02_ch02_pre-calculas11_wncp_tr.qxd
5/26/11
12:48 PM
Page 15
Home
Quit
Lesson 2.3 Exercises, pages 114–119
A
3. a) Simplify each radical, if possible.
√
√
√
3 2
27
27 ⴝ
√
9#3
√
√
√
8
3 2
√
32
√
8ⴝ
ⴝ3 3
√
6
√
6
√
2 3
√
2 3
√
4#2
√
√
32 ⴝ
ⴝ2 2
√
ⴝ4 2
√
18
√
48 ⴝ 16 # 3
√
√
48
√
16 # 2
√
√
18 ⴝ
ⴝ4 3
√
9#2
√
ⴝ3 2
b) Group the radicals in part a into sets of like radicals.
All radicals have index 2.
√ √ √ √
√ √ √ √
3 2, 8, 32, 18
√ √ √
Radicals with radicand 3: 3 3, 2 3, 4 3; that is,
√
√ √
Radicals with radicand 2: 3 2, 2 2, 4 2, 3 2; that is,
27, 2 3,
©P
DO NOT COPY.
48
2.3 Adding and Subtracting Radical Expressions—Solutions
15
02_ch02_pre-calculas11_wncp_tr.qxd
5/26/11
12:48 PM
Page 16
Home
Quit
4. Simplify by adding or subtracting like terms.
√
√
√
a) 3 2 + 2 2 - 5 2
√
b)
ⴝ (3 ⴙ 2 ⴚ 5) 2
ⴝ0
√
√
√
108 - 2 3 - 75
√
√
√
ⴝ 36 # 3 ⴚ 2 3 ⴚ 25 # 3
√
√
√
ⴝ6 3ⴚ2 3ⴚ5 3
√
ⴝⴚ 3
√
√
√
√
√
√
√
c) 5 7 + 2 5 - 28 + 45
d) 3 16 + 3 375 - 3 3 2
√
√
√
√
√
√
√
ⴝ5 7ⴙ2 5ⴚ 4#7ⴙ 9#5
ⴝ 3 8 # 2 ⴙ 3 125 # 3 ⴚ 3 3 2
√
√
√
√
√
√
√
ⴝ5 7ⴙ2 5ⴚ2 7ⴙ3 5
ⴝ 232 ⴙ 533 ⴚ 332
√
√
√
√
3
3
ⴝ3 7ⴙ5 5
ⴝⴚ 2ⴙ5 3
5. Simplify.
√
√
√
√
a) 6 x - 4 x + 2 x + x, x ≥ 0
√
ⴝ (6 ⴚ 4 ⴙ 2 ⴙ 1) x
√
ⴝ5 x
b)
√
√
√
4a + 16a - 9a, a ≥ 0
√
√
√
ⴝ 4 # a ⴙ 16 # a ⴚ 9 # a
√
√
√
ⴝ2 aⴙ4 aⴚ3 a
√
ⴝ3 a
c)
√
3
√
√
27x2 - 3 8x2 + 3 64x2 , x H ⺢
√
√
√
ⴝ 3 27 # x2 ⴚ 3 8 # x2 ⴙ 3 64 # x2
√
√
√
ⴝ 3 3 x2 ⴚ 2 3 x2 ⴙ 4 3 x2
√
3 2
ⴝ5 x
B
6. Explain why it is necessary to write
√
4
√
4
x4 as |x|.
x4 is defined
√ for x ç ⺢. The radical sign indicates the principal root, so
the value of 4 x4 cannot be negative. Although√x4 is always positive or zero,
x can be negative. So, it is necessary to write 4 x4 as x .
7. Identify the values of the variables for which each radical is defined,
then simplify.
√
√
√
a) 7 -x + 15 - x - 13 -x
The radicand cannot be negative, so ⴚx
√
√
√
» 0; that is, x ◊ 0.
√
7 ⴚx ⴙ 15 ⴚx ⴚ 13 ⴚx ⴝ (7 ⴙ 15 ⴚ 13) ⴚx
√
ⴝ 9 ⴚx
16
2.3 Adding and Subtracting Radical Expressions—Solutions
DO NOT COPY.
©P
02_ch02_pre-calculas11_wncp_tr.qxd
5/26/11
12:48 PM
Page 17
Home
√
Quit
√
b) 28m4n + m2 63n
The radicand cannot be negative.
