4ach04(more About Equations)

Certificate Mathematics in Action Full Solutions 4A

4 More about Equations • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

(x + 2)(x – 6) = 0 x = –2 or

Follow-up Exercise

Checking:

p. 162 1.

2.

3.

4.

4

2

By substituting x = u into the equation x – 10x + 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.

When x = 6,

x2 − 4x − 8 = 62 − 4 × 6 − 8 = 2 ∴ The real roots of the equation are –2 and 6. 2.

By substituting x2 = u into the equation 4x4 – 17x2 + 4 = 0, we have 4u2 – 17u + 4 = 0 (4u – 1)(u – 4) = 0 1 u= or u = 4 4 ∵ x2 = u 1 ∴ x2 = or x2 = 4 4 1 x=± or x = ±2 2 1 1 ∴ The real roots of the equation are –2, − , and 2. 2 2 3

6

(

Checking:

Alternative Solution By substituting x = u into the equation x + 3 x = 4, we have u2 + 3u = 4 u2 + 3u – 4 = 0 (u – 1)(u + 4) = 0 u = 1 or u = –4 ∵ = u x x = 1 or x = –4 (rejected) x=1 ∴ The real root of the equation is 1. ∴

3.

x –3 x−2 = 0 x =3 x−2

(

 2  2  x − 4x − 8  = 2   x2 – 4x – 8 = 4 x2 – 4x – 12 = 0

)

x2 = 3 x − 2 2 x2 = 9x – 18 x2 – 9x + 18 = 0 (x – 3)(x – 6) = 0 x = 3 or x = 6

3

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. x + 2 x − 2 = 10

2

2

When x = 1, x + 3 x = 1 + 3 1 = 4

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

4. x − 4x − 8 = 2

)

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

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

p. 163

x +3 x = 4 x – 4 = –3 x

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.

83

x 2 − 4 x − 8 = ( −2) 2 − 4 × ( −2) − 8 = 2

When x = –2, 2

1.

x=6

(2

2 x − 2 = 10 – x

)

2 2 x − 2 = (10 – x) 4x – 8 = 100 – 20x + x2 2 x – 24x + 108 = 0

4 More about Equations (x – 6)(x – 18) = 0 x=6 or x = 18 Checking: When x = 6, x + 2 x − 2 = 6 + 2 6 − 2 = 10 When x = 18, x + 2 x − 2 = 18 + 2 18 − 2 = 26 ≠ 10 ∴ The real root of the equation is 6.

p. 175 1.

p.168

(a) ∵ x2 – 4x = 0 x2 = 4x ∴ The equation of the required straight line is y = 4x. (b) ∵ x2 + 3x – 2 = 0 x2 = –3x + 2 ∴ The equation of the required straight line is y = –3x + 2.

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

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

1.

y = 3x – 5 x y

1 –2

2 1

3 4

(b) ∵ y = 3x2 y = 4x + 1 ∴ 3x2 = 4x + 1 2 3x – 4x – 1 = 0 ∴ The quadratic equation that can be solved is 3x2 – 4x – 1 = 0.

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

y = –2x – 2 x y

–2 2

–1 0

(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.

0 –2

∵ The two graphs do not intersect. ∴ The simultaneous equations have no real solutions. 3.

(a) ∵ y = x2 y=x–3 ∴ x2 = x – 3 2 x –x+3=0 ∴ The quadratic equation that can be solved is x2 – x + 3 = 0.

–x + y = 2 3. 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). 4.

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.

84

Certificate Mathematics in Action Full Solutions 4A

x2 – 2x – 3 = 0 x2 – 2x = 3 ∴ The corresponding simultaneous equations are  y = x 2 − 2x .  y = 3 Draw the straight line y = 3 on the graph of y = x2 – 2x. From the graphs, the roots of x2 – 2x – 3 = 0 are –1 and 3.

By substituting (3) into (2), we have 3y2 = 9y – 6 2 3y – 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).

(a) ∵

3.

x2 + 2x + 1 = 0 x – 2x + 4x +1 = 0 x2 – 2x = –4x –1 ∴ The corresponding simultaneous equations are  y = x 2 − 2x .   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.

(b) ∵

2

(c) ∵

2

2x + x – 6 = 0 2x2 + x – 5x – 6 = – 5x 2x2 – 4x – 6 = – 5x 5 x2 – 2x – 3 = − x 2 5 x2 – 2x = − x + 3 2 ∴ The corresponding simultaneous equations are  y = x 2 − 2x  y = − 5 x + 3 .  2 5 Draw the straight line y = − x + 3 on the graph of 2 y = x2 – 2x. From the graphs, the roots of 2x2 + x – 6 = 0 are –2 and 1.5.

4.

2.

 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).  x − 9 y + 6 = 0  (1)  2  ( 2) 3 y = x From (1), we have x = 9y – 6

85

……(3)

 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 x2 + 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.

5.

p.183 1.

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

p.185 1.

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 2 x – 23x + 120 = 0

4 More about Equations

2.

(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  (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 2 x + 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 x = 6 x = 8 ∴ The solutions are  or  . y = 8 y = 6

3.

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) ∵ The factors of f(x) are (x – 2), (x +1) and (x + 5). ∴ f(x) = ( x − 2)( x + 1)( x + 5) (c) ∵ f(x) = 0 ∴ (x – 2)(x + 1)(x + 5) = 0 x – 2 = 0 or x + 1 = 0 or x + 5 = 0 x = 2 or x = −1 or x = −5 2.

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 x2 + 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 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

4

−8

x + x − 6 x + x − 14 x 2 − 8 x + 48

3

x4 + x3 − 6x 2 − 8 x 2 − 8 x + 48 − 8 x 2 − 8 x + 48 ∴ 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 x = 2 or x = −3 or x =±2 2

p.192 1.

