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Lesson 2.4 Exercises, pages 126–133
A
3. Expand and simplify.
a)
c)
√ √
61 5 + 22
√ √
√
ⴝ 6( 5) ⴙ 6(2)
√
√
ⴝ
30 ⴙ 2 6
√
√
21 -7 + 22
√
√ √
ⴝ 2(ⴚ7) ⴙ 2 ( 2 )
√
ⴝ ⴚ7 2 ⴙ 2
b)
√ √
51 2 - 42
√ √
√
ⴝ 5( 2 ) ⴚ 5(4)
√
√
ⴝ
10 ⴚ 4 5
√
√
d) - 313 + 82
√
√ √
ⴝ ⴚ 3(3) ⴚ 3( 8)
√
√
ⴝ ⴚ 3 3 ⴚ 24
√
√
ⴝ ⴚ3 3 ⴚ 2 6
4. Expand and simplify.
a)
√ √
√
61 3 + 22
√ √
√ √
ⴝ 6( 3) ⴙ 6( 2)
√
√
ⴝ 18 ⴙ 12
√
√
ⴝ3 2ⴙ2 3
©P
DO NOT COPY.
√
√
√
b) -2 713 2 - 72
√ √
√
√
ⴝ ⴚ 2 7(3 2) ⴚ 2 7( ⴚ 7 )
√
ⴝ ⴚ 6 14 ⴙ 2(7)
√
ⴝ ⴚ 6 14 ⴙ 14
2.4 Multiplying and Dividing Radical Expressions—Solutions
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5. Expand and simplify.
√
√
a) 1 3 + 221 3 - 22
√ √
√
ⴝ 3( 3 ⴚ 2) ⴙ 2( 3 ⴚ 2)
√ √
√
ⴝ 3( 3) ⴚ 3(2)
√
ⴙ 2( 3 ) ⴚ 2(2)
√
√
ⴝ3ⴚ2 3ⴙ2 3ⴚ4
ⴝ ⴚ1
Quit
√
b) 1 5 + 222
√
√
ⴝ ( 5 ⴙ 2)( 5 ⴙ 2)
√ √
√
ⴝ 5( 5 ⴙ 2) ⴙ 2( 5 ⴙ 2)
√ √
√
√
ⴝ 5( 5 ) ⴙ 5(2) ⴙ 2( 5 ) ⴙ 2(2)
√
√
ⴝ5ⴙ2 5ⴙ2 5ⴙ4
√
ⴝ9ⴙ4 5
√
√
√
√ √
√
c) 1 3 + 222
d) 12 3 + 3 521 3 - 2 52
√ √
√
√ √ √ √
ⴝ 2 3( 3 ⴚ 2 5)
ⴝ ( 3 ⴙ 2)( 3 ⴙ 2 )
√
√
√
√ √ √
√ √
√
ⴙ 3 5( 3 ⴚ 2 5)
ⴝ 3( 3 ⴙ 2) ⴙ 2( 3 ⴙ 2 )
√
√
√ √
√ √
√ √
ⴝ 2(3) ⴚ 4( 15) ⴙ 3 15 ⴚ 6(5)
ⴝ 3( 3 ) ⴙ 3( 2) ⴙ 2( 3)
√
√
√ √
ⴝ 6 ⴚ 4 15 ⴙ 3 15 ⴚ 30
ⴙ 2( 2)
√
√
√
ⴝ ⴚ24 ⴚ 15
ⴝ3ⴙ 6ⴙ 6ⴙ2
√
ⴝ5ⴙ2 6
6. Rationalize the denominator.
√
3
2
a) √ ⴝ √2 # √
3
3
3
√
√
1
1
b) √ ⴝ √
2 3
ⴝ
3
5 3
5 3
√
ⴝ
ⴝ
# √3
3
3
5(3)
