1. Complete each equation. 1. a 3 = a2 • a 2. b 7 = b6 • b
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(For help, go to Lessons 8-3 and 10-3.)
Complete each equation.
1. a 3 = a2 • a 2. b 7 = b6 • b
3. c 6 = c3 • c 4. d 8 = d4 • d
Find the value of each expression.
11-1

2. 1. a3 = a(2 + 1) = a2 • a1 2. b7 = b(6 + 1) = b6 • b1
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Solutions
1. a3 = a(2 + 1) = a2 • a1 2. b7 = b(6 + 1) = b6 • b1
3. c6 = c(3 + 3) = c3 • c3 4. d8 = d(4 + 4) = d4 • d4
= = 13
= = 7
11-1

3. 243 = 81 • 3 81 is a perfect square and a factor of 243.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify
243 = • is a perfect square and a factor of 243.
= • Use the Multiplication Property of Square Roots.
= Simplify
11-1

4. 28x7 = 4x6 • 7x 4x6 is a perfect square and a factor of 28x7.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify x7.
28x7 = 4x6 • 7x 4x6 is a perfect square and a factor of 28x7.
= 4x6 • 7x Use the Multiplication Property of Square Roots.
= 2x3 7x Simplify 4x6.
11-1

5. Simplify each radical expression.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression.
a •
12 • = • 32 Use the Multiplication Property of
Square Roots.
= Simplify under the radical.
= • 6 64 is a perfect square and a factor of 384.
= • 6 Use the Multiplication Property of
Square Roots.
= Simplify
11-1

6. 7 5x • 3 8x = 21 40x2 Multiply the whole numbers and
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(continued)
b x • x
7 5x • x = x2 Multiply the whole numbers and
use the Multiplication Property of
Square Roots.
= x2 • 10 4x2 is a perfect square and a
factor of 40x2.
= x2 • Use the Multiplication Property of
Square Roots.
= 21 • 2x Simplify 4x2.
= 42x Simplify.
11-1

7. The distance you can see is 9 miles.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Suppose you are looking out a fourth floor window 54 ft above the ground. Use the formula d = h to estimate the distance you can see to the horizon.
d = h
= • 54 Substitute 54 for h.
= Multiply.
= 9 Simplify
The distance you can see is 9 miles.
11-1

8. Simplify each radical expression.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression. a. 13 64 b. 49 x4
= Use the Division Property of Square Roots. 13 64
= Simplify 13 8
= Use the Division Property of Square Roots. 49 x4 7 x2
= Simplify and x4.
11-1

9. Simplify each radical expression.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression. a.
120 10
= Divide.
120 10
= 4 • 3 4 is a perfect square and a factor of 12.
= 4 • 3 Use the Multiplication Property of Square Roots.
= Simplify
11-1

10. = Divide the numerator and denominator by 3x.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(continued) b.
75x5
48x
= Divide the numerator and denominator by 3x.
75x5
48x
25x4 16
= Use the Division Property of Square Roots.
25x4 16
= Use the Multiplication Property of
Square Roots.
25 • x4 16
= Simplify , x4, and
5x2 4
11-1

11. Simplify each radical expression.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression. a. 3 7 3 7
= • Multiply by to make the denominator a
perfect square.
= Use the Multiplication Property of Square Roots.
3 7 49
= Simplify
3 7 7
11-1

12. Simplify the radical expression.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(continued)
Simplify the radical expression. b. 11
12x3
= • Multiply by to make the denominator a
perfect square. 3x 11
12x3
= Use the Multiplication Property of Square Roots.
33x
36x4
= Simplify x4.
33x
6x2
11-1

13. Simplifying Radicals pages 581–583 Exercises 1. 10 2 2. 7 2 3. 5 3
ALGEBRA 1 LESSON 11-1
pages 581–583  Exercises
4. –4 5
5. –
7. 2n 7
8. 6b2 3
9. 6x 3
10. 2n n
11. 2a2 5a mi mi
28.
29.
30.
31.
32.
33.
34.
12. –4b2 3
13. 20
14. 18
18. –
19. 6n 2
20. 14t 2
21. 3x
22. 80t 3
23. 3a3 2
24. –6a2 2
25. 3 mi 21 7
3 3 2 5 2 11 5 3a 7 4c
5 3a 7
11-1

14. 52. not simplest form; radical in the denominator of a fraction
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
46.
47.
48.
50.
51.
52. not simplest form; radical in
the denominator of a fraction
53. not simplest form; radical in
35.
37.
39. –2 5
40. 2x 7
41.
42.
43.
44.
2n 2n 9
54. Simplest form; radicand has
no perfect-square factors
other than 1.
55. Simplest form; radicand has
56. a • 10 = =
36 • 5 =
b. Answers may vary.
Sample: a = 36,
b = 5; a = 9, b = 20
57. 30
59.
21x 7x 3 2 n 5n
9 2 4 s 3 2b b
a2 3 2
55y 2y 3 y
3 2 2
3 2 4
11-1