(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.

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)

86

Certificate Mathematics in Action Full Solutions 4A = (x + 1)(x – 1)(2x – 1)(x + 1) = (x + 1)2(x – 1)(2x – 1) 4 3 2 ∵ 2x + x – 3x – x + 1 = 0 ∴ (x + 1)2(x – 1)(2x – 1) = 0 (x + 1)2 = 0 or x – 1 = 0 or 2x – 1 = 0 1 x = −1 or 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) ∵ 8(x + 1)3 – (x + 2)3 = 0 ∴ x(7x2 + 18x + 12) = 0 ∴ x = 0 or 7x2 + 18x + 12 = 0 x=0

4.

or

x = ±3

or

∴ The real roots of the equation are –3, − 3 , 3 and 3. 5.

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.

6.

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.

7.

By substituting x3 = u into the equation x6 + 9x3 + 8 = 0, 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

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

− 18 ± 18 2 − 4(7)(12) x= 2( 7 ) =

∴

∴ The real roots of the equation are –4 and 4.

Exercise Exercise 4A (p.164) Level 1 1.

2.

3.

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. 2

4

2

By substituting x = u into the equation x – 11x – 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

87

8.

x2 + 2x +1 = 3 2

 2  2  x + 2x + 1  = 3   x2 + 2x + 1 = 9 x2 + 2x – 8 = 0 (x + 4)(x – 2) = 0 x = –4 or

x=2

Checking: x 2 + 2 x + 1 = ( −4) 2 + 2( −4) + 1 = 3

When x = –4, When x = 2,

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

∴ The real roots of the equation are –4 and 2. 9.

x – x – 12 = 0 x – 12 = x (x – 12)2 =

( x)

x2 – 24x + 144 = x x2 – 25x + 144 = 0 (x – 9)(x – 16) = 0 x=9

2

or

x = 16

Checking: When x = 9, x – x – 12 = 9 – 9 – 12 = –6 ≠ 0 When x = 16, x – x – 12 = 16 – 16 – 12 = 0

4 More about Equations x = 0 or

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

10.

x –5 x + 6 = 0 x + 6 =5 x

( )

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

2

or

x=9

Checking: When x = 4, x – 5 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. Alternative Solution By substituting x = u into the equation x – 5 x + 6 = 0, we have u2 – 5u + 6 = 0 (u – 2)(u – 3) = 0 u=2 or u = 3 ∵ x =u x = 2 or x =3 x = 4 or x=9 ∴ The real roots of the equation are 4 and 9. ∴

11.

x + 2 + x = 10

(

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

12.

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

x=3

Checking: When x = 0, x – x + 1 = 0 – 0 + 1 = –1 ≠ 1 When x = 3, x – x + 1 = 3 – 3 + 1 = 1 ∴ The real root of the equation is 3. 13. ∵ 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 x−2 = k

14.

2 x − 2 = –x + k

(2

)

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

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 9 u=1 or u = − 4 ∵ x2 = u 9 ∴ x2 = 1 or x2 = − (rejected) 4 x = ±1 ∴ The real roots of the equation are –1and 1. 16. By substituting x2 = u into the equation 9x4 – 37x2 + 4 = 0, we have 9u2 – 37u + 4 = 0 (9u – 1)(u – 4) = 0 1 u= or u = 4 9

88

Certificate Mathematics in Action Full Solutions 4A ∵ x2 = u 1 ∴ x2 = 9 x=±

x=1 or

x2 = 4

1 or 3

x = ±2 1 1 , and 2. 3 3

(

3

4 .

3x + 3 = 3x – 3

)

Checking: When x =

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

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

)

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

Checking: When x =

7 , 4

13 7 3 x − 3 + 5 = 3  − 3 + 5 = 2 4  7  7 13 2x = 2   = ≠ 2 4 2

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

x=± 3

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

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

When x = 2,

(

1 1 and . 5 5

1 , 3

1 3x – 1 = 3   – 1 = 0 ≠ 4 3

22.

19. By substituting x2 = u into the equation 4x4 – 19x2 + 21 = 0, we have 4u2 – 19u + 21 = 0 (4u – 7)(u – 3) = 0 7 u= or u = 3 4 2 ∵ x =u 7 ∴ x2 = or x2 = 3 4

7 7 , 2 2

and 3 . 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

89

4

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 1 x= or x = 2 3

1 1 ∴ The real roots of the equation are –5, − , and 5. 2 2

or

3

3x + 3 + 2 = 3x – 1

21.

17. By substituting x2 = u into the equation 4x4 – 101x2 + 25 = 0, we have 4u2 – 101u + 25 = 0 (4u – 1)(u – 25) = 0 1 u= or u = 25 4 ∵ x2 = u 1 ∴ x2 = or x2 = 25 4 1 x = ± or x = ±5 2

7 x=± 2

x=

∴ The real roots of the equation are 1 and

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

∴ The real roots of the equation are −

or

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 x 2 + 5 x + 1 = 3x2 + 15x + 3 – 3 + 2 x 2 + 5 x + 1

4 More about Equations = 3(x2 + 5x + 1) + 2 x 2 + 5 x + 1 – 3 By substituting

x=

− 7 ± 7 2 − 4(1)(14) 2(1)

or

(x – 2)(x – 7) = 0

x=

−7± −7 (rejected) 2

or

x=2

x 2 + 5 x + 1 = u into the expression

3x2 + 15x + 2 x 2 + 5 x + 1 , we have 2

3x + 15x + 2 x 2 + 5 x + 1

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 5 u=1 or u = − 3

x=7

Checking: When x = 2, x2 + x + 2 x x + 2 = 22 + 2 + 2(2) 2 + 2 = 14 2 When x = 7, x + x + 2 x x + 2 = 72 + 7 + 2(7) 7 + 2 = 98 ≠ 14 ∴ The real root of the equation is 2.

