Calculus 9e Purcell Varberg Rigdon (Solution)

CHAPTER 0 Preliminaries

0.1 Concepts Review 1. rational numbers

2. dense

3. If not Q then not P.

4. theorems Problem Set 0.1

1. 4 2(8 11) 6 4 2( 3) 6

4 6 6 16

− − + = − − +

= + + =

2.

( ) [ ]

[ ]

3 2 4 7 12 3 2 4( 5)

3 2 20 3(22) 66

− − = − −

⎡ ⎤

⎣ ⎦

= + = =

3. –4[5(–3 12 – 4) 2(13 – 7)]

–4[5(5) 2(6)] –4[25 12]

–4(37) –148

+ +

= + = +

= =

4.

[ ]

[ ] ( )

( )

5 1(7 12 16) 4 2

5 1(3) 4 2 5 3 4 2

5 1 2 5 2 7

− + − + +

= − + + = − + +

= + = + =

5. ⁵– ^{1 65}– ^{7 58} 7 13=91 91=91

6. ^{3 3 1 3 3 1}

4 7 21 6 3 21 6

42 6 7 43

42 42 42 42

+ − = + −

− −

= − + − = −

7. ^{1 1 1}–^{1 1 1 1 3 – 4 1}

3 2 4 3 6 3 2 12 6

1 1 1 1

3 2 –12 6

1 1 4

3 –24 24

1 3 1

3 24 24

⎡ ⎛⎜ ⎞⎟+ ⎤= ⎡ ⎛⎜ ⎞⎟+ ⎤

⎢ ⎝ ⎠ ⎥ ⎢ ⎝ ⎠ ⎥

⎣ ⎦ ⎣ ⎦

⎡ ⎛ ⎞ ⎤

= ⎢⎣ ⎜⎝ ⎟⎠+ ⎥⎦

⎡ ⎤

= ⎢⎣ + ⎥⎦

⎛ ⎞

= ⎜ ⎟=

⎝ ⎠

8. 1 2 1 1 1 1

3 5 2 3 5 2 1 5 3

3 5 2 15 15

1 2 1 2 1 2 1

3 5 2 15 3 5 15

1 6 1 1 5 1

3 15 15 3 15 9

⎡ ⎛ ⎞⎤

− ⎢⎣ − ⎜⎝ − ⎟⎠⎥⎦= − ⎡⎢⎣ − ⎛⎜⎝ − ⎞⎟⎠⎤⎥⎦

⎡ ⎛ ⎞⎤ ⎡ ⎤

= − ⎢⎣ − ⎜⎝ ⎟⎠⎥⎦= − ⎢⎣ − ⎥⎦

⎛ ⎞ ⎛ ⎞

= − ⎜⎝ − ⎟⎠= − ⎜⎝ ⎟⎠= −

9.

2 2 2

1 14

3 3

2

14 2 14 2 14 6

21 5 21 21 14

14 3 2 9 6

21 7 3 49 49

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟

⎜ − ⎟ ⎜ ⎟ ⎝ ⎠

⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞

= ⎜ ⎟⎝ ⎠ = ⎜⎝ ⎟⎠=

10.

2 2 35 33

5 33 11

7 7 7 7

1 7 1 6 6 2

1 7 7 7 7

⎛ − ⎞ ⎛ − ⎞ ⎛− ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝= ⎠ ⎝= ⎠= − = −

⎛ − ⎞ ⎛ − ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

11.

7 11 127 –21 117 –47 7 7

11 12 11 4 15 15

7 21 7 7 7

= = =

+ +

12.

1 3 7 4 6 7 5

2 4 8 8 8 8 8 5

1 3 7 4 6 7 3 3

2 4 8 8 8 8 8

− + − +

= = =

+ − + −

13. 1 3

2 2

1 1 2 3 2 1

1 – 1 – 1 – –

3 3 3 3

1 = = = =

+

14. 3 3 3

2 2 2

5 2 5 7

1 2 2 2 2

6 14 6 20

2 7 7 7 7

+ = + = +

+ −

= + = + =

15.

(

^{5 3}

)(

^{5 – 3}

) ( ) ( )

^{5 2– 3 2}

5 – 3 2

+ =

= =

16.

(

^{5 3}

) ( ) ( )( ) ( )

^{2 5 2 2 5 3 3 2}

5 2 15 3 8 2 15

− = − +

= − + = −

17. ²

2

(3 4)( 1) 3 3 4 4

3 4

x x x x x

x x

− + = + − −

= − −

18. ²

2 2

(2 3) (2 3)(2 3)

4 6 6 9

4 12 9

x x x

x x x

x x

− = − −

= − − +

= − +

19.

2 2

(3 – 9)(2 1) 6 3 – 18 – 9 6 – 15 – 9

x x x x x

x x

+ = +

=

20. ²

2

(4 11)(3 7) 12 28 33 77

12 61 77

x x x x x

x x

− − = − − +

= − +

21. ^{2 2 2 2}

4 3 2 3 2 2

4 3 2

(3 1) (3 1)(3 1)

9 3 3 3 3 1

9 6 7 2 1

t t t t t t

t t t t t t t t

t t t t

− + = − + − +

= − + − + − + − +

= − + − +

22. ³

2

3 2 2

3 2

(2 3) (2 3)(2 3)(2 3)

(4 12 9)(2 3)

8 12 24 36 18 27

8 36 54 27

t t t t

t t t

t t t t t

t t t

+ = + + +

= + + +

= + + + + +

= + + +

23.

2– 4 ( – 2)( 2)

– 2 – 2 2

x x x

x x x

= + = + , x≠2

24.

2 6 ( 3)( 2)

3 ( 3) 2

x x x x

x x x

− − − +

= = +

− − , x≠3

25.

2– 4 – 21 ( 3)( – 7)

3 3 – 7

t t t t

t t t

= + =

+ + , t≠ −3

26.

2

3 2 2

2 2 2 (1 )

2 ( 2 1)

2 ( 1) ( 1)( 1)

2 1

x x x x

x x x x x x

x x x x x

x

− −

− + = − +

− −

= − −

= − −

27.

2

12 4 2

2 x x 2

x x

+ + + +

12 4( 2) 2

( 2) ( 2) ( 2)

12 4 8 2 6 20

( 2) ( 2)

2(3 10) ( 2)

x x

x x x x x x

x x x

x x x x

x x x

= + + +

+ + +

+ + + +

= =

+ +

= + +

28.

