Solution Manual - Calculus 8th Edition Varberg, Purcell, Rigdon Ch00 PDF

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CHAPTER

0

0.1 Concepts Review

Preliminaries

8. −

1. rational numbers 2. dense

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

3. If not Q then not P. 4. theorems

2

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.

6.

7.

1 ⎡2 1 ⎛ 1 1 ⎞ ⎤ 1 − ⎜ − ⎟ =− 3 ⎢⎣5 2 ⎝ 3 5 ⎠ ⎥⎦ ⎡ 2 1 ⎛ 5 3 ⎞⎤ 3 ⎢ − ⎜ − ⎟⎥ ⎣ 5 2 ⎝ 15 15 ⎠⎦

5 1 65 7 58 – = – = 7 13 91 91 91 3 3 1 3 3 1 + − = + − 4 − 7 21 6 − 3 21 6 42 6 7 43 =− + − =− 42 42 42 42 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

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9.

2

2 14 ⎛ 2 ⎞ 14 ⎛ 2 ⎞ 14 6 ⎜ ⎟ = ⎜ ⎟ = ⎛⎜ ⎞⎟ ⎟ 21 ⎜ 5 − 13 ⎟ 21 ⎜ 14 21 ⎝14 ⎠ ⎝ ⎠ ⎝ 3⎠ 2

14 ⎛ 3 ⎞ 2⎛ 9 ⎞ 6 ⎜ ⎟ = ⎜ ⎟= 21 ⎝ 7 ⎠ 3 ⎝ 49 ⎠ 49

=

⎛ 2 ⎞ ⎛ 2 35 ⎞ ⎛ 33⎞ ⎜ − 5⎟ ⎜ − ⎟ ⎜ − ⎟ 33 11 7 ⎠ ⎝7 7⎠ ⎝ 7⎠ = = =− =− 10. ⎝ 6 2 ⎛ 1 1⎞ ⎛ 7 1⎞ ⎛ 6⎞ ⎜ −7⎟ ⎜ 7 −7⎟ ⎜ 7⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 11 – 12 11 – 4 7 7 7 21 7 7 7 = = = 11. 11 + 12 11 + 4 15 15 7 21 7 7 7 1 3 7 4 6 7 5 − + − + 2 4 8 =8 8 8 =8 =5 12. 1 3 7 4 6 7 3 3 + − + − 2 4 8 8 8 8 8

13. 1 –

1 1 + 12

14. 2 +

15.

(

= 1–

3 5 1+ 2

5+ 3

1 3 2

=1–

2 3

=

3 3

–

2 3

=

1 3

3 3 =2+ 2 5 7 − 2 2 2 6 14 6 20 = 2+ = + = 7 7 7 7 =2+

)(

) ( 5) – ( 3)

5– 3 =

2

2

= 5–3= 2

Section 0.1

1

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16.

(

5− 3

)2 = ( 5 )2 − 2 ( 5 )( 3 ) + ( 3 )2

27.

= 5 − 2 15 + 3 = 8 − 2 15

17. (3x − 4)( x + 1) = 3 x 2 + 3 x − 4 x − 4

= 3x 2 − x − 4

12

4 2 + +2 x x x + 2x 12 4( x + 2) 2x = + + x ( x + 2) x ( x + 2) x( x + 2) 12 + 4 x + 8 + 2 x 6 x + 20 = = x( x + 2) x( x + 2) 2

=

18. (2x − 3)2 = (2x − 3)(2x − 3)

+

2(3 x + 10) x (x + 2)

= 4 x2 − 6 x − 6 x + 9 = 4 x 2 − 12 x + 9

19.

28.

(3x – 9)(2x + 1) = 6 x 2 + 3 x – 18 x – 9 = 6x 2 – 15x – 9

20. (4x − 11)(3x − 7) = 12x 2 − 28x − 33x + 77 = 12x 2 − 61x + 77

21. (3t 2 − t + 1)2 = (3t 2 − t + 1)(3t 2 − t + 1) = 9t 4 − 3t 3 + 3t 2 − 3t 3 + t 2 − t + 3t 2 − t + 1

y 2 + 6 y − 2 9 y2 − 1 y 2 = + − + 2(3 y 1) (3 y 1)(3 y − 1) 2(3 y +1) 2y = + 2(3 y + 1)(3 y −1) 2(3 y +1)(3 y −1) 6y + 2 + 2 y 8y + 2 = = 2(3 y + 1)(3 y −1) 2(3 y +1)(3 y −1) 2(4 y +1) 4 y +1 = = 2(3 y + 1)(3 y − 1) (3 y +1)(3 y −1)

= 9t 4 − 6t 3 + 7t 2 − 2t + 1

0⋅ 0 = 0

b.