= 3(x2 + 5x + 1) + 2 x 2 + 5 x + 1 – 3 = 3u2 + 2u – 3 (b) By substituting

or

Exercise 4B (p.176)

∵

x2 + 5x + 1 = u

Level 1

∴

5 x 2 + 5 x + 1 = 1 or x 2 + 5 x + 1 = − (rejected) 3

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

( x 2 + 5 x + 1 ) 2 = 12 x 2 + 5x + 1 = 1 x 2 + 5x = 0 x( x + 5) = 0 x = 0 or Checking: When x = 0,

x = −5

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.

(

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

)

2

1.

x y

–2

= x2 + 2 x x + 2 + x + 2 – 2 = x2 + x + 2 x x + 2 = LHS ∴

(

x2 + x + 2 x x + 2 ≡ x + x + 2

y=6

)

2

1 6

2 6

3 6

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

–2 2.

y=x+3

x + x + 2 x x + 2 = 14 2

(b)

(x +

(x +

x+2 x+2

)

2

)

2

– 2 = 14

x y

(By (a))

– 16 = 0

(

)

2

2 x + 2 = (–x – 4) 2 x + 2 = x + 8x + 16 x2 + 7x + 14 = 0

or or or or

x+ x+2 – 4 = 0

(

)

2 2 x + 2 = (–x + 4) 2 x + 2 = x – 8x + 16 x2 – 9x + 14 = 0

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 + x + 2 + 4)( x + x + 2 – 4) = 0 x+ x+2 + 4 = 0

–1 2

3.

y = 2x – 1

x

0

1

2

90

Certificate Mathematics in Action Full Solutions 4A y

–1

1

3

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

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

9.

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

10. ∵ x2 – 4x – 4 = 0 x2 = 4x + 4 ∴ The equation of the required straight line is y = 4x + 4. 11. ∵ x2 + 5x + 1 = 0 x2 = –5x – 1 ∴ The equation of the required straight line is y = –5x – 1. 4.

y=x–1

x y

–1 0 1 –2 –1 0 ∵ 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). 5.

Level 2

y = 3x x y

–1 –3

0 0

1 3

∵ 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.

y = 2x + 3 x y

12. ∵ 2x2 + 3x – 4 = 0 2x2 = –3x + 4 3 x2 = − x + 2 2 ∴ The equation of the required straight line is 3 y = − x + 2. 2

–1 1

0 3

13. y = x2 + 1 x y

–2 5

–1 2

0 1

2 6

3 10

1 2

2 5

3 10

y = 4x – 2 x y

1 2

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). 7.

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

8.

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

91

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

4 More about Equations (1.0, 2.0) and (3.0, 10.0). 14. y = x2 – 1 x y

–2 3

–1 0

0 –1

1 0

2 3

3 8

y = 2x – 2 x y

1 0

2 2

16. y = x2 – x – 1 x y

2 1

3 5

2x + 3y = –1 x y

3 4

–2 –1 0 1 5 1 –1 –1

–2 –0.5 1 1 0 –1

∵ The two graphs intersect at only one point (1.0, 0). ∴ The solution of the simultaneous equations is (1.0, 0). 15. y = x2 x y

–3 –2 –1 9 4 1

0 0

1 1

2 4

4x + y + 5 = 0 x Y

–3 –2 –1 7 3 –1

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

17. ∵

– 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.

18. ∵

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

19. ∵

2

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

20. ∵

x2 + 5x + 4 = 0

92

Certificate Mathematics in Action Full Solutions 4A 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 y

–2 –1 0 1 2 3 7 1 –3 –5 –5 –3

4 1

5 7

2x2 – 5x – 10 = 0 2x – 5x – x – 6 – 4 = – x 2x2 – 6x – 6 = – x + 4 1 x2 – 3x – 3 = − x + 2 2 ∴ The corresponding simultaneous equations are  y = x 2 − 3x − 3  y = − 1 x + 2 .  2 1 Draw the straight line y = − x + 2 on the graph of 2 y = x2 – 3x – 3. From the graphs, the roots of 2x2 – 5x – 10 = 0 are approximately –1.3 and 3.8.

23. ∵

2

24. (a) y = x2 – 4x x –1 0 y 5 0

1 2 3 –3 –4 –3

4 0

5 5

(b)

x2 – 2x + 1 = 0 x – 2x – x – 4 + 1 = – x – 4 x2 – 3x – 3 = –x – 4 ∴ The corresponding simultaneous equations are  y = x 2 − 3x − 3 .  y = −x − 4

21. ∵

2

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. x2 – 4x + 5 = 0 x – 3x – x + 8 – 3 = 0 x2 – 3x – 3 = x – 8 ∴ The corresponding simultaneous equations are  y = x 2 − 3x − 3 .  y = x − 8

22. ∵

2

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.

93

(c) (i) ∵ x2 – 4x + 1 = 0 x2 – 4x = –1 ∴ The corresponding simultaneous equations  y = x 2 − 4x are  .  y = −1 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. x2 – 3x – 3 = 0 x – 3x – x + x – 3 = 0 x2 – 4x = – x + 3 ∴ The corresponding simultaneous equations  y = x 2 − 4x are  . y = −x + 3

(ii) ∵

2

Draw the straight line y = – x + 3 on the graph of y = x2 – 4x.

4 More about Equations From the graphs, the roots of x2 – 3x – 3 = 0 are approximately –0.8 and 3.8. 2x2 – 7x – 2 = 0 2x2 – 7x – x – 2 = – x 2x2 – 8x = – x + 2 1 x2 – 4x = − x + 1 2 ∴ The corresponding simultaneous equations  y = x 2 − 4x  1 are  .  y = − 2 x + 1 1 Draw the straight line y = − x + 1 on the graph 2 of y = x2 – 4x. From the graphs, the roots of 2x2 – 7x – 2 = 0 are approximately –0.3 and 3.8.

