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■ Section P.1 Real Numbers
Quick Review P.
5. (a) 1187.75 (b) –4. 6. (a) 20.65 (b) 0. 7. (–2) 3 -2(–2)+1=–3; (1.5) 3 -2(1.5)+1=1. 8. (–3) 2 +(–3)(2)+2^2 = 9. 0, 1, 2, 3, 4, 5, 6 10. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Section P.1 Exercises
1. –4.625 (terminating) 2. (repeating) 3. (repeating) 4. (repeating) 5.
all real numbers less than or equal to 2 (to the left of and including 2) 6.
all real numbers between –2 and 5, including –2 and excluding 5 7.
all real numbers less than 7 (to the left of 7) 8.
all real numbers between –3 and 3, including both –3 and 3 9.
all real numbers less than 0 (to the left of 0) 10.
all real numbers between 2 and 6, including both 2 and 6
11. –1 x<1; all numbers between –1 and 1 including – and excluding 1 12. , or ; all numbers less than or equal to 4 13. 45. (a) Associative property of multiplication (b) Commutative property of multiplication (c) Addition inverse property (d) Addition identity property (e) Distributive property of multiplication over addition 46. (a) Multiplication inverse property (b) Multiplication identity property, or distributive property of multiplication over addition, followed by the multiplication identity property. Note that we also use the multiplicative commutative property to say that. (c) Distributive property of multiplication over subtraction (d) Definition of subtraction; associative property of addition; definition of subtraction (e) Associative property of multiplication; multiplicative inverse; multiplicative identity
47.
53. 3.6930338μ10 10 54. 2.21802107μ10^11 55. 1.93175805μ10^11 56. 4.51908251μ10^11 57. 4.839μ10^8 58. –1.6μ10 – 59. 0.000 000 033 3 60. 673,000,000, 61. 5,870,000,000, 62. 0.000 000 000 000 000 000 000 001 674 7 (23 zeros between the decimal point and the 1)
63.
65. (a) When n=0, the equation aman=a m+n^ becomes ama^0 =am+0. That is, ama^0 =am. Since , we can divide both sides of the equation by am. Hence a^0 =1. (b) When n=–m, the equation aman=a m+n^ becomes ama–m=am+(–m). That is am-m=a^0. We know from part (a) that a^0 =1. Since , we can divide both sides of the equation ama–m^ =1 by am. Hence
.
66. (a)
(b) When the remainder is repeated, the quotients generated in the long division process will also repeat. (c) When any remainder is first repeated, the next quotient will be the same number as the quotient resulting after the first occurrence of the remainder, since the decimal representation does not terminate.
67. False. If the real number is negative, the additive inverse is positive. For example, the additive inverse of –5 is 5. 68. False. If the positive real number is less than 1, the reciprocal is greater than 1. For example, the reciprocal
of is 2.
69. [–2, 1) corresponds to. The answer is E. 70. (–2)^4 =(–2)(–2)(–2)(–2)=16. The answer is A. 71. In –7¤=–(7 2 ), the base is 7. The answer is B. 72.. The answer is D. 73. The whole numbers are 0, 1, 2, 3,.. ., so the whole numbers with magnitude less than 7 are 0, 1, 2, 3, 4, 5, 6. 74. The natural numbers are 1, 2, 3, 4,.. ., so the natural numbers with magnitude less than 7 are 1, 2, 3, 4, 5, 6. 75. The integers are.. ., –2, –1, 0, 1, 2,.. ., so the integers with magnitude less than 7 are –6, –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5, 6.
x^6 x^2
x^2 #^ x^4
x^2
= x^4
a-m^ =
am
a Z 0
a Z 0
* 10 -^1 -^7 = 6.364 * 10 -^8
1 3.72 14.3 2 * 10 -^7 +^6
15.91 * 10 -^1
* 101 -^9 = 2.6028 * 10 -^8
1 1.352 12.41 2 * 10 -^7 +^8
a
4 a^3 b a^2 b^3
b a
3 b^2 2 a^2 b^4
b = a
4 a b^2
b a
2 a^2 b^2
b =
12 a 2 a^2 b^4
ab^4
1 x-^3 y^2 2 -^4 1 y^6 x-^4 2 -^2
x^12 y-^8 y-^12 x^8
x^4 y-^4
= x^4 y^4
a
xy
b
xy 2
b
3
x^3 y^3 23
x^3 y^3 8
a
x^2
b
2
1 x^2
x^4
13 x^2 22 y^4 3 y^2
32 1 x^2 22 y^4 3 y^2
9 x^4 y^4 3 y^2
= 3 x^4 y^2
x^2 y^2
1 #^ u = u #^1 = u
2 Chapter P Prerequisites
Step Quotient Remainder 1 0 1 2 0 10 3 5 15 4 8 14 5 8 4 6 2 6 7 3 9 8 5 5 9 2 16 10 9 7 11 4 2 12 1 3 13 1 13 14 7 11 15 6 8 16 4 12 17 7 1
21. A parallelogram
This is a parallelogram with base 8 units and height 4 units. Perimeter ; Area
22. A rectangle
This is a rectangle with length 6 units and height 5 units. Perimeter=22; Area=
23.