√
3
15
B
7. Expand and simplify.
√
√ √
√
a) 1 3 + 8212 3 - 12 - 317 3 - 42
√ √
√
√ √
√
ⴝ 3(2 3 ⴚ 1) ⴙ 8(2 3 ⴚ 1) ⴚ 3(7 3) ⴚ 3(ⴚ4)
√
√
√
ⴝ 2(3) ⴚ 3 ⴙ 16 3 ⴚ 8 ⴚ 7(3) ⴙ 4 3
√
√
√
ⴝ 6 ⴚ 3 ⴙ 16 3 ⴚ 8 ⴚ 21 ⴙ 4 3
√
ⴝ ⴚ23 ⴙ 19 3
√
√ √
√
√
b) - 13 2 - 521 2 + 72 - 12 2 - 522
√
√ √
√
√ √
√
ⴝ (ⴚ3 2 ⴙ 5)( 2 ⴙ 7) ⴚ (2 2 ⴚ 5)(2 2 ⴚ 5)
√ √
√ √
√ √
√
√ √
√
ⴝ ⴚ3 2( 2 ⴙ 7) ⴙ 5( 2 ⴙ 7) ⴚ [2 2(2 2 ⴚ 5) ⴚ 5(2 2 ⴚ 5 ) ]
√
√
√
√
√
ⴝ ⴚ3(2) ⴚ 21 2 ⴙ 10 ⴙ 7 5 ⴚ [4(2) ⴚ 2 10 ⴚ 2 10 ⴙ 5]
√
√
√
√
ⴝ ⴚ6 ⴚ 21 2 ⴙ 10 ⴙ 7 5 ⴚ 8 ⴙ 4 10 ⴚ 5
√
√
√
ⴝ ⴚ19 ⴚ 21 2 ⴙ 5 10 ⴙ 7 5
22
2.4 Multiplying and Dividing Radical Expressions—Solutions
DO NOT COPY.
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√
√
√
√
√
√
c) 16 7 - 5 5216 7 + 5 52 + 13 7 + 4 522
√ √
√
√ √
√
√
√ √
√
ⴝ 6 7(6 7 ⴙ 5 5 ) ⴚ 5 5(6 7 ⴙ 5 5 ) ⴙ (3 7 ⴙ 4 5)(3 7 ⴙ 4 5 )
√ √
√ √
√ √
√ √
ⴝ 6 7(6 7) ⴙ 6 7(5 5) ⴚ 5 5(6 7) ⴚ 5 5(5 5)
√ √
√
√ √
√
ⴙ 3 7(3 7 ⴙ 4 5) ⴙ 4 5(3 7 ⴙ 4 5)
√
√
√ √
ⴝ 36(7) ⴙ 30 35 ⴚ 30 35 ⴚ 25(5) ⴙ 3 7(3 7)
√ √
√ √
√ √
ⴙ 3 7(4 5) ⴙ 4 5(3 7) ⴙ 4 5(4 5 )
√
√
ⴝ 252 ⴚ 125 ⴙ 9(7) ⴙ 12 35 ⴙ 12 35 ⴙ 16(5)
√
ⴝ 252 ⴚ 125 ⴙ 63 ⴙ 24 35 ⴙ 80
√
ⴝ 270 ⴙ 24 35
8. Identify the values of the variable for which each expression is
defined, then expand and simplify.
√
√
a) w12 w + 32
The radicands cannot be negative, so w
√
√
w(2 w ⴙ 3) ⴝ
√
√
w(2 w ) ⴙ
√
√
» 0.
w(3)
ⴝ 2w ⴙ 3 w
√
√
b) 13 x - 2212 x + 52
» 0.
√ √
√
(3 x ⴚ 2)(2 x ⴙ 5) ⴝ 3 x (2 x ⴙ 5) ⴚ 2(2 x ⴙ 5)
√ √
√
√
ⴝ 3 x (2 x ) ⴙ 3 x(5) ⴚ 2(2 x) ⴚ 2(5)
√
√
ⴝ 6x ⴙ 15 x ⴚ 4 x ⴚ 10
√
The radicands cannot be negative, so x
√
√
ⴝ 6x ⴙ 11 x ⴚ 10
√
√
c) 1 c + d22
The radicands cannot be negative, so c
» 0 and d » 0.
