chapter 1 ssm

SECTION 1-1 17

CHAPTER 1

Section 1-1

1. To solve an equation is to find the solution set, that is, to find the set of all elements in the domain of the variable that make the equation true.

3. An equation is linear if it can be written, through simplification, in the form ax + b = 0, a ≠ 0.

5. To check a solution, substitute the value obtained into the original equation to see whether a true statement is obtained.

7. If an equation, like 2 2P w , contains more than one variable, it only makes sense to solve it if the variable to be solved for is specified.

9. 10x – 7 = 4x – 25

6x – 7= – 25

6x = – 18

x = – 3

11. 3(x + 2) = 5(x – 6)

3x + 6 = 5x – 30

–2x + 6 = –30

–2x = –36

x = 18

13. 5 + 4(t – 2) = 2(t + 7) +1

5 + 4t – 8 = 2t + 14 + 1

4t – 3 = 2t + 15

2t – 3 = 15

2t = 18

t = 9

15. 5 – 3 4

5

a =

7 2

2

a LCD = 10

10 · 5 – 10 · (3 4)

5

a = 10 ·

(7 2 )

2

a

50 – 2(3a – 4) = 5(7 – 2a) 50 – 6a + 8 = 35 – 10a –6a + 58 = 35 – 10a 4a = –23

a = –234

Common Error: After line 2, students often write

2

50 103 4

5a

50 6 4a

forgetting to distribute the –2. Put compound numerators in parentheses to avoid this.

17. 3

4

x –

4

2

x =

3

8 LCD = 8

8 · ( 3)

4

x – 8 ·

( 4)

2

x = 8·

3

8

2(x + 3) – 4(x – 4) = 3

2x + 6 – 4x + 16 = 3 – 2x + 22 = 3 –2x = –19

x = 19

2

19. 0.1(t + 0.5) + 0.2t = 0.3(t – 0.4)

0.1t + 0.05 + 0.2t = 0.3t – 0.12

0.3t + 0.05 = 0.3t – 0.12 0.05 = –0.12 No solution

21. 0.35(s + 0.34) +0.15s = 0.2s – 1.66

0.35s + 0.119 + 0.15s = 0.2s – 1.66

0.5s + 0.119 = 0.2s – 1.66 0.3s = –1.779 s = –5.93 Solution: –5.93

18 CHAPTER 1 EQUATIONS AND INEQUALITIES

23. 2

y +

5

2 = 4 –

2

3y

Excluded value: y ≠ 0 LCD = 6y

6y · 2

y + 6y ·

5

2 = 6y(4) – 6y ·

2

3y

12 + 15y = 24y – 4 –9y = –16

y = 16

9

25. 1

z

z =

1

1z + 2

Excluded value: z ≠ 1 LCD = z – 1

(z – 1)1

z

z = (z – 1)

1

1z + 2(z – 1)

z = 1 + 2z – 2 –z = –1 z = 1 No solution: 1 is excluded

27. 3

y +

10

5

y =

2 23

4

y LCD = 60

60 · 3

y + 60 ·

( 10)

5

y = 60 ·

(2 2)60 3

4

y

20y + 12(y – 10) = 15(2y – 2) – 180 20y + 12y – 120 = 30y – 30 – 180 32y – 120 = 30y – 210 2y – 120 = –210 2y = –90 y = –45

29. 1 – 3

2

x

x

= 2 3

2

x

x

Excluded value: x ≠ 2 LCD = x – 2

(x – 2)1 – (x – 2)( 3)

2

x

x

= (x – 2)(2 3)

2

x

x

x – 2 – (x – 3) = 2x – 3

x – 2 – x + 3 = 2x – 3

1 = 2x – 3

4 = 2x

2 = x No solution: 2 is excluded

31.6

4y + 1 =

5

2 8y

Excluded value: y ≠ –4 LCD 2(y + 4)

2(y + 4)6

4y + 2(y + 4)1 = 2(y + 4)

5

2( 4)y

12 + 2y + 8 = 5

2y + 20 = 5

2y = –15

y = –15

2

33. 2

3 1

4 4

a

a a

– 2

3

2a a =

3

a

2

3 1

( 2)

a

a

– 3

( 2)a a =

3

a Excluded values: a ≠ 0, –2

a(a + 2)2 2

(3 1)

( 2)

a

a

– a(a + 2)2 3

( 2)a a = a(a + 2)2 3

a

a(3a – 1) – (a + 2)3 = (a + 2)23

3a2 – a – 3a – 6 = 3(a2 + 4a + 4)

3a2 – 4a – 6 = 3a2 + 12a + 12

–16a = 18

a = –9

8

SECTION 1-1 19

35. 3.142x – 0.4835(x – 4) = 6.795 3.142x – 0.4835x + 1.934 = 6.795 2.6585x + 1.934 = 6.795 2.6585x = 4.861

x = 4.861

2.6585

x = 1.83 to 3 significant digits

37. 2.32

2

x

x –

3.76

x = 2.32

Excluded values: x ≠ 0, 2 LCD = x(x – 2)

x(x – 2)2.32

2

x

x – x(x – 2)

3.76

x = 2.32x(x – 2)

2.32x2 – 3.76(x – 2) = 2.32x(x – 2) 2.32x2 – 3.76x + 7.52 = 2.32x2 – 4.64x –3.76x + 7.52 = –4.64x 7.52 = –0.88x x = –8.55

39. an = a1 + (n – 1)d a1 + (n – 1)d = an (n – 1)d = an – a1

d = 1

1na a

n

41. 1

f =

1

1

d +

2

1

d LCD = d1d2f

d1d2f1

f = d1d2f

1

1

d + d1d2f

2

1

d

d1d2 = d2f + d1f

d2f + d1f = d1d2

(d2 + d1)f = d1d2

f = 1 2

2 1

d d

d d

43. A = 2ab + 2ac + 2bc 2ab + 2ac + 2bc = A

2ab + 2ac = A – 2bc a(2b + 2c) = A – 2bc

a = 2

2 2

A bc

b c

45. y = 2 3

3 5

x

x

(3x + 5)y = 2x – 3 3xy + 5y = 2x – 3 5y + 3 = 2x – 3xy 5y + 3 = x(2 – 3y)

5 3

2 3

y

y

= x

x = 5 3

2 3

y

y

47. The "solution" is incorrect. Although 3 is a solution of the two last equations, they are not equivalent to the first equation because both sides have been multiplied by x – 3, which is zero when x = 3. It is not permitted to multiply both sides of an equation by zero. When x = 3, the first equation involves division by zero. Since 3, the only possible solution, is not a solution, the given (first) equation has no solution.


49. 1

11x

x

x

= 3 Excl. val.: x ≠ 0

1

11x

x

x x

x

= 3

2 1

1

x

x

= 3 Excl. val.: x ≠ –1

( 1) ( 1)x x

1

1x 1

= 3

x – 1 = 3 x = 4

51. 2

1

1

1x

x

x

= x + 2 Excl. val.: x ≠ 0

2

1

1

1x

x

x x

x

= x + 2

2 2

1

x x

x

= x + 2 Excl. val.: x ≠ 1

( 1)x

1

( 2)

1

x

x

1

= x + 2

x + 2 = x + 2

Solution: All real numbers except the excluded numbers 0 and 1.

53. y = 1 b

x c

a

y =

( )

( ) 1 bx c

a x c

x c

y = ( )a x c

x c b

y = ax ac

x c b

y(x + c + b) = ax + ac xy + cy + by = ax + ac cy + by – ac = ax – xy cy + by – ac = x(a – y)

cy by ac

a y

= x

x = cy by ac

a y

55. Let x = the number, Then 10 less than two thirds the number is one fourth the number.

2

3x – 10 =

1

4x

2

3x – 10 =

1

4x

122

3x

– 12(10) = 121

4x

8x – 120 = 3x –120 = –5x x = 24

The number is 24.

57. Let x = first of the consecutive even numbers x + 2 = second of the numbers x + 4 = third of the numbers x + 6 = fourth of the numbers first + second + third = 2 more than twice fourth x + x + 2 + x + 4 = 2 + 2(x + 6) 3x + 6 = 2 + 2x + 12 3x + 6 = 2x + 14 x = 8

The four consecutive numbers are 8, 10, 12, 14.

59. Let P = perimeter of triangle, 16 = length of one

side, 2

7P = length of second side,

1

3P = length of

third side. We use the perimeter formula P = a + b + c

P = 16 + 2

7P +

1

3P

21P = 21(16) + 212

7P

+ 211

3P

21P = 336 + 6P + 7P 21P = 336 + 13P 8P = 336 P = 42 feet

SECTION 1-1 21

60. Let w = width of rectangle 2w – 3 = length of rectangle

We use the perimeter formula P = 2a + 2b. 54 = 2w + 2(2w – 3) 54 = 2w + 4w – 6

54 = 6w – 6 60 = 6w 10 = w

17 = 2w – 3 dimensions: 17 meters 10 meters

63. Let P = price before discount

0.30P = 30 percent discount on P Then price before discount – discount = price after discount P – 0.30P = 140 0.7P = 140

P = 140

0.7

P = $200

65. Let x = sales of employee

Then x – 7,000 = sales on which 8% commission is paid 0.08(x – 7,000) = (rate of commission) (sales) = (amount of commission) 2,150 + 0.08(x – 7,000) = (base salary) + (amount of commission) = earnings Earnings = 3,170 2,150 + 0.08(x – 7,000) = 3,170 2,150 + 0.08x – 560 = 3,170 0.08x + 1,590 = 3,170 0.08x = 1,580

x = 1,580

0.08

x = $19,750

67. (A) We note: The temperature increased 2.5°C for each additional 100 meters of depth. Hence, the temperature increased 25 degrees for each additional kilometer of depth.

Let x = the depth (in kilometers), then x – 3 = the depth beyond 3 kilometers. 25(x – 3) = the temperature increase for x – 3 kilometers of depth. T = temperature at 3 kilometers + temperature increase. T = 30 + 25(x – 3)

(B) We are to find T when x = 12. We use the above relationship as a formula T = 30 + 25(12 – 3) = 255˚C

(C) We are to find x when T = 200. We use the above relationship as an equation. 200 = 30 + 25(x – 3) 200 = 30 + 25x – 75 200 = –45 + 25x 245 = 25x x = 9.8 kilometers


69. Let x = total population Using the assumption given for the ratios, we have

number tagged in second sample total number tagged

number in second sample total population

22

80 =

80

x LCD: 80x

80x22

80

= 80x80

x

22x = 6,400

x = 6, 400

29122

x = 291 kangaroo rats total (approximately)

71. Let x = amount of distilled water 50 = amount of 30% solution

Then 50 + x = amount of 25% solution acid in 30% solution + acid in distilled water = acid in 25% solution

0.3(50) + 0 = 0.25(50 + x) 0.3(50) = 0.25(50 + x) 15 = 12.5 + 0.25x 2.5 = 0.25x x = 10 gallons

73. Let x = amount of 50% solution 5 = amount of distilled water Then x – 5 = amount of 90% solution acid in 90% solution + acid in distilled water = acid in 50% solution 0.9(x – 5) + 0 = 0.5x 0.9x – 4.5 = 0.5x –4.5 = –0.4x x = 11.25 liters

75. Let t = time for both computers to finish the job

Then t + 1 = time worked by old computer t = time worked by new computer Since the old computer can do 1 job in 5 hours, it works at a rate

(1 job) ÷ (5 hours) = 1

5 job per hour

Similarly the new computer works at a rate of 1

3 job per hour.

Part of job completed by Part of job completed by old computer in + new computer in = 1 whole job. t + 1 hours t hours

(Rate of old)(time of old) + (Rate of new)(Time of new) = 1 1

5(t + 1) +

1

3(t) = 1

1 115 ( 1) 15

5 3t t

= 15

3(t + 1) + 5t = 15 3t + 3 + 5t = 15 8t + 3 = 15 8t = 12 t = 1.5 hours

SECTION 1-1 23

77. Let d = distance flown north

(A) Using t = d

r, we note:

rate flying north = 150 – 30 = 120 miles per hour rate flying south = 150 + 30 = 180 miles per hour

time flying north + time flying south = 3 hours

120

d +

180

d = 3

360120

d + 360

180

d = 3(360)

3d + 2d = 1080 5d = 1080 d = 216 miles

(B) We still use the above ideas, except that rate flying north = rate flying south = 150 miles per hour.

150

d +

150

d = 3

2

150

d = 3

75

d = 3

d = 225 miles 79. Let x = the speed of the current. Then distance upstream = 1,000 rate upstream = 3 – x distance downstream = 1,200 rate downstream = 3 + x We can use the formula d = r · t to find an expression for each time:

1,000 = (3 – x) · (time upstream); time upstream = 1,000

3 x

1,200 = (3 + x) · (time downstream); time downstream = 1,200

3 x

The problem tells us that these two times are equal, so we set them equal to each other and solve.