2

2

6 2 9 1

y

y y

− + −

2

2(3 1) (3 1)(3 1)

2(3 1) 2

2(3 1)(3 1) 2(3 1)(3 1) y

y y y

y y

y y y y

= +

− + −

= + +

+ − + −

6 2 2

2(3 1)(3 1)

y y

y y

= + +

+ −

8 2

2(3 1)(3 1) y

y y

= +

+ −

2(4 1) 4 1

2(3 1)(3 1) (3 1)(3 1)

y y

y y y y

+ +

= =

+ − + −

29. a. 0 0⋅ =0 b. ⁰

0 is undefined.

c. 0

17=0 d. 3

0 is undefined.

e. 0⁵=0 f. 17⁰ =1 30. If ⁰

0=a, then 0= ⋅0 a, but this is meaningless because a could be any real number. No single value satisfies 0

0=a. 31. _.083

12 1.000 96 40

36 4

32. .285714 7 2.000000

1 4 60 56 40 35 50 49 10

7 30 28 2

33. .142857

21 3.000000 2 1

90 84 60 42 180 168 120 105 150 147 3 34. .294117...

17 5.000000... 0.2941176470588235 3 4

160 153 70 68 20 17 30 17 130 119 11

→

35. 3.6 3 11.0 9

20 18 2

36. .846153

13 11.000000 10 4

60 52 80 78 20 13 70 65 50 39 11 37. x = 0.123123123...

1000 123.123123...

0.123123...

999 123

123 41 999 333 x

x x x

=

=

=

= =

38. x=0.217171717… 1000 217.171717...

10 2.171717...

990 215

215 43 990 198 x

x x x

=

=

=

= =

39. x = 2.56565656...

100 256.565656...

2.565656...

99 254

254 99 x x x x

=

=

=

=

40. x=3.929292… 100 392.929292...

3.929292...

99 389

389 99 x x x x

=

=

=

=

41. x = 0.199999...

100 19.99999...

10 1.99999...

90 18

18 1 90 5 x

x x x

=

=

=

= =

42. x=0.399999… 100 39.99999...

10 3.99999...

90 36

36 2 90 5 x

x x x

=

=

=

= =

43. Those rational numbers that can be expressed by a terminating decimal followed by zeros.

44. 1

p , q p q

= ⎜ ⎟⎛ ⎞

⎝ ⎠ so we only need to look at 1 q. If 2ⁿ 5 ,^m

q= ⋅ then

1 1 1

(0.5) (0.2) .

2 5

n m

n m

q

⎛ ⎞ ⎛ ⎞

=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⋅ = The product of any number of terminating decimals is also a terminating decimal, so (0.5) and (0.2) ,^{n m} and hence their product, 1

q, is a terminating decimal. Thus p

q has a terminating decimal expansion.

45. Answers will vary. Possible answer: 0.000001,

12

1 0.0000010819...

π ^≈

46. Smallest positive integer: 1; There is no smallest positive rational or irrational number.

47. Answers will vary. Possible answer:

3.14159101001...

48. There is no real number between 0.9999…

(repeating 9's) and 1. 0.9999… and 1 represent the same real number.

49. Irrational

50. Answers will vary. Possible answers:

and ,

π π

− − 2 and 2

51. ( 3 1)+ ³≈20.39230485

52.

(

²− ³

)

⁴≈0.0102051443 53. ⁴1.123 – 1.09³ ≈0.00028307388 54.

(

^3.1415

)

^{−1/ 2}≈0.5641979034 55. 8.9π +² 1 – 3π ≈0.000691744752 56. ⁴(6π²−2)π ≈3.661591807

57. Let a and b be real numbers with a<b. Let n be a natural number that satisfies

a b n< − /

1 . Let S ={k:k n>b}. Since a nonempty set of integers that is bounded below contains a least element, there is a

S

k

₀

∈

such that

k

₀

/ n > b

but b

n k −1)/ ≤

( ₀ . Then

n a n b

n k n

k₀ −1= ₀ −1 > −1 >

Thus, a< ^k0_n⁻¹ ≤b. If ^k0_n⁻¹ <b, then choose

n

r = k⁰⁻¹. Otherwise, choose r= ^k0_n⁻².

Note that 1

a b r

< − <n .

Given a<b, choose r so that a< <r₁ b. Then choose r r₂,₃ so that a<r₂< < <r₁ r₃ b, and so on.

58. Answers will vary. Possible answer: ≈120 in³ 59. 4000 mi 5280 ^ft 21,120, 000 ft

r= × mi=

equator 2 2 (21,120, 000) 132, 700,874 ft

πr π

= =

≈

60. Answers will vary. Possible answer:

beats min hr day

70 60 24 365 20 yr

min hr day year

735,840, 000 beats

× × × ×

=

61.

2 2

3

16 12 (270 12) 2

93,807, 453.98 in.

V = πr h= π⎛⎜ ⋅ ⎞⎟ ⋅

⎝ ⎠

≈

volume of one board foot (in inches):

1 12 12× × =144 in.3

number of board feet:

93,807, 453.98

651, 441 board ft

144 ≈

62. V =π(8.004) (270)² −π(8) (270)² ≈54.3 ft.³ 63. a. If I stay home from work today then it

rains. If I do not stay home from work, then it does not rain.

b. If the candidate will be hired then she meets all the qualifications. If the candidate will not be hired then she does not meet all the qualifications.

64. a. If I pass the course, then I got an A on the final exam. If I did not pass the course, thn I did not get an A on the final exam.

b. If I take off next week, then I finished my research paper. If I do not take off next week, then I did not finish my research paper.

65. a. If a triangle is a right triangle, then

2 2 2

.

a +b =c If a triangle is not a right triangle, then a²+b² ≠c².

b. If the measure of angle ABC is greater than 0^o and less than 90^o, it is acute. If the measure of angle ABC is less than 0^o or greater than 90^o, then it is not acute.

66. a. If angle ABC is an acute angle, then its measure is 45^o. If angle ABC is not an acute angle, then its measure is not 45^o. b. If a² <b² then a<b. If a²≥b² then

. a≥b

67. a. The statement, converse, and contrapositive are all true.

b. The statement, converse, and contrapositive are all true.

68. a. The statement and contrapositive are true.

The converse is false.

b. The statement, converse, and contrapositive are all false.

69. a. Some isosceles triangles are not equilateral. The negation is true.

b. All real numbers are integers. The original statement is true.

c. Some natural number is larger than its square. The original statement is true.

70. a. Some natural number is not rational. The original statement is true.

b. Every circle has area less than or equal to 9π. The original statement is true.

c. Some real number is less than or equal to its square. The negation is true.