35. (a) about $183,000 (b) about $277, 36. (a) 1996: about $144,000; 1997: about $183,
An increase of about 27% (b) 2000: about $277,000; 2001: about $251,
A decrease of about 9% (c) 1995: about $120,000; 2004: about $311,
An increase of about 159%
37. The three side lengths (distances between pairs of points) are
Since two sides of the triangle formed have the same length, the triangle is isosceles.
L 1.
L -0.
L 0.
[1995, 2005] by [0, 150]
[1995, 2005] by [0, 50]
[1995, 2005] by [0, 100]
[1995, 2005] by [0, 150]
[1995, 2005] by [0, 5]
[1995, 2005] by [0, 10]
= a
a b
b
= a -
= b
b
2
a
b = a
b
a
b = a
b = 1 2, 6 2
y
x
7
7
= 8 #^4 = 32
= 2120 + 16 L 24.
233 - 1 - 1 2 4^2 + 3 - 1 - 1 - 3 2 4^2 = 242 + 22 = 220
y
x
5
7
4 Chapter P Prerequisites
38. (a) Midpoint of diagonal from (–7, –1) to (3, –1) is
=(–2, –1)
Midpoint of diagonal from (–2, 4) to (–2, –6) is
=(–2, –1)
Both diagonals have midpoint (–2, –1), so they bisect each other. (b) Midpoint of diagonal from (–2, –3) to (6, 7) is
=(2, 2)
Midpoint of diagonal from (0, 1) to (4, 3) is
=(2, 2)
Both diagonals have midpoint (2, 2), so they bisect each other.
39. (a) Vertical side: length=6-(–2)=8; horizontal side: length=3-(–2)=5; diagonal side: length
(b) 82 +5 2 =64+25=89= , so the Pythagorean Theorem implies the triangle is a right triangle.
40. (a)
(b) Since , the triangle is a right triangle.
41. (x-1)^2 +(y-2)^2 =5^2 , or (x-1)^2 +(y-2)^2 = 42. [x-(–3)]^2 +(y-2)^2 =1 2 , or (x+3)^2 +(y-2)^2 = 43. [x-(–1)]^2 +[y-(–4)]^2 =3 2 , or (x+1)^2 +(y+4)^2 = 44. (x-0)^2 +(y-0)^2 = , or x^2 +y^2 = 45. (x-3)^2 +(y-1)^2 =6 2 , so the center is (3, 1) and the radius is 6. 46. [x-(–4)]^2 +(y-2)^2 =11 2 , so the center is (–4, 2) and the radius is 11. 47. (x-0)^2 +(y-0)^2 = , so the center is (0, 0) and the radius is. 48. (x-2)^2 +[(y-(–6)]^2 =5 2 , so the center is (2, –6) and the radius is 5.
49.
50. , or
51.
52. The distance between y and c is greater than d, so |y-c|>d. 53. and
1+a=8 2+b= a=7 b=
54. Show that two sides have the same length, but not all three sides have the same length:
55. The midpoint of the hypotenuse is . The distances from this point to the vertices are:
56. |x-2|<3 means the distance from x to 2 must be less than 3. So x must be between – 1 and 5. That is, –1 4. (c): (10-2) 1/3^ =81/3^ =2. Meanwhile, substituting x=–6 gives –2 rather than 2; substituting x=8 gives 6 1/3^ ≠1.82 rather than 2. 5. Yes: –3x+5=0. 6. No: There is no variable x in the equation. 7. No: Subtracting x from both sides gives 3=–5, which is false and does not contain the variable x. 8. No: The highest power of x is 2, so the equation is quadratic and not linear. 9. No: The equation has a root in it, so it is not linear. 10. No. The equation has in it, so it is not linear. 11. 3x=24 12. 4x=– x=8 x=– 13. 3t=12 14. 2t= t=4 t=
15.
29. (a) The figure shows that is a solution of the equation.
(b) The figure shows that is a solution of the
equation.
30. (a) The figure shows that is not a solution of the equation. (b) The figure shows that is a solution of the equation. 31. (a): 2(0)-3=0-3=–3<7. Meanwhile, substituting x=5 gives 7 (which is not less than 7); substituting x=6 gives 9. 32. (b) and (c): 3(3)-4=9-4=5 5 , and 3(4)-4=12-4=8 5.