15. perfect-square factors other than 1. 73. Answers may vary.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
77. 30a4
seconds
79. C
80. F
81. B
82. I
83. [2] A = 96 ft2
s = =
16 • = ft
[1] correct answer,
without work shown
84. quadratic; y = 0.2x2
85. exponential; y = 4(2.5)x
86. linear; y = –4.2x + 7
69. –3 ±
70. 1 ± 5
71.
72. a = • 2 =
25 • =
b. The radicand has no
perfect-square factors
other than 1.
73. Answers may vary.
Sample: 12, 27, 48.
74. a in.
b in.
75. 12x
76. 10b2
60.
62.
63. – 3
65. 2ab 5b
66. ab2c abc
67.
68.
–2 a a2
2 ± 10 3
x y
y 2 3m 4m
8 6a 3a
11-1

16. Simplifying Radicals 89. 87. 90. 3n2 + 5n + 5 88. 91. 3v2 – v – 9
ALGEBRA 1 LESSON 11-1
89.
90. 3n2 + 5n + 5
91. 3v2 – v – 9
92. 5t 3 + 8t 2 – 14t – 11
93. –3b2 – 23b – 21
87.
88.
11-1

17. Simplify each radical expression.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression. • 12 36 3
8 2 48 2 a5
2 a a3 3x
15x3 5 5x
11-1

18. The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
(For help, go to Lesson 10-4.)
Simplify each expression.
– (3t)2 + (4t)2
Solve each equation.
4. c2 = b2 = a = 65
b2 = = c = a2 + 52
11-2

19. The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
Solutions
= (5 • 5) + (6 • 6) = = 61
2. 92 – 42 = (9 • 9) – (4 • 4) = 81 – 16 = 65
3. (3t)2 + (4t)2 = (32 • t2) + (42 • t2) = 9t2 + 16t2 = 25t 2
4. c2 = 36
c = ± = ±6
b2 = a = 65
b2 = 49 – a2 = 65 – 16
b2 = a2 = 49
b = ± = ±5 a = ± = ±7
b2 = = c2
b2 = 32 – 12 c = ± = ± • 5 = ± • 5 = ±4 5
b2 = 20
b = ± = ± 4 • 5 = ± • 5 = ±
= a2 + 52
100 – 52 = a2
48 = a2
a = ± = ± • 3 = ± • = ±
11-2

20. The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
What is the length of the hypotenuse
of this triangle?
a2 + b2 = c2 Use the Pythagorean Theorem.
= c2 Substitute 8 for a and 15 for b.
= c2 Simplify.
289 = c2 Find the principal square root of each side.
17 = c Simplify.
The length of the hypotenuse is 17 m.
11-2

21. The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
A toy fire truck is near a toy building on a table such that the base of the ladder is 13 cm from the building. The ladder is extended 28 cm to the building. How high above the table is the top of the ladder?
Define: Let b = height (in cm) of the ladder from
a point 9 cm above the table.
Relate: The triangle formed is a right triangle.
Use the Pythagorean Theorem.
11-2

22. The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
(continued)
Write: a2 + b2 = c2
132 + b2 = 282 Substitute.
169 + b2 = 784 Simplify.
b2 = 615 Subtract 169 from each side.
b2 = Find the principal square root of each side.
b Use a calculator and round to the
nearest tenth.
The height to the top of the ladder is 9 cm higher than 24.8 cm,
so it is about 33.8 cm from the table.
11-2

23. The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
Determine whether the given lengths are sides of a right triangle.
a. 5 in., 5 in., and 7 in.
Determine whether a2 + b2 = c2,
where c is the longest side.
Simplify.
50 = 49 /
This triangle is not a right triangle.
b. 10 cm, 24 cm, and 26 cm
Determine whether a2 + b2 = c2,
where c is the longest side.
Simplify.
676 = 676
This triangle is a right triangle.
11-2

24. The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
If two forces pull at right angles to each other, the
resultant force is represented as the diagonal of a rectangle,
as shown in the diagram. The diagonal forms a right triangle
with two of the perpendicular sides of the rectangle.
For a 50–lb force and a 120–lb force, the resultant force is
130 lb. Are the forces pulling at right angles to each other?
Determine whether a2 + b2 = c2
where c is the greatest force.
, ,900
16,900 = 16,900
The forces of 50 lb and 120 lb are pulling at right angles to each other.
11-2

25. The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
pages 587–590  Exercises
1. 10
2. 25
3. 17
4. 26
5. 2.5
6. 1
7. 4
8. 5
9. 12 m
14. about 15.5 ft
15. about 5.8 km
16. yes
17. no
18. no
19. yes
20. no
21. yes
22. yes
23. no
24. no
25. yes
or 0.3
28. 3
29. 6
32. a ft
b ft2
33. yes
34. no 4 15
11-2

26. The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
35. yes
36. yes cm lb
42. a. These lengths could be
2 legs or one leg and
the hypotenuse.
b. about 12.8 in. or 6 in.
43. a = = 100 = 102
b. 5; 12; 7; 41
c. Answers may vary. Sample: 10, 24, 26
44. a. 6.9 ft
b ft2
c. 981 watts ft
46. a. Answers may vary.
Sample: 5, , 5
b. 5 units2
47. a ft
b ft
c ft
d. No; the hypotenuse d must
be longer than each leg.
48. An integer has 2 as a factor;
the integer is even; if an integer
is even, then it has 2 as a factor; true.
11-2