71. a. True; If x is positive, then x² is positive.

b. False; Take x= −2. Then x²>0 but 0

x< .

c. False; Take 1

x= 2. Then

x

²

=

₄¹

< x

d. True; Let x be any number. Take

2 1

y=x + . Then y>x².

e. True; Let y be any positive number. Take 2

x= y. Then 0< <x y.

72. a. True; ^x+ − < + + −

( )

^{x x 1}

( )

^{x : 0}<¹ b. False; There are infinitely many prime

numbers.

c. True; Let x be any number. Take 1 1

y= +x . Then y ¹

> x.

d. True; 1/n can be made arbitrarily close to 0.

e. True; 1/ 2ⁿ can be made arbitrarily close to 0.

73. a. If n is odd, then there is an integer k such that n=2k+1. Then

2 2 2

2

(2 1) 4 4 1

2(2 2 ) 1

n k k k

k k

= + = + +

= + +

b. Prove the contrapositive. Suppose n is even. Then there is an integer k such that

2 .

n= k Then n² =(2 )k ² =4k²=2(2k²). Thus n² is even.

74. Parts (a) and (b) prove that n is odd if and only if n² is odd.

75. a. 243= ⋅ ⋅ ⋅ ⋅3 3 3 3 3

b. 124= ⋅4 31= ⋅ ⋅2 2 31 or 2²⋅31

c.

2 2

5100 2 2550 2 2 1275 2 2 3 425 2 2 3 5 85 2 2 3 5 5 17 or 2 3 5 17

= ⋅ = ⋅ ⋅

= ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 76. For example, let A= ⋅b c²⋅d³; then

2 2 4 6

A =b ⋅c ⋅d , so the square of the number is the product of primes which occur an even number of times.

77.

2 2 2

2 p; 2 p2; 2 ;

q p

q q

= = = Since the prime

factors of p² must occur an even number of times, 2q² would not be valid and p 2

q = must be irrational.

78.

2 2 2

3 p; 3 p2; 3 ;

q p

q q

= = = Since the prime

factors of p²must occur an even number of times, 3q² would not be valid and p 3

q = must be irrational.

79. Let a, b, p, and q be natural numbers, so a b and ^p

q are rational. ^{a p aq bp}

b q bq

+ = + This

sum is the quotient of natural numbers, so it is also rational.

80. Assume a is irrational, p 0

q ≠ is rational, and p r

a⋅ =q s is rational. Then a ^{q r} p s

= ⋅

⋅ is rational, which is a contradiction.

81. a. – 9=–3; rational b. 0.375 ³;

=8 rational

c. (3 2)(5 2)=15 4=30; rational d. (1+ 3)²= +1 2 3+ = +3 4 2 3;

irrational

82. a. –2 b. –2 c. x = 2.4444...;

10 24.4444...

2.4444...

9 22

22 9 x x x x

=

=

=

=

d. 1

e. n = 1: x = 0, n = 2: ³,

x=2 n = 3: –², x= 3 n = 4: ⁵

x=4

The upper bound is ³. 2 f. 2

83. a. Answers will vary. Possible answer: An example

isS ={ :x x²<5, a rational number}.x Here the least upper bound is 5, which is real but irrational.

b. True

0.2 Concepts Review 1. [ 1, 5); (− −∞ −, 2]

2. b > 0; b < 0 3. (b) and (c) 4. − ≤ ≤1 x 5

Problem Set 0.2 1. a.

b.

c.

d.

e.

f.

2. a. (2, 7) b. [ 3, 4)− c. (−∞ −, 2] d. [ 1, 3]−

3. 7 2 5 2 ; ( 2, )

x x

x

− < −

− < − ∞

4.

( )

3 5 4 6

1 ; 1,

x x

x

− < −

< ∞

5.

7 – 2 9 3

–5 2

5 5

– ; – ,

2 2

x x

x x

≤ +

≤

⎡ ⎞

≥ ⎢⎣ ∞ ⎟⎠

6. 5 3 6 4 1 ; ( ,1)

x x

x

− > −

> −∞

7. 4 3 2 5

6 3 3

2 1; ( 2, 1) x

x x

− < + <

− < <

− < < − −

8. 3 4 9 11

6 4 20

3 3

5; , 5

2 2

x x x

− < − <

< <

⎛ ⎞

< < ⎜⎝ ⎟⎠

9.

–3 1 – 6 4

–4 –6 3

2 1 1 2

– ; – ,

3 2 2 3

x x x

< ≤

< ≤

⎡ ⎞

> ≥ ⎢⎣ ⎟⎠

10. 4 5 3 7

1 3 2

1 2 2 1

; ,

3 3 3 3

x x x

< − <

− < − <

⎛ ⎞

> > − ⎜⎝− ⎟⎠

11. x² + 2x – 12 < 0;

–2 (2) – 4(1)(–12)2 –2 52

2(1) 2

–1 13

x ± ±

= =

= ±

( ) ( )

– –1 13 – –1 – 13 0;

x x

⎡ + ⎤ ⎡ ⎤<

⎣ ⎦ ⎣ ⎦

(

–1 – 13, – 1+ 13

)

12. ² 5 6 0 ( 1)( 6) 0;

x x

x x

− − >

+ − >

(−∞ − ∪, 1) (6, )∞

13. 2x² + 5x – 3 > 0; (2x – 1)(x + 3) > 0;

( , 3) 1, 2

⎛ ⎞

−∞ − ∪⎜ ∞⎟

⎝ ⎠

14. 4 ² 5 6 0 (4 3)( 2) 0; 3, 2

4

x x

x x

− − <

⎛ ⎞

+ − < ⎜− ⎟

⎝ ⎠

15. 4 – 3 0;

x x

+ ≤ [–4, 3)

16. ^{3 2} 0; ,² (1, )

1 3

x x

− ≥ ⎛⎜−∞ ⎤⎥∪ ∞

− ⎝ ⎦

17. 2 5

2 5 0

2 5 0;

x x

x x

<

− <

− <

(– , 0) 2, 5

⎛ ⎞

∞ ∪⎜ ∞⎟

⎝ ⎠

18. 7

4 7

7 7 0

4 7 28

4 0;

x x

x x

≤

− ≤

− ≤

(

^{, 0}

)

^1,

4

⎡ ⎞

−∞ ∪⎢⎣ ∞ ⎟⎠

19. 1 3 2 4

1 4 0

3 2

1 4(3 2) 3 2 0

9 12 2 3

0; , ,

3 2 3 4

x x

x x

x x

− ≤

− − ≤

− −

− ≤

−− ≤ ⎛⎜⎝−∞ ⎞ ⎡⎟⎠ ⎣∪⎢ ∞⎞⎠⎟

20. 3

5 2

3 2 0

5

3 2( 5)

5 0

2 7 7

0; 5,

5 2

x x

x x

x x

+ >

− >

+

− +

+ >

− −+ > ⎛⎜⎝− − ⎞⎟⎠

21. (x+2)(x−1)(x− >3) 0; ( 2,1)− ∪(3,8)

22. 3 1

(2 3)(3 1)( 2) 0; , , 2

2 3

x+ x− x− < ⎛⎜−∞ − ⎞ ⎛⎟ ⎜∪ ⎞⎟

⎝ ⎠ ⎝ ⎠

23. (2 - 3)( -1) ( - 3)x x ² x ≥0; ^{– , 3, 3}

[ )

2

⎛ ∞ ⎤∪ ∞

⎜ ⎥

⎝ ⎦

24.