Since m4 » 0, m ç ⺢.
n cannot be negative, so n » 0.
√
√
√
√
28 m4n ⴙ m2 63n ⴝ 4 # 7 # m4 # n ⴙ m2 9 # 7 # n
√
√
2
√
ⴝ 2m2 7n ⴙ 3m2 7n
ⴝ 5m
7n
√
√
c) 4 3 2p4q - 6p 3 2pq
The cube root of a number is defined for all real numbers. So, each
radical is defined for p, q ç ⺢.
√
√
√
√
√
√
ⴝ 4p 3 2pq ⴚ 6p 3 2pq
√3
4 3 2p4q ⴚ 6p 3 2pq ⴝ 4 3 2 # p3 # pq ⴚ 6p 3 2pq
ⴝ ⴚ2p 2pq
8. Simplify.
a)
√
√
√
√
5b + 4 5b - 3 3 5b - 2 5b, b ≥ 0
√
√
√
√
ⴝ 5b ⴙ 4 5b ⴚ 2 5b ⴚ 3 3 5b
√
√3
ⴝ 3 5b ⴚ 3 5b
√
√
√
b) 3 x3 + 5 2x - 4x3, x ≥ 0
√
√
√
ⴝ 3 x2 # x ⴙ 5 2x ⴚ 4 # x2 # x
√
√
√
ⴝ 3x x ⴙ 5 2x ⴚ 2x x
√
√
ⴝ x x ⴙ 5 2x
√
√
ⴝ ⴚ10e
6e ⴙ 6e
√
c) 5e 24e3 - 7 54e5 + e2 6e + 6e, e ≥ 0
√
√
√
ⴝ 5e 4 # 6 # e2 # e ⴚ 7 9 # 6 # e4 # e ⴙ e2 6e ⴙ 6e
√
√
√
ⴝ 5e(2e) 6e ⴚ 7(3e2) 6e ⴙ e2 6e ⴙ 6e
√
√
√
ⴝ 10e2 6e ⴚ 21e2 6e ⴙ e2 6e ⴙ 6e
√
2
√
√
√
√
d) 3 16v5 + 3 3w4 + 2w 3 24w - 5v 3 54v2, v, w H ⺢
√
√
√
√
ⴝ 3 8 # 2 # v3 # v2 ⴙ 3 3 # w3 # w ⴙ 2w 3 8 # 3 # w ⴚ 5v 3 27 # 2 # v2
√
√
√
√
ⴝ 2v 3 2v2 ⴙ w 3 3w ⴙ 2w(2) 3 3w ⴚ 5v(3) 3 2v2
√
√
√
√
ⴝ 2v 3 2v2 ⴙ w 3 3w ⴙ 4w 3 3w ⴚ 15v 3 2v2
√
√
3
2
3
ⴝ ⴚ13v 2v ⴙ 5w 3w
©P
DO NOT COPY.
2.3 Adding and Subtracting Radical Expressions—Solutions
17
02_ch02_pre-calculas11_wncp_tr.qxd
5/26/11
12:48 PM
Page 18
Home
Quit
9. A square with area 24 square units is placed
beside a square with area 50 square units.
In simplest form, write a radical expression
for the perimeter of the shape formed.
The side length of a square is the square root of its area.
Small square:
Large square:
Its area is√24, so its side
Its area is 50, so its side length is
√
50.
length is 24 .
The perimeter of the shape formed consists of 3 sides of each square and
the length that is the difference √
in their side
√ lengths.
√
√
Perimeter of shape formed ⴝ 3 24 ⴙ 3 50 ⴙ ( 50 ⴚ 24 )
√
√
√
√
50 ⴚ 24
√
√
ⴝ 2 24 ⴙ 4 50
√
√
ⴝ 2 4 # 6 ⴙ 4 25 # 2
√
√
ⴝ 3 24 ⴙ 3 50 ⴙ
ⴝ 4 6 ⴙ 20 2
10. Two squares are enclosed in a large square
as shown. The area of the smallest square
is A square units. The area of the middle
square is 4A square units. Determine the
area and perimeter of the shaded region
in terms of A.
4A
A
The side length of a square is the square root
√ of its area.