1,000 1,200

3 3x x

LCD: (3 + x)(3 – x)

1,000 1, 200(3 )(3 ) (3 )(3 )

3 3(3 )1,000 (3 )1, 200

3,000 1,000 3,600 1, 200

2, 200 600

6000.27

2, 200

x x x xx x

x x

x x

x

x

The speed of the current is about 0.27 meters per second.


81. Let x = frequency of second note y = frequency of third note 264

4 =

5

x =

6

y

264

4 =

5

x

264

4 =

6

y

66 = 5

x 66 =

6

y

x = 330 hertz y = 396 hertz

83. We are to find d when p = 40.

40 = –1

5d + 70

5(40) = 51

5d

+ 5(70)

200 = –d + 350 –150 = –d d = 150 centimeters

Section 1-2 1. To solve an inequality is to find the solution set, that is, to find the set of all values of the variables that make the inequality a true statement. 3. The sense of an inequality reverses if we multiply or divide both sides by a negative number. There is no corresponding distinction in solving an equation.

5. –8 ≤ x ≤ 7 7. –6 ≤ x < 6 9. x ≥ –6

11. (–2, 6] 13. (–7, 8) 15. (–∞, –2]

-10 -5 0 5 10

x]

17. [–7, 2); –7 ≤ x < 2 19. (–∞, 0]; x ≤ 0 21. 12 > 6, 12 + 5 > 6 + 5

23. –6 > –8, –6 – 3 > –8 – 3 25. 2 > –1, –2(2) < –2(–1) 27. 2 < 6, 2

2 <

6

2

29. 7x – 8 < 4x + 7 3x < 15 x < 5 or (–∞, 5)

31. 12 – y ≥ 2(9 – 2y) 12 – y ≥ 18 – 4y 12 + 3y ≥ 18

3y ≥ 6 y ≥ 2

or [2, ∞)

33. 2

N

> 4

N < –8 or (–∞,–8)

)-8

N

35. –5t < –10 t > 2 or (2, ∞)

37. 3 – m < 4(m – 3) 3 – m < 4m – 12 –5m < –15

m > 3 or (3, ∞)

Common Error:

Neglecting to reverse the order after division by –5

39. –2 – 4

B ≤

1

3

B 41. –4 < 5t + 6 ≤ 21

–10 < 5t ≤ 15 –2 < t ≤ 3 or (–2, 3]

SECTION 1-2 25

12 24

B

≤ 12(1 )

3

B

–24 – 3B ≤ 4(1 + B) –24 – 3B ≤ 4 + 4B –7B ≤ 28 B ≥ –4 or [–4, ∞)

-2 3t( ]

43.

[4, 7] (–5, 5) ( 5,5) [4,7] ( 5,7]

5 7x

45.

[–1, 4) (2, 6]

[ 1, 4) (2,6] (2, 4)

2 4x

47.

( ,1)

( 2, ) ( ,1) ( 2, ) ( , )

x

49.

( , 1)

[3,7) ( , 1) [3,7)

1 or 3 7x x

51.

[2,3] (1,5) [2,3] (1,5) (1,5)

1 5x

53.

( , 4)

( 1,6]

( ,4) ( 1,6] ( ,6]

6x

55. 7

q – 3 >

4

3

q + 1

21 37

q

> 21( 4)

13

q

3q – 63 > 7(q – 4) + 21 3q – 63 > 7q – 28 + 21 3q –63 > 7q – 7 –4q > 56 q < –14 or (–∞,–14)

)-14

q

57. 2

5

x –

1

2(x – 3) ≤

2

3

x –

3

10(x + 2) LCD = 30

12x – 15(x – 3) ≤ 20x – 9(x + 2) 12x – 15x + 45 ≤ 20x – 9x – 18 –3x + 45 ≤ 11x – 18 –14x ≤ –63 x ≥ 4.5 or [4.5, ∞)


59. –4 ≤ 9

5x + 32 ≤ 68

–36 ≤ 9

5x ≤ 36

5

9(–36) ≤ x ≤

5

9(36)

–20 ≤ x ≤ 20 or [–20, 20]

[ ]-20 20

x

61. –20 < 5

2(4 – x) < –5

2

5(–20) < 4 – x <

2

5 (–5)

–8 < 4 – x < –2 –12 < –x < –6 6 < x < 12 or (6, 12)

63. 16 < 7 – 3x ≤ 31 9 < –3x ≤ 24 –3 > x ≥ –8 –8 ≤ x < –3 or [–8, –3)

[ )-8 -3

x

65. –8 ≤ –1

4(2 – x) + 3 < 10

(–4)( –8) ≥ (–4) 1

4

(2 – x) + (–4)3 > –4(10)

32 ≥ 2 – x – 12 > –40

32 ≥ – x – 10 > –40 42 ≥ –x > –30 –42 ≤ x < 30 or [–42, 30)

67. 0.1(x – 7) < 0.8 – 0.05x 0.1x – 0.7 < 0.8 – 0.05x 0.1x + 0.05x – 0.7 < 0.8 0.15x – 0.7 < 0.8 0.15x < 0.8 + 0.7 0.15x < 1.5 x < 10 or (–∞, 10) )

10 x

69. 0.3x – 2.04 ≥ 0.04(x + 1) 0.3x – 2.04 ≥ 0.04x + 0.04 0.3x – 0.04x – 2.04 ≥ 0.04 0.26x – 2.04 ≥ 0.04 0.26x ≥ 0.04 + 2.04 0.26x ≥ 2.08 x ≥ 8 or [8, ∞)

[8

x

71. 1 x represents a real number exactly when 1 – x is positive or zero. We can write this as an inequality statement and solve for x.

1 – x ≥ 0 –x ≥ –1 x ≤ 1

73. 3 5x represents a real number exactly when 3x + 5 is positive or zero. We can write this as an inequality statement and solve for x.

3x + 5 ≥ 0 3x ≥ –5

x ≥ –5

3

75.4

1

2 3x represents a real number exactly when 2x + 3 is positive. (not zero).

We can write this as an inequality statement and solve for x. 2x + 3 > 0 2x > –3

x > –3

2

SECTION 1-2 27

77. (A) For ab > 0, ab must be positive, hence a and b must have the same sign. Either 1. a > 0 and b > 0 or 2. a < 0 and b < 0

(B) For ab < 0, ab must be negative, hence a and b must have opposite signs. Either 1. a > 0 and b < 0 or 2. a < 0 and b > 0

(C) For a

b > 0,

a

b must be positive, hence a and b must have the same sign. Answer as in (A).

(D) For a

b < 0,

a

b must be negative, hence a and b must have opposite signs. Answer as in (B).

79. (A) If a – b = 1, then a = b + 1. Therefore, a is greater than b. >

(B) If u – v = –2, then v = u + 2. Therefore, u is less than v. <

81. If b

a is greater than 1

b

a > 1

a · b

a < a · 1. (since a is negative.)

b < a 0 < a – b a – b is positive

83. (A) F (B) T (C) T

85. If a < b, then by definition of <, there exists a positive number p such that a + p = b. Then, adding c to both sides, we obtain (a + c) + p = b + c, where p is positive. Hence, by definition of <, we have a + c < b + c

87. (A) If a < b, then by definition of <, there exists a positive number p such that a + p = b. If we multiply both sides of this by the positive number c, we obtain (a + p)c = bc, or ac + pc = bc, where pc is positive. Hence, by definition of <, we have ac < bc.

(B) If a < b, then by definition of <, there exists a positive number p such that a + p = b. If we multiply both sides of this by the negative number c, we obtain (a + p)c = bc, or ac + pc = bc, where pc is negative. Hence, by definition of <, we have ac > bc.

89. 150 ≤ T ≤ 250 150 ≤ 30 + 25(x – 3) ≤ 250 150 ≤ 30 + 25x – 75 ≤ 250 150 ≤ 25x – 45 ≤ 250 195 ≤ 25x ≤ 295 7.8 ≤ x ≤ 11.8 Depth from 7.8 km to 11.8 km.

91. Let x = number of calculators sold

Then Revenue = (price per calculator) (number of calculators sold) = 63x Cost = fixed cost + variable cost = 650,000 + (Cost per calculator) (number sold) = 650,000 + 47x

(A) We want Revenue > Cost 63x > 650,000 + 47x 16x > 650,000 x > 40,625 More than 40,625 calculators must be sold for the company to make a profit.

(B) We want

Revenue = Cost 63x = 650,000 + 47x 16x = 650,000 x = 40,625

(C) 40,625 calculators sold represents the break-even point, the boundary between profit and loss.


93. (A) The company might try to increase sales and keep the price the same (see part B). It might try to increase the price and keep the sales the same (see part C). Either of these strategies would need further analysis and implementation that are out of place in a discussion here.

(B) Here the cost has been changed to 650,000 + 50.5x, but the revenue is still 63x. Revenue > Cost 63x > 650,000 + 50.5x 12.5x > 650,000 x > 52,000 calculators

(C) Let p = the new price. Here the cost is still 650,000 + 50.5x as in part (B) where x is now known to be 40,625. Thus, cost = 650,000 + 50.5(40,625). The revenue is (price per calculator) (number of calculators) = p(40,625).

Revenue > Cost p(40,625) > 650,000 + 50.5(40,625)

p > 650,000 50.5(40,625)

40,625

p > 66.50 The price could be raised by $3.50 to $66.50.

95. We want 220 ≤ W ≤ 2,750. We are given W = 110I. Substituting, we must solve 220 ≤ 110I ≤ 2,750 2 ≤ I ≤ 25 or [2, 25].

Section 1-3

1. The absolute value of a positive number is equal to the number. The absolute value of 0 is 0. If the number is negative, find its absolute value by changing its sign. This is Definition 1 in words.

3. The equation 5 10x states that the distance of x from 5 is 10. Then either x = 5 + 10, thus x = 15,

or x = 5 – 10 , thus x = –5.

5. The symbol denotes the nonnegative square root. Thus the left side of this statement is nonnegative, while the right side need not be.

7. 5 9. |(–6) – (–2)| = |–4| = 4 11. |5 – 5 | = 5 – 5 since 5 – 5 is positive.

13. | 5 – 5| = – ( 5 – 5) = 5 – 5 since

5 – 5 is negative.

15. d(B,O) = |0 – (–4)| = |4| = 4

17. d(O,B) = |–4 – 0| = |–4| = 4

19. d(B,C) = |5 – (–4)| = |9| = 9

21. The distance between x and 3 is equal to 4. |x – 3| = 4

23. The distance between m and –2 is equal to 5. |m – (–2)| = 5 |m + 2| = 5

25. The distance between x and 3 is less than 5. |x – 3| < 5

27. The distance between p and –2 is more than 6. |p – (–2)| > 6 |p + 2| > 6

29. The distance between q and 1 is not less than 2. |q – 1| ≥ 2

31. y is 3 units from 5. 33. y is less than 3 units from 5. 35. y is more than 3 units from 5.

SECTION 1-3 29

|y – 5| = 3 y – 5 = ±3 y = 5 ± 3 y = 2, 8

y 2 8

|y – 5| < 3 –3 < y – 5 < 3 2 < y < 8 (2, 8)

|y – 5| > 3 y – 5 < –3 or y – 5 > 3 y < 2 or y > 8

– , 2 8,

) ( y 2 8

37. |u – (–8)| = 3

u is 3 units from –8. |u + 8| = 3 u + 8 = ± 3 u = –8 ± 3 u = –11 or –5

u -11 -5

39. |u – (–8)| ≤ 3 u is no more than 3 units from –8. |u + 8| ≤ 3 –3 ≤ u + 8 ≤ 3 –11 ≤ u ≤ –5 [–11, –5]

[ ] u -11 -5

41. |u – (–8)| ≥ 3 u is at least 3 units from -8. |u + 8| ≥ 3 u + 8 ≤ –3 or u + 8 ≥ 3 u ≤ –11 or u ≥ –5 ( , 11] [ 5, )

] [ u-11 -5

43. |2x – 11| ≤ 13 –13 ≤ 2x – 11 ≤ 13 –2 ≤ 2x ≤ 24 –1 ≤ x ≤ 12

1,12

45. |100 – 40t| > 60 100 – 40t > 60 or 100 – 40t < –60 –40t > –40 or –40t < –160 t < 1 or t > 4

( , 1) (4, )

47. |4x – 7| = 13 4x – 7 = 13 or 4x –7 = –13 4x = 20 or 4x = –6

x = 5 or x = 3

2

49. 1 3

2 4w < 2

–2 < 1

2w –

3

4 < 2

–8 < 2w – 3 < 8 –5 < 2w < 11

–5

2 < w <

11

2

–2.5 < w < 5.5 (–2.5, 5.5)

51. |0.2u + 1.7| ≥ 0.5

0.2u + 1.7 ≥ 0.5 or 0.2u + 1.7 ≤ –0.5 0.2u ≥ –1.2 or 0.2u ≤ –2.2

u ≥ –6 or u ≤ –11 ( , 11] [ 6, )

53. 9

325

C < 31

–31 < 9

5C + 32 < 31

–63 < 9

5C < –1

–35 < C < –5

9

5

35,9

55. 2x < 2 |x| < 2

–2 < x < 2 (–2, 2)

57. 2(1 3 )t ≤ 2

|1 – 3t| ≤ 2 –2 ≤ 1 – 3t ≤ 2 –3 ≤ –3t ≤ 1

1 ≥ t ≥ –1

3

–1

3≤ t ≤ 1

1,1

3

59. 2(2 3)t > 3

|2t – 3| > 3 2t – 3 < –3 or 2t – 3 > 3 2t < 0 2t > 6 t < 0 t > 3 (–∞, 0) (3, ∞)


61. |2.25 – 1.02x| ≤ 1.64 –1.64 ≤ 2.25 – 1.02x ≤ 1.64 –3.89 ≤ –1.02x ≤ –0.61 3.81 ≥ x ≥ 0.598 to three significant digits .598 ≤ x ≤ 3.81

63. |21.7 – 11.3x| = 15.2 21.7 – 11.3x = 15.2 or 21.7 – 11.3x = –15.2 –11.3x = –6.5 or –11.3x = –36.9 x = 0.575 or x = 3.27 to three significant digits

65. 0 < |x – 3| < 0.1 The distance between x and 3 is between 0 and 0.1, that is, less than 0.1 but x ≠ 3. –0.1 < x – 3 < 0.1 except x ≠ 3 2.9 < x < 3.1 but x ≠ 3 2.9 < x < 3 or 3 < x < 3.1 (2.9,3) (3,3.1)

(

2.9 3 3.1

) x

67. 0 < |x – a| < 1

10

The distance between x and a is between 0 and 1

10,

that is, less than 1

10 but x ≠ a.