7 x + 5 = 4 x - 7
x = - 4
7 x + 5 = 4 x - 7
x = 2
2 x^2 + x - 6 = 0
x =
2 x^2 + x - 6 = 0
x = - 2
t = -
7 t = - 5
7 t + 11 = 6
4 t - 4 + 3 t + 15 = 6
41 t - 12 + 31 t + 52 = 6
12 a
t - 1 3
t + 5 4
b = 12 a
b
t =
3 t + 15 - 12 t + 24 = 8
31 t + 52 - 121 t - 22 = 8
24 a
t + 5 8
t - 2 2
b = 24 a
b
x =
2 x = 7
6 x = 4 x + 7
6 x - 12 = 4 x - 5
312 x - 42 = 3 a
4 x - 5 3
b
x =
17 = 10 x
2 x + 17 = 12 x
2 x - 3 + 20 = 12 x
4 a
2 x - 3 4
z =
2 z = 11
7 z = 5 z + 11
7 z - 13 = 5 z - 2
15 z - 9 - 8 z - 4 = 5 z - 2
z =
- 19 z = - 8
- 18 z = z - 8
- 18 z - 9 = z - 17
6 - 8 z - 10 z - 15 = z - 17
x =
x +
3 a
x +
b = 3112
x =
x +
2 a
x +
b = 2112
x =
x =
2 x =
3 a
x b = 3 a
b
x =
2 a
x b = 2 a
b
y = - 8
4 y = 5 y + 8
y = -
4 - 3 y = 2 y + 8
x = 2
- 5 x = - 10
- 2 x = 3 x - 10
4 - 2 x = 3 x - 6
x = 1
2 x = 4 x - 2
2 x - 3 = 4 x - 5
x
= x-^1
Section P.3 Linear Equations and Inequalities 7
33. (b) and (c): 4(2)-1=8-1=7 and –1<7 11 , and also 4(3)-1=12-1=11 and –1<11 11. Meanwhile, substituting x=0 gives –1 (which is not greater than –1). 34. (a) , (b) , and (c): 1-2(–1)=1+2=3 and – 3 3 ; 1-2(0)=1-0=1 and –3 1 3 ; and –3 –3 3. 35.
2x-1 4x+ 2x 4x+ –2x 4 x – 38.
3x-1 6x+ 3x 6x+ –3x 9 x – 39.
x < 40.
3x <
x <
41.
10-6x+6x-3 2x+ 7 2x+ 6 2x 3 x x 3 42.
4-4x+5+5x>3x- 9+x>3x- 10+x>3x 10>2x 5>x
3x-2>– 3x>– x>–
45.
12 2y-5 – 17 2y –
y
y
4> 3y-1 >– 5> 3y >–
> y >–
–1< y <
47. 0 2z+5< –5 2z <
z <
48. –6<5t-1< –5< 5t <
–1< t <
3(x-5)+4(3-2x)<– 3x-15+12-8x<– –5x-3<– –5x<–
3(3-x)+2(5x-2)<– 9-3x+10x-4<– 7x+5<– 7x<–
5(2y-3)+2(3y-1)<10y- 10y-15+6y-2<10y- 16y-17<10y- 16y<10y+ 6y<
y 6
10 a
2 y - 3 2
3 y - 1 5
b 6 101 y - 12
x 6 -
6 a
3 - x 2
5 x - 2 3
b 6 61 - 12
x 7
12 a
x - 5 4
3 - 2 x 3
b 6 121 - 22
4112 7 4 a
3 y - 1 4
b 7 41 - 12
3142 3 a
2 y - 5 3
b 31 - 22
5 a
3 x - 2 5
b 7 51 - 12
x -
5 x - 19
5 x + 7 - 12
4 a
5 x + 7 4
b 41 - 32
x 6 5
1 0 1 2 3 4 5 6 7 8 9
2 1 0 1 2 3 4 5 6 7 8
5 4 3 2 1 0 1 2 3 4 5
2 x + 6 6 9
5 4 3 2 1 0 1 2 3 4 5
5 4 3 2 1 0 1 2 3 4 5
5 4 3 2 1 0 1 2 3 4 5
1 0 1 2 3 4 5 6 7 8 9
1 0 1 2 3 4 5 6 7 8 9
8 Chapter P Prerequisites
■ Section P.4 Lines in the Plane
Exploration 1
1. The graphs of y=mx+b and y=mx+c have the same slope but different y-intercepts. 2.
The angle between the two lines appears to be 90°. 3.
In each case, the two lines appear to be at right angles to one another.
Quick Review P.
1. –75x+25= –75x=
x=
2. 400-50x= –50x=– x= 3. 3(1-2x)+4(2x-5)= 3-6x+8x-20= 2x-17= 2x= x= 4. 2(7x+1)=5(1-3x) 14x+2=5-15x 29x+2= 29x=
x=
5. 2x-5y= –5y=–2x+
y=
4x+3y= 3y=–4x+
y=
7. 2x+y=17+2(x-2y) 2x+y=17+2x-4y y=17-4y 5y=
y=
8. x^2 +y=3x-2y y=3x-2y-x^2 3y=3x-x^2
y=
Section P.4 Exercises
1. m=–2 2.
7. , so x= 8. , so y=– 9. , so y= 10. , so x= 11. y-4=2(x-1)
12.