(iii) ∵

25. (a) y = 2x2 – x x y

–1 3

–0.5 1

0 0

0.5 0

1 1

1.5 3

(b)

 y = 2x2 − x  7 2. are   y = 3 x − 3 Draw the straight line y =

7 2 x − on the graph 3 3

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 3 2x2 – x = 2 ∴ The corresponding simultaneous equations  y = 2x2 − x  3 are  .  y = 2 3 Draw the straight line y = 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.

2x2 – 3x + 1 = 0 2x – x – 2x + 1 = 0 2x2 – x = 2x – 1 ∴ The corresponding simultaneous equations  y = 2x2 − x are  .  y = 2x −1 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.

(c) (i) ∵

2

3x2 – 5x + 1 = 0 10 2 2x2 – x + = 0 3 3 7 2 2x2 – x – x + = 0 3 3 7 2 2x2 – x = x − 3 3 ∴ The corresponding simultaneous equations

(ii) ∵

(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 b = −6 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 m = −1 By substituting m = –1 into (1), we have n = 7

94

Certificate Mathematics in Action Full Solutions 4A Exercise 4C (p.186)

∴ The simultaneous equations have no real solutions.

Level 1 1.

2.

3.

5.

 y = x 2 + 1  (1)   y = 2 x + 4  ( 2) By substituting (2) into (1), we have 2x + 4 = x2 + 1 2 x – 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 (–1, 2) and (3, 10).  (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).  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= or 4

6.

By substituting (1) into (2), we have x2 – x = 1 – 2x 2 x +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.

8. y–4=0

By substituting y =

4.

 x 2 + y 2 = 4  (1)   y = 5 − x  ( 2) 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= =

95

− (−10) ± (−10) 2 − 4( 2)(21) 2(2) 10 ± − 68 (rejected) 4

3 x + y = 2 x 2 + 10  (1)   (2)  y = 9x − 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 + 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.

y=4

5 into (1), we have 4 13 5 x = 3  − 7 = − 4 4 By substituting y = 4 into (1), we have x = 3(4) – 7 = 5 ∴ The solutions of the simultaneous equations are  13 5   − ,  and (5, 4).  4 4

 y = 2 x 2 + 1  (1)   (2)  y = 2x By substituting (1) into (2), we have 2x2 + 1 = 2x 2 2x – 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 = x 2 − x  (1)   y = 1 − 2 x  (2)

9.

 (1) y = x  2  y = x − 3 x + 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

 (1) 10.  y = 4 x − 5 2  ( 2) y = x − 6 x + k  By substituting (2) into (1), we have

4 More about Equations x2 – 6x + k = 4x – 5 x2 – 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  (1) 11. 7 = 6 x − y 2  y = x − 4 x − k  ( 2) By substituting (2) into (1), we have 7 = 6x – (x2 – 4x – k) x2 – 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 k = −18  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 2 x – 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 2 x – 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  ( 2) y = x − 2 x + k  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.

96

Certificate Mathematics in Action Full Solutions 4A ∴ (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. any other reasonable answers)

x=

(or

 (1) 18.  x − y − 1 = 0 2 2  ( 2) x − xy + y = 7  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 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  (1) 19.  x + y = 25  ( 2) 3 x − 4 y + 25 = 0  3 x + 25 From (2), we have y = ……(3) 4 By substituting (3) into (1), we have

2

 3 x + 25  x2 +   = 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 3( −3) + 25 y= =4 4 ∴ The solution of the simultaneous equations is (–3, 4).  (1)  ( 2)

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

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

97

12 into (3), we have 13 3  12  5 x =  −  +1 = − 2  13  13 ∴ The solutions of the simultaneous equations are  5 12  (–2,–2) and  − ,−  .  13 13  2 2  21. 2 x − 3 xy − 2 y − 12 = 0  (1)  (2) − 2 x + 3 y + 4 = 0 By substituting y = −

Level 2

2 2 20.  x + y + 3x + 2 y = −2 2 x − 3 y = 2

3 (−2) + 1 = –2 2

12 13

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

3  3  2 y + 2  − 3 y + 2  y − 2 y 2 − 12 = 0 2  2  9 2 9 2 y + 12 y + 8 − y − 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 3 7 By substituting y = 1 into (3), we have x = (1) + 2 = 2 2 3 By substituting y = 2 into (3), we have x = (2) + 2 = 5 2 ∴ The solutions of the simultaneous equations are 7   , 1 and (5, 2). 2   (1) 22. (a)  y = mx 2  ( 2) y = x − 2 x + 4  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 m = −6 or m = 2 (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

4 More about Equations y = 2(2) = 4 ∴ The coordinates of P are (2, 4).  (1) 23.  y + k = 2 x 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. ∴ (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.  (1) 24. (a)  y = x + 4 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 1 k ≥− 2 1 (b) By substituting the minimum value of k, i.e. − , 2 into (3), we have 1 − x2 + 2x – 2 = 0 2 – x2 + 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). 2 25.  y = − x + kx + 4  (1) 4 x + y − 3 = 0  ( 2) 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.

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 2 x + 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. 27. (a) ∵ 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  xy = 30  (1) .   x + y = 11  (2) (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

Exercise 4D (p.193) Level 1 1.