√
√ 2 √
√ √
√
( c ⴙ d ) ⴝ ( c ⴙ d )( c ⴙ d )
√ √
√
√ √
√
ⴝ c( c ⴙ d) ⴙ d( c ⴙ d)
√
√
ⴝ c ⴙ cd ⴙ cd ⴙ d
√
ⴝ c ⴙ 2 cd ⴙ d
√
√
√
√
d) 12 x - 3 y212 x + 3 y2
The radicands cannot be negative, so x » 0 and y » 0.
The expression is the binomial factors of a difference of squares.
√
√ √
√
√
√
(2 x ⴚ 3 y )(2 x ⴙ 3 y ) ⴝ (2 x )2 ⴚ (3 y )2
ⴝ 4x ⴚ 9y
©P
DO NOT COPY.
2.4 Multiplying and Dividing Radical Expressions—Solutions
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√
√
√
√
√
√
e) 12 a - b213 a - 4 b2 - 1 a - 3 b22
The radicands cannot be negative, so a
» 0 and b » 0.
√ √
√
√
√ 2
(2 a ⴚ b)(3 a ⴚ 4 b) ⴚ ( a ⴚ 3 b)
√ √
√
√ √
√
√
√ √
√
ⴝ 2 a(3 a ⴚ 4 b) ⴚ b(3 a ⴚ 4 b) ⴚ ( a ⴚ 3 b)( a ⴚ 3 b)
√ √
√ √
√ √
√
√
ⴝ 2 a(3 a ) ⴚ 2 a(4 b) ⴚ b(3 a ) ⴚ b (ⴚ4 b)
√ √
√
√ √
√
ⴚ [ a( a ⴚ 3 b) ⴚ 3 b( a ⴚ 3 b) ]
√
√
√ √
√ √
ⴝ 6a ⴚ 8 ab ⴚ 3 ab ⴙ 4b ⴚ [ a( a ) ⴚ a(3 b )
√ √
√
√
ⴚ 3 b( a ) ⴚ 3 b( ⴚ3 b ) ]
√
√
√
ⴝ 6a ⴚ 11 ab ⴙ 4b ⴚ [a ⴚ 3 ab ⴚ 3 ab ⴙ 9b]
√
√
√
ⴝ 6a ⴚ 11 ab ⴙ 4b ⴚ a ⴙ 3 ab ⴙ 3 ab ⴚ 9b
√
√
ⴝ 5a ⴚ 5 ab ⴚ 5b
9. Simplify. Describe the strategy you used in part c.
√
√
√
2 5 + 4
5 8 - 2 5
√
a) √
b)
5
6
√
√
√ √
√
(2 5 ⴙ 4) # 5
(5 8 ⴚ 2 5) # 6
√
√
√
√
ⴝ
ⴝ
5
√
5
√
√
2 5# 5ⴙ4# 5
√ √
ⴝ
5
√
ⴝ
#
5
10 ⴙ 4 5
5
√
6
6
√
√ √
5 8# 6ⴚ2 5# 6
√ √
ⴝ
ⴝ
#
6
√
6
√
5 48 ⴚ 2 30
6
√
√
20 3 ⴚ 2 30
ⴝ
6
√
10 3 ⴚ
ⴝ
3
√
√
- 3 12 + 2 3
√
c)
18
√
√
ⴝ
ⴚ6 3 ⴙ 2 3
√
3 2
√ √
ⴚ4 3 # 2
√
ⴝ √
3 2
2
√ √
ⴚ4 3 # 2
ⴝ √ √
3
2#
√
ⴚ4 6
ⴝ
6
√
ⴝ
2
√
30
√
√
4 2 - 6 5
√
d)
2 3
√
√
ⴝ
4 2ⴚ6 5
√
√
2 3
√
# √3
3
√
√ √
4 2# 3ⴚ6 5# 3
√ √
ⴝ
ⴝ
√
2 3
√
√
√
#
3
4 6 ⴚ 6 15
6
2 6 ⴚ 3 15
ⴝ
3
ⴚ2 6
3
In part c, I first simplified the radicals, then collected like terms.