0 is undefined. 0

c.

0 =0 17

d.

3 is undefined. 0

e.

05 = 0

29. a. 22. (2t + 3)3 = (2t + 3)(2t + 3)(2t + 3) = (4t2 + 12t + 9)(2t + 3) = 8t 3 + 12t 2 + 24t 2 + 36t + 18t + 27 = 8t 3 + 36t 2 + 54t + 27

23.

x 2 – 4 ( x – 2)( x + 2) = =x+2, x ≠ 2 x–2 x–2

24.

x 2 − x − 6 ( x − 3)( x + 2) = = x+2 , x ≠ 3 (x − 3) x −3

25.

t2 – 4 t – 21 (t + 3)( t – 7) = = t – 7 , t ≠ −3 t +3 t +3

f. 17 0 = 1

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

30. If

31.

.083 12 1.000 96

2

26.

2 x− 2 x

x3 − 2 x2 + x

=

=−

2

2 x(1− x)

x( x2 − 2 x + 1) −2 x( x −1) = x( x −1)( x −1)

Section 0.1

40 36 4

2 x −1

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32.

.285714 7 2.000000 14 60 56 40 35 50 49 10 7 30 28 2

33.

.142857 21 3.000000 21 90 84 60 42 180 168 120 105 150 147 3

34.

.294117... 17 5.000000... → 0.2941176470588235 34 160 153 70 68 20 17 30 17 130 119 11 

Instructor’s Resource Manual

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... 1000x = 123.123123... x = 0.123123... 999 x = 123 123 41 x= = 999 333 38. x = 0.217171717… 1000x = 217.171717... 10x = 2.171717... 990 x = 215 215 43 = x= 990 198 39. x = 2.56565656... 100x = 256.565656... x = 2.565656... 99 x = 254 254 x= 99 40. x = 3.929292… 100x = 392.929292... x = 3.929292... 99 x = 389 389 x= 99 Section 0.1

3

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41. x = 0.199999... 100x = 19.99999... 10 x = 1.99999... 90 x = 18 18 1 x= = 90 5

52.

54.

56.

p 1 ⎛1⎞ = p ⎜ ⎟ , so we only need to look at . If q q ⎝q ⎠ q = 2 n ⋅ 5 m, then m

1 ⎛ 1 ⎞ ⎛ 1⎞ = ⋅ = (0.5)n (0.2)m . The product q ⎜⎝ 2 ⎟⎠ ⎜⎝ 5 ⎟⎠ of any number of terminating decimals is also a terminating decimal, so (0.5) n and (0.2) m,

and hence their product, decimal. Thus

1 , is a terminating q

p has a terminating decimal q

expansion. 45. Answers will vary. Possible answer: 0.000001, 1 ≈ 0.0000010819...

π 12

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: −π and π , − 2 and 2 51. ( 3 + 1)3 ≈ 20.39230485

4

Section 0.1

)

4

≈ 0.0102051443

(3.1415 )−1/ 2

≈ 0.5641979034

8.9π2 + 1 – 3π ≈ 0.000691744752

55.

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

n

2− 3

53. 4 1.123 – 3 1.09 ≈ 0.00028307388

42. x = 0.399999… 100x = 39.99999... 10 x = 3.99999... 90 x = 36 36 2 = x= 90 5

44.

(

4

(6π 2 − 2)π ≈ 3.661591807

57. Let a and b be real numbers with a < b . Let n be a natural number that satisfies 1 / n < b − a . Let S = {k : k n > b}. Since a nonempty set of integers that is bounded below contains a least element, there is a k0 ∈ S such that k 0 / n > b but

( k0 −1) / n ≤ b . Then

k0 −1 k0 1 1 = − >b− > a n n n n k0 −1 k 0 −1 < b , then choose . If < ≤ Thus, a b n n r=

k0 − 1 n .

Otherwise, choose r =

k0 −2 n

.

1 0 but x < 0.

c.

False; Take x =

e.

a 2 + b2 = c 2 . If a triangle is not a right 2

2

b.

66. a.

72. a.

b. 68. a. b. 69. a. b.

True; Let x be any number. Take 1 1 y = + 1 . Then y > . x x

d.

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

e.

True; 1/ 2 n 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

c. 70. a.

n2 = (2k + 1)2 = 4 k 2 + 4k +1 = 2(2 k 2 + 2 k ) + 1

The statement and contrapositive are true. The converse is false.

b.

The statement, converse, and contrapositive are all false.