27. The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
54. 10
56. 5
57. n2 + (n + 1)2 = (n + 2)2; 3, 4, 5
58. a. b
59. a. a2 + 2ab + b2
b. c2 c.
d. (a + b)2 = c ab ; a2 + 2ab + b2 =
2ab + c2; a2 + b2 = c2
e. This equation is the same as the
Pythagorean Theorem.
49. A figure is a square;
the figure is a rectangle;
if a figure is a rectangle
then the figure is a square;
false.
50. You are in Brazil; you are
south of the equator; if you
are south of the equator you
are in Brazil; false.
51. An angle is a right angle;
its measure is 90°; if the
measure of an angle is 90°,
then it is a right angle; true.
units2
53. 6 in. ab 2 1 2
11-2

28. The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
60. D
61. H
62. B
63. C
64. A
65. [2] It is a right triangle. Substitute 17,
the longest side, for c and substitute
the other lengths for a and b in the
Pythagorean Theorem
= 289
[1] incorrect equation OR incorrect explanation
67.
69. 2b b
70.
71.
72. 3 and 4
73. 8 and 9
74. –8 and –7
and 12
76. rational
77. irrational
78. irrational
79. rational 2
2x2
2 6v
v 4 –
> –
> 6 3
11-2

29. The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
80. 8x2 – 4x
81. 12a2 + 15a
82. 18t 3 – 6t 2
83. –10p4 + 26p3
84. 15b3 + 5b2 – 45b
85. –7v4 + 42v2 – 7v
11-2

30. The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
1. Find the missing length 2. Find the missing length
to the nearest tenth. to the nearest tenth.
3. A triangle has sides of lengths
12 in., 14 in., and 16 in. Is the
triangle a right triangle?
4. A triangular flag is attached to a post. The bottom of the flag is 48 in.
above the ground. How far from the ground is the top of the flag?
16.6
5.7 no
57 in.
11-2

31. The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
(For help, go to Lessons 11-2 and 2-7.)
Find the length of the hypotenuse with the given leg lengths.
If necessary, round to the nearest tenth.
1. a = 3, b = 4 2. a = 2, b = 5
3. a = 3, b = 8 4. a = 7, b = 5
For each set of values, find the mean.
5. x1 = 6, x2 = y1 = –4, y2 = 8
7. x1 = –5, x2 = –7 8. y1 = –10, y2 = –3
11-3

32. The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
Solutions
1. a2 + b2 = c2 2. a2 + b2 = c2
= c = c2
= c = c2
25 = c = c2
c = = c =
The length of the hypotenuse is 5. The length of the hypotenuse is about 5.4.
3. a2 + b2 = c a2 + b2 = c2
= c = c2
= c = c2
73 = c = c2
c = c =
The length of the hypotenuse is The length of the hypotenuse is
about about 8.6.
5. mean = = = 10
6. mean = = = 2
6 + 14 2 20
–4 + 8 4
7. mean = = = –6
8. mean = = = –6.5
–5 + (–7)
–10 + (–3)
–12
–13
11-3

33. The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
Find the distance between F(6, –9) and G(9, –4).
d = ( x2 – x1)2 + (y2 – y1)2 Use the distance formula.
d = (9 – 6)2 + [–4 – (–9)]2 Substitute (9, –4) for (x2, y2)
and (6, –9) for (x1, y1).
d = Simplify within parentheses.
d = Simplify to find the exact
distance.
d Use a calculator. Round to
the nearest tenth.
The distance between F and G is about 5.8 units.
11-3

34. The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
Find the exact lengths of each side of quadrilateral EFGH. Then find the perimeter to the nearest tenth.
EF = [4 – (–1)]2 + (3 + 5)2
= 52 + (–2)2 =
= 29
FG = (3 – 4)2 + (–2 – 3)2
= (–1)2 + (–5)2 =
= 26
GH = |–2 – 3| = 5
EH = [–2 – (–1)]2 + (–2 – 5)2
= (–1)2 + (–7)2 =
= 50
The perimeter = units.
11-3

35. The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
Find the midpoint of CD.
, = , Substitute (–3, 7) for
(x1, y1) and (5, 2) for
(x2, y2).
7 + 2 2
x1+ x2
y1+ y2
(–3) + 5
= , Simplify each numerator. 2 9
= 1, 4 Write as a mixed
number. 9 2 1
The midpoint of CD is M 1, 1 2
11-3

36. The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
A circle is drawn on a coordinate plane. The endpoints
of the diameter are (–3, 5) and (4, –3). What are the coordinates of the center of the circle?
, = , Substitute (–3, 5) for
(x1, y1) and (4, –3) for
(x2, y2).
5 + (–3) 2
x1+ x2
y1+ y2
(–3) + 4
= , = , 1 1 2
The center of the circle is at , 1 . 1 2
11-3

37. The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
pages 594–597  Exercises
1. 15
2. 14
3. 10
5. 2.8
6. 8.1
7. 16
9. (1, 6)
10. (–1, 12)
11. (0, 0)
22. MN ; NP ; MP = 5
23. JK = 4; KL ; LJ
24. TU ; UV ; VT
25. a. OR = , ST =
b. ;
c. yes
26. a. (20, 80), (–40, 30)
b ft
c. (–10, 55) or 55L10
12. (–2, 3)
13. 5, –5
14. – 2 , –1
15. (–4, 4)
, –9
19. AB ; BC ; AC = 5
20. DE ; EF ; DF
21. RS ; ST ; RT 1 2 1 2 5 2 5 2 1 2
11-3