I was left with an expression with a monomial in the numerator and in the
denominator. I rationalized the denominator, then removed a common factor.
24
2.4 Multiplying and Dividing Radical Expressions—Solutions
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10. Simplify.
√
- 5
a) √
7 - 3
√
√
√
5 3 + 2
√
b) √
3 - 2
√
√
√
√
(5 3 ⴙ 2) # ( 3 ⴙ 2)
√
√
√
ⴝ √
( 3 ⴚ 2) ( 3 ⴙ 2)
√ √
√
√ √
√
5 3( 3 ⴙ 2) ⴙ 2( 3 ⴙ 2)
√
√
ⴝ
( 3)2 ⴚ ( 2)2
√ √
√ √
√ √
√ √
5 3( 3) ⴙ 5 3( 2) ⴙ 2( 3) ⴙ 2( 2)
√
# (√7 ⴙ 3)
ⴚ 5
ⴝ √
( 7 ⴚ 3) ( 7 ⴙ 3)
√ √
√
ⴚ 5( 7) ⴚ 5(3)
√
ⴝ
( 7)2 ⴚ (3)2
√
√
ⴝ
ⴝ
ⴝ
ⴚ 35 ⴚ 3 5
7ⴚ9
ⴝ
ⴚ2
ⴝ 17 ⴙ 6 6
√
35 ⴙ 3 5
2
√
√
5 8 - 3 2
√
c) √
32 + 3 2
Simplify
first.
√
√
√
√ ⴝ √
√
√
5 8ⴚ3 2
10 2 ⴚ 3 2
32 ⴙ 3 2
4 2ⴙ3 2
√
3ⴚ2
√
√
ⴝ 15 ⴙ 5 6 ⴙ 6 ⴙ 2
√
√
√
ⴚ ( 35 ⴙ 3 5 )
√
Quit
√
ⴝ √
7 2
7 2
√
6 3 - 2
√
d)
5 + 4 2
√
√
(6 3 ⴚ 2) # (5 ⴚ 4 2)
√
√
ⴝ
(5 ⴙ 4 2) (5 ⴚ 4 2)
√
√
√
6 3(5 ⴚ 4 2) ⴚ 2(5 ⴚ 4 2)
√
ⴝ
(5)2 ⴚ (4 2)2
√
√ √
√
6 3(5) ⴚ 6 3(4 2) ⴚ 2(5) ⴚ 2(ⴚ4 2)
ⴝ
ⴝ1
ⴝ
√
25 ⴚ 32
√
√
30 3 ⴚ 24 6 ⴚ 10 ⴙ 8 2
ⴚ7
√
√
√
ⴚ 30 3 ⴙ 24 6 ⴙ 10 ⴚ 8 2
ⴝ
7
11. Rationalize each denominator. Describe any patterns in each list of
quotients. Predict the next three quotients in each pattern.
1
1
1
√ ,√
√ ,√
√ ,...
5 + 2 6 + 2 7 + 2
a) √
√ 1 √
5ⴙ
√ 1 √
2
2
√
√
( 6 ⴚ 2)
1
#
√
√
√
√
ⴝ
( 6 ⴙ 2) ( 6 ⴚ 2)
√
√
ⴝ
ⴝ
5ⴚ
3
2
√ 1 √
7ⴙ
2
ⴝ √
1
√
√
# (√7 ⴚ √2)
√
( 7 ⴙ 2) ( 7 ⴚ 2)
√
√
ⴝ
©P
6ⴙ
√
√
( 5 ⴚ 2)
1
#
√
√
√
√
ⴝ
( 5 ⴙ 2) ( 5 ⴚ 2)
√
√
7ⴚ
5
DO NOT COPY.
