4(1)(k + 8)= 0 16 – 4k – 32 = 0 k =− 4 ∴

Answer: D

y = x 2 + ax + b  y = mx + c

(1,

11. Answer: A

From the graphs, the simultaneous equations have two real solutions. ∴ The equation x2 + ax + b = c has two real roots. ∴ III is correct. 7.

k =4

 (1)  ( 2)

By substituting (1) into (2), we have x2 + ax + b = mx + c x2 + (a – m)x + (b – c) = 0 ……(3) ∴ The simultaneous equations have two real solutions. ∴ (3) has two real roots. ∴ ∆ >0 (a – m)2 – 4(1)(b – c) > 0 (a – m)2 – 4(b – c) > 0 ∴ (a – m)2 > 4(b – c)

12. Answer: D Let f(x) = x3 – 3x + 2. ∵ f(1) = 13 – 3(1) + 2 =1–3+2 =0 ∴ x – 1 is a factor of x3– 3x + 2. By long division, x2 + x − 2 x − 1 x3 + 0 x 2 − 3 x + 2 x3 − x 2 x 2 − 3x x2 − x − 2x + 2 − 2x + 2

8.

Answer: D ∵ 2x2 + x – 3 = 0 2x2 = – x + 3 x2 = ∴

−x + 3 2

x3 – 3x + 2 = (x – 1)(x2 + x – 2) = (x – 1)(x – 1)(x + 2) = (x – 1)2(x + 2) f(x) = 0 (x – 1)2(x + 2) = 0 (x – 1)2 = 0 or x + 2 = 0 2 x = 1 or x= −

.

The equation of the required straight line is

−x +3 y= . 2 9.

∵ ∴

The corresponding simultaneous equations are

y = x2  − x+3  y=  2  ∴

∴

Answer: C

 (1) y = 4 x  2 y = x + k  ( 2) By substituting (2) into (1), we have x2 + k = 4x 2 x – 4x + k = 0 ……(3) ∵ The simultaneous equations have only one solution. ∴ (3) has only one real root. ∴ ∆ =0 (–4)2 – 4(1)k = 0 16 – 4k = 0

HKMO (p. 201) Let f(x) = 2x3 + 7x2 – 29x – 70. ∵ f(–2) = 2(–2)3 + 7(–2)2 – 29(–2) – 70 = –16 + 28 + 58 – 70 =0 ∴ x + 2 is a factor of 2x3 + 7x2 – 29x – 70. By long division, 2 x2 + 3 x − 3 5 x + 2 2 x3 + 7 x 2 − 2 9x − 70 2 x3 + 4 x 2 3x 2 − 29x 3x 2 + 6 x − 35 x − 70 − 35 x − 70

∴ ∵ ∴ ∴

2x3 + 7x2 – 29x – 70 = (x + 2)(2x2 + 3x – 35) = (x + 2)(x + 5)(2x – 7) f(x) = 0 (x + 2)(x + 5)(2x – 7) = 0 x+2=0 or x + 5 = 0 or 2x – 7 = 0 x = –2

or

x = –5

or

x=

7 2 112

Certificate Mathematics in Action Full Solutions 4A ∵ ∴

p is the positive real root. 7 p= 2

Let’s Discuss p. 170 Both methods are correct. p. 180 Ken’s method is more tedious as it involves squaring of the expression y + 1.

113