–1

10 < x – a <

1

10 except x ≠ a

a – 1

10 < x < a +

1

10 but x ≠ a

a – 1

10 < x < a or a < x < a +

1

10

1 1, ,

10 10a a a a

69. We consider two cases for |x – 2| = 2x – 7.

Case 1: x – 2 ≥ 0 For this case, the possible values of x are in the set x ≥ 2. Then |x – 2| = x – 2 We solve x – 2 = 2x – 7

–x = –5 x = 5 A solution, since 5 is

among the possible values of x. Case 2: x – 2 < 0 For this case, the possible values of x are in the set x < 2. Then |x – 2| = 2 – x We solve 2 – x = 2x – 7

–3x = –9 x = 3 Not a solution, since 3 is not

among the possible values of x. Solution: x = 5

71. We consider two cases for |3x + 5| = 2x + 6 Case 1: 3x + 5 ≥ 0 For this case, the possible values of x are in the

set x ≥ –5

3.

Then |3x + 5| = 3x + 5.

We solve 3x + 5 = 2x + 6 x = 1 A solution, since 1 is

among the possible values of x.

Case 2: 3x + 5 < 0

For this case, the possible values of x are in the

set x < –5

3.

Then |3x + 5| = –(3x + 5)

We solve –(3x + 5) = 2x + 6 –3x – 5 = 2x + 6 –5x = 11 x = –2.2 A solution, since –2.2 is

among the possible values of x.Solution x = 1, –2.2

SECTION 1-3 31

73. We consider four cases for |x| + |x + 3| = 3. Case 1: x ≥ 0 and x + 3 ≥ 0, that is, x ≥ 0 and x ≥ –3 (or simply x ≥ 0)

|x| = x and |x + 3| = x + 3 Hence: x + x + 3 = 3 2x + 3 = 3 x = 0 which is a possible value for x in this case.

Case 2: x < 0 and x + 3 ≥ 0, that is, x < 0 and x ≥ –3 (or simply –3 ≤ x < 0) |x| = –x and |x + 3| = x + 3 Hence: –x + x + 3 = 3 3 = 3 This is satisfied by all x, but the condition –3 ≤ x < 0 must be imposed.

Case 3: x ≥ 0 and x + 3 < 0, that is, x ≥ 0 and x < –3. These are mutually contradictory, so no solution

is possible in this case. Case 4: x < 0 and x + 3 ≤ 0, that is, x < 0 and x ≤ –3 (or simply x ≤ –3)

|x| = –x and |x + 3| = –(x + 3) Hence: –x + [–(x + 3)] = 3 –x – x – 3 = 3 –2x = 6 x = –3 Combining the results of the four cases, –3 ≤ x ≤ 0 is the solution.

75. We consider two cases for |3 – x| = 2(4 + x) Case 1: 3 – x ≥ 0 For this case, the possible values of x are in the set

x ≤ 3.

Then |3 – x | = 3 – x.

We solve 3 – x = 2(4 + x) 3 – x = 8 + 2x

–3x = 5

x = 5

3

A solution, since 5

3 is among the possible

values of x. Case 2: 3 – x < 0 For this case, the possible values of x are in the

set x > 3.

Then |3 – x | = x – 3 We solve x – 3 = 2(4 + x) x – 3 = 8 + 2x –x = 11 x = –11 Not a solution, since –11 is not

among the possible values of x.

Solution x = 5

3

77. Case 1: x > 0. Then |x| = x. Hence x

x =

x

x = 1.

Case 2: x = 0. Then |x| = 0. Hence x

x is not defined.

Case 3: x < 0. Then |x| = –x. Hence x

x =

x

x = –1.

Thus, the possible values of x

x are 1 and –1.

79. The absolute value of no number is negative, thus can never be less than –3.


81. There are three possible relations between real numbers a and b; either a = b, a > b, or a < b. We examine each case separately.

Case 1: a = b |b – a| = |0| = 0; |a – b| = |0| = 0

Case 2: a > b |b – a| = – (b – a) = a – b | a – b| = a – b

Case 3: b > a |b – a| = b – a |a – b| = – (a – b) = b – a Thus in all three cases |b – a| = |a – b|.

83. If m < n, then m + m < m + n (adding m to both sides) Also, m + n < n + n (adding n to both sides). Hence, m + m < m + n < n + n 2m < m + n < 2n

m < 2

m n < n

85. Case 1. m > 0. Then |m| = m; |–m| = – (–m) = m. Hence |m| = |–m| Case 2. m < 0. Then |m| = –m; |–m| = –m. Hence |m| = |–m| Case 3. m = 0. Then 0 = m = –m, hence |m| = |–m| = 0.

87. If n ≠ 0, n > 0 or n < 0. Case 1.

n > 0. If m ≥ 0 |m| = m, m

n≥ 0;

m

n =

m

n; |n| = n.

Hence: m

n =

m

n =

m

n

If m < 0 |m| = –m, m

n < 0;

m

n = –

m

n; |n| = n.

Hence: m

n = –

m

n =

m

n

=

m

n

Case 2.

n < 0. If m > 0 |m| = m, m

n< 0;

m

n = –

m

n; |n| = –n

Hence: m

n = –

m

n =

m

n =

m

n

If m ≤ 0 |m| = –m, m

n > 0;

m

n =

m

n; |n| = –n

Hence: m

n =

m

n =

m

n

= m

n

89. First note that a ≤ b is true if a < b or if a = b. Hence a < b implies a ≤ b. Also a = b implies a ≤ b. Now consider three cases (m > 0, m = 0, m < 0).

Case 1. m > 0. Then |m| = m. Also –|m| < 0 Hence –|m| < 0 < m = |m| –|m| < m = |m| –|m| ≤ m ≤ |m|

Case 2. m = 0. Then –|m| = m = |m| = 0. Hence –|m| ≤ m ≤ |m|

Case 3. m < 0. Then |m| = –m, hence –|m| = m. Also |m| > 0. Hence –|m| = m < 0 < |m| –|m| = m < |m| –|m| ≤ m ≤ |m|

SECTION 1-4 33

91. 45.4

3.2

x < 1

–1 < 45.4

3.2

x < 1

–3.2 < x – 45.4 < 3.2 42.2 < x < 48.6

93. The difference between P and 500 has an absolute value of less than 20. |P – 500| < 20 –20 < P – 500 < 20 480 < P < 520 Production is between 480 and 520 units.

95. The difference between A and 12.436 has an absolute value of less than the error of 0.001. |A – 12.436| < 0.001 –0.001 < A – 12.436 < 0.001 12.435 < A < 12.437 or, in interval notation, (12.435, 12.437)

97. The difference between N and 2.37 has an absolute value of no more than 0.005. |N – 2.37| ≤ 0.005.

Section 1-4

1. In the complex number system, every negative real number has an (imaginary) square root.

3. Yes. The square of any pure imaginary number is a negative real number. For example, the square of 3i is –9.

5. (A) is true. Every real number a can be written as a complex number a + 0i.

(B) is false. For example, i is a complex number that is not a real number.

7. 2 – 9i = 2 + (–9)i (A) real part: 2 (B) imaginary part: –9i (C) conjugate: 2 – (–9)i = 2 + 9i

9. –3

2 +

5

6i

(A) real part: –3

2

(B) imaginary part: 5

6i

(C) conjugate: –3

2 –

5

6i

11. 6.5 + 2.1i (A) real part: 6.5 (B) imaginary part: 2.1i (C) conjugate: 6.5 – 2.1i

13. iπ = 0 + πi (A) real part: 0 (B) imaginary part: πi (C) conjugate: 0 – πi = –πi

15. 4π = 4π + 0i (A) real part: 4π (B) imaginary part: 0i = 0 (C) conjugate: 4π – 0i = 4π

17. –5 + i 2 (A) real part: –5

(B) imaginary part: i 2

(C) conjugate: –5 – i 2

19. (3 + 5i) + (2 + 4i) = 3 + 5i + 2 + 4i = 5 + 9i 21. (8 – 3i) + (–5 + 6i) = 8 – 3i – 5 + 6i = 3 + 3i

23. (9 + 5i) – (6 + 2i) = 9 + 5i – 6 – 2i = 3 + 3i 25. (3 – 4i) – (–5 + 6i) = 3 – 4i + 5 – 6i = 8 – 10i

27. 2 + (3i + 5) = 2 + 3i + 5 = 7 + 3i 29. (2i)(4i) = 8i2 = 8(–1) = –8

31. –2i(4 – 6i) = –8i + 12i2 = –8i + 12(–1)

= –12 – 8i

33. (1 + 2i)(3 – 4i) = 3 – 4i + 6i – 8i2 = 3 + 2i – 8(–1)

= 3 + 2i + 8 = 11 + 2i

35. (3 – i)(4 + i) = 12 + 3i – 4i – i2 = 12 – i – (–1) = 12 – i + 1 = 13 – i

37. (2 + 9i)(2 – 9i) = 4 – 81i2 = 4 + 81 = 85 or 85 + 0i


39. 1

2 4i =

1 (2 4 )

(2 4 ) (2 4 )

i

i i

= 2

2 4

4 16

i

i

= 2 4

4 16

i

= 2 4

20

i = 0.1 – 0.2i

41. 4 3

1 2

i

i

= (4 3 ) (1 2 )

(1 2 ) (1 2 )

i i

i i

= 2

2

4 5 6

1 4

i i

i

= 4 5 6

1 4

i

= 10 5

5

i = 2 – i

43. 7

2

i

i

= (7 ) (2 )

(2 ) (2 )

i i

i i

= 2

2

14 5

4

i i

i

= 14 5 1

4 1

i

= 15 5

5

i = 3 – i

45. 2 8 = 16 = 4

47. 2 8 = 2 8 1 = 2 8 i

= 16 i = 4i

49. 2 8 = 1 2 8 = i 2 8 = i 16 = 4i

51. 2 8 = 1 2 1 8 = i 2 i 8

= i2 16 = –1 · 4 = –4

53. (2 – 4 ) + (5 – 9 ) = (2 – i 4 ) + (5 – i 9 )

= 2 – 2i + 5 – 3i = 7 – 5i

55. (9 – 9 ) – (12 – 25 )

= (9 – i 9 ) – (12 – i 25 ) = (9 – 3i) – (12 – 5i)

= 9 – 3i – 12 + 5i = –3 + 2i

57. (3 – 4 )(–2 + 49 ) = (3 – i 4 )(–2 + i 49 ) = (3 – 2i)(–2 + 7i)

= –6 + 25i – 14i2

= –6 + 25i + 14 = 8 + 25i

59. 5 4

7

=

5 4

7

i =

5 2

7

i =

5

7 –

2

7i 61.

1

2 9 =

1

2 9i =

1

2 3i =

1 (2 3 )

(2 3 ) (2 3 )

i

i i

= 2

2 3

4 9

i

i

= 2 3

4 9

i

= 2 3

13

i

= 2

13 +

3

13i

63. 5

i

=

5

i ·

i

i =

2

5i

i

=

5

1

i

= 5i or 0 + 5i 65. 2(2 ) 5(2 ) 6i i = 24 10 6i i

= 4 10 6 2 10i i 67. (5 + 2i)2 – 4(5 + 2i) – 1

= 25 +20i + 4i2 – 20 – 8i – 1 = 25 +20i – 4 – 20 – 8i – 1 = 12i or 0 + 12i

69. x2 – 2x + 2 = (1 – i)2 – 2(1 – i) + 2 = 1 – 2i + i2 – 2 + 2i + 2

= 1 – 2i – 1 – 2 + 2i + 2 = 0 or 0 + 0i

71. 3 x represents an imaginary number when 3 – x is negative.