2
6ⴚ
4
2
The numerator of the quotient is the
conjugate of the denominator in the
original expression. The denominator
is the absolute value of the difference
of the radicands. The next three
quotients are:
√
8ⴚ
6
√ √
√ √
,
,
2
9ⴚ
7
2
10 ⴚ
8
√
2
2.4 Multiplying and Dividing Radical Expressions—Solutions
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√
√
3
Quit
√
5
√ ,√
√ ,√
√ ,...
3 + 2 4 + 3 5 + 4
√
√
b) √
3
√
3ⴙ
4
√
√
2
√
√
ⴝ √
4ⴙ
√
# (√3 ⴚ √2)
3
5
√
5ⴙ
4
√
√
5
√
# (√5 ⴚ √4)
√
( 5 ⴙ 4) ( 5 ⴚ 4)
ⴝ5ⴚ
√
20
√
√
# (√4 ⴚ √3)
4
√
ⴝ4ⴚ
6
√
ⴝ √
3
√
√
( 4 ⴙ 3) ( 4 ⴚ 3)
√
√
√
ⴝ √
√
( 3 ⴙ 2) ( 3 ⴚ 2)
ⴝ3ⴚ
4
12
The first term of the quotient is the
square of the numerator in the original
expression. The second term is the
square root of the product of the
radicands in the denominator of the
original expression. The next three
quotients are:
√
√
√
6 ⴚ 30, 7 ⴚ 42, 8 ⴚ 56
12. The dimensions of the Parthenon of ancient Greece are thought to
be based on the golden rectangle. The length of a golden rectangle is
2
times as long as its width.
√
5 - 1
2
to 3 decimal places.
5 - 1
a) Write the value of √
Use a calculator. √
2
5ⴚ1
⬟ 1.618
2
to 3 decimal places.
5 - 1
b) Write the value of the reciprocal of √
√
5ⴚ1
⬟ 0.618
2
Reciprocal:
c) Compare the values in parts a and b. How are they related?
1.618 ⴚ 0.618 ⴝ 1; the values have a difference of 1.
2
5 - 1
d) Write an expression to represent the difference between √
and its reciprocal. Simplify the expression. What do you notice?
√
A common denominator is: 2( 5 ⴚ 1)
√
√
2
5ⴚ1
ⴚ
√
√
( 5 ⴚ 1)( 5 ⴚ 1)
√
2( 5 ⴚ 1)
√
√
4 ⴚ (5 ⴚ 5 ⴚ 5 ⴙ 1)
√
ⴝ
2 5ⴚ2
√
2 5ⴚ2
ⴝ √
2(2)
5ⴚ1
ⴝ √
ⴚ
2
2( 5 ⴚ 1)
2 5ⴚ2
ⴝ1
The expression simplifies to 1, which is the same as the difference
between the estimated decimal values in part c.
26
2.4 Multiplying and Dividing Radical Expressions—Solutions
DO NOT COPY.
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C
13. a) Simplify.
√
√
5 3
ii) √ - √
12
8
2
1
1
i) √ - √
5
3
A common denominator is:
√
Simplify first.
15
√
√
√
√
2
5 3
2
5 3
√ ⴚ √ ⴝ √ ⴚ √
√
√
1# 3
1# 5
1
1
√ ⴚ√ ⴝ√ √ ⴚ√ √
5
3
5
√
ⴝ
#
3
3
√
3ⴚ
#
12
5
5
√
√
√ √
( 3 ⴚ 5) # 15
√
√
ⴝ
ⴝ
15
ⴝ
2 3
2 2
ⴝ
15
√
√ √
3 # 15 ⴚ 5 # 15
45 ⴚ
15
2 3# 2
2 ⴚ 15
2 2
#
3
√
2 6
2 6
√
ⴝ
3 5ⴚ5 3
15
√
√
ⴚ13
75
√
2 2
√
6
ⴝ √ #√
√
( 15)2
√
√
ⴝ
2 3
√
√
√ √
√ √
2
5 3
2# 2
5 3# 3
√ ⴚ √ ⴝ √ √ ⴚ √ √
15
√
8
A common denominator is: 2 6
√
6
ⴚ13 6
12
√
6
2
√ - √
√
2 5 + 3 3
7 - 2 3
iii) √
Rationalize the denominator of each term, then subtract.