All real numbers are integers. The original statement is true. Some natural number is larger than its square. The original statement is true.

True; x + ( − x) < x + 1 + ( − x) : 0 < 1

c.

The statement, converse, and contrapositive are all true.

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

True; Let y be any positive number. Take y x = . Then 0 < x < y . 2

False; There are infinitely many prime numbers.

If angle ABC is an acute angle, then its measure is 45o. If angle ABC is not an acute angle, then its measure is not 45o.

The statement, converse, and contrapositive are all true.

x2 .

If a triangle is a right triangle, then 2

1 2 . Then x = 2

d. True; Let x be any number. Take

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.

triangle, then a + b ≠ c .

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

Prove the contrapositive. Suppose n is even. Then there is an integer k such that n = 2 k. Then n2 = (2k )2 = 4 k2 = 2(2k2 ) . Thus n 2 is even.

74.

Parts (a) and (b) prove that n is odd if and only if n 2 is odd.

75. a.

243 = 3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3

b.

124 = 4 ⋅ 31 = 2 ⋅ 2 ⋅ 31 or 22 ⋅ 31

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

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Section 0.1

5

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5100 = 2 ⋅ 2550 = 2 ⋅ 2 ⋅ 1275

c.

82. a.

= 2⋅ 2⋅ 3⋅ 425 = 2⋅ 2⋅ 3⋅ 5⋅ 85 = 2⋅ 2⋅ 3⋅ 5⋅ 5⋅ 17 or 2 2 ⋅ 3⋅ 52 ⋅ 17

c.

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

A2 = b 2 ⋅ c 4 ⋅ d 6 , so the square of the number is the product of primes which occur an even number of times. 77.

p 2 = ;2 = q

p2

3 =

b. –2

x = 2.4444...; 10 x = 24.4444... x = 2.4444... 9x = 22 22 x= 9

d. 1 ; 2q2

p2 ;

= Since the prime q2 factors of p2 must occur an even number of p times, 2q2 would not be valid and = 2 q must be irrational.

78.

–2

p p2 ;3= ; 3q 2 = p 2; Since the prime 2 q q

factors of p2 must occur an even number of times, 3q 2 would not be valid and

e.

n = 4: x =

f. 83. a.

3 2 , n = 3: x = – , 3 2

5 4

The upper bound is

p = 3 q

3 . 2

2

Answers will vary. Possible answer: An example is S = {x : x2 < 5, x a rational number}. Here the least upper bound is 5, which is real but irrational.

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

n = 1: x = 0, n = 2: x =

a b

p a p aq +bp This are rational. + = b q bq q sum is the quotient of natural numbers, so it is also rational.

and

b. True

0.2 Concepts Review 1. [−1, 5); (−∞, −2] 2. b > 0; b < 0

p ≠ 0 is rational, and 80. Assume a is irrational, q p r q ⋅r is = is rational. Then a = p ⋅s q s rational, which is a contradiction. a⋅

81. a.

– 9 = –3; rational

b.

3 0.375 = ; rational 8

c.

(3 2)(5 2) = 15 4 = 30; rational

d.

(1 + 3) 2 = 1 + 2 3 + 3 = 4 + 2 3; irrational

3. (b) and (c) 4. −1 ≤ x ≤ 5

Problem Set 0.2 1. a.

b.

c.

d.

6

Section 0.2

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–3 < 1 – 6 x ≤ 4 9. –4 < –6 x ≤ 3

e.

2 1 ⎡ 1 2⎞ > x ≥ – ; ⎢– , ⎟ 3 2 ⎣ 2 3⎠

f.

2. a. c.

(2, 7) (−∞, −2]

b. d.

[−3, 4)

[−1, 3]

10.

3. x − 7 < 2 x − 5 − 2 < x;( − 2, ∞ )

4 < 5−3x < 7 −1 < −3 x < 2 1 2 2 1 > x > − ; ⎛⎜ − , ⎞⎟ 3 3 ⎝ 3 3⎠

4. 3x − 5 < 4 x − 6

1 < x ;( 1, ∞ ) 11. x2 + 2x – 12 < 0; x= 7x – 2 ≤ 9x + 3 5. –5 ≤ 2 x

= –1 ± 13

(

7. −4 < 3 x + 2 < 5 − 6 < 3x < 3 − 2 < x < 1; (− 2, − 1)

( –1 –

)

13, – 1 + 13

)

13. 2x2 + 5x – 3 > 0; (2x – 1)(x + 3) > 0; ⎛1 ⎞ ( −∞, −3) ∪⎜ , ∞⎟ ⎝2 ⎠

14.