3 – x < 0 – x < –3 x > 3

73. 2 3x represents an imaginary number when 2 – 3x is negative. 2 – 3x < 0 – 3x < –2

x > 2

3

SECTION 1-4 35

75. (2x – 1) + (3y + 2)i = 5 – 4i We note: a + bi = c + di if and only if a = c and b = d. Thus 2x – 1 = 5 and 3y + 2 = –4 2x = 6 3y = –6 x = 3 y = –2

77. (1 ) ( 2)

1

x y i

i

= 2 – i

(1 + i) (1 ) ( 2)

1

x y i

i

= (2 – i)(1 + i)

(1 + x) + (y – 2)i = 2 – i + 2i – i2 (1 + x) + (y – 2)i = 3 + i

We note: a + bi = c + di if and only if a = c and b = d. Thus 1 + x = 3 and y – 2 = 1 x = 2 y = 3

79. (10 – 2i)z + (5 + i) = 2i (10 – 2i)z = –5 + i

z = 5

10 2

i

i

z = 5

10 2

i

i

· 10 2

10 2

i

i

z = 2

2

50 10 10 2

100 4

i i i

i

z = 50 2

100 4

= 1

2 or

1

2 + 0i

81. (4 + 2i)z + (7 – 2i) = (4 – i)z + (3 + 5i) 3iz + (7 – 2i) = (3 + 5i) 3iz = –4 + 7i

z = 4 7

3

i

i

z = 4 7

3

i

i

·

i

i =

2

2

4 7

3

i i

i

z = 4 7

3

i

= 7

3 +

4

3i

83. x is a square root of y if x2 = y. Thus,

(2 – i)2 = 4 – 4i + i2 and (–2 + i)2 = 4 – 4i + i2

= 4 – 4i – 1 = 4 – 4i – 1

= 3 – 4i = 3 – 4i

Hence, 2 – i and –2 + i are square roots of 3 – 4i.

85. The error arises when equating 1 1

with ( 1)( 1) .

For positive real numbers a and b,

a b ≠ ( )( )a b .

87. i4k = (i4)k = (i2·i2)k = [(–1)(–1)]k = 1k = 1

89. 1. Definition of addition

2. Commutative property for addition of real numbers.

3. Definition of addition (read from right to left).

91. The product of a complex number and its conjugate is a real number.

z z = (x + yi)(x –yi) = x2 – (yi)2 = x2 – y2i2 = x2 + y2 or (x2 + y2) + 0i. This is a real number.


93. The conjugate of a complex number is equal to the complex number if and only if the number is real. To prove a theorem containing the phrase "if and only if", it is often helpful to prove two parts separately.

Thus: z = z if z is real;

z = z only if z is real

Hypothesis: z is real Hypothesis: z = z

Conclusion: z = z Conclusion: z is real

Proof: Assume z is real, then Proof: Assume z = z, z = x + 0i = x that is, x – yi = x + yi

z = x – 0i = x Then by the definition of equality x = x –y = y

Hence z = z . –2y = 0 y = 0 Hence z = x + 0i, that is, z is real.

95. The conjugate of the sum of two complex numbers is equal to the sum of their conjugates.

z w = ( ) ( )x yi u vi

= x yi u vi

= ( )x u y v i

= (x + u) – (y + v)i = x + u – yi – vi = (x – yi) + (u – vi)

= z + w

97. The conjugate of the product of two complex numbers is equal to the product of their conjugates.

zw = ( )( )x yi u vi

= 2xu xvi yui yvi

= ( )xu xv yu i yv

= xu – yv – (xv + yu)i = xu – xvi – yv – yui = x(u – vi) – yui + yv(–1)

= x(u – vi) – yui + yvi2 = x(u – vi) – yi(u – vi) = (x – yi)(u – vi)

= z w Section 1-5 1. A quadratic equation can be written in the standard form ax2 + bx + c = 0, where a ≠ 0. 3. The product of two numbers can only be zero if one or both of the numbers is 0. 5. One would have to choose the quadratic formula, because factoring does not always work and completing the square can require somewhat laborious arithmetic.

7. 2x2 = 8x 2x2 – 8x = 0 2x(x – 4) = 0 2x = 0 or x – 4 = 0 x = 0 x = 4

9. –8 = 22t – 6t2 6t2 – 22t – 8 = 0 2(3t2 – 11t – 4) = 0 2(3t + 1)(t – 4) = 0 3t + 1 = 0 or t – 4 = 0

t = 1

3 t = 4

SECTION 1-5 37

11. 3w2 + 13w = 10 3w2 + 13w – 10 = 0 (3w – 2)(w + 5) = 0 3w – 2 = 0 or w + 5 = 0 3w = 2 w = –5

w = 2

3

13. m2 – 25 = 0 m2 = 25

m = ± 25 m = ±5

15. c2 + 9 = 0 c2 = –9

c = ± 9 c = ±3i

17. 4y2 + 9 = 0 4y2 = –9

y2 = –9

4

y = ±9

4

y = ±9

4

i

y = ±3

2

i

19. 25z2 – 32 = 0 25z2 = 32

z2 = 32

25

z = ±32

25

z = ±32

25

z = ±4 2

5

21. (2k – 5)2 = 16

2k – 5 = ± 16 2k – 5 = ± 4

2k = 5 ± 4

k = 5 4

2

k = 9

2 or k =

1

2

Common Errors: It is incorrect to cancel this way: 2 2 2

2

≠ ±2 2

or this way 2 2 22

≠ 1 ± 2 2

23. (n – 3)2 = –4

n – 3 = ± 4 n – 3 = ±2i n = 3 ± 2i

25. x2 – 2x – 1 = 0 a = 1, b = –2, c = –1 b2 – 4ac = (–2)2 – 4(1)(–1) = 8

The discriminant is positive; there are two real roots.

x = 2 4

2

b b ac

a

x = ( 2) 8

2(1)

=

2 2 2

2

x = 1 ± 2

27. x2 – 2x + 3 = 0 a = 1, b = –2, c = 3 b2 – 4ac = (–2)2 – 4(1)(3) = –8 The discriminant is negative; there are no real roots.

x = 2 4

2

b b ac

a

x = ( 2) 8

2(1)

x = 2 2 2

2

i= 1 ± i 2

29. 2t2 + 8 = 6t 2t2 – 6t + 8 = 0 t2 – 3t + 4 = 0 a = 1, b = –3, c = 4

b2 – 4ac = (–3)2 – 4(1)(4) = –7 The discriminant is negative; there are no real roots.

t = 2 4

2

b b ac

a

t = ( 3) 7

2(1)

=

3 7

2

i

31. 2t2 + 1 = 6t

2t2 – 6t + 1 = 0 a = 2, b = –6, c = 1

b2 – 4ac = (–6)2 – 4(2)(1) = 28 The discriminant is positive; there are two real roots.

t = 2 4

2

b b ac

a

t = ( 6) 28

2(2)

=

6 28

4

t =6 2 7

4

=

3 7

2


33. x2 – 4x – 1 = 0

x2 – 4x = 1 x2 – 4x + 4 = 5 (x – 2)2 = 5

x – 2 = ± 5

x = 2 ± 5

35. 2r2 + 10r + 11 = 0

r2 + 5r + 11

2 = 0

r2 + 5r = –11

2

r2 + 5r + 25

4 = –

11

2 +

25

4

5

2r

2 =

3

4

r + 5

2 = ±

3

4

r = –5

2 ±

3

4

r = –5

2 ±

3

2

r = 5 3

2

37. 4u2 + 8u + 15 = 0

u2 + 2u + 15

4 = 0

u2 + 2u = –15

4

u2 + 2u + 1 = –15

4 + 1

(u + 1)2 = –11

4

u + 1 = ±11

4

u = –1 ± 11

4

i

u = –2

2 ±

11

2

i

u = 2 11

2

i

39. 3w2 + 4w + 3 = 0

w2 + 4

3w + 1 = 0

w2 + 4

3w = –1

w2 + 4

3w +

4

9 = –1 +

4

9

2

3w

2 = –

5

9

w + 2

3 = ±

5

9

w = –2

3 ±

5

9

i

w = –2

3 ±

5

3

i

w = 2 5

3

i

41. 12x2 + 7x = 10 12x2 + 7x – 10 = 0 (4x + 5)(3x – 2) = 0 Polynomial is factorable. 4x + 5 = 0 or 3x – 2 = 0 4x = –5 3x = 2

x = –5

4 x =

2

3

43. (2y – 3)2 = 5 Format for the square root method.

2y – 3 = ± 5

2y = 3 ± 5

y = 3 5

2

SECTION 1-5 39

45. x2 = 3x + 1 x2 – 3x – 1 = 0 Polynomial is not factorable, use quadratic formula.

x = 2 4

2

b b ac

a

a = 1, b = –3, c = –1

x = 2( 3) ( 3) 4(1)( 1)

2(1)

x = 3 13

2

47. 7n2 = –4n 7n2 + 4n = 0 n(7n + 4) = 0 Polynomial is factorable. n = 0 or 7n + 4 = 0 7n = –4

n = –4

7

49. 1 + 2

8

x =

4

x

Excluded value: x ≠ 0 x2 + 8 = 4x x2 – 4x + 8 = 0

Polynomial is not factorable, use quadratic formula, or

complete the square. x2 – 4x = –8 x2 – 4x + 4 = –4 (x – 2)2 = –4

x – 2 = ± 4

x – 2 = ±i 4 x – 2 = ±2i x = 2 ± 2i

51. 24

10 m + 1 =

24

10 m Excluded value: m ≠ –10, 10:

LCD is (10 + m)(10 – m)

(10 + m)(10 – m)24

10 m + (10 + m)(10 – m) = (10 + m)(10 – m)

24

10 m 24(10 – m) + 100 – m2 = 24(10 + m) 240 – 24m + 100 – m2 = 240 + 24m 340 – 24m – m2 = 240 + 24m 0 = m2 + 48m – 100 m2 + 48m – 100 = 0 Polynomial is

factorable. (m + 50)(m – 2) = 0 m + 50 = 0 or m – 2 = 0 m = –50 m = 2

53. 2

2x =

4

3x –

1

1x Excluded values: x ≠ 2, 3, –1

(x – 2)(x -–3)(x + 1) 2

2x = (x – 2)(x – 3)(x + 1)

4

3x – (x – 2)(x – 3)(x + 1)

1

1x

2(x – 3)(x + 1) = 4(x – 2)(x + 1) – (x – 2)(x – 3) 2(x2 – 2x – 3) = 4(x2 – x – 2) – (x2 – 5x + 6) 2x2 – 4x – 6 = 4x2 – 4x – 8 – x2 + 5x – 6 2x2 – 4x – 6 = 3x2 + x – 14 0 = x2 + 5x – 8 x2 + 5x – 8 = 0 Polynomial is not factorable, use quadratic formula.

x = 2 4

2

b b ac

a

a = 1, b = 5, c = –8

x = 25 (5) 4(1)( 8)

2(1)

=

5 57

2


55. 2

3

x

x

– 2

2 9

x

x = 1 –

1

3

x

x

Excluded values: x ≠ 3, –3

(x – 3)(x + 3) ( 2)

3

x

x

– (x – 3)(x + 3)2

2 9

x

x = (x – 3)(x + 3) – (x – 3)(x + 3)

1

3

x

x

(x – 3)(x + 2) – x2 = x2 – 9 + (x – 1)(x + 3) x2 – x – 6 – x2 = x2 – 9 + x2 + 2x – 3 –x – 6 = 2x2 + 2x – 12 0 = 2x2 + 3x – 6

2x2 + 3x – 6 = 0 Polynomial is not factorable, use quadratic formula.

x = 2 4

2

b b ac

a

a = 2, b = 3, c = –6

x = 23 (3) 4(2)( 6)

2(2)

x = 3 57

4

57. s = 1

2gt2

1

2gt2 = s

gt2 = 2s

t2 = 2s

g

t = 2s

g

59. P = EI – RI2 RI2 – EI + P = 0

I = 2 4

2

b b ac

a

a = R, b = –E, c = P

I = 2( ) ( ) 4( )( )

2( )

E E R P

R

I = 2 4

2

E E RP

R

(positive square root)

61. In this problem, a = 1, b = 4, c = c. Thus, the discriminant b2 – 4ac = (4)2 – 4(1)(c) = 16 – 4c. Hence, if 16 – 4c > 0, thus 16 > 4c or c < 4, there are two distinct real roots. if 16 – 4c = 0, thus c = 4, there is one real double root, and if 16 – 4c < 0, thus 16 < 4c or c > 4, there are two distinct imaginary roots.