(a) ∵ 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, x 2 + 3x + 2 x − 1 x3 + 2x 2 − x − 2 x3 − x 2 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

98

Certificate Mathematics in Action Full Solutions 4A ∴ (x – 1)(x + 1)(x + 2) = 0 x – 1 = 0 or x + 1 = 0 or x = 1 or x = −1 or 2.

x2 − x − 2 x + 3 x3 + 2x 2 − 5x − 6

x+2=0 x = −2

x 3 + 3x 2 − x 2 − 5x

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

− x 2 − 3x − 2x − 6 − 2x − 6

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

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

x3 + x 2

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

− 2 x 2 − 10 x − 2x 2 − 2x − 8x − 8 − 8x − 8 3

2

5. 2

∴ x – x – 10x – 8 = (x + 1)(x – 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 or x = −1 or x = −2 or 3.

3

x–4=0 x=4

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

3x 3 + 3x 2 − 6 x

(b) By long division, x2 + x − 2 x − 2 x3 − x2 − 4x + 4

x2 + x − 2 x2 + x − 2

x3 − 2x 2

∴ 3x3 + 4x2 – 5x – 2 = (3 x + 1)( x − 1)( x + 2)

x 2 − 4x x 2 − 2x

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

− 2x + 4 − 2x + 4 ∴ x3 – x2 – 4x + 4 = (x – 2)(x2 + x – 2) = ( x − 2)( x − 1)( x + 2) 6.

4.

or or

(a) ∵ f(–3) = (–3)3 + 2(–3)2 – 5(–3) – 6 = –27 + 18 + 15 – 6 =0 ∴ x + 3 is a factor of f(x). (b) By long division,

99

(a) ∵ f (1) = 3(1)3 + 4(1)2 – 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). (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. 3x + 1 x 2 + x − 2 3x 3 + 4 x 2 − 5 x − 2

2

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

x–2=0 x=2

x+2=0 x = −2

or

x+2=0

or

x = −2

(a) ∵ 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 = 16 + 24 – 8 – 24 – 8 =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.

4 More about Equations x2 − 4x + 4 x + x − 2 x 4 − 3 x 3 − 2 x 2 + 12 x − 8

(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

x 4 + x3 − 2x 2 − 4x3

+ 12 x

2x2 − x + 3

− 4 x 3 − 4 x 2 + 8x

2

x + x − 2 2 x 4 + x3 − 2 x 2 + 5x − 6

4x + 4x − 8 2

2x 4 + 2x3 − 4x2

4x2 + 4x − 8

− x3 + 2x 2 + 5x − x3 − x 2 + 2x

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

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

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

(c) ∵ f(x) = 0 ∴ (x – 1)(x + 2)(2x2 – x + 3) = 0 x – 1 = 0 or x + 2 = 0 or

(x – 2)2 = 0 x=2

x = 1 or x = –2 or x = 7.

(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).

=

9.

x = –1

or

x= =

4x 3 + 7 x 2 − 5x

∴

4x 3 + 4 x 2 − 8x 10.

3x + 3x − 6 2

∴ 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 x = 1 or x = −2 or x = −1 or x = −3

−5± −3 (rejected) 2

8(x – 3)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 9 2

or

x= =

∴ 11.

− 5 ± 5 2 − 4(1)(7) 2(1)

x = −1

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).

1 ± − 23 (rejected) 4

(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 4 + x3 − 2x 2

8.

− ( −1) ± (−1) 2 − 4(2)(3) 2( 2)

∴ x = 1 or x = −2

(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 + 4x + 3 2 x + x − 2 x 4 + 5x 3 + 5x 2 − 5x − 6

3x 2 + 3x − 6

2x2 – x + 3 = 0

x=

− ( −18) ± (−18) 2 − 4( 4)(27) 2( 4) 18 ± − 108 (rejected) 8

9 2

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

100

Certificate Mathematics in Action Full Solutions 4A

x=

2 − ( −110) ± ( −110) 2 − 4(100)(31) or x = 5 2(100) =

∴ 12.

x=

110 ± − 300 (rejected) 200

x 3 − 3x 2 2 x 2 − 41x 2x 2 − 6x

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 ∴ 3x + 1 = 0 or 3x2 + 3x + 1 = 0 x=−

1 or 3

x=

− 3 ± 32 − 4(3)(1) 2(3) =

∴

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

x=−

−3± −3 (rejected) 6

1 3

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

x+7=0 x = −7

x 4 + x 3 − 20 x 2

x3 − 2x2

− 6 x 2 − 6 x + 120

− x 2 − 4x

− 6 x 2 − 6 x + 120

− x 2 + 2x − 6 x + 12 − 6 x + 12

2 ∴ x4 + x3 – 26x2 – 6x + 120 = ( x − 4)( x + 5)( x − 6)

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

By long division,

(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

Divide f(x) by (x – 4)(x + 5), i.e. x2 + x – 20. x2 −6 2 x + x − 20 x 4 + x 3 − 26 x 2 − 6 x + 120

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

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

∴ x3 – x2 – 41x + 105 = (x – 3)(x2 + 2x – 35) = ( x − 3)( x − 5)( x + 7)

15. (a) ∵ 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).

Level 2

(b) ∵ f(x) = 0 ∴ (x – 2)(x – 3)(x + 2) = 0 x – 2 = 0 or x – 3 = 0 or x = 2 or x = 3 or

− 35 x + 105 − 35 x + 105

x+2=0 x = −2

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

x2 – 6 = 0 x=± 6

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.