√
6
√
√
2 5ⴙ3 3
ⴝ
√
7ⴚ2 3
√
√
ⴝ
2
√ ⴚ√
√
# (2√5 ⴚ 3√3) ⴚ
6
√
√
(2 5 ⴙ 3 3) (2 5 ⴚ 3 3)
√
√
√
√
2 30 ⴚ 3 18
ⴚ
20 ⴚ 27
√
√
2 30 ⴚ 9 2
ⴝ
ⴚ
ⴚ7
√
√
2
√
√
# (√7 ⴙ 2√3)
√
√
( 7 ⴚ 2 3) ( 7 ⴙ 2 3)
14 ⴙ 2 6
7 ⴚ 12
√
14 ⴙ 2 6
ⴚ5
√
ⴚ 2 30 ⴙ 9 2
ⴝ
ⴙ
7
√
√
√
√
√
14 ⴙ 2 6
5
Common denominator: 35
√
√
5( ⴚ 2 30) ⴙ 5(9 2) ⴙ 7 14 ⴙ 7(2 6)
ⴝ
35
√
√
√
√
ⴚ 10 30 ⴙ 45 2 ⴙ 7 14 ⴙ 14 6
ⴝ
35
b) How are the strategies you used in part a similar to those used to
add and subtract quotients of integers? How are they different?
The strategies are similar in that I still have to find a common
denominator to add or subtract the terms. The strategies are different
in that I have to rationalize the denominators.
©P
DO NOT COPY.
2.4 Multiplying and Dividing Radical Expressions—Solutions
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14. Rationalize the denominator.
3
a) √
3
4
√
3
(
4)2
#
√
√
√
ⴝ
3
3
4
4 ( 3 4)2
√
3( 3 4)2
3
ⴝ
4
3
b) √
4
To make the denominator
an integer,
√
√ I multiply
3
4 by ( 3 4 )2.
3
Quit
To make the denominator
an
integer,
√
√ I multiply
4
3 by ( 4 3 )3.
√
4
(
3)3
#
√
√
√
ⴝ
4
4
3
3 ( 4 3)3
√
4( 4 3)3
4
4
ⴝ
4
3
√
√
2 x + 3 y
√ , x, y ≥ 0 in simplest form.
15. Write √
x - y
√
√
√
√
√
√
2 xⴙ3 y
(2 x ⴙ 3 y) # ( x ⴙ y )
√
√
√ ⴝ √
√
√
xⴚ y
( x ⴚ y) ( x ⴙ y )
√ √
√
√ √
√
2 x( x ⴙ y) ⴙ 3 y( x ⴙ y)
√
√
ⴝ
( x)2 ⴚ ( y)2
√
√
ⴝ
2x ⴙ 2 xy ⴙ 3 xy ⴙ 3y
xⴚy
√
2x ⴙ 5 xy ⴙ 3y
ⴝ
xⴚy
16. Decide if it is possible to determine two whole numbers a and b that
satisfy each condition. Justify your answers.
√
√ √
a
#
b and √ are rational.
a) a
b
Yes, it is possible. When a and b are perfect squares, b ⴝ 0,
√
a
√
a
is a product of natural numbers and √ is a quotient of natural
b
√
√ √
4
2
numbers. For example, 4 # 9 ⴝ 2 # 3, or 6, and √ ⴝ .
9
b)
# √b
3
√
√
a
a # b and √ are irrational.
√
b
Yes, it
√ is possible. When a and b are different prime numbers,
a
and √ are irrational. For example,
b
√
2
√
a
# √b
# √3 ⴝ √6 or 2.4494 Á , and
√
2
√ ⴝ 0.8164 Á .
3
28
2.4 Multiplying and Dividing Radical Expressions—Solutions
DO NOT COPY.
©P