63. x2 + 3ix – 2 = 0

x = 2 4

2

b b ac

a

a = 1, b = 3i, c = –2

x = 23 (3 ) 4(1)( 2)

2(1)

i i =

3 9 8

2

i

x = 3 1

2

i =

3

2

i i

x = –i, –2i

SECTION 1-5 41

65. x2 + 2ix = 3 x2 + 2ix – 3 = 0

x = 2 4

2

b b ac

a

a = 1 b = 2i c = –3

x = 22 (2 ) 4(1)( 3)

2(1)

i i

x = 2 4 12

2

i

x = 2 8

2

i

x = 2 2 2

2

i

x = 2( 2)

2

i

x = –i ± 2

x = 2 – i, – 2 – i

67. x3 – 1 = 0 (x – 1)(x2 + x + 1) = 0 x – 1 = 0 or x2 + x + 1 = 0

x = 1 x = 2 4

2

b b ac

a

a = 1, b = 1, c = 1

x = 21 (1) 4(1)(1)

2(1)

x = 1 1 4

2

x = 1 3

2

x = 1 3

2

i or –

12 ±

12 i 3

69. The solutions of ax2 + bx + c = 0 are given by 2 4

2

b b ac

a

. If b2 – 4ac is negative, it can be written as

–(4ac – b2), where 4ac – b2 is positive. Then

22

2

2

44

2 2

4

2

4

2 2

b ac bb b ac

a a

b i ac b

a

b i ac b

a a

The last expression clearly represents two imaginary numbers, the two imaginary solutions of the equation.

71. If a quadratic equation has two roots, they are 2 4

2

b b ac

a

and

2 4

2

b b ac

a

. If a, b, c are rational,

then so are –b, 2a, and b2 – 4ac. Then, either 2 4b ac is rational, hence 2 4

2

b b ac

a

and

2 4

2

b b ac

a

are both rational, or, 2 4b ac is irrational, hence

2 4

2

b b ac

a

and

2 4

2

b b ac

a

are both irrational, or, 2 4b ac is imaginary, hence 2 4

2

b b ac

a

and

2 4

2

b b ac

a

are both

imaginary. There is no other possibility; hence, one root cannot be rational while the other is irrational.


73. r1 = 2 4

2

b b ac

a

r2 =

2 4

2

b b ac

a

r1r2 = 2 2( 4 ) ( 4 )

2 2

b b ac b b ac

a a

= 2 2 2

2

( ) ( 4 )

4

b b ac

a

=

2 2

2

( 4 )

4

b b ac

a

= 2 2

2

4

4

b b ac

a

=

2

4

4

ac

a =

c

a

75. The ± in front still yields the same two numbers even if a is negative.

77. Let x = one number. Since their sum is 21, 21 – x = other number Then, since their product is 104, x(21 – x) = 104 21x – x2 = 104 0 = x2 – 21x + 104 x2 – 21x + 104 = 0 (x – 13)(x – 8) = 0 x – 13 = 0 or x – 8 = 0 x = 13 x = 8 The numbers are 8 and 13.

79. Let x = first of the two consecutive even integers.

Then x + 2 = second of these integers

Since their product is 168, x(x + 2) = 168 x2 + 2x = 168 x2 + 2x – 168 = 0 (x – 12)(x + 14) = 0 x – 12 = 0 or x + 14 = 0 x = 12 x = –14 If x = 12, the two consecutive positive even integers must be 12 and 14. We discard the other solution, since the numbers must be positive.

81. The per person consumption in 1960 is found by setting the number of years after 1960 = x = 0. Then y = 122 ounces. To find the year when consumption is again 122, set y = 122 and solve.

122 = –0.0665x2 +3.58x + 122

–0.0665x2 +3.58x = 0

x( –0.0665x +3.58) = 0 x = 0 or –0.0665x +3.58 = 0

(year 1960) x = 3.58

0.0665≈ 54

The model predicts consumption at 122 ounces per person in 1960+54 or 2014.

83. From the diagram, the dimension of the planting area are 30 – 2x and 20 – 2x. Then

Area = 400 = (30 – 2x)( 20 – 2x)

400 = 600 – 100x + 4x2

0 = 4x2 – 100x + 200

x2 – 25x + 50 = 0

x = 2 4

2

b b ac

a

a = 1, b = –25, c = 50

x = 2( 25) ( 25) 4(1)(50)

2(1)

; x = 2.19 or x = 22.81

Since x must be less than 20, the second solution is discarded. The walkway should be 2.19 ft wide.

SECTION 1-5 43

85. From the area formula, Area = 1,200 = w . From the perimeter formula, Perimeter = 150 = 2 2w .Solve for w and substitute the result in for w in the first equation.

2 150 2

75

w

w

Then

2

2

1, 200 (75 )

1,200 75

75 1,200 0

2 4

2

b b ac

a

a = 1, b = –75, c = 1,200

2( 75) ( 75) 4(1)(1, 200)

2(1)

= 23.1 or = 51.9

If = 23.1, then w = 75 − 23.1 = 51.9, so there is actually only one solution, 23.1 feet by 51.9 feet.

87. If p = 3, the demand is given by q = 1,600 – 200p = 1,600 – 200(3) = 1,000 hamburgers. The revenue is given by R = qp = 1,000 · 3 = $3,000

89. The revenue equation is R = qp = (1,600 – 200p)p.

Solve (A) 2,800 = (1,600 – 200p)p 200p2 – 1,600p + 2,800 = 0 p2 – 8p + 14 = 0

p = 2( 8) ( 8) 4(1)(14)

2(1)

=

8 8

2

p = $2.59 or p = $5.41

(B) 3,200 = (1,600 – 200p)p 200p2 – 1,600p + 3,200 = 0 p2 – 8p + 16 = 0 (p – 4)2 = 0 p = $4

(C) 3,400 = (1,600 – 200p)p 200p2 – 1,600p + 3,400 = 0 p2 – 8p + 17 = 0 Since the discriminant b2 – 4ac = (–8)2 – 4(1)(17) = –4 is negative, there is no solution.

91. Let r = rate of slow plane. Then r + 140 = rate of fast plane.

After 1 hour r(1) = r = distance traveled by slow plane.

(r + 140)(1) = r + 140 = distance traveled by fast plane.

Applying the Pythagorean theorem, we have r2 + (r + 140)2 = 2602 r2 + r2 + 280r + 19,600 = 67,600 2r2 + 280r – 48,000 = 0 r2 + 140r – 24,000 = 0 (r + 240)(r – 100) = 0 r + 240 = 0 or r – 100 = 0 r = –240 r = 100

Discarding the negative solution, we have r = 100 miles per hour = rate of slow plane r + 140 = 240 miles per hour = rate of fast plane

Slow Plane after 1 hour

260 miles

Fast Plane after 1 hourAirport


93. Let t = time to travel 500 miles.

Then 200t = distance travelled by plane going north

170t = distance travelled by plane going east

Applying the Pythagorean theorem, we have

(200t)2 + (170t)2 = 5002

40,000t2 + 28,900t2 = 250,000

68,900t2 = 250,000

2 250,000

68,900

250,000

68,900

1.91 hours

t

t

t

Discarding the negative solution, we have t = 1.91 hours or 1.91(60) = 114 minutes. 114 minutes after 6:00AM is 7:54AM.

95. Let t = time for smaller pipe to fill tank alone t – 5 = time for larger pipe to fill tank alone 5 = time for both pipes to fill tank together

Then 1

t = rate for smaller pipe

1

5t = rate for larger pipe

Part of job

completed by

smaller pipe

+

Part of job

completed by

larger pipe

= 1 whole job

1

t(5) +

1

5t (5) = 1

5

t +

5

5t = 1 Excluded values: t ≠ 0, 5

t(t – 5)5

t + t(t – 5)

5

5t = t(t – 5)

5(t – 5) + 5t = t(t – 5) 5t – 25 + 5t = t2 – 5t 10t – 25 = t2 – 5t 0 = t2 – 15t + 25 t2 – 15t + 25 = 0

t = 2 4

2

b b ac

a

a = 1, b = –15, c = 25

t = 2( 15) ( 15) 4(1)(25)

2(1)

=

15 125

2

t = 13.09, 1.91 t – 5 = 8.09, –3.09

Discarding the answer for t which results in a negative answer for t – 5, we have 13.09 hours for smaller pipe alone, 8.09 hours for larger pipe alone.

97. Let v = speed of car. Applying the given formula, we have 165 = 0.044v2 + 1.1v 0 = 0.044v2 + 1.1v – 165 0.044v2 + 1.1v – 165 = 0

v = 2 4

2

b b ac

a

a = 0.044, b = 1.1, c = –165

v = 21.1 (1.1) 4(0.044)( 165)

2(0.044)

=

1.1 5.5

0.088

v = –75 or 50

Discarding the negative answer, we have v = 50 miles per hour.

SECTION 1-5 45

99. (A) Let = length of building w = width of building. Then, using the hint, in the similar triangles ABC and AFE, we have

200 w

= 400

200

200 w

= 2

= 2(200 − w) = 400 − 2w

Since the cross-sectional area of the building is given as 15,000 ft2, we have w = 15,000

Substituting, we get, (400 − 2w)w = 15,000 400w − 2w2 = 15,000 0 = 2w2 − 400w + 15,000 0 = w2 − 200w + 7,500 0 = (w − 50)(w − 150) w − 50 = 0 or w − 150 = 0 w = 50 w = 150 = 400 − 2w = 400 − 2w = 400 − 2(50) = 400 − 2(150) = 300 = 100

Thus, there are two solutions: the building is 50 ft wide and 300 ft long or 150 ft wide and 100 ft long.

(B) Preceding as in (A), but with the cross-sectional area given as 25,000 ft2, we have w = 25,000.

2

2

2

(400 2 ) 25,000

400 2 25,000

0 2 400 25,000

0 200 12,500

w w

w w

w w

w w

However, since 2 24 ( 200) 4(1)(12,500) 10,000b ac is negative, there are no real

solutions; the builder cannot meet this condition.

101. Let x = distance from the warehouse to Factory A. Since the distance from the warehouse to Factory B via Factory A is known (it is the difference in odometer readings: 52937 — 52846) to be 91 miles, then 91 − x = distance from Factory A to Factory B. The distance from Factory B to the warehouse is known (it is the difference in odometer readings: 53002—52937) to be 65 miles. Applying the Pythagorean theorem, we have


x2 + (91 − x)2 = 652 x2 + 8281 − 182x + x2 = 4225 2x2 − 182x + 4056 = 0 x2 − 91x + 2028 = 0 (x − 52)(x − 39) = 0 x − 52 = 0 or x − 39 = 0 x = 52 mi x = 39 mi

Factory AFactory B

Warehouse

x 65

91 - x

Since we are told that the distance from the warehouse to Factory A was greater than the distance from Factory A to Factory B, we discard the solution x = 39, which would lead to 91 − x = 52 miles, a contradiction. 52 miles.

Section 1-6

1. If an equation is solved by raising both sides to the same power, the resulting equation may have solutions that are not solutions of the original equation; these are called extraneous solutions.

3. Since 2x x , an absolute value equation can be regarded as a radical equation; these can often be solved

by squaring both sides.

5. This statement is true, since ± 5 are the only two solutions of x2 = 5. T 7. This statement is false. The left side, (x + 5)2, is not equal to x2 + 10x + 25. This only equals x2 + 25 when x = 0. F 9. This statement is false.

( 1x + 1)2 is not equal to x − 1 + 1 or x, in general. In fact

( 1x + 1)2 = x − 1 + 2 1x + 1 = x + 2 1x . This is only equal to x in case x = 1. F

11. This statement is false. If x3 = 2 then x = 3 2 or x is equal to one of two non-real complex numbers whose cube is 2. If x = 8, x3 must equal 512. F

13. 2x = 4 x + 2 = 16 x = 14 Check:

14 2 ?

4

16 ?

4

4

4 Solution: 14

15. 3 5y + 10 = 0

3 5y = −10

This equation has no solution,

since the left side can never be

a negative.

17. 3 2y = y − 2

3y − 2 = y2 − 4y + 4 0 = y2 − 7y + 6

y2 − 7y + 6 = 0 (y − 1)(y − 6) = 0

y = 1, 6

Check: 1: 3(1) 2 ?

1 − 2 6: 3(6) 2 ?

6 − 2

1 ?

−1 16 ?

4

1 ≠ −1 4

4 Not a solution A solution Solution: 6

SECTION 1-6 47

19. 5 6w − w = 2

5 6w = w + 2 5w + 6 = w2 + 4w + 4 0 = w2 − w − 2 w2 − w − 2 = 0 (w − 2)(w + 1) = 0 w = 2, −1

2

2

Common Error : 5 6 4

or 5 6 4

are not equivalent to the

original equation.

w w

w w

Check:

2: 5(2) 6 − 2 ?

2 −1: 5( 1) 6 −(−1) ?