101

4 More about Equations 81) = 0 (4x + 3)(4x2 + 6x + 333) = 0 ∴ 4x + 3 = 0 or 4x2 + 6x + 333 = 0

x 2 − 2x −1 x + 4 x − 21 x 4 + 2 x 3 − 30 x 2 + 38 x + 21 2

x 4 + 4 x 3 − 21x 2

x=−

− 2 x 3 − 9 x 2 + 38 x − 2 x 3 − 8 x 2 + 42 x − x2

− 4 x + 21

− x2

− 4 x + 21

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

20.

x2 – 2x – 1 = 0

− ( −2) ± (−2) 2 − 4(1)(−1) 2(1)

x=−

x= = ∴ 18.

x=−

x=−

12 5

66 ± − 768 (rejected) 42 22.

or x = =

∴

x=

− ( −93) ± ( −93) 2 − 4( 43)(63) 2(43) 93 ± − 2187 (rejected) 86

12 5

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 +

x=

− 58 ± 58 2 − 4(157)(13) 2(157) − 58 ± − 4800 (rejected) 314

7 11

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 or x =1 or

x–2=0 x=2

x4 – 3x3 – 2x2 + 12x – 8 = 0 (x4 – 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 1 x= or x= or

(x – 2)2 = 0 x=2

4

(x + 3)3 – 27(2x – 3)3 = 0 (x + 3)3 – [3(2x – 3)]3 = 0 [(x + 3) – 3(2x – 3)][(x + 3)2 + 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 ∴ –5x + 12 = 0 or 43x2 – 93x + 63 = 0 x=

7 or 11

x=−

∴ 21.

− 6 ± − 5292 (rejected) 8

3 4

=

− ( −66) ± ( −66) 2 − 4(21)(61) 2(21)

1 3

− 6 ± 6 2 − 4( 4)(333) 2(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

= 1± 2 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 1 x = − or 3

x=

= ∴

∴ x4 + 2x3 – 30x2 + 38x + 21 = ( x − 3)( x + 7)( x 2 − 2 x − 1)

3 or 4

23.

x4 – 3x3 – 3x2 + 3x + 2 = 0 (x – 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 x = −1 or x = 1 or 4

x=

− ( −3) ± ( −3) 2 − 4(1)(−2) 2(1)

102

Certificate Mathematics in Action Full Solutions 4A

=

(

3 ± 17 2

Revision Exercise 4 (p.196)

Checking:

Level 1 1.

)

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

When x = 4, 3 x + 4 + x = 3( 4) + 4 + 4= 8 2

4

2

By substituting x = u into the equation x – 9x + 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

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:

x=± 5

∴ 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 8 u=− or u = 64 27 ∵ x3 = u 8 ∴ x3 = − or x3 = 64 27 2 x=− or x = 4 3 ∴ The real roots of the equation are −

3.

5. 2 and 4. 3

x y

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

∵

x =u

∴

x =

u=

− 1 ± 12 − 4(1)(−3) 2(1)

u=

− 1 + 13 or 2

− 1+ 13 2

 − 1 + 13   x =   2  

or

u=

x =

1 − 2 13 + 13 4

=

7 − 13 2

∴ The real root of the equation is 3x + 4 + x = 8

4.

x – 8 = − 3x + 4

103

–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). 6.

y = –3x + 1

− 1− 13 2

x y

− 1− 13 (rejected) 2

–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).

2

=

y = –x – 1

7.

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

7 − 13 . 2

8.

2x2 – 7x + 6 = 0 2x2 + 6x – 13x – 2 + 8 = 0 2x2 + 6x – 2 = 13x– 8 13 x2 + 3x – 1 = x − 4 2 ∴ The equation of the required straight line is ∵

4 More about Equations

y=

9.

13 x−4 . 2

k =±2 2

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

4 5 By substituting x = 0 into (2), we have y = 2(0) + 1 = 1 4  4 By substituting x = − into (2), we have y = 2 −  + 1 5  5 3 =− 5 x=−

∴ The solutions of the simultaneous equations are  4 3 (0, 1) and  − ,−  .  5 5 2 11.  y = 2 x − x + 1  (1)  ( 2) 2 x − y = 3 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 12.  x + y = 4  (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 2 x + 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

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 k = ±5

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. 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 2 x + 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 49 x=5 or x = − (rejected) 5 By substituting x = 5 into (1), we have y = 2(5) = 10 ∴ The present ages of the brothers are 5 and 10. 16. (a) Let y cm be the length of a side of the square.

104

Certificate Mathematics in Action Full Solutions 4A ∵ 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.

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

x 4 − x3 − 2x 2 − 2x 3 + x 2 + 5x − 2x3 + 2x 2 + 4x

(b) By long division, x 2 + 3x − 5 x − 3 x 3 + 0 x 2 − 14 x + 15

+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 x = −1 or x=2 or

(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.

− x2

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

x3 − 3x 2

x=

3 x 2 − 14 x

3 2

3x 2 − 9 x

or

x= =

− 5 x + 15 − 5 x + 15

∴

x=

− ( −6) ± ( −6) 2 − 4(4)(3) 2( 4) 6 ± − 12 (rejected) 8

3 2

2 ∴ x3 – 14x + 15 = ( x − 3)( x + 3x − 5)

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

or

x= =

− 3 ± 32 − 4(1)(−5) 2(1) − 3 ± 29 2

18. (a) Let f(x) = x4 – 3x3 – x2 + 5x + 2. ∵ f(–1) = (–1)4 – 3(–1)3 – (–1)2 + 5(–1) + 2 =1+3–1–5+2 =0 ∴ 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. (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.

105

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 (x2 – 1)(x2 + 2) = 0 x2 = 1 or x2 = –2 (rejected) x = –1 or x = 1 ∴ The equation has only two real roots. ∴ a = 0, b = –1 or a = 1, b = –2 (or any other reasonable answers)  (1) 21.  y = c 2  ( 2) y = x + 4 x + 2  By substituting (2) into (1), we have x2 + 4x + 2 = c 2 x + 4x + (2 – c) = 0……(3) Let the two solutions of the simultaneous equations be h

4 More about Equations 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 1 ∴ A possible value of c is –1 or . (or any other 4 reasonable answers)

x= =

Level 2 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 x2 + 2x + 1 = 0 or x2 + 2x + 4 = 0 (x + 1)2 = 0

− 2 ± 2 2 − 4(1)(4) or x = 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 (u – 1)2 = 0 u=1 ∵ x2 – 3x = u ∴ x2 – 3x = 1 2 x – 3x – 1 = 0

3 ± 13 2

∴ The real roots of the equation are

3 − 13 and 2

3 + 13 . 2 3x – 2 x + 1 = 9

25.