( ) ( )

(2 3)( 1) (2 3) 0;

,1 1,3 3, 2

x− x− x− >

⎛ ⎞

−∞ ∪⎜⎝ ⎟⎠∪ ∞

25. ³– ²– 2–

5 6 0

( 5 – 6) 0 ( 1)( – 6) 0;

x x x

x x x x x x

<

<

+ <

(−∞ − ∪, 1) (0, 6)

26. ^{3 2}

2

2

1 0 ( 1)( 1) 0 ( 1)( 1) 0;

x x x

x x

x x

− − + >

− − >

+ − >

( 1,1)− ∪ ∞(1, )

27. a. False. b. True.

c. False.

28. a. True. b. True.

c. False.

29. a. ⇒Let a<b, so ab<b². Also, a²<ab. Thus, a²<ab<b² and a²<b². ⇐ Let

2 2

a <b , so a≠b Then

( )

( )

2 2 2

2 2

0 2

2 2

a b a ab b

b ab b b b a

< − = − +

< − + = −

Since b>0, we can divide by 2b to get 0

b a− > .

b. We can divide or multiply an inequality by any positive number.

1 1

a 1 a b

b b a

< ⇔ < ⇔ < . 30. (b) and (c) are true.

(a) is false: Take a= −1,b=1. (d) is false: if a≤b, then − ≥ −a b. 31. a. 3x + 7 > 1 and 2x + 1 < 3

3x > –6 and 2x < 2 x > –2 and x < 1; (–2, 1) b. 3x + 7 > 1 and 2x + 1 > –4

3x > –6 and 2x > –5 x > –2 and –⁵;

x> 2

(

^{− ∞2,}

)

c. 3x + 7 > 1 and 2x + 1 < –4 x > –2 and –⁵;

x< 2 ∅ 32. a. 2 7 1 or 2 1 3

2 8 or 2 2 4 or 1

x x

x x

x x

− > + <

> <

> <

(−∞ ∪,1) (4, )∞ b. 2 7 1 or 2 1 3

2 8 or 2 2 4 or 1 ( , 4]

x x

x x

x x

− ≤ + <

≤ <

≤ <

−∞

c. 2 7 1 or 2 1 3 2 8 or 2 2

4 or 1 ( , )

x x

x x

x x

− ≤ + >

≤ >

≤ >

−∞ ∞

33. a. ^{2 2}–

3 2 2

– –

3 2

( 1)( 2 – 7) 1

3 5 – 7 1

2 – 5 – 6 0 ( 3)( 1)( – 2) 0

x x x x

x x x x

x x x

x x x

+ + ≥

+ ≥

+ ≥

+ + ≥

[ 3, 1]− − ∪[2, )∞

b. ^{4 2}

4 2

2 2

2

2 8

2 8 0

( 4)( 2) 0

( 2)( 2)( 2) 0

x x

x x

x x

x x x

−

− −

−

≥

≥ + ≥

+ + − ≥

(−∞ − ∪, 2] [2, )∞

c. ^{2 2 2}

2 2

2 2

( 1) 7( 1) 10 0

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

( 4)( 1) 0

( 2)( 1)( 1)( 2) 0

x x

x x

x x

x x x x

−

− −

+ + + <

+ − + − <

<

+ + − − <

( 2, 1)− − ∪(1, 2) 34. a. _1.99<¹_<2.01

x

x x 1 2.01 99

.

1 < <

1 99 .

1 x< and 1<2.01x 99

. 1

< 1

x and

01 . 2

> 1 x

99 . 1

1 01

. 2

1 <x<

⎟⎠

⎜ ⎞

⎝

⎛

99 . 1 , 1 01 . 2

1

b. _3.01

2 99 1 .

2 <

< + x

) 2 ( 01 . 3 1 ) 2 ( 99 .

2 x+ < < x+ 1

98 . 5 99 .

2 x+ < and 1<3.01x+6.02

99 . 2

98 .

−4

<

x and

01 . 3

02 .

−5

>

x

99 . 2

98 . 4 01

. 3

02 .

5 < <−

− x

⎟⎠

⎜ ⎞

⎝

⎛− −

99 . 2

98 . , 4 01 . 3

02 . 5

35. 2 5;

2 5 or 2 5 3 or 7 x

x x

x x

− ≥

− ≤ − − ≥

≤ − ≥

(−∞ − ∪, 3] [7, )∞

36. x+ <2 1;

–1 2 1

–3 –1

x x

< + <

< <

(–3, –1) 37. 4 5 10;

10 4 5 10

15 4 5

15 5 15 5

; ,

4 4 4 4

x x x x + ≤

− ≤ + ≤

− ≤ ≤

⎡ ⎤

− ≤ ≤ ⎢⎣− ⎥⎦

38. 2 – 1x >2;

2x – 1 < –2 or 2x – 1 > 2 2x < –1 or 2x > 3;

1 3 1 3

– or , – , – ,

2 2 2 2

x< x> ⎛⎜⎝ ∞ ⎞ ⎛⎟ ⎜⎠ ⎝∪ ∞⎞⎟⎠

39. ² 5 7 7

2 2

5 7 or 5 7

7 7

2 2

2 or 12

7 7

7 or 42;

x

x x

x x

x x

− ≥

− ≤ − − ≥

≤ − ≥

≤ − ≥

(−∞ − ∪, 7] [42, )∞ 40. 1 1

4 x+ <

1 1 1

4

2 0;

4 x x

− < + <

− < <

–8 < x < 0; (–8, 0) 41. 5 6 1;

5 6 1 or 5 6 1 5 5 or 5 7

7 7

1 or ; ( ,1) ,

5 5

x

x x

x x

x x

− >

− < − − >

< >

⎛ ⎞

< > −∞ ∪⎜ ∞⎟

⎝ ⎠

42. 2 – 7x >3;