2

4 – 2

2 1 + 1

2 A solution A solution Solution: 2, −1

21. |2x + 1| = x + 2 |2x + 1|2 = (x + 2)2 4x2 + 4x + 1 = x2 + 4x + 4 3x2 = 3 x2 = 1 x = 1 or x = –1

Check: x = 1 x = –1 |2x + 1| = x + 2 |2x + 1| = x + 2

|2·1 + 1| ?

1 + 2 |2(–1) + 1| ?

(–1) + 2

|3| ?

3 |–1| ?

1

3

3 1

1 A solution A solution Solution: 1, –1

23. |x – 5| = 7 – 2x |x – 5|2 = (7 – 2x)2

x2 –10x + 25 = 49 – 28x + 4x2 0 = 3x2 – 18x + 24

3(x – 2)(x – 4) = 0 x – 2 = 0 or x – 4 = 0

x = 2 x = 4

Check: x = 2 x = 4 |x – 5| = 7 – 2x |x – 5| = 7 – 2x

|2 – 5| ?

7 – 2·2 |4 – 5| = 7 – 2·4

|–3| ?

3 |–1| ?

–1

3

3 1 ≠ –1 A solution Not a solution Solution: 2

25. |3x – 4| = 2x – 5 |3x – 4|2 = (2x – 5)2 9x2 – 24x + 16 = 4x2 – 20x + 25 5x2 – 4x – 9 = 0

(5x – 9)(x + 1) = 0 5x – 9 = 0 or x + 1 = 0

x = 9

5 x = –1

Check: x = 9

5 x = –1

|3x – 4| = 2x – 5 |3x – 4| = 2x – 5

9

3 45

?

29

5

– 5 |3(–1) – 4| ?

2(–1) – 5

7

5 ?

–7

5 |–7|

? –7

7

5 ≠ –

7

5 7 ≠ –7

Not a solution Not a solution No solution

27. If we set u = 3x then u2 = 6x and the equation would become

2u2 – 4u = 0; hence, the equation is of quadratic type.

29. Since x3 ≠ (x)2 this is not an equation of quadratic type.


31. If we set u = 2

1

xthen u2 =

4

1

xand the equation would become

10

9+ 4u – 7u2= 0; hence, the equation is of quadratic type.

33. 3 2t = 1 – 2 t

3t – 2 = 1 – 4 t + 4t

– t – 3 = – 4 t

t2 + 6t + 9 = 16t

t2 – 10t + 9 = 0

(t – 9)(t – 1) = 0 t = 9, 1

Check: 9: 3 9 2 ?

1 – 2 9 1: 3 1 2 ?

1 – 2 1

25

1 – 6 1 ?

1 – 2 5 ≠ – 5 1 ≠ – 1 Not a solution Not a solution

No solution

35. m4 + 2m2 – 15 = 0 Let u = m2, then u2 + 2u – 15 = 0 (u + 5)(u – 3) = 0 u = –5, 3 m2 = –5 m2 = 3

m = ±i 5 m = ± 3

37. 3x = 2 2x 9x2 = x2 – 2 8x2 = –2

x2 = –1

4

x = ±1

4

x = ±1

2i

Check: 12 i: 3( 1

2 i) ?

212( ) 2i – 1

2 i: 3(– 12 i)

? 21

2( ) 2i

32 i

? 1

4 2 – 32 i

? 1

4 2

32 i

? 9

4 – 32 i

? 9

4

32 i

3

2 i – 32 i ≠ 3

2 i

A solution Not a solution Solution: 1

2 i

39. 2y2/3 + 5y1/3 – 12 = 0 Let u = y1/3, then 2u2 + 5u – 12 = 0 (2u – 3)(u + 4) = 0 2u – 3 = 0 u + 4 = 0

u = 3

2 u = –4

y1/3 = 3

2 y1/3 = –4

y = 27

8 y = –64

41. (m2 – 2m)2 + 2(m2 – 2m) = 15 Let u = m2 – 2m, then u2 + 2u = 15 u2 + 2u – 15 = 0 (u + 5)(u – 3) = 0 u = –5 u = 3 m2 – 2m = –5 m2 – 2m = 3 m2 – 2m + 1 = –4 m2 –2m – 3 = 0 (m – 1)2 = –4 (m – 3)(m + 1) = 0 m – 1 = ±2i m – 3 = 0 m + 1 = 0 m = 1 ± 2i m = 3 m = –1

SECTION 1-6 49

43. 2 3t + 2 = 2t

2t + 3 + 4 2 3t + 4 = t – 2

2t + 7 + 4 2 3t = t – 2

4 2 3t = –t – 9 16(2t + 3) = t2 + 18t + 81 32t + 48 = t2 + 18t + 81 0 = t2 – 14t + 33 0 = (t – 11)(t – 3) t = 11 t = 3

2

Common Error : 2 3 4 2

is not an equivalent equation

to the given equation

( 2 3 2) ?2 3 4

t t

t t

Check: 11: 2(11) 3 + 2 ?

11 2

25 + 2 ?

9 5 + 2 ≠ 3 Not a solution

3: 2(3) 3 + 2 ?

3 2

9 + 2 ?

1 3 + 2 ≠ 1 Not a solution No solution

45. 3w + 2 w = 3

3w = 3 – 2 w

w + 3 = 9 – 6 2 w + 2 – w

w + 3 = 11 – w – 6 2 w

2w – 8 = –6 2 w

w – 4 = –3 2 w w2 – 8w + 16 = 9(2 – w) w2 – 8w + 16 = 18 – 9w w2 + w – 2 = 0 (w + 2)(w – 1) = 0 w = –2 w = 1

Check: –2: ( 2) 3 + 2 ( 2) ?

3

1 + 4 ?

3

1 + 2

3 A solution

1: 1 3 + 2 1 ?

3

4 + 1

3 A solution Solution: –2, 1

47. 8 z = 1 + 5z

8 – z = 1 + 2 5z + z + 5

8 – z = z + 6 + 2 5z

2 – 2z = 2 5z

1 – z = 5z z2 – 2z + 1 = z + 5 z2 – 3z – 4 = 0 (z – 4)(z + 1) = 0 z = 4 z = –1

Check: 4: 8 4 ?

1 + 4 5

4 ?

1 + 9 2 ≠ 1 + 3 Not a solution

–1: 8 ( 1) ?

1 + ( 1) 5

9 ?

1 + 4

3

1 + 2 A solution

Solution: –1


49. 24 12 1x x – 6x = 9 x = 2 4

2

b b ac

a

a = 2 b = 6 c = 5

24 12 1x x = 6x + 9 x = 26 6 4(2)(5)

2(2)

4x2 + 12x + 1 = 36x2 + 108x + 81 x = 6 36 40

4

0 = 32x2 + 96x + 80 x = 6 4

4

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

4

i

x = –3

2 ±

1

2i

Check: –3

2 +

1

2i:

23 1 3 1

4 12 12 2 2 2

i i

– 63 1

2 2i

?

9

2( 3 ) 18 6 1i i + 9 – 3i ?

9

29 6 18 6 1i i i + 9 – 3i ?

9

9 6 1 18 6 1i i + 9 – 3i ?

9

9 + 9 – 3i ?

9

3i + 9 – 3i

9 A solution

–3

2 –

1

2i:

23 1 3 1

4 12 12 2 2 2

i i

– 63 1

2 2i

?

9

2( 3 ) 18 6 1i i + 9 + 3i ?

9

29 6 18 6 1i i i + 9 + 3i ?

9

9 6 1 18 6 1i i + 9 + 3i ?

9

9 + 9 + 3i ?

9

3i + 9 + 3i ?

9 9 + 6i ≠ 9 Not a solution

Solution: –3

2 +

1

2i

SECTION 1-6 51

51. y–2 – 2y–1 + 3 = 0 Let u = y–1, then u2 – 2u + 3 = 0 u2 – 2u = –3 u2 – 2u + 1 = –2 (u – 1)2 = –2

u – 1 = ±i 2

u = 1 ± i 2

y–1 = 1 ± i 2

1

y = 1 ± i 2

y = 1

1 2i =

1 (1 2)

(1 2) (1 2)

i

i i

y = 1 2

1 ( 2)

i

= 1 2

3

i

or alternatively, write

2

1

y –

2

y + 3 = 0 y ≠ 0 LCD = y2

1 – 2y + 3y2 = 0 3y2 – 2y + 1 = 0

y = 2 4

2

b b ac

a

a = 3, b = – 2, c = 1

y = 2( 2) ( 2) 4(3)(1)

2(3)

y = 2 8

6

y = 2 2 2

6

i=

1 2

3

i

53. 2t–4 – 5t–2 + 2 = 0 Let u = t–2, then 2u2 – 5u + 2 = 0 (2u – 1)(u – 2) = 0

2u – 1 = 0 or u – 2 = 0

u = 1

2 u = 2

t–2 = 1

2 t–2 = 2

t2 = 2 t2 = 1

2

t = ± 2 t = ±1

2

t = ±1

2

t = ±2

2

55. 3z–1 – 3z–1/2 + 1 = 0 Let u = z–1/2, then 3u2– 3u + 1 = 0

u = 2 4

2

b b ac

a

a = 3, b = – 3, c = 1

u = 2( 3) ( 3) 4(3)(1)

2(3)

u = 3 3

6

i

z–1/2 = 3 3

6

i

z–1 =

2

3 3

6

i

= 29 2(3) 3 ( 3)

36

i i

z–1 = 9 6 3 3

36

i

z–1 = 6 6 3

36

i

z–1 = 1 3

6

i

z = 6

1 3i=

6 (1 3)

(1 3) (1 3)

i

i i

z = 6(1 3)

1 ( 3)

i

= 6(1 3)

4

i or

3 3 3

2

i


57. By squaring: m – 7 m + 12 = 0

m + 12 = 7 m m2 + 24m + 144 = 49m m2 – 25m + 144 = 0 (m – 9)(m – 16) = 0 m – 9 = 0 or m – 16 = 0 m = 9, 16

Check: m = 9 m = 16

9 – 7 9 + 12 ?

0 16 – 7 16 + 12 ?

0

0

0 0

0

A solution A solution Solution: m = 9, 16

By substitution:

m – 7 m + 12 = 0

Let u = m , then u2 – 7u + 12 = 0 (u – 4)(u – 3) = 0 u – 4 = 0 or u – 3 = 0 u = 3, 4

m = 3 m = 4

m = 9 m = 16 These answers have already been checked.

59. t – 11 t + 18 = 0 By squaring:

t – 11 t + 18 = 0

t + 18 = 11 t t2 + 36w + 324 = 121t t2 – 85w + 324 = 0 (t – 4)(t – 81) = 0 t – 4 = 0 t – 81 = 0 t = 4, 81

Check:

4 – 11 4 + 18 ?

0

0

0

81 – 11 81 + 18 ?

0

0

0 Solution: t = 4, 81

By substitution: t – 11 t + 18 = 0

Let u = t , then

u2 – 11u + 18 = 0 (u – 9)(u – 2) = 0 u = 2, 9

u = 2 u = 9

t = 2 t = 9

t = 4 t = 81 These answers have already been checked.

61. 7 2x – 2x = 5x

7 – 2x – 2 7 2x 2x + x + 2 = x + 5

–2 7 2x 2x – x + 9 = x + 5

–2 7 2x 2x = 2x – 4

– 7 2x 2x = x – 2

(7 – 2x)(x + 2) = x2 – 4x + 4

7x + 14 – 2x2 – 4x = x2 – 4x + 4

0 = 3x2 – 7x – 10

0 = (3x – 10)(x + 1) 3x – 10 = 0 x + 1 = 0

x = 10

3 x = –1

Check: x = 10

3: 10

37 2 – 103 2

? 10

3 5

20213 3 – 10 6

3 3 ?

10 153 3

13 – 16

3 ?

253

13

– 43≠ 5

3

Not a solution

x = –1: 7 2( 1) – ( 1) 2 ?

( 1) 5

9 – 1 ?

4

3 – 1

2 A solution

Solution: –1

SECTION 1-6 53

63. 3 + x-4 = 5x-2 x ≠ 0, LCD = x4 3x4 + 1 = 5x2 3x4 – 5x2 + 1 = 0

Let u = x2, then 3u2 – 5u + 1 = 0

u = 2 4

2

b b ac

a

a = 3, b = –5, c = 1

u = 2( 5) ( 5) 4(3)(1)

2(3)

u = 5 13

6

x2 = 5 13

6

x = ±5 13

6

(four roots)

65. 2 5x = 0.01x + 2.04

200 5x = x + 204 40,000(x + 5) = (x + 204)2 40,000x + 200,000 = x2 + 408x + 41,616 0 = x2 – 39,592x – 158,384

Although this is factorable in the integers, one is unlikely to notice this or to detect the factors, so the quadratic formula is used

x = 2 4

2

b b ac

a

a = 1

b = –39,592 c = –158,384

x = 2( 39,592) ( 39,592) 4(1)( 158,384)

2(1)

x = 39,592 39,600

2

x = –4, x = 39,596

Check:

–4: 2 4 5 ?