13. y+4=–2(x-5) 14. y-4=3(x+3) 15. Since m=1, we can choose A=1 and B=–1. Since x=–7, y=–2 solves x-y+C=0, C must equal 5: x-y+5=0. Note that the coefficients can be multiplied by any non-zero number, e.g., another answer would be 2x-2y+10=0. This comment also applies to the following problems.
y - 3 = -
1 x + 42
x + 8
x + 8
y + 5 4 + 3
y + 5 7
y - 3 4 + 2
y - 3 6
5 - x
5 - x
m =
m =
m =
m =
m =
x -
x^2
x + 8
12 a
x +
y b
x +
y
x -
[–4.7, 4.7] by [–3.1, 3.1] m=
[–4.7, 4.7] by [–3.1, 3.1] m=
[–4.7, 4.7] by [–3.1, 3.1] m=
[–4.7, 4.7] by [–3.1, 3.1] m=
[–4.7, 4.7] by [–3.1, 3.1] m=
10 Chapter P Prerequisites
16. Since m=1, we can choose A=1 and B=–1. Since x=–3, y=–8 solves x-y+C=0, C must equal –5: x-y-5=0. See comment in #15. 17. Since m=0, we can choose A=0 and B=1. Since x=1, y=–3 solves 0x+y+C=0. C must equal 3: 0x+y+3=0, or y+3=0. See comment in #15. 18. Since m=–1, we can choose A=1 and B=1. Since x=–1, y=–5 solves x+y+C=0, C must equal 6 : x+y+6=0. See comment in #15. 19. The slope is m=1=–A/B, so we can choose A= and B=–1. Since x=–1, y=2 solves x-y+C=0, C must equal 3: x-y+3=0. See comment in #15. 20. Since m is undefined we must have B=0, and we can choose A=1. Since x=4, y=5 solves x+0y+C=0, C must equal –4: x-4=0. See comment in #15. 21. Begin with point-slope form: y-5=–3(x-0), so y=–3x+5. 22. Begin with point-slope form: , so
23. , so in point-slope form, ,
and therefore.
24. , so in point-slope form, , and
therefore.
25. Solve for y :. 26. Solve for y :. 27. Graph y=49-8x; window should include (6.125, 0) and (0, 49), for example, [–5, 10][–10, 60]. 28. Graph y=35-2x; window should include (17.5, 0) and (0, 35), for example , [–5, 20][–10, 40]. 29. Graph y=(429-123x)/7; window should include (3.488, 0) and (0, 61.29), for example, [–1, 5][–10, 80]. 30. Graph y=(3540-2100x)/12=295-175x; window should include (1.686, 0) and (0, 295), for example, [–1, 3](–50, 350]. 31. (a): The slope is 1.5, compared to 1 in (b). 32. (b): The slopes are and 4, respectively. 33. Substitute and solve: replacing y with 14 gives , and replacing x with 18 gives. 34. Substitute and solve: replacing y with 14 gives , and replacing x with 18 gives. 35. Substitute and solve: replacing y with 14 gives , and replacing x with 18 gives. 36. Substitute and solve: replacing y with 14 gives , and replacing x with 18 gives. 37. 38. 39. 40.
In #41–44, use the fact that parallel lines have the same slope; while the slopes of perpendicular lines multiply to give –1.
41. (a) Parallel: , or.
(b) Perpendicular: , or
42. (a) Parallel: , or.
(b) Perpendicular: , or
43. (a) Parallel: , or.
(b) Perpendicular: , or.
44. (a) Parallel: , or.
(b) Perpendicular: , or y = -.
5 x + 3 y = 33 x + 11
y =
x -
3 x - 5 y = 13
y =
x -
3 x - 2 y = 7
y = -
2 x + 3 y = 9 x + 3
y =
y - 3 = x + 4.
1 x + 22
y - 3 = - 21 x + 22 y = - 2 x - 1
y = -
x +
y - 2 = -
1 x - 12
y - 2 = 31 x - 12 y = 3 x - 1
Ymin = -12.5, Ymax = 12.5, Yscl = 1.
Ymin = - 20 >3, Ymax = 20 >3, Yscl = 2 > 3
Ymin = -50, Ymax = 50, Yscl = 5
Ymin = -30, Ymax = 30, Yscl = 3
y = 20
x = 14
y = - 7
x = - 10
y = - 18
x = 2
y = 21
x = 4
[–1, 3] by [–50, 350]
[–1, 5] by [–10, 80]
[–5, 20] by [–10, 40]
[–5, 10] by [–10, 60]
y =
x - 8
y = -
x +
y =
x +
y - 2 =
m = 1 x - 42
y = -
x + 4
y - 5 = -
m = - 1 x + 42
y =
x +
y - 2 =
1 x - 12
Section P.4 Lines in the Plane 11
(b) Slope of the line between the points (6, 67.6) and
(13, 52.1) is
Using the point-slope form equation for the line, we have y-67.6=–2.2143(x-6), so y=–2.2143x+80.8857.
(c) The year 2006 is represented by x=16. Using y=2.2143x+80.8857 and x=16, the model predicts U.S. exports to Japan in 2006 will be approximately $45.5 billion.
55.
24=4(a-3) 6=a- 9=a
56.
a=
57. ß ;
ß
58. ß ;
ß
59. (a) No, it is not possible for two lines with positive slopes to be perpendicular, because if both slopes are positive, they cannot multiply to – 1. (b) No, it is not possible for two lines with negative slopes to be perpendicular, because if both slopes are negative, they cannot multiply to – 1. 60. (a) If b=0, both lines are vertical; otherwise, both have slope m=–a/b, and are, therefore, parallel. If c=d, the lines are coincident. (b) If either a or b equals 0, then one line is horizontal and the other is vertical. Otherwise, their slopes are –a/b and b/a, respectively. In either case, they are perpendicular. 61. False. The slope of a vertical line is undefined. For example, the vertical line through (3, 1) and (3, 6) would
have a slope of , which is undefined.