38. The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
28. Check students’ work.
29. a.
b. (0, – )
c. one half mile south of the substation
30. about 9.5 km apart
31. a mi
b. 20 mi, 21.2 mi
c. 15 min, 16 min
32. Yes; all sides are congruent.
27. Answers may vary. Sample:
Suppose you have (1, 1) and
(4, –3). To find the distance,
square the difference between
the x-coordinates. Square the
difference between the
y-coordinates. Find the sum
and take the square root,
so = 5. To find the
midpoint, add x-coordinates
together and divide by 2.
Repeat for y-coordinates.
So, 1 2
1 + 4 2 ,
1 – 3 2 5 2 =
, –1
11-3

39. The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
33. a. R(–27, –5)
b. PR =
RQ =
34. a. about 4.3 mi
b. about 17.4 mi
35. a. M(–0.5, 3); N(5.5, 3)
b. They are equal.
36. (x, y)
37. They are opposites.
38. a. 5 units
b. Answers may vary. Sample:
(5, 0), (0, 5), (–5, 0), (0, –5),
(3, –4), (–3, 4), (–3, –4)
c. circle 4 3
50. 26
52. 8
54. –14, 14
55. –10, 10
56. no solution
57. –5, 5
58. no solution
39. a. y = – x + 1 b.
c. 3
40. Yes; the distance from
each point to the
center is 5.
42. 5
44. –5
45. 3
46. –8 9 5 7 5
, –
11-3

40. The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
59. – ,
60. k2 + 11k + 24
61. v2 + 2v – 35
62. 2p2 – 17p – 9
63. 8w4 + 19w2 + 11
64. 7t3 + 5t2 + 5t – 2
65. 6c3 – 27c2 + 33c + 24 2 3 2 3
11-3

41. The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
1. Find the distance between M(2, –1) and N(–4, 3) to the nearest tenth.
2. Find the distance between P(–2.5, 3.5) and R(–7.5, 8.5) to the nearest tenth.
3. Find the midpoint of AB, A(3, 6) and B(0, 2).
4. Find the midpoint of CD, C(6, –4) and D(12, –2).
5. Find the perimeter of triangle RST to the
nearest tenth of a unit.
7.2
7.1
(1 , 4) 1 2
(9, – 3)
9.5 units
11-3

42. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
(For help, go to Lesson 11-1.)
Simplify each radical expression. x2
Rationalize each denominator. 3 11 5 8 15 2x
11-4

43. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Solutions
5. = • = =
6. = • = = = = = =
7. = • = = 3 11
3 •
11 • 5 8
5 • 8
8 • 8
4 • 10
4 • 10 40
2 10 10 4 15 2x
15 • 2x
2x • 2x
30x
= • 13 = • =
= • 2 = • 2 =
= • 6 = 4 • • = 4 • 3 • 6 =
x2 = • 5 • x2 = • • x2 = 5x 5
11-4

44. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify
= Both terms contain
= (4 + 1) 3 Use the Distributive Property to
combine like radicals.
= Simplify.
11-4

45. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify –
– = • 5 9 is a perfect square and a factor of 45.
= – 9 • 5 Use the Multiplication Property of
Square Roots.
= – Simplify
= (8 – 3) 5 Use the Distributive Property to
combine like terms.
= Simplify.
11-4

46. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify 5( ).
5( ) = Use the Distributive Property.
= 4 • Use the Multiplication Property
of Square Roots.
= Simplify.
11-4

47. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify ( 6 – )( ).
( 6 – )( )
= – – Use FOIL.
= 6 – – 3(21) Combine like radicals and
simplify and
= 6 – • 14 – 63 9 is a perfect square factor of 126.
= 6 – • – 63 Use the Multiplication Property of
Square Roots.
= 6 – – 63 Simplify
= –57 – Simplify.
11-4

48. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify . 8
7 – 3
= • Multiply the numerator and
denominator by the conjugate
of the denominator. 8
7 – 3
= Multiply in the denominator.
8( )
7 – 3
= Simplify the denominator.
8( ) 4
= 2( ) Divide 8 and 4 by the common
factor 4.
= Simplify the expression.
11-4

49. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
A painting has a length : width ratio approximately equal to the golden ratio ( ) : 2. The length of the painting is 51 in. Find the exact width of the painting in simplest radical form. Then find the approximate width to the nearest inch.
Define: 51 = length of painting
x = width of painting
Relate: ( ) : 2 = length : width
Write: =
x ( ) = 102 Cross multiply.
= Solve for x.
102
( )
x( ) 51 x 2
11-4

50. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
(continued)
x = • Multiply the numerator and the
denominator by the conjugate
of the denominator.
(1 – 5)
102
( )
x = Multiply in the denominator.
102(1 – 5)
1 – 5
x = Simplify the denominator.
102(1 – 5) –4
x = Divide 102 and –4 by the
common factor –2.
– 51(1 – 5) 2
x = Use a calculator.
x 32
The exact width of the painting is inches.
The approximate width of the painting is 32 inches.
– 51(1 – 5) 2
11-4

51. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
pages 603–606  Exercises
3. –2 5
6. –8 3
7. yes
8. yes
9. no
11. –3 3
13. –2 5
16. 4 –
18. 6 –
21. 6 –
22. –9 – – – –
29. –6 2
30. – –
31.
32. – –
; 24.1
3( ) 5
11-4

52. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
47.
units
units
units
51. 4x + x units
52. Answers may vary. Sample:
, ,
53. a. The student simplified
as instead of or b
54. a or 2.8 ft
b. s 2
35. – ; –1.3
36. 6 – ; 0.3 ft
– 6
42.
44. –24
45. – 2 4 3 10 5 8
11-4

53. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4 % % %
58. a. x
b. x x
59.
60. about 251 years
61. They are unlike radicals.
62. a. 1, 0, 1, 1;
4, 1, 5, ;
5, 3, 8, ;
8, 6, 14, 10;
10, 9, 19,
b. No; the only values it
worked for were 0 and 1.
a + b = a b
64.
65.
66.
67. 2
69. 70
– –
71. a b
c (p + q)
72. B /
9 2 2 21 n 2
n – 1 2 15 ab b
11-4

54. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
73. I
74. [2] ( – 2)( ) =
– – =
3(5) – – 5(2) =
– – 10 =
[1] correct technique, but with a computational error
75. [4]
[3] correct steps but answer not completely simplified
[2] correct technique, but with a computational error
[1] correct answer but no work shown
7 – 21 5
• Multiply the numerator and
denominator by the conjugate
of the denominator. Simplify the
denominator.
11-4

55. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
units
units
units
79. (3, 5)
80. (–2, 6.5)
81. 0, 7
82. –2, 9
83. –9, –3
84. –4, 6
85. –15, –2
86. –3, –
87. b2 + 22b + 121
88. 4p2 + 28p + 49
89. 25g2 – 49
90. 9x2 – 1
k2 – 81
92. d 2 – 2.2d 1 9 1 2
11-4

56. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4 16
5 – 7
Simplify each expression.
– – ( )
4. ( 3 – )( ) 5. 40
–2 5 –
– –
11-4

57. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
(For help, go to Lesson 10-3.)
Evaluate each expression for the given value.
x – 3 for x = x + 7 for x = x + 3 for x = 1
Simplify each expression.
4. ( 3)2 5. ( x + 1) ( 2x – 5) 2
11-5

58. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solutions
x – 3 for x = 16: – 3 = 4 – 3 = 1
x + 7 for x = 9: = = 4
x + 3 for x = 1: = = 2 • 2 = 4
4. ( 3)2 = 3
5. ( x + 1)2 = x + 1
6. ( 2x – 5)2 = 2x – 5
11-5

59. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve each equation. Check your answers.
a x – 5 = 4
x = 9 Isolate the radical on the left side
of the equation.
( x)2 = 92 Square each side.
x = 81
Check: x – 5 = 4
– Substitute 81 for x.
9 –
4 = 4
11-5

60. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
(continued)
b x – 5 = 4
( x – 5)2 = 42 Square each side.
x – 5 = 9 Solve for x.
x = 21
Check: x – 5 = 4
21– 5 = 4 Substitute 21 for x.
16 = 4
4 = 4
11-5

61. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
On a roller coaster ride, your speed in a loop depends on the height of the hill you have just come down and the radius of the loop in feet. The equation v = 8 h – 2r gives the velocity v in feet per second
of a car at the top of the loop.
11-5

62. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
(continued)
The loop on a roller coaster ride has a radius of 18 ft.
Your car has a velocity of 120 ft/s at the top of the loop.
How high is the hill of the loop you have just come
down before going into the loop?
Solve v = 8 h – 2r for h when v = 120 and r = 18.
120 = 8 h – 2(18) Substitute 120 for v and 18 for r.
= Divide each side by 8 to isolate the radical.
15 = h – 36 Simplify.
8 h – 2(18) 8
120
(15)2 = ( h – 36)2 Square both sides.
225 = h – 36
261 = h
The hill is 261 ft high.
11-5

63. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve 3x – 4 = 2x + 3.
( 3x – 4)2 = ( 2x + 3)2 Square both sides.
3x – 4 = 2x + 3 Simplify.
3x = 2x + 7 Add 4 to each side.
x = 7 Subtract 2x from each side.
Check: x – 4 = 2x + 3
3(7) – (7) + 3 Substitute 7 for x.
17 = 17
The solution is 7.
11-5

64. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve x = x + 12.
(x)2 = ( x + 12)2 Square both sides.
x2 = x + 12
x2 – x – 12 = 0 Simplify.
(x – 4)(x + 3) = 0 Solve the quadratic equation by factoring.
(x – 4) = 0 or (x + 3) = 0 Use the Zero–Product Property.
x = 4  or   x = –3 Solve for x.
Check: x = x + 12
– –3 + 12
4 = 4 –3 = 3 /
The solution to the original equation is 4.
The value –3 is an extraneous solution.
11-5

65. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve 3x + 8 = 2.
3x = –6
( 3x)2 = (–6) Square both sides.
3x = 36
x = 12
Check: x + 8 = 2
3(12)    Substitute 12 for x.
= 2   x = 12 does not solve the original equation. /
3x + 8 = 2 has no solution.
11-5

66. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
pages 610–612  Exercises
1. 4
2. 49
3. 36
4. 137
5. 15
6. 16 ft
watts
9. 4.5
10. 3
11. 7
12. –2
13. 4
15. 2
16. –4
17. none
18. –
19. –7
20. none
21. 3
22. 5
23. no solution
24. 2
25. no solution
26. 4 or
, 1
29. a. 25 b
30. about 2.5 in.
31. An extraneous solution is
a solution of a new equation
that does not satisfy the
original equation.
32. Answers may vary. Sample:
x – 2 = – 2x , 3x = 3 5 4 1 4 1 2
11-5

67. Solving Radical Equations
ALGEBRA 1 LESSON 11-5 ft
34. 3
35. no solution
36. no solution
37. 1, 6
39. 11
40. 0, 12
41. 3, 6
42. 44
43. no solution
44. a. 68 ft
b mi/h
c. As radius increases,
velocity decreases.
As height decreases,
d. Velocity depends upon
the difference of the
height and the radius.
45. a.
b. approximately (6, 3.6)
c. 6; it is the x-coordinate of
the point of intersection.
46. a. V = 10x2
b. x =
c. 2, 3, 4, 5, 6, 7
47. a. – 7, 7
b. 49
c. In both cases 3 is
added to each side.
To solve the first
equation you find the
square roots of each
side, and in the second
square of each side.
48. –2, 8
49. 0 V 10
11-5

68. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
51. –1
52. Subtract 2x from each side.
Square both sides. Solve for x.
Check the solution if there is one.
53. The square of x – 1 will have
only 2 terms while x – 1 squared
will have 3 terms.
54. a. about 2.0 m
b. about 32.4 m
55. C
56. G
57. B
58. A
59. B
60. C
61. [2] – 5x = 4x – 3
15 – 5x = 4x – 3
–9x = –18
x = 2
Check: – 5(2) (2) – 3
5 = 5
The solution is 2.
[1] correct technique with a minor error
OR correct answer, no work shown
65. 32
11-5

69. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
66. 4( – 3)
, –4.5
, –0.4
70. –0.2, –4.8
71. –10.7, 0.7
72. –11.7, 1.7
73. –1.6, 3.1
74. (x + 12)(x – 2)
75. (m – 13)(m – 1)
76. (b + 18)(b – 2)
77. (2p + 1)(p + 7)
78. 3(d – 1)(d + 5)
79. (4v – 5)(v – 5)
11-5

70. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve each radical equation.
x – 3 = x – 2 = x + 2
x + 7 = 5x – x = 2x + 8
x = 3 2 5 7 2 5 4
no solution
11-5

71. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
(For help, go to Lessons 10-1 and 10-3.)
Graph each pair of quadratic functions on the same graph.
1. y = x2, y = x y = x2, y = x2 – 4
Evaluate each expression for the given value of x.
x for x = 4
x + 7 – 3 for x = 2
x + 2 for x = 9
11-6

72. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
Solutions
1. y = x2, y = x2 + 3  
2. y = x2, y = x2 – 4  
x for x = 4: = 2
x – 7 – 3 for x = 2: – 3 = – 3 = – 3 – 3 = 0
x + 2 for x = 9: = 3(3) + 2 = = 11
11-6

73. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
Find the domain of each function.
a. y = x + 5
x Make the radicand 0.
> –
x –5
> –
The domain is the set of all numbers greater than or equal to –5.
b. y = x – 12  
4x – Make the radicand 0.
> –
4x 12
> –
x 3
The domain is the set of all numbers greater than or equal to 3.
11-6

74. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
The size of a television screen is the length of the screen’s diagonal d in inches.The equation d = 2A estimates the length of a diagonal of a television with screen area A.
Graph the function.
Domain
2A 0
A 0
> – 50
100
200
300
400 10
14.1 20
24.5
28.3
Screen
Area
(sq. in.)
Length of
Diagonal
(in.)
11-6

75. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
Graph y = x + 4 by translating the graph of
y = x .
For the graph y = x + 4,
the graph of y = x is shifted 4 units up.
11-6

76. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
Graph ƒ(x) = x + 3 by translating the graph of
y = x .
the graph of y = x is shifted to the left 3 units.
For the graph ƒ(x) = x + 3,
11-6

77. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
pages 616–619  Exercises
1. x 2
2. x
3. x 0
4. x –7
5. x –3
6. x 5
7. x –
8. x –2
9. x
13. x y
0 0
3 3
5.3 4
14. x ƒ(x)
1 3
4 6
15. x y
1 –3
4 –6
> –
10. x y
0 0
2 2
4.5 3
11. x ƒ(x)
1 2
4 4
12. x y
2 0
3 2
6 4
> – 3 4
> –
> –
> –
> – 5 3
> –
> –
> – 4 3
11-6

78. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
22.
23.
24.
25.
26.
27.
16. h v
0  0
1  8
4 16
17. D
18. A
19. C
20. B
21.
11-6

79. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
28.
29.
30. x 4; y 0
31. x 4; y 0
32. Form an inequality setting
the radicand Solve for x.
Answers may vary. Sample:
y = x – 2
Domain: x –
x 2
33. a-d. Answers may vary. Samples:
a. y = x + 2
b. y = x + 2
c. y = 2 x
d. Check students’ work.
34. Translate the graph of
y = x  8 units to the left.
35. Translate the graph of
y = x  10 units down.
36. Translate the graph of
y = x  12 units up.
37. Translate the graph of
y = x  9 units right.
> –
> –
< –
> –
> –
> –
> –
11-6

80. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
38. x y
2.5 0
3.5 1
6.5 2
39. x ƒ(x)
0 0
1 4
2 5.7
4 8
40. x y
–6 0
–5 1
–2 2
0 2.4
41. x y
0 0
2 1
4 1.4
8 2
42. x y
2 3
3 4
6 5  
43. x ƒ(x)
–2 –4
–1 –3
2 –2
11-6

81. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
44. x y
0 3
1 4.4
2 5
3 5.4 
45. x y
–3 1
–2 2.4
–1 3
0 3.4
46. x y
1 –2
2 –0.3
3   0.4
4   1
47. B
48. D
49. A
50. C
51. a. p > 0 b.
c. about 45 lb/in.2
52. a. no
b. Answers may vary.
Sample: The graph
of y = x is the first
quadrant of the graph
of x = y2.
c. y = – x
53. y = 3 x rises more
steeply because
3 x > 3x for all
positive values of x.
54. False; x must equal 81.
55. False; only combine
like terms.
56. true
57. False; x = –1.
11-6

82. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
58. a. about 213 cameras
b. month 4
59. a.
b. y = |x| + 5
60. Translate the graph of
y = x right 2 units and up 3 units.
61. a. i.  ii. 
iii.  iv. 
b. The greater the absolute value of n,
the steeper the graph. If n < 0, then the
graph lies in Quadrant II. If
n > 0, the graph lies in Quadrant I.
11-6

83. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
62. Check students’ work.
63. B
64. H
65. C
66. H
67. B
68. H
69. [2] x y
6 0
7 1
8 1.4
9 1.7
10 2
[1] incorrect coordinates on graph
70. 16
71. 7
74. no solution
75.
76. ,
77. 4 – ,
78. no solution
79. , 2 3
–2 – 2 – 2
–1 – 3 – 3
11-6

84. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
80. no solution
81. ,
82. (2x + 1)(x – 4)
83. (3x – 5)(x + 2)
84. (2x + 1)(2x + 9)
85. 2(x – 8)(x + 3)
86. 4(x2 – x – 15)
87. x(x – 13)(x + 1)
–13 – 18 – 18
11-6

85. Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
1. Find the domain of the function ƒ(x) = 2x – 4.
2. Graph y = 3 x.
3. Graph y = x – 3.
4. Describe how to translate the graph of y = x
to obtain the graph of the function y = x – 15.
x > 2
Shift the graph to the right 15 units.
11-6

86. Let c = , s = , t = . Calculate c, s, and t for the given values.
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
(For help, go to Lessons 1-6 and 4-2.)
Let c = , s = , t = . Calculate c, s, and t for the given values.
1. A = 3, O = 4, H = 5 2. A = 5, O = 12, H = 13
Solve each equation.
3. = =
= = A H O 15 x
0.75 1
0.84 21 20
0.52
0.34 14
11-7

87. Trigonometric Ratios Solutions 1. For A = 3, O = 4, H = 5:
ALGEBRA 1 LESSON 11-7
Solutions
1. For A = 3, O = 4, H = 5:
c = = , s = = , t = =
2. For A = 5, O = 12, H = 13:
c = = , s = = , t = =
= =
0.75x = x = 20(0.34)
x = x = 6.8
= =
0.84 = x = 14(0.52)
x = x = 7.28 A H O 3 5 4 13 12 15 x
0.75 1 20
0.34
0.84 21
0.52 14
1. For A = 3, O = 4, H = 5:
c = = , s = = , t = =
2. For A = 5, O = 12, H = 13:
c = = , s = = , t = =
= =
0.75x = x = 20(0.34)
x = x = 6.8
= =
0.84 = x = 14(0.52)
x = x = 7.28 A H O 3 5 4 13 12 15 x
0.75 1 20
0.34
0.84 21
0.52 14
11-7

88. Use the triangle. Find sin A, cos A, and tan A.
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
Use the triangle. Find sin A, cos A, and tan A.
sin A = = =
opposite leg
hypotenuse 6 10 3 5
 cos A = = = 8 10 4 5
adjacent leg
hypotenuse
  tan A = = =
opposite leg
adjacent leg 6 8 3 4
11-7

89. Find sin 40° by using a calculator.
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
Find sin 40° by using a calculator.
Use degree mode when finding  
trigonometric ratios.
To find sin 40°, press
Rounded to the nearest ten-thousandth, the sin 40° is
11-7

90. Find the value of x in the triangle.
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
Find the value of x in the triangle.
Step 1: Decide which trigonometric ratio to use.
You know the angle and the length of the hypotenuse.
You are trying to find the adjacent side. Use the cosine.
Step 2: Write an equation and solve.
cos 30° =
adjacent leg
hypotenuse
cos 30° = Substitute x for adjacent leg
and 15 for hypotenuse. x 15
x = 15(cos 30°) Solve for x.
Use a calculator.
x Round to the nearest tenth.
The value of x is about 13.0.
11-7