0.01(–4) + 2.04 39,596: 2 39,596 5 ?

0.01(39,596) + 2.04

2

2 398

398

Solution: –4, 39596


67. 2x-2/5 – 5x-1/5 + 1 = 0 Let u = x-1/5, then 2u2 – 5u + 1 = 0

u = 2 4

2

b b ac

a

a = 2, b = –5, c = 1

u = 2( 5) ( 5) 4(2)(1)

2(2)

u = 5 17

4

x-1/5 = 5 17

4

x-1/5 =

5 17

4

x1/5 = 4

5 17 x1/5 =

4

5 17

x = 4

5 17

5 x =

4

5 17

5

x ≈ 0.016203 x ≈ 1974.98

69. The “solution” is incorrect because an incorrect attempt at squaring both sides was made. The

square of 3 5x is 3 10 3 25x x not 3 25x .

71. Substitute t = 14 into the given formula and solve.

144 1,100

x x

Let u x , then

2

2

2

144 1,100

15,400 275

0 275 15,400

u u

u u

u u

2 4

2

b b acu

a

a = 1, b = 275, c = –15,400

2275 (275) 4(1)( 15, 400)

2(1)u

u = 47.72 or –322.72

The negative “solution” for x is discarded

x = 47.72 x = (47.72)2

x = 2,277 feet

73. Let x = length and y = width, then A = xy = 45. From the Pythagorean theorem, since the diagonal is 10 inches,

2 2 210x y . Solve xy = 45 for y in terms of x to

obtain 45

yx

, then substitute this for y into

2 2 210x y .

22 2

22

4 2

4 2

4510

2,025100

2,025 100

100 2,025 0

xx

xx

x x

x x

Let 2u x , then 2 100 2,025 0u u

2 4

2

b b acu

a

a = 1, b = –100, c = 2,025

2( 100) ( 100) 4(1)(2,025)

2(1)u

u = 71.79 or u = –28.21 x2 = 71.79 x2 = –28.21 x = 8.5 in impossible

45

yx

=5.3 in Dimensions: 5.3 in by 8.5 in

SECTION 1-6 55

75. Let x = width of cross-section of the beam y = depth of cross-section of the beam From the Pythagorean theorem x2 + y2 = 162

Thus, y = 2256 x

Since the area of the rectangle is given by xy, we have xy = 120

x 2256 x = 120 x2(256 – x2) = 14,400 256x2 – x4 = 14,400 –x4 + 256x2 – 14,400 = 0 (x2)2 – 256x2 + 14,400 = 0

x2 = 2 4

2

b b ac

a

a = 1, b = –256, c = 14,400

x2 = 2( 256) ( 256) 4(1)(14, 400)

2(1)

x2 = 256 65,536 57,600

2

x2 = 256 7,936

2

x2 = 128 ± 1,984

x = 128 1,984

If x = 128 1,984 ≈ 13.1 then

y = 2256 x = 256 (128 1,984) = 128 1,984 ≈ 9.1

Thus the dimensions of the rectangle are 13.1 inches by 9.1 inches. Notice that if x = 128 1,984 , then

y = 128 1,984 and the dimensions are still 13.1 inches by 9.1 inches.


77.

Let w = width of trough h = altitude of triangular end

Examining the triangular end of the trough sketched above, we see that

h2 + 2

1

2w

= 22. The area of this end, A = 1

2wh. Since the volume of the trough

V is given by V = A·6, we have

9 = 6A

9 = 61

2wh

9 = 3wh 3 = wh

Since h2 = 22 – 2

1

2w

h2 = 22 – 1

4w2

h = 214

4w

Hence we solve

3 = w 214

4w

9 = w2 214

4w

9 = 4w2 – 1

4(w2)2

36 = 16w2 – (w2)2 (w2)2 – 16w 2 + 36 = 0

w2 = 2 4

2

b b ac

a

a = 1, b = –16, c = 36

w2 = 2( 16) ( 16) 4(1)(36)

2(1)

w2 = 16 256 144

2

w2 = 16 112

2

w2 = 8 ± 2 7

w = 8 2 7

w = 1.65 ft or 3.65 ft

CHAPTER 1 REVIEW 57

CHAPTER 1 REVIEW

1. 8x + 10 = 4x – 30 4x + 10 = –30 4x = –40 x = –10 (1-1)

2. 4 – 3(x + 2) = 5x – 7(4 – x)

4 3 6 5 28 7

3 2 12 28

15 2 28

15 26

26

15

x x x

x x

x

x

x

(1-1)

3. 10

15

y –

1

5 =

1

6

y –

1

10 LCD: 30

30( 10)

15

y – 30

1

5 = 30

( 1)

6

y – 30

1

10

2(y + 10) – 6 = 5(y + 1) – 3 2y + 20 – 6 = 5y + 5 – 3

2y + 14 = 5y + 2 –3y + 14 = 2

–3y = –12 y = 4 (1-1)

4. 3(2 – x) – 2 ≤ 2x – 1 6 – 3x – 2 ≤ 2x – 1 –3x + 4 ≤ 2x – 1 –5x ≤ –5 x ≥ 1 [1, ∞)

(1-2)

5. |y + 9| < 5 –5 < y + 9 < 5 –14 < y < –4 (–14, –4)

( ) y-14 -4 (1-3)

6. |3 – 2x| ≤ 5 –5 ≤ 3 – 2x ≤ 5 –8 ≤ –2x ≤ 2 4 ≥ x ≥ –1 –1 ≤ x ≤ 4

[–1, 4] [ ] x -1 4

(1-3)

7. (A) 9 – 4i = 9 + (–4)i Real part: 9 Imaginary part: –4i Conjugate: 9 – (–4)i = 9 + 4i

(B) 5i = 0 + 5i Real part = 0 Imaginary part: 5i Conjugate: 0 + (–5)i = –5i

(C) –10 = –10 + 0i Real part: –10 Imaginary part: 0 Conjugate: –10 – 0i = –10 (1-4)

8. (A) (4 + 7i) + (–2 – 3i) = 4 + 7i – 2 – 3i = 2 + 4i

(B) (–3 + 5i) – (4 – 8i) = –3 + 5i – 4 + 8i = –7 + 13 i

(C) (1 – 2i)(3 + 4i) = 3 + 4i – 6i – 8i2 = 3 – 2i + 8 = 11 – 2i

(D) 21 9

5 2

i

i

= (21 9 ) (5 2 )

(5 2 ) (5 2 )

i i

i i

= 2

2

105 42 45 18

25 4

i i i

i

= 105 87 18

25 4

i

= 87 87

29

i = 3 + 3i (1-4)


9. 2x2 – 7 = 0 2x2 = 7

x2 = 7

2

x = ±7

2= ±

14

2

(1-5)

10. 5x2 + 20 = 0 5x2 = –20 x2 = –4

x = ± 4 x = ±2i (1-5)

11. 2x2 = 4x 2x2 – 4x = 0 2x(x – 2) = 0 2x = 0 x – 2 = 0 x = 0 x = 2 (1-5)

12. 2x2 = 7x – 3 2x2 – 7x + 3 = 0 (2x – 1)(x – 3) = 0 2x – 1 = 0 x – 3 = 0

x = 1

2 x = 3

(1-5)

13. m2 + m + 1 = 0

m = 2 4

2

b b ac

a

a = 1 b = 1 c = 1

m = 21 (1) 4(1)(1)

2(1)

=

1 3

2

=

1 3

2

i = –

1

2 ±

3

2i (1-5)

14. y2 = 3

2(y + 1)

2y2 = 3(y + 1) 2y2 = 3y + 3

2y2 – 3y – 3 = 0

y = 2 4

2

b b ac

a

a = 2 b = –3 c = –3

y = 2( 3) ( 3) 4(2)( 3)

2(2)

y = 3 33

4

(1-5)

15. 5 6x – x = 0

5 6x = x 5x – 6 = x2 0 = x2 – 5x + 6 x2 – 5x + 6 = 0 (x – 3)(x – 2) = 0 x = 2, 3

Check: 5(2) 6 – 2 ?

0

0

0

5(3) 6 – 3 ?

0

0

0 Solution: 2, 3 (1-6)

16. 15 6x represents a real number exactly when

15 + 6x is positive or zero. We can write this as an inequality statement and solve for x. 15 + 6x ≥ 0 6x ≥ –15

x ≥ 5

2

(1-2)

17. 7

2 x =

2

10 4

3 10

x

x x

7

2 x =

10 4

( 2)( 5)

x

x x

Excluded values: x ≠ 2, –5

1

– 2x

1

7( 5)

2x

x

= (x – 2)(x + 5)

10 4

( 2)( 5)

x

x x

–7(x + 5) = 10 – 4x –7x – 35 = 10 – 4x –3x = 45 x = –15 (1-1)

CHAPTER 1 REVIEW 59

18. 3

2 2

u

u

= 1

6 –

1

3 3

u

u

3

2( 1)

u

u

= 1

6 –

1

3( 1)

u

u

Excluded value: u ≠ 1

6(u – 1)( 3)

2( 1)

u

u

= 6(u – 1) 1

6 – 6(u – 1)

(1 )

3( 1)

u

u

3(u – 3) = u – 1 – 2(1 – u) 3u – 9 = u –1 – 2 + 2u 3u – 9 = 3u – 3 –9 = –3 No solution (1-1)

19. 3

8

x ≤ 5 –

2

3

x

24( 3)

8

x ≤ 120 – 24

(2 )

3

x

3(x + 3) ≤ 120 – 8(2 – x) 3x + 9 ≤ 120 – 16 + 8x 3x + 9 ≤ 8x + 104

–5x ≤ 95 x ≥ –19 [–19, ∞)

(1-2)

20. |3x – 8| > 2 3x – 8 < –2 or 3x – 8 > 2 3x < 6 3x > 10

x < 2 or x > 10

3

10– , 2 ,

3

) ( x2 10

3(1-3)

21. 2(1 2 )m ≤ 3

|1 – 2m| ≤ 3 –3 ≤ 1 – 2m ≤ 3 –4 ≤ –2m ≤ 2

2 ≥ m ≥ –1 –1 ≤ m ≤ 2

[–1, 2] (1-3)

22. (A) d(A,B) = |20 – 5| = |15| = 15 (B) d(A,C) = |(–8) – 5| = |–13| = 13 (C) d(B,C) = |(–8) – 20| = |–28| = 28 (1-3)

23. (A) (3 + i)2 – 2(3 + i) + 3 = 9 + 6i + i2 – 6 – 2i + 3 = 9 + 6i – 1 – 6 – 2i + 3 = 5 + 4i (B) i27 = i26i = (i2)13i = (–1)13i = (–1)i = –i (1-4)

24. (A) (2 – 4 ) – (3 – 9 ) = (2 – i 4 ) – (3 – i 9 ) = (2 – 2i) – (3 – 3i) = 2 – 2i – 3 + 3i = –1 + i

(B) 2 1

3 4

= 2 1

3 4

i

i

= 2

3 2

i

i

= (2 ) (3 2 )

(3 2 ) (3 2 )

i i

i i

= 2

2

6 7 2

9 4

i i

i

= 6 7 2

9 4

i

= 4 7

13

i =

4

13 –

7

13i

(C) 4 25

4

= 4 25

4

i

i

=

4 5

2

i

i

=

4 5

2

i i

i i

=

2

2

4 5

2

i i

i

=

4 5

2

i

= 5

2 – 2i (1-4)

25. 2

11

3y

= 20

y + 11

3 = ± 20

y = –11

3 ± 20 =

11 3 4 5

3 3

y = 11 3 2 5

3

=

11 6 5

3

(1-5)

26. 1 + 2

3

u =

2

u Excluded value: u ≠ 0

u2 + 3 = 2u u2 – 2u = –3 u2 – 2u + 1 = –2 (u – 1)2 = –2

u – 1 = ± 2

u = 1 ± 2

u = 1 ± i 2 (1-5)


27. 2 6

x

x x –

2

3x = 3

( 3)( 2)

x

x x –

2

3x = 3 Excluded values: x ≠ 3, –2

(x – 3)(x + 2) ( 3)( 2)

x

x x – (x – 3)(x + 2)

2

3x = 3(x – 3)(x + 2)

x – 2(x + 2) = 3(x – 3)(x + 2) x – 2x – 4 = 3(x2 – x – 6) –x – 4 = 3x2 – 3x – 18 0 = 3x2 – 2x – 14 3x2 – 2x – 14 = 0

x = 2 4

2

b b ac

a

a = 3, b = –2, c = –14

x = 2( 2) ( 2) 4(3)( 14)

2(3)

x = 2 172

6

x = 2 2 43

6

=

1 43

3

(1-5)

28. 2x2/3 – 5x1/3 – 12 = 0 Let u = x1/3, then 2u2 – 5u – 12 = 0 (2u + 3)(u – 4) = 0

u = –3

2, 4

x1/3 = –3

2 x1/3 = 4

x = –27

8 x = 64 (1-6)

29. m4 + 5m2 – 36 = 0 Let u = m2, then u2 + 5u – 36 = 0 (u + 9)(u – 4) = 0 u = –9, 4 m2 = –9 m2 = 4 m = ±3i m = ±2

Common Error: 2 5 1 9y y

is not equivalent to the equation formed by squaring both members of the given equation.