4 x2 − 5 x − 6 < 0 ⎛ 3 ⎞ (4 x +3)( x −2) < 0; ⎜ − , 2 ⎟ ⎝ 4 ⎠

15.

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(

12. x 2 − 5x − 6 > 0 ( x +1)( x − 6) > 0; ( −∞, −1) ∪ (6, ∞)

8. −3 < 4 x − 9 < 11 6 < 4 x < 20 3 3 < x < 5;⎜⎛ ,5⎟⎞ 2 ⎝2 ⎠

)

⎡ x – –1 + 13 ⎤ ⎡ x – –1 – 13 ⎤ < 0; ⎣ ⎦⎣ ⎦

5 ⎡ 5 ⎞ x ≥ – ; ⎢– , ∞ ⎟ 2 ⎣ 2 ⎠

6. 5 x − 3 > 6 x − 4 1 > x ;(−∞ ,1)

–2 ± (2) 2 – 4(1)(–12) –2 ± 52 = 2(1) 2

x +4 ≤ 0; [–4, 3) x –3

Section 0.2

7

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16.

3x − 2 2⎤ ⎛ ≥ 0; ⎜ −∞ , ⎥ ∪ (1, ∞) − x 1 3⎦ ⎝

3 >2 + x 5

20.

3 − 2> 0 x +5

17.

2 − 5< 0 x 2 − 5x < 0; x ⎛2 ⎞ (–∞ , 0) ∪ ⎜ , ∞ ⎟ ⎝5 ⎠

18.

7 ≤7 4x 7 −7≤ 0 4x 7 − 28x ≤ 0; 4x ⎡1 ⎞ (−∞ , 0 ) ∪ ⎢ , ∞ ⎟ ⎣4 ⎠

19.

3 − 2(x + 5) >0 x +5

2 0; ⎜ −5, − ⎟ x +5 2⎠ ⎝

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

3⎞ ⎛ 1 ⎞ ⎛ 22. (2x + 3)(3x − 1)(x − 2) < 0; ⎜−∞ , − ⎟ ∪ ⎜ , 2 ⎟ 2 ⎝ ⎠ ⎝3 ⎠

3⎤ ⎛ 23. (2x - 3)(x -1) 2 ( x -3) ≥ 0; ⎜ – ∞ , ⎥ ∪[ 3, ∞ ) 2⎦ ⎝

24. (2x − 3)(x − 1) 2 (x − 3) > 0; ⎛ 3⎞ (−∞,1 ) ∪ ⎜1, ⎟ ∪ (3, ∞ ) ⎝ 2⎠

1 −4 ≤ 0 3 x −2 1− 4(3x − 2) ≤0 3x − 2 ⎛ 9 −12 x 2 ⎞ ⎡3 ⎞ ≤ 0; ⎜ −∞, ⎟ ∪ ⎢ , ∞ ⎟ 3x − 2 3⎠ ⎣ 4 ⎠ ⎝

25.

x 3 – 5x 2 – 6x < 0 x ( x2 – 5 x – 6) < 0 x ( x + 1)( x – 6) < 0;

(−∞ , − 1) ∪ (0,6)

26. x 3 − x 2 − x + 1 > 0 ( x 2 − 1)( x − 1) > 0 ( x +1)( x −1) 2 > 0; (− 1,1) ∪ (1, ∞ )

8

Section 0.2

27. a.

False.

c.

False.

b.

True.

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28. a.

True.

c.

False.

29. a.

b.

True.

33. a.

( x +1)( x2 +2 x – 7) ≥ x2 –1

x3 + 3 x2 – 5 x – 7 ≥ x2 –1 x 3 + 2 x 2 – 5x – 6 ≥ 0 ( x + 3)( x + 1)( x – 2) ≥ 0 [ −3, −1] ∪ [2, ∞ )

⇒ Let a < b , so ab < b2 . Also, a 2 < ab .

Thus, a2 < ab < b 2 and a2 < b2 . ⇐ Let a 2 < b 2 , so a ≠ b Then 0 < ( a − b) = a2 − 2 ab + b2 2

x 4 − 2x 2 ≥ 8

b.

< b − 2ab + b = 2b ( b − a ) 2

2

x4

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

We can divide or multiply an inequality by any positive number. a 1 1 a 1 and 2x + 1 < 3 3x > –6 and 2x < 2 x > –2 and x < 1; (–2, 1)

32. a.

c.

b.

2 x −7 ≤1 or 2 x +1 < 3 2 x ≤ 8 or 2 x < 2

x ≤ 4 or x < 1 ( −∞, 4]

c.