91. Define: Let x = the distance from the boat to the lighthouse.
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
Suppose the angle of elevation from a rowboat to the top of a lighthouse is 708. You know that the lighthouse is 70 ft tall. How far from the lighthouse is the rowboat? Round your answer to the nearest foot.
Draw a diagram. 
Define: Let x = the distance from the boat to the lighthouse.
Relate: You know the angle of elevation and the opposite side.
You are trying to find the adjacent side. Use the tangent.
11-7

92. tan 70° = Substitute for the angle and the sides. opposite leg
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
(continued)
Write: tan A =
tan 70° = Substitute for the angle and the sides.
opposite leg
adjacent leg 70 X
x(tan 70°) = 70 Multiply each side by x.
x = Divide each side by tan 70°. 70
tan 70
x Use a calculator.
x Round to the nearest unit.
The rowboat is about 25 feet from the lighthouse.
11-7

93. Define: Let x = the ground distance from the start of the runway.
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
A pilot is flying a plane 15,000 ft above the ground. The pilot begins a 3° descent to an airport runway. How far is the airplane from the start of the runway (in ground distance)?
Draw a diagram.
Define: Let x = the ground distance from the start of the runway.
Relate: You know the angle of depression and the opposite side.
You are trying to find the adjacent side. Use the tangent.
11-7

94. tan 3° = Substitute for the angle and the sides. opposite leg
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
(continued)
Write: tan A =
tan 3° = Substitute for the angle and the sides.
15,000 x
opposite leg
adjacent leg
x(tan 3°) = 15,000 Multiply each side by x.
x = Divide each side by tan 3°.
15,000
tan 3
x Use a calculator.
x ,000 Round to the nearest 10,000 feet.
The airplane is about 290,000 feet (or about 55 miles)
from the start of the runway.
11-7

95. Trigonometric Ratios pages 625–627 Exercises 1. 2. 3. 4. 5. 6.
ALGEBRA 1 LESSON 11-7
pages 625–627  Exercises 1. 2. 3. 4. 5. 6.
18. about 2.0 mi
19. about 172 ft
20. about 816 ft
21. about 0.4 mi
22. c = 29; sin A = ;
cos A = ; tan A =
23. b = 15; sin A = ;
cos A = ; tan A =
24. a = 24; sin A = ;
25. AC 6; AB 8
26. AC ; BC
27. BC 6; AB
28. AC 4; AB
29. about 55 m
30. about 4.4 m
31. a. 1,720,000 ft
b. 326 mi 8 17 3 5 15 17 8 15 4 5 12 13 5 13 12 5 3 4 4 5 3 5 4 3 21 29 20 29 21 20
11-7

96. [3] correct equation, but minor computational error
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
40. Answers may vary.
Sample:
about 8.2 cm
41. about 203 ft
42. about 5938 m
43. about 57 ft ft
46. a. about 229 km
b. about 4 km
47. C
48. G
49. [2] sin A ,
cos A ,
tan A
[1] at least one correct
equation
50. [4]
tan 56° = x
573 ft
[3] correct equation, but
minor computational error
[2] incorrect equation used
[1] no work shown m
33. about 6.8 m
37. q ; r
38. a. about 177 ft
b. about 86 ft
39. a. about 252 ft
b. about 377 ft
850 x
11-7

97. Trigonometric Ratios 54. 2 51. 55. 0 56. 1 57. (n – 20)(n + 20)
ALGEBRA 1 LESSON 11-7
54. 2
55. 0
56. 1
57. (n – 20)(n + 20)
58. (x – 15)2
59. (10p – 7)(10p + 7)
d – d +
61. 2(7w – 8)(7w + 8)
62. (x + 13)2
51.
52.
53. 1 4 1 4 3 2 1 4 3 2
11-7

98. 1. Use the figure to find sin A, cos A, and tan A.
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
1. Use the figure to find sin A, cos A, and tan A.
2. Find the value of x to the nearest tenth.
3. A group of skateboarders wants to build a ramp with an
angle of incline of 14°. What should the rise be for every
10 meters of run?
4. A park ranger on a 220 ft tower spots a fire at an angle
of depression of 4°. How far is the fire from the base of
the tower?
5. A wheelchair ramp is to have an angle of 3.5° with the ground.
The deck at the top of the ramp is 18 in. above the ground.
How long should the ramp be? 21 29 20
, ,
6.9
about 2.5 m
about 3146 ft
about in.
11-7

99. Radical Expressions and Equations
ALGEBRA 1 CHAPTER 11
12. – , – 2
13. – , 5 (– , 5) ft mi
19.
1. yes
2. no
3. no
4. yes cm
units
units
units
units
, 2
, – 3 1 2 – –
– 6
28. Answers may vary. Sample: =
29. B
30. 4
31. 13
32. 6
33. 7 1 2 1 2 3 5 1 2 1 2 1 2
11-A

100. Radical Expressions and Equations
ALGEBRA 1 CHAPTER 11
34. ft
37. x 0; 9 5
38. x 0;
39. x 4;
40. x –9; cm
42. The graph of y = x
is shifted 3 units down.
43. r =
44. about 32 ft
45. about 8.1
46. about 12.1
47. about 1.106
48. about
49. BC , AC
50. about 5 ft V h
> –
11-A