So, the side length of the small square is: A units
√
√
The side length of the middle square is: 4A, or 2 A units
The side length of
√the large
√ square
√is the sum of the side lengths of the
other 2 squares: A ⴙ 2 A, or 3 A
Area of shaded region
ⴝ area
√ of large square ⴚ area of small square ⴚ area of middle square
ⴝ (3 A )2 ⴚ A ⴚ 4A
ⴝ 9A ⴚ 5A
ⴝ 4A
√
From the diagram, the length of 2 grid squares is A .
The perimeter of the shaded region is the length of 20 grid squares.
So, perimeter ⴝ 20 grid squares
ⴝ 10 # (2 grid squares)
√
ⴝ 10 A
18
2.3 Adding and Subtracting Radical Expressions—Solutions
DO NOT COPY.
©P
02_ch02_pre-calculas11_wncp_tr.qxd
5/26/11
12:48 PM
Page 19
Home
Quit
C
11. In right ^ABC, AB has length
A
3 units and AC has length 6 units.
A congruent triangle is placed adjacent
to ^ABC as shown. Determine the
perimeter of the shape formed.
3
B
6
D
E
C
Use the Pythagorean Theorem in ¢ABC to determine the length of BC.
(AC)2 ⴝ (BC)2 ⴙ (AB)2
62 ⴝ (BC)2 ⴙ 32
27 ⴝ (BC)2
√
27 ⴝ BC
√
3 3 ⴝ BC
√
√
The perimeter of each triangle is: 6 ⴙ 3 ⴙ 3 3 ⴝ 9 ⴙ 3 3
BE ⴝ AB ⴝ 3
So, the perimeter of the shape formed is:
2 times the√perimeter of ¢ABC ⴚ2 times BE
ⴝ 2(9 ⴙ 3 3) ⴚ 2(3)
√
√
ⴝ 18 ⴙ 6 3 ⴚ 6
ⴝ 12 ⴙ 6 3
√
So, the perimeter of the shape formed is (12 ⴙ 6 3 ) units.
12. Determine whether ^EDG is a
113
right triangle. How did you
find out?
E
F
312
G
65
D
Use the Pythagorean Theorem
in right ¢DFG to determine
the length of DF.
(DG)2 ⴝ (FG)2 ⴙ (DF)2
√
√
(6 5 )2 ⴝ (3 12 )2 ⴙ (DF)2
180 ⴝ 108 ⴙ (DF)2
(DF)2 ⴝ 72
√
DF ⴝ 72
√
Use the Pythagorean Theorem
in right ¢DEF to determine
the length of ED.
(ED)2 ⴝ (EF)2 ⴙ (DF)2
√
√
√
(ED)2 ⴝ (11 3 ⴚ 3 12 )2 ⴙ (6 2 )2
(ED)2 ⴝ 75 ⴙ 72
(ED)2 ⴝ 147
ED ⴝ
√
147
√
ED ⴝ 7 3
DF ⴝ 6 2
To determine whether ¢EDG is a right triangle, use the Pythagorean
Theorem to check whether (EG)2 ⴝ (ED)2 ⴙ (DG)2
L .S. ⴝ (EG)2
√
R.S . ⴝ (ED)2 ⴙ (DG)2
2
ⴝ (11 3 )
ⴝ 363
√
√
ⴝ (7 3 )2 ⴙ (6 5 )2
ⴝ 147 ⴙ 180
ⴝ 327
Since L.S. ⴝ R.S., ¢EDG is not a right triangle.
©P
DO NOT COPY.
2.3 Adding and Subtracting Radical Expressions—Solutions
19
02_ch02_pre-calculas11_wncp_tr.qxd
5/26/11
12:48 PM
Page 20
Home
Quit
13. Determine if there are any values of x and y such that
√
x +
√
√
x + y and
y are equal. Explain your reasoning.
»0
√
√
√
xⴙyⴝ xⴙ y
√
√
√
( x ⴙ y )2 ⴝ ( x ⴙ y )2
√
x ⴙ y ⴝ x ⴙ 2 xy ⴙ y
√
x, y
2 xy ⴝ 0
For xy ⴝ 0, x ⴝ 0, or y ⴝ 0, or both x ⴝ 0√and y ⴝ 0√
So, there are values of x and y such that x ⴙ y ⴝ x ⴙ
20
√
y.
2.3 Adding and Subtracting Radical Expressions—Solutions
DO NOT COPY.
©P