2 x − 7 ≤ 1 or 2 x +1 > 3 2 x ≤ 8 or 2 x > 2

x ≤ 4 or x > 1 ( −∞, ∞)

2) ≥ 0

(x 2 + 1)2 − 7(x 2 + 1) + 10 < 0 [( x 2 + 1) − 5][( x 2 +1) − 2] < 0 ( x2 − 4)( x2 − 1) < 0 ( x + 2)( x+ 1)( x− 1)( x− 2) < 0 (− 2, − 1) ∪ (1, 2)

34. a.

1.99 <

1 < 2. 01 x

1.99 x < 1 < 2.01x 1.99 x < 1 and 1 < 2.01x 1 and x > 1 x< 1.99 2.01 1 1 1 and 2x + 1 < –4 5 x > –2 and x < – ; ∅ 2 2 x − 7 > 1 or 2 x +1 < 3 2 x > 8 or 2 x < 2 x > 4 or x < 1 (−∞,1) ∪ (4, ∞ )

≥0

( x + 2)( x+ 2)( x− 2) ≥ 0 (− ∞, −2] ∪[2, ∞)

b. 3x + 7 > 1 and 2x + 1 > –4 3x > –6 and 2x > –5 5 x > –2 and x > – ; ( −2, ∞ ) 2 c.

2+

2

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.

2 − 2 x −8

1 ⎞ ⎛ 1 , ⎜ ⎟ ⎝ 2.01 1.99⎠

b.

2.99 <

1 < 3.01 x+2

2.99( x + 2) 2.99 5.02 4.98 − < x 2;

2x – 1 < –2 or 2x – 1 > 2 2x < –1 or 2x > 3; 1 3 ⎛ 1⎞ ⎛ 3 ⎞ x < – or x > , ⎜ – ∞, – ⎟ ∪ ⎜ , ∞ ⎟ 2 2 ⎝ 2⎠ ⎝ 2 ⎠ 39.

2x −5 ≥ 7 7 2x 2x − 5 ≤ −7 or −5 ≥ 7 7 7 2x 2x ≤ −2 or ≥ 12 7 7 x ≤ − 7 or x ≥ 42; (−∞, −7] ∪ [42, ∞)

40.

x +1 < 1 4 x +1 < 1 4 x −2 < < 0; 4 –8 < x < 0; (–8, 0) −1 <

41. 5x − 6 > 1; 5x − 6 < − 1 or 5x − 6 > 1 5x < 5 or 5 x > 7 7 ⎛7 ⎞ x < 1 or x > ;( −∞,1) ∪ ⎜ , ∞ ⎟ 5 ⎝5 ⎠

42.

2 x – 7 > 3; 2x – 7 < –3 or 2x – 7 > 3 2x < 4 or 2x > 10 x < 2 or x > 5; (−∞, 2) ∪ (5, ∞ )

1 − 3 > 6; x 1 1 − 3 < −6 or −3 > 6 x x 1 1 + 3 < 0 or − 9 > 0 x x 1+ 3x 1− 9x < 0 or > 0; x x ⎛ 1 ⎞ ⎛ 1⎞ ⎜ − 3, 0 ⎟ ∪ ⎜0, 9 ⎟ ⎝ ⎠ ⎝ ⎠ 5 > 1; x 5 5 2 + < –1 or 2 + > 1 x x 5 5 3+ < 0 or 1+ > 0 x x 3x + 5 x +5 < 0 or > 0; x x ⎛ 5 ⎞ (–∞ , – 5) ∪ ⎜ – , 0 ⎟ ∪ (0, ∞ ) ⎝ 3 ⎠

44. 2 +

45. x2 − 3 x − 4 ≥ 0; 3 ± (–3) 2 – 4(1)(–4)

3 ±5 = = –1, 4 2(1) 2 (x + 1)( x − 4) = 0;(−∞ , − 1]∪ [4, ∞ )

x=

4 ± (−4)2 − 4(1)(4)

2 46. x − 4 x + 4 ≤ 0; x =

2(1)

=2

(x − 2)(x − 2) ≤ 0; x = 2

47. 3x2 + 17x – 6 > 0; x=

–17 ± (17) 2 – 4(3)(–6) 2(3)

=

–17 ± 19 1 = –6, 6 3

⎛1 ⎞ (3x – 1)(x + 6) > 0; (– ∞, – 6) ∪ ⎜ , ∞ ⎟ ⎝3 ⎠

48. 14 x 2 + 11x −15 ≤ 0; x=

− 11±

(11) 2 − 4(14)(− 15) 2(14)

=

− 11± 31 28

3 ...