3x – 9 = 2 x + 1

(

)

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

22. By substituting x = u into the equation x + 2 x = k, we have u2 + 2u = k 2 u + 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)

− ( −3) ± (−3) 2 − 4(1)(−1) 2(1)

Checking: When x =

13 20  20   20  , 3x – 2 x + 1 = 3   – 2  + 1 = 9 9 3 9     ≠9

When x = 4, 3x – 2 x + 1 = 3(4) – 2(4) + 1 = 9 ∴ The real root of the equation is 4. 5x – 5 x + 6 = 6

26.

5x – 6 = 5 x + 6

(

)

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

3 3 3 , 5x – 5 x + 6 = 5   – 5  + 6 = 0 ≠ 6 5 5 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 y

–2 6

–1 2

0 0

–1 0.5 –1 0

2 1

1 0

2 2

3 6

2x – 3y = 1 x y

106

Certificate Mathematics in Action Full Solutions 4A

∵ 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).

107

4 More about Equations 28. y = x2 – 4x + 3 x 0 1 2 y 3 0 –1 x + 2 x y

y =1 3 0 2 3 0

3 0

4 3

4 –3

2x2 – 5x + 2 = 0 2x2 – 4x + 2 = x x x2 – 2x + 1 = 2 ∴ The corresponding simultaneous equations are  y = x 2 − 2x + 1  . y = x  2 x Draw the straight line y = 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.

29. ∵

∵ 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:

x y

–2 –1 9 4

2x2 + 4x + 1 = 0 2x + 4x – 8x + 1 + 1 = 1 – 8x 2x2 – 4x + 2 = 1 – 8x 1 x2 – 2x + 1 = − 4 x 2 ∴ The corresponding simultaneous equations are  y = x 2 − 2x + 1  .  y = 1 − 4x  2 1 Draw the straight line y = − 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.

30. ∵

y = x2 – 2x + 1

2

0 1

1 0

2 1

3 4

4 9

108

Certificate Mathematics in Action Full Solutions 4A 31. (a) y = x2 + 2x x –2 –1 y 0 –1

0 0

1 3

2 8

∴ The corresponding simultaneous equations are  y = x 2 + 2x .   y = 3x + 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)

2x2 + 3x – 2 = 0 2x + 4x – x – 2 = 0 2x2 + 4x = x + 2 x x2 + 2x = + 1 2 ∴ The corresponding simultaneous equations are  y = x 2 + 2x  y = x +1 .  2 x Draw the straight line y = + 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.

(iii) ∵

2

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 = − 2

x2 – 3x + 1 = 0 x + 2x – 5x + 1 = 0 x2 + 2x = 5x – 1 ∴ The corresponding simultaneous equations are  y = x 2 + 2x .   y = 5x −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.

(c) (i) ∵

2

(ii) ∵

109

x2 – x – 1 = 0 x2 + 2x – 3x – 1 = 0 x2 + 2x = 3x + 1

4 8 , b = − and c = 4 3 3

y = ax + bx + c, we have 4 2 8 y =− x − x+4 3 3 We are going to solve the following simultaneous equations: 4 8  y = − x2 − x + 4  (5)  3 3  x − 2 y + 8 = 0  (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 x(8x + 19) = 0 19 x = 0 (rejected) or x = − 8 19 By substituting x = − into (6), we have 8

into

4 More about Equations −

19 – 2y + 8 = 0 8 45 y= 16

∴ The coordinates of P are ( −

19 45 , ). 8 16

 2 33.  x + y − x = 1  (1) 2 x + 3 y = 1  (2) 1 − 2x ……(3) 3 By substituting (3) into (1), we have 1 − 2x x2 + –x=1 3 3x2 + 1 – 2x – 3x = 3 3x2 – 5x – 2 = 0 (3x + 1)(x – 2) = 0 1 x = − or x = 2 3 1 By substituting x = − into (3), we have 3  1 1 − 2 −   3 = 5 y= 3 9 From (2), we have y =

1− 2( 2) = –1 3 ∴ The solutions of the simultaneous equations are  1 5  − ,  and (2, –1).  3 9 By substituting x = 2 into (3), we have y =

x − y + 6 = 0  (1) 34.  3  x 2 + 3x − y + 4 = 0  (2)  4 From (1), we have y = x + 6 ……(3) By substituting (3) into (2), we have 3 2 x + 3x – (x + 6) + 4 = 0 4 2 3x + 12x – 4x – 24 + 16 = 0 3x2 + 8x – 8 = 0 Using the quadratic formula, we have x=

− 8 ± 8 2 − 4(3)(−8) 2(3)

− 4 − 2 10 − 4 + 2 10 or x = 3 3 x = –3.44(cor. to 2 d.p.) or x = 0.77(cor. to 2 d.p.) − 4 − 2 10 By substituting x = into (3), we have 3 x=

y=

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

By substituting x = y=

− 4 + 2 10 into (3), we have 3

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

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

35. (a) ∵ 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  (1)  x + y = 41 .  2 2  ( 2) x + y = 841  (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  x = 20  x = 21 ∴ The solutions are  or  . y = 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.  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 x2 + 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 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