62. True. If b=0, then and the graph of is a
vertical line. If , then the graph of
is a line with slope and y -intercept
. If and a=0, y= , which is a horizontal line.
An equation of the form ax+by=c is called linear for this reason.
63. With (x 1 , y 1 )=(– 2 , 3) and m=4, the point-slope form equation y-y 1 =m(x-x 1 ) becomes y- =4[x-(– 2 )] or y-3=4(x+ 2 ). The answer is A. 64. With m=3 and b=–2, the slope-intercept form equation y=mx+b becomes y=3x+(– 2 ) or y=3x-2. The answer is B. 65. When a line has a slope of m 1 =–2, a perpendicular line
must have a slope of. The answer is E.
66. The line through (x 1 , y 1 )=(– 2 , 1) and (x 2 , y 2 )=( 1 , – 4 )
has a slope of.
The answer is C.
67. (a)
(b)
(c)
(d) From the graphs, it appears that a is the x -intercept and b is the y -intercept when c=1. Proof: The x -intercept is found by setting y=0.
When c=1, we have. Hence so
x=a. The y -intercept is found by setting x=0.
When c=1, we have. Hence , so
y=b.
y b
a
y b
x a
x a
b
[–5, 5] by [–5, 5]
[–5, 5] by [–5, 5]
[–5, 5] by [–5, 5]
m =
y 2 - y 1 x 2 - x 1
m 2 = -
m 1
c b
b Z 0
c b
a b
y = -
a b
x +
c b
b Z 0
x =
c a
a Z 0
CD 1
a
AB 1 a = 3
BC AD 1 b = 4
DC 1
a - 4
AB 1 a = 6
AD BC 1 b = 5
a 2
a - 0 5 - 3
a - 3
a - 3
[0, 15] by [0, 100]
m =
Section P.4 Lines in the Plane 13
(e)
From the graphs, it appears that a is half the x -intercept and b is half the y -intercept when c=2. Proof: When c=2, we can divide both sides by 2 and we have. By part (d) the x -intercept is 2 a and the y -intercept is 2 b. (f) By a similar argument, when c=–1, a is the opposite of the x -intercept and b is the opposite of the y -intercept.
68. (a)
These graphs all pass through the origin. They have different slopes. (b) If m>0, then the graphs of y=mx and y=–mx have the same steepness, but one increases from left to right, and the other decreases from left to right. (c)
These graphs have the same slope, but different y-intercepts.
69. As in the diagram, we can choose one point to be the origin, and another to be on the x -axis. The midpoints of the sides, starting from the origin and working around counterclockwise in the diagram, are then
, and
D .The opposite sides are therefore parallel, since
the slopes of the four lines connecting those points are:
70. The line from the origin to (3, 4) has slope , so the
tangent line has slope , and in point-slope form ,
the equation is.
71. A has coordinates , while B is , so the
line containing A and B is the horizontal line y=c/2,
and the distance from A to B is.
■ Section P.5 Solving Equations Graphically, Numerically, and Algebraically
Exploration 1
2. Using the numerical zoom, we find the zeros to be 0. and 2.21. 3.
By this method we have zeros at 0.79 and 2.21. 4.
Zooming in and tracing reveals the same zeros, correct to two decimal places.
5. The answers in parts 2, 3, and 4 are the same. 6. On a calculator, evaluating 4x 2 -12x+7 when x=0. gives y=0.0164 and when x=2.21 gives y=0.0164, so the numbers 0.79 and 2.21 are approximate zeros. 7.
Zooming in and tracing reveals zeros of 0.792893 and 2.207107 accurate to six decimal places. If rounded to two decimal places, these would be the same as the answers found in part 3.
[2.17, 2.24] by [–0.12, 0.11] [0.75, 0.83] by [–0.11, 0.12]
[2.05, 2.36] by [–0.5, 0.43] [0.63, 0.94] by [–0.39, 0.55]
[–1, 4] by [–5, 10] [–1, 4] by [–5, 10]
[–1, 4] by [–5, 10]
2 a^ +^ b 2
b 2
(^2) = a 2
a
a + b 2
c 2
a b
b 2
c 2
b
y - 4 = -
1 x - 32
mCD =
c b
; mDA =
e d - a
mAB =.
c b
; mBC =
e d - a
a
d 2
e 2
b
C a
b + d 2
c + e 2
A a b
a 2
, 0 b , B a
a + b 2
c 2
b ,
[–8, 8] by [–5, 5]
[–8, 8] by [–5, 5]
x 2 a
y 2 b
[–10, 10] by [–10, 10]
[–10, 10] by [–10, 10] [–10, 10] by [–10, 10]
14 Chapter P Prerequisites
x=–5 or x=
Rewrite as 3x 2 +11x-20=0; the left side factors to (3x-4)(x+5)=0: 3x-4=0 or x+5=
x=–
7. Rewrite as (2x)^2 =5^2 ; then 2x=—5, or. 8. Divide both sides by 2 to get (x-5)^2 =8.5. Then and. 9. Divide both sides by 3 to get. Then
and.