2x – 7 < –3 or 2x – 7 > 3 2x < 4 or 2x > 10

x < 2 or x > 5; (−∞, 2)∪(5, )∞

43. ¹ 3 6;

1 1

3 6 or 3 6

1 1

3 0 or 9 0 x

x x

x x

− >

− < − − >

+ < − >

1 3 1 9

0 or 0;

1 1

, 0 0,

3 9

x x

x x

+ −

< >

⎛− ⎞ ⎛∪ ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

44. 2 ⁵ 1;

+x >

5 5

2 –1 or 2 1

5 5

3 0 or 1 0

3 5 5

0 or 0;

x x

x x

x x

x x

+ < + >

+ < + >

+ < + >

(– , – 5) – , 05 (0, ) 3

⎛ ⎞

∞ ∪⎜ ⎟∪ ∞

⎝ ⎠

45. x²−3x− ≥4 0;

3 (–3) – 4(1)(–4)2 3 5 –1, 4

2(1) 2

x ± ±

= = =

(x+1)(x−4)=0; (−∞ − ∪, 1] [4, )∞

46.

2

2 4 ( 4) 4(1)(4)

4 4 0; 2

2(1)

( 2)( 2) 0; 2

x x x

x x x

± − −

− + ≤ = =

− − ≤ =

47. 3x² + 17x – 6 > 0;

–17 (17) – 4(3)(–6)2 –17 19 1

2(3) 6 –6, 3

x= ± = ± =

(3x – 1)(x + 6) > 0; (– , – 6) ¹, 3

⎛ ⎞

∞ ∪⎜ ∞⎟

⎝ ⎠

48. 14x²+11x−15≤0;

11 (11)2 4(14)( 15) 11 31

2(14) 28

3 5, 2 7 x

x

− ± − − − ±

= =

= −

3 5 3 5

0; ,

2 7 2 7

x x

⎛ + ⎞⎛ − ⎞≤ ⎡− ⎤

⎜ ⎟⎜ ⎟ ⎢ ⎥

⎝ ⎠⎝ ⎠ ⎣ ⎦

49. x− <3 0.5⇒5x− <3 5(0.5)⇒5x−15 <2.5 50. x+ <2 0.3⇒4 x+ <2 4(0.3)⇒ 4x+18 <1.2

51. 2 6 2 6 12 x− < ⇒ε6 x− < ⇒ε x− <ε

52. 4 2 4 2 8

x+ < ⇒ε2 x+ < ⇒ε x+ <ε 53. 3 15 3( 5)

3 5

5 ;

3 3

x x

x x

ε ε

ε ε δ ε

− < ⇒ − <

⇒ − <

⇒ − < =

54. 4 8 4( 2)

4 2

2 ;

4 4

x x

x x

ε ε

ε ε δ ε

− < ⇒ − <

⇒ − <

⇒ − < =

55. 6 36 6( 6)

6 6

6 ;

6 6

x x

x x

ε ε

ε ε δ ε + < ⇒ + <

⇒ + <

⇒ + < =

56. 5 25 5( 5)

5 5

5 ;

5 5

x x

x x

ε ε

ε ε δ ε + < ⇒ + <

⇒ + <

⇒ + < =

57. C=πd

– 10 0.02 – 10 0.02 –10 0.02 10 0.02

– 0.0064

C d d

d

≤

π ≤

⎛ ⎞

π⎜⎝ π ⎟⎠ ≤

≤ ≈

π π

We must measure the diameter to an accuracy of 0.0064 in.

58. ^{50 1.5, 5}

(

³²

)

^{50 1.5;}

C− ≤ 9 F− − ≤

( )

5 32 90 1.5

9

122 2.7 F

F

− − ≤

− ≤

We are allowed an error of 2.7 F.

59.

2 2

2 2

– –

2–

– 1 2 – 3 – 1 2 – 6 ( – 1) (2 – 6)

2 1 4 24 36

3 22 35 0

(3 – 7)( – 5) 0;

x x

x x

x x

x x x x

x x

x x

<

<

<

+ < +

+ >

>

– , (5, 7 ) 3

⎛ ∞ ⎞∪ ∞

⎜ ⎟

⎝ ⎠

60.

( )

²

2

2 1 1

(2 1) 1

x x

x x

− ≥ +

− ≥ +

2 2

2

4 4 1 2 1

3 6 0

3 ( 2) 0

x x x x

x x

x x

− + ≥ + +

− ≥

− ≥ (−∞, 0]∪[2, )∞ 61.

2 2

2 2

2

2 2 3 10

4 6 10

(4 6) ( 10)

16 48 36 20 100

15 68 64 0

(5 4)(3 16) 0;

x x

x x

x x

x x x x

x x

x x

−

−

− < +

− < +

− < + + < + +

− <

+ − <

–4 16, 5 3

⎛ ⎞

⎜ ⎟

⎝ ⎠

62.

( )( )

2 2

2 2

2

3 1 2 6

3 1 2 12

(3 1) (2 12)

9 6 1 4 48 144

5 54 143 0

5 11 13 0

x x

x x

x x

x x x x

x x

x x

− < +

− < +

− < +

− + < + +

− − <

+ − <

11,13 5

⎛ ⎞

⎜− ⎟

⎝ ⎠

63.

( )

2 2

2 2 2 2

and Order property: when is positive.

Transitivity

x y x x x y x y y y x y xz yz z

x y

x y x x

< ⇒ ≤ < < ⇔ <

⇒ <

⇒ < =

Conversely,

( )

( )( )

2 2 2

2 2 2

2 2 2

– 0 Subtract from each side.

– 0 Factor the difference of two squares.

– 0 This is the only factor that can be negative.

Add t

x y x y x x

x y y

x y x y

x y

x y y

< ⇒ < =

⇒ <

⇒ + <

⇒ <

⇒ < o each side.

64. ⁰< < ⇒ =^{a b a}

( )

^{a 2 and b}=

( )

^{b 2,}so

( ) ( )

^{a 2< b 2,} and, by Problem 63, a < b ⇒ a< b.

65. a. a–b = +a (– )b ≤ a+ –b = a+b b. a–b ≥ a–b ≥ a –b Use Property 4

of absolute values.

c. a b c (a b) c a b c

a b c

+ + = + + ≤ + +

≤ + +

66. _{2 2}

2

2

2

1 1 1 1

2 2

3 3

1 1

3 2

1 1

3 2

1 1

3 2

x x

x x

x x

x x x x

⎛ ⎞

− + = + −⎜⎜ + ⎟⎟

+ + ⎝ ⎠

≤ + −

+ +

= +

+ +

= +

+ +

by the Triangular Inequality, and since

2 3 0, x + >

2

1 1

2 0 0, 0.