110

Certificate Mathematics in Action Full Solutions 4A (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 x = 4 ∴ The solutions are  or  . y = 5 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 39 − 5 x y= ……(1) 6 ∵ 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 x  6  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 39 − 5(3) y= =4 6 By substituting x = 4.8 into (1), we have 39 − 5( 4.8) y= = 2.5 6 ∴ The dimensions of each rectangle are 3 cm × 4 cm or 4.8 cm × 2.5 cm. 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 ∴ The length of the side of EFGH is

(b) ∵ The triangle has two equal sides and one right angle. ∴ It is an isosceles right-angled triangle. 40. (a)  y = 2 x + k − 2  (1) 2  y = x − 6 x + 6  (2) By substituting (2) into (1), we have x2 – 6x + 6 = 2x + k – 2 x2 – 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 k = −8 (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). 41. (a) ∵ f(–1) = (–1)3 + 7(–1)2 + 15(–1) + 9 = – 1 + 7 – 15 + 9 =0 ∴ x + 1 is a factor of f(x). By long division, x 2 + 6x + 9 x + 1 x 3 + 7 x 2 + 15 x + 9 x3 + x 2 6 x 2 + 15 x 6x2 + 6x 9x + 9 9x + 9

……(1) x + y cm. 2

2

2

∵ The area of EFGH is 18 cm . 2

 2 2  x + y  = 18   x2 + y2 = 18 ……(2) ∵ The perimeter of ABCD is 24 cm. ∴ 4(x + y) = 24 y=6–x ……(3) By substituting (3) into (2), we have x2 + (6 – x) 2 = 18 x2 + 36 – 12x + x 2 = 18 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.

111

∴ 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 x = −1 or x = −3 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,

4 More about Equations x 2 − x − 12 x − 2 x 3 − 3 x 2 − 10 x + 24

x2 − x − 6 x − x − 2 x 4 − 2 x 3 − 7 x 2 + 8 x + 12 2

x3 − 2x 2

x 4 − x3 − 2x 2

− x − 10 x

− x3 − 5x 2 + 8x

− x 2 + 2x

− x3 + x2 + 2x

2

− 12 x + 24 − 12 x + 24

− 6 x 2 + 6 x + 12 − 6 x 2 + 6 x + 12

∴ x3 – 3x2 – 10x + 24 = (x – 2)(x2 – x – 12) = ( x − 2)( x − 4)( x + 3) (b) ∵ f(x) = 0 ∴ (x – 2)(x – 4)(x + 3) = 0 x – 2 = 0 or x – 4 = 0 or x = 2 or x = 4 or

∴ 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 x = −1 or x = 2 or x = −2 or x = 3

x+3=0 x = −3

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

45.

By long division, x 2 + x − 20 x − 3 x 3 − 2 x 2 − 23 x + 60

x4 + 6x3 – 7x2 – 36x + 6 = 0 (x4 – 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 3 − 3x 2

x = ± 6 or

x 2 − 23 x

− 6 ± 6 2 − 4(1)(−1) 2(1)

= − 3 ± 10

x 2 − 3x − 20 x + 60 − 20 x + 60

46.

x4 + x3 – 5x2 – x + 4 = 0 (x – 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 2 ∴ x –1=0 or x2 + x – 4= 0 4

∴ 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

x=

x = ±1

x+5=0 x = −5

44. (a) ∵ 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).

or

x= =

− 1 ± 12 − 4(1)(−4) 2(1) − 1± 17 2

Multiple Choice Questions (p.200) 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) x = ±1

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

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

112

Certificate Mathematics in Action Full Solutions 4A u = –8 or u = –1 ∵ x3 = u ∴ x3 = –8 or x3 = –1 x = −2 or x = −1 3.

x = 1 or x =1

x = –2 (rejected)

Answer: D ∵ 2x2 + x – 3 = 0 2x2 = – x + 3 −x + 3 x2 = 2 ∴ The corresponding simultaneous equations are y = x2  y = − x + 3 .  2 ∴ The equation of the required straight line is −x + 3 y= . 2

9.

Answer: C  (1)  y = 4x  2  (2) y = x + k  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 k =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

Answer: B ∵ y = x2 y = 3x –

3 2

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

6.

x2 = 3x –

Answer: C ∵ The equation x2 + ax + b = 0 has no x-intercepts. ∴ It has no real roots. ∴ I is wrong.

10. Answer: A  y = x 2 − 5 x + 6  (1)   (2)  y = 2x

∵ x2 + (a – 1)x + b = 0 x2 + ax – x + b = 0 x2 + ax + b = x ∴ The corresponding simultaneous equations are  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  y = x 2 + ax + b + b = c are  . y = c

x2 + ax

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

113

 (1)  (2)

8.

From the graphs, the roots of ax2 + (b – m)x + (c – d) = 0 are 0 and 2. 5.

Answer: D  y = x 2 + ax + b   y = mx + c

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)

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

4.

7.

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 and (6, 12).

(1, 2)

11. Answer: A  y = 2 x + k + 2  (1)  2  y = − x + 6 x − 6  (2) By substituting (2) into (1), we have –x2 + 6x – 6 = 2x + k + 2 2 x – 4x + (k + 8) = 0 ……(3) ∵ The simultaneous equations have only one solution.

4 More about Equations ∴ (3) has only one real root. ∴ ∆=0 (–4)2 – 4(1)(k + 8)= 0 16 – 4k – 32 = 0 k = −4 ∴ 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 x 3 + 0 x 2 − 3x + 2

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

x3 − x 2 x 2 − 3x x2 − x − 2x + 2 − 2x + 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 x = 1 or x = −2 ∴

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 x 2 + 3 x − 35 x + 2 2 x 3 + 7 x 2 − 29 x − 70 2x3 + 4x 2 3 x 2 − 29 x 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 7 x = –2 or x = –5 or x= 2 ∵ p is the positive real root. 7 ∴ p= 2

Let’s Discuss 114