10. Divide both sides by 4 to get (u+1)^2 =4.5. Then and. 11. Adding 2y^2 +8 to both sides gives 4y^2 =14. Divide both
sides by 4 to get , so.
12. 2x+3=_13 so , which gives
x=–8 or x=5.
13. x^2 +6x+3^2 =7+3 2 (x+3)^2 = x+3= x=–3_ 4 x=–7 or x= 14. x^2 +5x=
(x+2.5)^2 =9+6. x+2.5= or
or
16. x^2 +6x=
(x+3)^2 =4+
or
17. 2x^2 -7x+9=x^2 -2x-3+3x 2x 2 -7x+9=x^2 +x- x^2 -8x=– x^2 -8x+(–4)^2 =–12+(–4)^2 (x-4)^2 = x-4=— x=4_ x=2 or x= 18. 3x 2 -6x-7=x^2 +3x-x 2 -x+ 3x 2 -8x=
or
19. a=1, b=8, and c=–2:
x≠–8.24 or x≠0.
20. a=2, b=–3, and c=1:
or x=
21. x^2 -3x-4=0, so a=1, b=–3, and c=–4:
x=–1 or x=
22. , so a=1, , and c=–5:
x≠–1.53 or x≠3.
x =
b = - 13
x^2 - 13 x - 5 = 0
x =
x =
x =
x =
x =
x = 146 L 3.
146 L -0.
x =
x -
B
a x -
b
2
x^2 -
x + a -
b
2
b
2
x^2 -
x =
x = - 3 - 113 L -6.61 x = - 3 + 113 L 0.
x = - 3 ; 113
x + 3 = ; 113
x^2 + 6 x + a
b
2 = 4 + a
b
2
x =
x = + 111 L 6.
- 111 L 0.
x =
x -
a x -
b
2 = 11
x^2 - 7 x + a -
b
2 = -
b
2
x^2 - 7 x = -
x = -2.5 + 1 15.25 L 1.
x = -2.5 - 1 15.25 L -6.
x^2 + 5 x + a
b
2 = 9 + a
b
2
x =
y = ; B
y^2 =
u + 1 = ; 1 4.5 u = - 1 ; 1 4.
x = - 4 ; B
x + 4 = ; B
1 x + 422 =
x - 5 = ; 1 8.5 x = 5 ; 1 8.
x = ;
x =
[–10, 10] by [–30, 30]
16 Chapter P Prerequisites
23. x^2 +5x-12=0, so a=1, b=5, c=–
x≠–6.77 or x≠1.
24. x^2 -4x-32=0, so a=1, b=–4, c=–32:
x=–4 or x=
25. x -intercept: 3; y -intercept: – 2 26. x -intercepts: 1, 3; y -intercept: 3 27. x -intercepts: – 2, 0, 2; y -intercept: 0 28. no x -intercepts; no y -intercepts 29.
35. x^2 +2x-1=0; x≠0. 36. x^3 -3x=0; x≠–1. 37. Using TblStart =1.61 and Tbl=0.001 gives a zero at 1.62. Using TblStart =–0.62 and Tbl=0.001 gives a zero at – 0.62. 38. Using TblStart=1.32 and Tbl=0.001 gives a zero at 1.32. 39. Graph and y=2: t=6 or t= 40. Graph and y=4: x=–5 or x= 41. Graph and y=7: x=1 or x=– 42. Graph and y=4: or 43. Graph and y=x^2 : x=–3 or x= 44. Graph and y=2x-3: x= 45. (a) The two functions are (the one that begins on the x -axis) and y 2 =x^2 -1. (b) This is the graph of. (c) The x -coordinates of the intersections in the first picture are the same as the x -coordinates where the second graph crosses the x -axis. 46. Any number between 1.324 and 1.325 must have the digit 4 in its thousandths position. Such a number would round to 1.32. 47. The left side factors to (x+2)(x-1)=0: x+2=0 or x-1= x=–2 x= 48. Graphing y=x^2 -18 in (e.g.) [–10, 10]*[–20, 10] and looking for x -intercepts gives x≠–4.24 or x≠4.24. x^2 -3x=12-3x+ x^2 -18= 49. 2x-1=5 or 2x-1=– 2x=6 2x=– x=3 x=– 50. x^2 +4x+4=4(x+3) x^2 = or
- 18 is an extraneous solution, x = 18 L 2.
x = - 18 x = 18
x + 2 = 21 x + 3
y = 31 x + 4 - x^2 + 1
y 1 = 31 x + 4
y = ƒ x + 1 ƒ
y = ƒ 2 x - 3 ƒ
x =
x = -
y = ƒ 3 - 5 x ƒ
y = ƒ 2 x + 5 ƒ
y = ƒ x + 1 ƒ
y = ƒ x - 8 ƒ
[–5, 5] by [–5, 5]
[–5, 5] by [–5, 5]
[–5, 5] by [–5, 5]
[–5, 5] by [–5, 5] [–5, 5] by [–5, 5]
[–5, 5] by [–5, 5]
[–5, 5] by [–5, 5] [–5, 5] by [–5, 5]
[–5, 5] by [–5, 5] [–5, 5] by [–5, 5]
x =
x =
Section P.5 Solving Graphically, Numerically, and Algebraically 17
(d) Let c=–1. The graph suggests y=–1 does not intersect. Since absolute value is never negative, =–1 has no solutions. (e) There is no other possible number of solutions of this equation. For any c , the solution involves solving two quadratic equations, each of which can have 0, 1, or 2 solutions.