3 2

x x x

+ > ⇒ > >

+ +

2 3 3

x + ≥ and x+ ≥2 2, so

2

1 1

3 3 x

+ ≤ and 1 1

2 2,

x ≤

+ thus,

2

1 1 1 1

2 3 2

3 x

x

+ ≤ +

+ +

67.

2 2

– 2 (–2)

9 9

x x

x x

= +

+ +

2 2 2

– 2 –2

9 9 9

x x

x x x

≤ +

+ + +

2 2 2 2

– 2 2 2

9 9 9 9

x x

x

x x x x

≤ + = +

+ + + +

Since ²

2

1 1

9 9, 9 9 x

x

+ ≥ ≤

+ 2

2 2

9 9

x x

x

+ +

+ ≤

2

– 2 2 9 9 x x x

≤ + +

68. 2 ² 2 7 ² 2 7

4 4 7 15 x ≤ ⇒ x + x+ ≤ x + x+

≤ + + = and x²+ ≥1 1 so

2

1 1.

1 x

+ ≤

Thus,

2 2

2 2

2 7 1

2 7

1 1

15 1 15

x x

x x

x x

+ +

= + +

+ +

≤ ⋅ = 69. ^{4 1 3 1 2 1 1}

2 4 8 16

x + x + x + x+

4 3 2

4 3 2

1 1 1 1

2 4 8 16

1 1 1 1

1 since 1.

2 4 8 16

1 1 1 1

So 1.9375 2.

2 4 8 16

x x x x

x

x x x x

≤ + + + +

≤ + + + + ≤

+ + + + ≤ <

70. a. ²

2 0

(1 ) 0 0 or 1 x x x x

x x

x x

<

− <

− <

< >

b. ²

2 0

( 1) 0

0 1

x x

x x

x x x

<

− <

− <

< <

71. a≠ ⇒0

2 2

2

1 1

0 a– a – 2

a a

⎛ ⎞

≤⎜⎝ ⎟⎠ = +

so, ^{2 2}

2 2

1 1

2 a or a 2

a a

≤ + + ≥ .

72.

and

2 2

2 a b

a a a b a b b b

a a b b

a b

a b

<

+ < + + < +

< + <

< + <

73.

2 2

2 2

0

and a b

a ab ab b

a ab b

a ab b

< <

< <

< <

< <

74.

( ) ( )

( )

2 2

2 2 2 2

2

1 1

2 4 2

1 1 1 1

0 2

4 2 4 4

0 1( ) which is always true.

4

ab a b ab a ab b

a ab b a ab b

a b

≤ + ⇔ ≤ + +

⇔ ≤ − + = − +

⇔ ≤ −

75. For a rectangle the area is ab, while for a square the area is

2

2 .

2 a b a _{= ⎜}^{⎛ + ⎞}_⎟

⎝ ⎠ From

Problem 74,

1 2

( )

2 2

a b ab≤ a+b ⇔ab≤ ⎜^{⎛ + ⎞}⎟

⎝ ⎠

so the square has the largest area.

76. 1 ^{2 3 99} 0;

( , 1]

x x x x

+ + + + + ≤

−∞ −

…

77. 1 1 1 1 10 20 30

R≤ + +

1 6 3 2

60 1 11

60 60 11 R R R

≤ + +

≤

≥

1 1 1 1

20 30 40

1 6 4 3

120 120

13 R R R

≥ + +

≥ + +

≤

Thus, 60 120 11≤ ≤R 13

78. ^{2 2}

2 2 2

4 ; 4 (10) 400

4 400 0.01

4 100 0.01

100 0.01 4

A r A

r r r

π π π

π π

π

π

= = =

− <

− <

− <

0.01 2 0.01

4 100 4

0.01 0.01

100 100

4 4

0.00004 in r

r

π π

π π

δ

− < − <

− < < +

≈

0.3 Concepts Review

1. (x+2)²+(y−3)² 2. (x + 4)² + (y – 2)² = 25 3. ^{2 5 3 7}, (1.5, 5)

2 2

− + +

⎛ ⎞ =

⎜ ⎟

⎝ ⎠

4. ^{d b} c a

−

−

Problem Set 0.3 1.

2 2

(3 – 1) (1 – 1) 4 2

d = + = =

2.

2 2

( 3 2) (5 2) 74 8.60

d = − − + + = ≈

3.

2 2

(4 – 5) (5 8) 170 13.04

d = + + = ≈

4.

2 2

( 1 6) (5 3) 49 4 53

7.28

d = − − + − = + =

≈

5. d₁= (5 2)+ ²+(3 – 4)² = 49 1+ = 50

2 2

(5 10) (3 8) 25 25 50

2

2 2

( 2 10) (4 8) 3

144 16 160

so the triangle is isosceles.

1 2

d d

d d

= − + − = + =

= − − + −

= + =

=

6. a= (2 4)− ²+ − −( 4 0)² = 4 16+ = 20

2 2

(4 8) (0 2) 16 4 20

b= − + + = + =

2 2

(2 8) ( 4 2) 36 4 40

c= − + − + = + =

2 2 2,

a +b =c so the triangle is a right triangle.

7. (–1, –1), (–1, 3); (7, –1), (7, 3); (1, 1), (5, 1) 8.

( )

2 2 2 2

2 2

( 3) (0 1) ( 6) (0 4) ;

6 10 12 52

6 42

7 7, 0

x x

x x x x

x x

− + − = − + −

− + = − +

=

= ⇒

9. ^{–2 4 –2 3}, 1, ¹ ;

2 2 2

+ +

⎛ ⎞ ⎛= ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

1 2 25

(1 2)2 – 3 9 3.91

2 4

d= + +⎛⎜ ⎞⎟ = + ≈

⎝ ⎠

10. midpoint of ^{1 2 3 6}, ^{3 9},

2 2 2 2

AB=⎛⎜ + + ⎞ ⎛⎟ ⎜= ⎞⎟

⎝ ⎠ ⎝ ⎠

4 3 7 4 7 11

midpoint of , ,

2 2 2 2

CD=⎛⎜⎝ + + ⎞ ⎛⎟ ⎜⎠ ⎝= ⎞⎟⎠

2 2

3 7 9 11

2 2 2 2

4 1 5 2.24 d = ⎛⎜⎝ − ⎞⎟⎠ +⎛⎜⎝ − ⎞⎟⎠

= + = ≈

11 (x – 1)² + (y – 1)² = 1

12. ^{2 2 2}

2 2

( 2) ( 3) 4

( 2) ( 3) 16

x y

x y

+ + − =

+ + − =

13. ^{2 2 2}

2 2 2

2

2 2

( 2) ( 1)