30. 2y – 5 1y = –3

– 5 1y = –3 – 2y

5y + 1 = 9 + 6 2y + y – 2

5y + 1 = y + 7 + 6 2y

4y – 6 = 6 2y

2y – 3 = 3 2y

4y2 – 12y + 9 = 9(y – 2) 4y2 – 12y + 9 = 9y – 18 4y2 – 21y + 27 = 0 (4y – 9)(y – 3) = 0 y = 9

4 , 3

Check: 94 2 – 9

45 1 ?

–3

14 – 49

4 ?

–3

–3

–3

3 2 – 5(3) 1 ?

–3

–3

–3 Solution: 9

4 , 3 (1-6)

31. 2.15x – 3.73(x – 0.930) = 6.11x 2.15x – 3.73x + 3.4689 = 6.11x –1.58x + 3.4689 = 6.11x 3.4689 = 7.69x x = 0.451 (1-1)

32. –1.52 ≤ 0.770 – 2.04x ≤ 5.33 –2.29 ≤ –2.04x ≤ 4.56 1.12 ≥ x ≥ –2.24 –2.24 ≤ x ≤ 1.12 or [–2.24, 1.12] (1-2)

CHAPTER 1 REVIEW 61

33. |9.71 – 3.62x| > 5.48 9.71 – 3.62x > 5.48 or 9.71 – 3.62x < –5.48 –3.62x > –4.23 –3.62x < –15.19 x < 1.17 x > 4.20

(1-3)

34. 8 4

3 5t ≤

1

2

1 8 4 1

2 3 5 21 8 4 1

30 30 30 302 3 5 2

15 80 24 15

95 24 65

95 65

24 2465 95

24 24

t

t

t

t

t

t

(1-3)

35. 6.09x2 + 4.57x – 8.86 = 0

x = 2 4

2

b b ac

a

a = 6.09, b = 4.57, c = –8.86

x = 24.57 (4.57) 4(6.09)( 8.86)

2(6.09)

x = 4.57 236.7145

12.18

x = 4.57 15.3855

12.18

x = –1.64, 0.888 (1-5)

36. P = M – Mdt

M – Mdt = P

M(1 – dt) = P

M = 1

P

dt (1-1)

37. P = EI – RI2 RI2 – EI + P = 0

I = 2 4

2

b b ac

a

a = R, b = –E, c = P

I = 2( ) ( ) 4( )( )

2( )

E E R P

R

I = 2 4

2

E E PR

R

(1-5)

38. x = 4 5

2 1

y

y

x(2y + 1) = (2y + 1)4 5

2 1

y

y

2xy + x = 4y + 5 2xy + x – 4y = 5 2xy – 4y = 5 – x y(2x – 4) = 5 – x

y = 5

2 4

x

x

(1-1)


39. The original equation can be rewritten as

4

( 1)( 3)x x =

3

( 1)( 2)x x

Thus, x = 1 cannot be a solution of this equation. This extraneous solution was introduced when both sides were multiplied by x – 1 in the second line. x = 1 must be discarded and the only correct solution is x = –1. (1-1)

40. In this problem, a = 1, b = –8, c = c. Thus, the discriminant b2 – 4ac = (–8)2 – 4(1)(c) = 64 – 4c. Hence, if c = –16, the discriminant is 64 – 4(–16)=128 > 0. Therefore there are two real solutions. if c = 16, the disciminant is 64 – 4(16) = 0. Therefore, there is one real solution. if c = 32, the discriminant is 64 – 4(32) = –128 < 0. Therefore, there are two imaginary solutions. In general if 64 – 4c > 0, thus 64 > 4c or c < 16, there are two real solutions. if 64 – 4c = 0, thus c = 16, there is one real solution. if 64 – 4c < 0, thus 64 < 4c or c > 16, there are imaginary solutions. (1-5)

41. The given inequality a + b < b – a is equivalent to, successively, a < –a 2a < 0 a < 0 Thus its truth is independent of the value of b, and dependent on a being negative. True for all real b and all negative a. (1-2)

42. If a > b and b is negative, then a

b <

b

b, that is,

a

b < 1, since dividing both sides by b reverses the

order of the inequality. a

b is less than 1. (1-2)

43. y = 1

1

1

1 x

y = 1

1

1(1 )

(1 )1 (1 ) x

x

x x

y = 1

1 1

x

x

y = 1 x

x

–xy = 1 – x x – xy = 1 x(1 – y) = 1

x = 1

1 y (1-1)

44. 0 < |x – 6| < d means: the distance between x and 6 is between 0 and d, that is, less than d but x≠ 6. –d < x – 6 < d except x ≠ 6 6 – d < x < 6 + d but x ≠ 6 6 – d < x < 6 or 6 < x < 6 + d (6 – d, 6) (6, 6 + d)

(1-3)

45. 2x2 = 3 x – 1

2

4x2 = 2 3 x – 1

4x2 – 2 3 x + 1 = 0

x = 2 4

2

b b ac

a

a = 4, b = –2 3 , c = 1

x = 2( 2 3) ( 2 3) 4(4)(1)

2(4)

46. 4 = 8x-2 – x-4 4x4 = 8x2 – 1 x ≠ 0 LCD = x4 4x4 – 8x2 + 1 = 0 Let u = x2, then 4u2 – 8u + 1 = 0

u = 2 4

2

b b ac

a

a = 4, b = –8, c = 1

u = 2( 8) ( 8) 4(4)(1)

2(4)

CHAPTER 1 REVIEW 63

x = 2 3 4

8

x = 2 3 2

8

i

x = 3

4

i or

3

4 ±

1

4i (1-5)

u = 8 48

8

=

8 4 3

8

Common Error: It is incorrect

to "cancel" the 8's at this point

u = 4(2 3)

8

=

2 3

2

x2 = 2 3

2

x = ±2 3

2

(four real roots) (1-6)

47. 2ix2 + 3ix – 5i = 0

2(2 3 5) 0

(2 5)( 1) 0

2 5 0 or 1 0

2 5 1

5

2

i x x

i x x

x x

x x

x

(1-5)

48. (a + bi)2 2 2 2

a bi

a b a b

= ( )

1

a bi2 2 2 2

a bi

a b a b

= 2 2

( )a a bi

a b

– 2 2

( )bi a bi

a b

= 2 2 2

2 2

a abi abi b i

a b

= 2 2

2 2

a b

a b

= 1 (1-4)

49. Let x = the number

1

x = its reciprocal

Then x – 1

x =

Excluded value:16

015 x

15x2 –15 = 16x 15x2 – 16x – 15 = 0 (5x + 3)(3x – 5) = 0

5x + 3 = 0 or 3x – 5 = 0

x = –3

5 x =

5

3 (1-5)

50. (A) H = 0.7(220 – A)

(B) We are to find H when A = 20. H = 0.7(220 – 20) H = 140 beats per minute. (C) We are to find A when H = 126. 126 = 0.7(220 – A) 126 = 154 – 0.7A –28 = –0.7A A = 40 years old. (1-1)

51. Let x = amount of 80% solution Then 50 – x = amount of 30% solution since 50 = amount of 60% solution

acid in acid in acid in

80% solution + 30% solution = 60% solution 0.8(x) + 0.3(50 – x) = 0.6(50) 0.8x + 15 – 0.3x = 30 15 + 0.5x = 30 0.5x = 15

x = 15

0.5

52. Let x = the wind speed Then 300 + x = the speed flying with the wind 300 – x = the speed flying against the wind

Solving d = rt for t, we have t = d

r. We use this formula,

together with

time down (with the wind) = time back (against the wind) – 3

2

time flying down = distance flown

rate down =

1,200

300 x

time flying back = distance back

rate back =

1,200

300 x


x = 30 milliliters of 80% solution

50 – x = 20 milliliters of 30% solution (1-2)

So, 1, 200

300 x =

1, 200

300 x –

3

2Excluded values: x ≠ –300, 300

2(300 – x)(1,200) = 2(300 + x)(1,200) – 3(300 – x) (300 + x)

720,000 – 2,400x = 720,000 + 2,400x – 270,000+ 3x2

0 = 3x2 + 4,800x – 270,000

0 = x2 + 1,600x – 90,000

x = 2 4

2

b b ac

a

a = 1, b = 1,600, c = –90,000

x = 21,600 (1,600) 4(1)( 90,000)

2(1)

x = –1,654.4 or x = 54.4

Discarding the negative answer, we have

wind speed = 54.4 miles per hour. (1-5) 53. (A) Let x = distance rowed then 15 – 3 = 12 km/hr = the rate rowed upstream 15 + 3 = 18 km/hr = the rate rowed downstream

Using t = d

r as in the previous problem yield

time upstream = 12

x time downstream =

18

x

So 12

x +

18

x =

25

60 LCD = 180

180

1 ·

12

x +

180

1 ·

18

x =

180

1 ·

25

60

15x + 10x = 75 25x = 75 x = 3 km

(B) Now let x = still-water speed x – 3 = the rate rowed upstream x + 3 = the rate rowed downstream

time upstream = 3

3x time downstream =

3

3x

So 3

3x +

3

3x =

23

60 Excluded values: x = 3, –3

60(x + 3)3 + 60(x – 3)3 = 23(x + 3)(x – 3) 180x + 540 + 180x – 540 = 23x2 – 207 0 = 23x2 – 360x – 207

x = 2 4

2

b b ac

a

a = 23, b = –360, c = –207

x = 2( 360) ( 360) 4(23)( 207)

2(23)

x = 16.2 or –0.6 Discarding the negative answer, we have x = 16.2 km/hr.

(C) Now 18 – 3 = 15 km/hr = the rate rowed upstream 18 + 3 = 21 km/hr = the rate rowed downstream

So 3

15 +

3

21 = round trip time

= 0.343 hr = 0.343 60 min = 20.6 min (1-1, 1-5)

CHAPTER 1 REVIEW 65

54. (A) Apply the given formula with C = 15. 15 = x2 – 10x + 31 0 = x2 – 10x + 16 x2 – 10x + 16 = 0 (x – 8)(x – 2) = 0 x = 2 or 8 Thus the output could be either 2,000 or 8,000 units.

(B) Apply the given formula with C = 6. 6 = x2 – 10x + 31 0 = x2 – 10x + 25 x2 – 10x + 25 = 0 (x – 5)2 = 0 x = 5 Thus the output must be 5,000 units. (1-5)

55. The break-even points are defined by C = R (cost = revenue). Applying the formulas in this problem and the previous one, we have x2 – 10x + 31 = 3x x2 – 13x + 31 = 0

x = 2 4

2

b b ac

a

a = 1, b = –13, c = 31

x = 2( 13) ( 13) 4(1)(31)

2(1)

x = 13 45

2

thousand or approximately

3,146 and 9,854 units (1-5)

56. Let P = the percentage required. Then the distance of P from 54 is less than or equal to 1.2, thus

54 1.2P

Solving, we obtain

1.2 54 1.2

52.8 55.2

P

P

(1-3)


58.

B

2

8 feeth

B

2

In the isosceles triangle we note:

1

2Bh = A = 24

Hence

h = 48

B

57. Let x = width of page

y = height of page

Then xy = 480, thus y = 480

x.

Since the printed portion is surrounded by margins of 2 cm on each side, we have

x – 4 = width of printed portion y – 4 = height of printed portion Hence (x – 4)(y – 4) = 320, that is

(x – 4)480

4x

= 320

Solving this, we obtain:

x480

x

– 4x – 4480

x

+ 16 = 320

480 – 4x – 1,920

x + 16 = 320

–4x – 1,920

x = –176 LCD: x x ≠ 0

–4x2 – 1,920 = –176x 0 = 4x2 – 176x + 1,920 0 = x2 – 44x + 480

0 = (x – 20)(x – 24) x – 20 = 0 or x – 24 = 0

x = 20 x = 24

480

x = 24

480

x = 20

Thus, the dimensions of the page are 20 cm by 24 cm. (1-5)

Applying the Pythagorean theorem, we have

h2 + 2

2

B

= 82

2

48

B

+ 2

2

B

= 82

2

2,304

B +

2

4

B = 64

4B22

2,304

B

+ 4B22

4

B

= 4B2(64)

9,216 + B4 = 256B2 (B2)2 – 256B2 + 9,216 = 0

B2 = 2 4

2

b b ac

a

a = 1, b = – 256, c = 9,216

B2 = 2( 256) ( 256) 4(1)(9, 216)

2(1)

B2 = 256 65,536 36,864

2

B2 = 256 28,672

2

B2 = 128 ± 32 7

B = 128 32 7

B = 14.58 ft or 6.58 ft (1-6)

chapter 1 ssm

Documents