70. (a) Let D=b 2 -4ac. The two solutions are ;
adding them gives
(b) Let D=b^2 -4ac. The two solutions are ;
multiplying them gives
71. From #70(a),. Since a=2, this means
b=–10. From #70(b), ; since a=2, this
means c=6. The solutions are ; this
reduces to , or approximately 0.697 and 4.303.
■ Section P.6 Complex Numbers
Quick Review P.
1. x+ 2. x+2y 3. a+2d 4. 5z- 5. x^2 -x- 6. 2 x^2 +5x- 7. x^2 - 8. x^2 - 9. x^2 -2x- 10. x^2 -4x+
Section P.6 Exercises
In #1–8, add or subtract the real and imaginary parts separately.
1. (2-3i)+(6+5i)=(2+6)+(–3+5)i=8+2i 2. (2-3i)+(3-4i)=(2+3)+(–3-4)i=5-7i 3. (7-3i)+(6-i)=(7+6)+(–3-1)i=13-4i 4. (2+i)-(9i-3)=(2+3)+(1-9)i=5-8i 5. (2-i)+ =(2+3)+ =5- i 6. + = + (–3+3)i = 7. (i^2 +3)-(7+i^3 )=(–1+3)-(7-i) =(2-7)+i=–5+i 8. - = -(6-9i)= +9i= (^) +9i In #9–16, multiply out and simplify, recalling that i^2 =–1. 9. (2+3i)(2-i)=4-2i+6i-3i^2 =4+4i+3=7+4i 10. (2-i)(1+3i)=2+6i-i-3i^2 =2+5i+3=5+5i 11. (1-4i)(3-2i)=3-2i-12i+8i^2 =3-14i-8=–5-14i 12. (5i-3)(2i+1)=10i^2 +5i-6i- =–10-i-3=–13-i 13. (7i-3)(2+6i)=14i+42i^2 -6-18i =–42-6-4i=–48-4i 14. (6-5i)=(3i)(6-5i)=18i-15i^2 = 15+18i 15. (–3-4i)(1+2i)=–3-6i-4i-8i^2 =–3-10i+8=5-10i 16. (6+5i)= i(6+5i) = i+ i^2 =– + i **17.
20.** In #21–24, equate the real and imaginary parts. 21. x=2, y= 22. x=3, y=– 23. x=1, y= 24. x=7, y=–7/ In #25–28, multiply out and simplify, recalling that i^2 =–1. 25. (3+2i)^2 =9+12i+4i^2 =5+12i 26. (1-i) 3 =(1-2i+i^2 )(1-i)=(–2i)(1-i) =–2i+2i 2 =–2-2i 27. = (1+i)^4
= (1+2i+i^2 )^2 = = (–4)=–
= = =i
In #29–32, recall that (a+bi)(a-bi)=a^2 +b 2.
29. 22 +3^2 = 30. 52 +6^2 = 31. 32 +4^2 = 32. 12 + 1 1222 =
14 i 2
1 13 + i + 3 i + 13 i^2
1 13 + i 2
11 + 13 i 2
1 13 + i 2
13 + 213 i + i^2
a 1 13 + i 23
b
3 a
i b
3
12 i 22
a
b
4 a
i b
4
1 - 5 = 15 i
1 - 3 = 13 i
1 - 25 = 5 i
1 - 16 = 4 i
1 1 - 2 + 2 i 2 1 12 + 22
1 1 - 4 + i 2
1 17 + i^2 216 - 1 - 812 1 17 - 12
1 15 - 3 i 2 1 - 2 + 1 - 92 1 15 - 22
13 - 1 - 32 1 - 1 - 132 i
x 1 #^ x 2 =
c a
x 1 + x 2 = -
b a
b^2 - 1 b^2 - 4 ac 2 4 a^2
c a
1 - b 22 - 1 1 D 22 4 a^2
# - b^ -^1 D
2 a
b a
- 2 b + 1 D - 1 D 2 a
- b + 1 D 2 a
- b - 1 D 2 a - b ; 1 D 2 a
ƒ x^2 - 4 ƒ
y = ƒ x^2 - 4 ƒ
Section P.6 Complex Numbers 19
In #33–40, multiply both the numerator and denominator by the complex conjugate of the denominator, recalling that (a+bi)(a-bi)=a^2 +b 2.
33.
In #41–44, use the quadratic formula.
41. x=–1 2i
42.