(5 2) (3 1) 9 16 25

( 2) ( 1) 25

x y r

r r

x y

− + + =

− + + =

= + =

− + + =

14. (x−4)²+(y−3)²=r²

2 2 2

2

2 2

(6 4) (2 3) 4 1 5

( 4) ( 3) 5

r r

x y

− + − =

= + =

− + − =

15. center ^{1 3 3 7}, (2, 5)

2 2

+ +

⎛ ⎞

=⎜⎝ ⎟⎠=

1 2 2 1

radius (1 – 3) (3 – 7) 4 16

2 2

1 20 5

2

= + = +

= =

2 2

( – 2)x +( – 5)y =5

16. Since the circle is tangent to the x-axis, r=4.

2 2

(x−3) +(y−4) =16 17. x²+2x+10+y²– 6 – 10y =0

2 2

2 2

2 2

2 – 6 0

( 2 1) ( – 6 9) 1 9

( 1) ( – 3) 10

x x y y

x x y y

x y

+ + =

+ + + + = +

+ + =

center (–1, = 3); radius= 10

18. ^{2 2}

2 2

2 2

6 16

( 6 9) 16 9

( 3) 25

x y y

x y y

x y

+ − =

+ − + = +

+ − =

center (0, = 3); radius=5 19. x²+y²– 12x+35=0

2 2

2 2

2 2

– 12 –35

( – 12 36) –35 36

( – 6) 1

x x y

x x y

x y

+ =

+ + = +

+ =

center=(6, 0); radius=1

20. ^{2 2}

2 2

2 2

10 10 0

( 10 25) ( 10 25) 25 25

( 5) ( 5) 50

x y x y

x x y y

x y

+ − + =

− + + + + = +

− + + =

( )

center 5, 5 ; = − radius 50= =5 2

21. 4x²+16x+15 4+ y²+6y=0

2 2 3 9 9

4( 4 4) 4 15 16

2 16 4

x + x+ + ⎛⎜⎝y + y+ ⎞⎟⎠= − + +

2

2 3 13

4( 2) 4

4 4

x+ + ⎛⎜⎝y+ ⎞⎟⎠ =

2

2 3 13

( 2)

4 16

x+ +⎛⎜⎝y+ ⎞⎟⎠ =

center = 3

2, ;

4

⎛− − ⎞

⎜ ⎟

⎝ ⎠ radius = 13 4

22. ^{2 2}

2 2

4 16 105 4 3 0

16

3 9

4( 4 4) 4

4 64

105 9

16

16 16

x x y y

x x y y

+ + + + =

⎛ ⎞

+ + + ⎜ + + ⎟

⎝ ⎠

= − + +

2 2

2 2

4( 2) 4 3 10

8

3 5

( 2)

8 2

x y

x y

⎛ ⎞

+ + ⎜ + ⎟ =

⎝ ⎠

⎛ ⎞

+ +⎜⎝ + ⎟⎠ =

3 5 10

center 2, ; radius

8 2 2

⎛ ⎞

= − −⎜ ⎟ = =

⎝ ⎠

23. ^{2 – 1} 1

2 – 1= 24. ^{7 5} 2

4 3

− =

−

25. –6 – 3 9

–5 – 2=7 26. 6 4

0 2 1

− + =

− 27. 5 – 0 5

0 – 3=–3 28. 6 0 0 6 1

− = +

29. 2 1( 2)

2 2

4 0

y x

y x

x y

− = − −

− = − + + − =

30. 4 1( 3)

4 3

7 0

y x

y x

x y

− = − −

− = − + + − =

31. ^{2 3}

2 – 3 0

y x

x y

= +

+ =

32. 0 5

0 5 0

y x

x y

= +

+ − =

33. ^{8 – 3 5}; 4 – 2 2

m= =

– 3 5( – 2) 2 2 – 6 5 – 10 5 – 2 – 4 0

y x

y x

x y

=

=

=

34. ^{2 1 1};

8 4 4

m −

= =

−

1 1( 4) 4

4 4 4

4 0 0

y x

y x

x y

− = −

− = −

− + =

35. 3y = –2x + 1; 2 1

– ;

3 3

y= x+ 2

slope – ;

= 3 -intercept 1

y =3

36. 4 5 6

5 3

4 2

y x

y x

− = −

= − +

5 3

slope ; -intercept

4 y 2

= − =

37. 6 – 2 10 – 2 –2 10 – 8 –5 4;

y x

y x

y x

=

=

= +

slope = –5; y-intercept = 4 38. 4 5 20

5 4 20

4 4

5

x y

y x

y x

+ = −

= − −

= − −

slope ; 4 -intercept = 4 5 y

= − −

39. a. m = 2;

3 2( – 3) 2 – 9

y x

y x

+ =

=

b. –¹; m= 2

3 –1( – 3) 2

1 3

– –

2 2

y x

y x

+ =

=

c. 2 3 6

3 –2 6

–2 2;

3

x y

y x

y x

+ =

= +

= +

–2; m= 3

3 –2( – 3) 3 –2 – 1

3

y x

y x

+ =

=

d. ³; m=2

3 3( – 3) 2

3 15

2 – 2

y x

y x

+ =

=

e. ^{–1 – 2} –³;

3 1 4

m= =

+

3 –3( – 3) 4

3 3

– –

4 4

y x

y x

+ =

=

f. x = 3 g. y = –3

40. a. 3 5

3(3) (1) 5 4 x cy

c c

+ =

+ =

= − b. c=0

c. 2 1

2 1

x y

y x

+ = −

= − − m= −2;

3 5

3 5

3 5

x cy

cy x

y x

c c

+ =

= − +

= − + 2 3

3 2

c c

− = −

=

d. c must be the same as the coefficient of x, so c=3.

e. y− =2 3(x+3);

perpendicular slope 1;

= −3

1 3

3 9

c c

− = −

= 41. ³;

m= 2

1 3( 2) 2

3 2

2

y x

y x

+ = +

= +

42. a. m=2;

3 10

3 10

10

3 3

kx y y kx y kx

− =

− = − +

= −

2; 6 3

k = k=

b. ¹;

m= −2 1

3 2

3 2 k

k

= −

= − c. 2 3 6

3 2 6

2 2;

3

x y

y x

y x

+ =

= − +

= − +

3 3 9

; ;

2 3 2 2

m= k = k=

43. y = 3(3) – 1 = 8; (3, 9) is above the line.