45. False. When a=0, z=a+bi becomes z=bi, and then =–(–bi)=bi=z. 46. True. Because i^2 =–1, i^3 =i(i^2 )=–i, and i^4 =(i^2 )^2 =1, we obtain i+i^2 +i^3 +i^4 =i+(–1)+(–i)+1=0. 47. (2+3i)(2-3i) is a product of conjugates and equals 22 +3 2 =13. The answer is E. 48.. The answer is D. 49. Complex, nonreal solutions of polynomials with real coefficients always come in conjugate pairs. So another solution is 2+3i, and the answer is A. 50. The answer is C. 51. (a) i=i i^5 = =i i^2 =–1 i^6 = =– i^3 =(–1)i=–i i^7 = =–i i^4 =(–1) 2 =1 i^8 = = =
(b) i–1^ = = =–i i–5^ = = =–i
i–2^ = =–1 i–6^ = =–
i–3^ = = =i i–7^ = = =i
i–4= =(–1)(–1)=1 i–8^ = = =
(c) i^0 = (d) Answers will vary.
52. Answers will vary. One possibility: the graph has the shape of a parabola, but does not cross the x-axis when plotted in the real plane, beacuse it does not have any real zeros. As a result, the function will always be positive or always be negative. 53. Let a and b be any two real numbers. Then (a+bi) -(a-bi)=(a-a)+(b+b)i=0+2bi=2bi. 54. imaginary part is zero. 55. and are equal. 56. = and are equal. 57. but Because the coefficient of x in x^2 -ix+2=0 is not a real number, the complex conjugate, i , of –i, need not be a solution.
■ Section P.7 Solving Inequalities
Algebraically and Graphically
Quick Review P.
1. –7<2x-3< –4< 2x < –2< x < 2. 5x-2 7x+ –2x 6 x – 3. |x+2|= x+2=3 or x+2=– x=1 or x=– 4. 4x^2 -9=(2x-3)(2x+3) 5. x^3 -4x=x(x 2 -4)=x(x-2)(x+2) 6. 9x^2 -16y 2 =(3x-4y)(3x+4y)
1 - i 22 - i 1 - i 2 + 2 = 0 1 i 22 - i 1 i 2 + 2 Z 0.
1 a - bi 2 + 1 c - di 2 = 1 a + c 2 - 1 b + d 2 i
1 a + c 2 - 1 b + d 2 i 1 a + bi 2 + 1 c + di 2 =
1 a + bi 2 + 1 c + di 2 1 a + c 2 + 1 b + d 2 i =
1 a - bi 2 #^1 c - di 2 = 1 ac - bd 2 - 1 ad + bc 2 i
1 ac - bd 2 - 1 ad + bc 2 i 1 a + bi 2 #^1 c + di 2 =
1 a + bi 2 #^1 c + di 2 = 1 ac - bd 2 + 1 ad + bc 2 i =
1 a + bi 2 #^1 a + bi 2 = 1 a + bi 2 #^1 a - bi 2 = a^2 + b^2 ,
1 #^1
i^4
i^4
i^2
i^2
i
i^3
i^4
i
i
i^2
i^2
i^4
i^2
i
i
i^4
i
# i
i
i
i^4 #^ i^41 #^1
i^3 #^ i^4
i^2 #^ i^4
i #^ i^4
11 - i 23 = 1 - 2 i2 1 1 - i 2 = - 2 i + 2 i^2 = - 2 - 2 i.
i
i
# - i
= -i
x = 2 ; 115 i
x =
i
x = -
i
i
1 + 12 + 12 - 2 + 1 - 212 - 12 i 3
1 + 12 - 1 12 + 22 i + 11 - 122 i - 1 12 - 22 i^2 3
31 + 12 + 11 - 122 i4 1 1 - 12 i 2 3
11 + i - 12 i - 12 i^2 2 1 1 - 12 i 2 3
11 - 12 i2 1 1 + i 2 1 + 12 i
# 1 -^12 i
1 - 12 i
11 - 3 i2 1 1 + 2 i 2 5
1 + 2 i - 3 i - 6 i^2 5
i
11 - i2 1 2 - i 2 1 - 2 i
# 1 +^2 i
1 + 2 i
12 - i - 2 i + i^2 2 1 1 + 2 i 2 5
i
14 + 3 i2 1 5 - 2 i 2 29
20 - 8 i + 15 i - 6 i^2 29
12 - i2 1 1 + 2 i 2 5 + 2 i
# 5 -^2 i
5 - 2 i
12 + 4 i - i - 2 i^2 2 1 5 - 2 i 2 29
i
13 + 4 i2 1 - 1 - i 2 2
12 + i 22 1 - i 2 1 + i
# 1 -^ i
1 - i
14 + 4 i + i^2 2 1 - i + i^2 2
2 + i 3 i
# -^3 i
i
2 + i 2 - i
# 2 +^ i
2 + i
4 + 4 i + i^2 5
i
i 2 - i
# 2 +^ i
2 + i
2 i + i^2 5
i
2 + i
# 2 -^ i
2 - i
2 - i 5
i
20 Chapter P Prerequisites