44. ( , 0), (0, ); ⁰ 0

; ; 1

b b

a b m

a a

b bx x y

y x b y b

a a a b

= − = −

−

= − + + = + =

45. 2 3 4

–3 5

x y x y

+ =

+ =

2 3 4

9 – 3 –15 11 –11 –1 x y x y

x x

+ =

=

=

= –3(–1) 5

2 y y + =

=

Point of intersection: (–1, 2)

3 –2 4

2 4

–3 3

y x

y x

= +

= +

3 m=2

2 3( 1) 2

3 7

2 2

y x

y x

− = +

= +

46. 4 5 8

2 10

x y

x y

− =

+ = −

4 5 8

4 2 20

7 28 4

x y

x y

y y

− =

− − =

− =

= − 4 5( 4) 8

4 12

3 x

x x

− − =

= −

= −

Point of intersection:

(

^{− −3, 4 ;}

)

4 5 8

5 4 8

4 8

5 5

x y

y x

y x

− =

− = − +

= −

5 m= −4

4 5( 3) 4

5 31

4 4

y x

y x

+ = − +

= − −

47. 3 – 4 5

2 3 9

x y

x y

=

+ =

9 – 12 15

8 12 36

17 51 3

x y

x y

x x

=

+ =

=

= 3(3) – 4 5

–4 –4

1 y y y

=

=

=

Point of intersection: (3, 1); 3x – 4y = 5;

–4 –3 5

3 5

4 –4

y x

y x

= +

= –4 m= 3

– 1 –4( – 3) 3 –4 5

3

y x

y x

=

= +

48. 5 – 2 5

2 3 6

x y

x y

=

+ =

15 – 6 15

4 6 12

19 27 27 19

x y

x y

x x

=

+ =

=

=

2 27 3 6

19 3 60

19 20 19 y y y

⎛ ⎞ + =

⎜ ⎟

⎝ ⎠

=

=

Point of intersection: ^{27 20}, ; 19 19

⎛ ⎞

⎜ ⎟

⎝ ⎠

5 2 5

–2 –5 5

5 5

2 –2

x y

y x

y x

− =

= +

= –2 m= 5

20 2 27

– –

19 5 19

2 54 20

–5 95 19

2 154

5 95

y x

y x

y x

⎛ ⎞

= ⎜ − ⎟

⎝ ⎠

= + +

= − +

49. center: , ^{2 6 –1 3} (4, 1)

2 2

+ +

⎛ ⎞ =

⎜ ⎟

⎝ ⎠

2 6 3 3

midpoint , (4, 3)

2 2

+ +

⎛ ⎞

=⎜ ⎟=

⎝ ⎠

2 2

inscribed circle: radius (4 – 4) (1 – 3) 4 2

= +

= =

2 2

( – 4)x +( – 1)y =4 circumscribed circle:

2 2

radius= (4 – 2) +(1 – 3) = 8

2 2

( – 4)x +( – 1)y =8

50. The radius of each circle is 16=4. The centers are

(

^{1, 2 and −}

) (

^{−9,10 .}

)

The length of the belt is the sum of half the circumference of the first circle, half the circumference of the second circle, and twice the distance between their centers.

2 2

1 1

2 (4) 2 (4) 2 (1 9) ( 2 10)

2 2

8 2 100 144 56.37

L π π

π

= ⋅ + ⋅ + + + − −

= + +

≈

51. Put the vertex of the right angle at the origin with the other vertices at (a, 0) and (0, b). The midpoint of the hypotenuse is , .

2 2

⎛a b⎞

⎜ ⎟

⎝ ⎠ The distances from the vertices are

2 2 2 2

2 2

– 0 –

2 2 4 4

1 ,

2

a b a b

a

a b

⎛ ⎞ +⎛ ⎞ = +

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= +

2 2 2 2

2 2

0 – –

2 2 4 4

1 , and

2

a b a b

b

a b

⎛ ⎞ +⎛ ⎞ = +

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= +

2 2 2 2

2 2

0 – 0 –

2 2 4 4

1 ,

2

a b a b

a b

⎛ ⎞ +⎛ ⎞ = +

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= +

which are all the same.

52. From Problem 51, the midpoint of the hypotenuse,

(

^{4, 3, ,}

)

is equidistant from the vertices. This is the center of the circle. The radius is 16 9+ =5. The equation of the circle is

2 2

(x−4) +(y−3) =25.

53. x²+y²– 4 – 2 – 11x y =0

2 2

2 2

2 2

2 2

2 2

( – 4 4) ( – 2 1) 11 4 1 ( – 2) ( – 1) 16

20 – 12 72 0

( 20 100) ( – 12 36)

–72 100 36 ( 10) ( – 6) 64

x x y y

x y

x y x y

x x y y

x y

+ + + = + +

+ =

+ + + =

+ + + +

= + +

+ + =

center of first circle: (2, 1) center of second circle: (–10, 6)

2 2

(2 10) (1 – 6) 144 25 169 13

d= + + = +

= =

However, the radii only sum to 4 + 8 = 12, so the circles must not intersect if the distance between their centers is 13.

54. ^{2 2}

2 2

2 2

2 2

0

4 4

4 4

x ax y by c

a b

x ax y by

a b

c

+ + + + =

⎛ ⎞ ⎛ ⎞

+ + + + +

⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= − + +

2 2 2 2

2 2

2 2

4

2 2 4

4 0 4

4

a b a b c

x y

a b c

a b c

+ −

⎛ + ⎞ +⎛ + ⎞ =

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

+ − > ⇒ + >

55. Label the points C, P, Q, and R as shown in the figure below. Let d = OP , h= OR, and

a= PR . Triangles ΔOPR and ΔCQR are similar because each contains a right angle and they share angle∠QRC. For an angle of

30 , ³

2 d

h = and 1

2 2

a h a

h= ⇒ = . Using a property of similar triangles, QC / RC = 3 / 2,

2 3 4

2

2 2 a 3

a = → = +

−

By the Pythagorean Theorem, we have

2 2 3 2 3 4 7.464

d = h −a = a= + ≈

56. The equations of the two circles are

2 2 2

2 2 2

( ) ( )

( ) ( )

x R y R R

x r y r r

− + − =

− + − =

Let

( )

^{a a,} denote the point where the two circles touch. This point must satisfy

2 2 2

2 2

( ) ( )

( )

2