11.2 Simplify Radical Expressions
Warm Up
Lesson Presentation
Lesson Quiz
11.2 Warm-Up
Use the distributive property to write an equivalent
expression.
1. 2(x + 6)
ANSWER
2x + 12
2. x(x2 + 2)
ANSWER
x3 + 2x
3. (x + x2)(–3)
ANSWER
–3x – 3x2
4. 0.1(6 + 10x)
ANSWER
0.6 + x
11.2 Warm-Up
5. The area of a square field is 148 square meters.
What is the side length of the field?
ANSWER
about 12.2 m
11.2 Example 1
Simplify.
a.
32 =
=
16 2
Factor using perfect square factor.
16
2
Product property of radicals
= 4 2
b.
9 x3 =
=
Simplify.
9 x2 x
9
= 3x x
x2
Factor using perfect square factors.
x
Product property of radicals
Simplify.
11.2 Guided Practice
1.
Simplify (a)
24 and (b)
ANSWER
(a) 2 6 and (b) 5x
25x2 .
11.2 Example 2
Simplify.
a.
b.
6
6
3x 4 x
=
6 6
Product property of radicals
=
36
Multiply.
=
6
=
4 3x
Simplify.
x
Product property of radicals
= 4 3 x2
= 4
=
3
4x 3
Multiply.
x2
Product property of radicals
Simplify.
11.2 Example 2
Simplify.
c.
7x y2 3 x = 3 7x y2 x
= 3 7x 2 y2
= 3
7
= 3xy 7
Product property of radicals
Multiply.
x2
y2 Product property of radicals
Simplify.
11.2 Example 3
Simplify.
a.
b.
13
100
7
x2
=
13
100
Quotient property of radicals
=
13
10
Simplify.
=
7
x2
Quotient property of radicals
=
7
x
Simplify.
11.2 Guided Practice
2. Simplify (a) 2 x3
x and (b)
ANSWER
(a) x
2
2 and (b) 1
y
1 .
y2
11.2 Example 4
Simplify the expression.
a.
5
7
=
5
7
=
5 7
49
5 7
=
7
7
7
Multiply by
7 .
7
Product property of radicals
Simplify.
11.2 Example 4
b.
2
3b
3b
3b
Multiply by
3b .
3b
=
2
3b
=
6b
9b2
Product property of radicals
=
6b
9
b2
Product property of radicals
=
6b
3b
Simplify.
11.2 Guided Practice
Simplify the expression.
3.
1
3
4.
1
x
5.
3
2x
=
3
3
=
x
x
=
3 2x
2x
11.2 Example 5
Simplify the expression.
a.
4 10 + 13 – 9 10 = 4 10 – 9 10 + 13 Commutative property
= (4 – 9) 10 + 13 Distributive property
= –5 10 + 13
b.
5 3 + 48 = 5 3 + 16 3
= 5 3 + 16
= 5 3 +4 3
= (5 + 4) 3
= 9 3
Simplify.
Factor using perfect
square factor.
3 Product property of radicals
Simplify.
Distributive property
Simplify.
11.2 Guided Practice
Simplify the expression.
6.
2 7 + 3 63
= 11 7
11.2 Example 6
Simplify the expression.
a. 5 (4 – 20 ) = 4 5 – 5
= 4 5 –
20 Distributive property
100 Product property of radicals
= 4 5 – 10
b.
Simplify.
( 7 + 2 )( 7 – 3 2 )
2
= ( 7 ) + 7 (–3 2 ) + 2 7 + 2 (–3 2 )
Multiply.
2
= 7 – 3 7 2 + 7 2 – 3( 2 ) Product property of radicals
= 7 – 3 14 + 14 – 6
Simplify.
= 1 – 2 14
Simplify.
11.2 Guided Practice
7. Simplify the expression (4 – 5 ) (1 – 5 ) .
ANSWER
9– 5 5
11.2 Example 7
ASTRONOMY
The orbital period of a planet is
the time that it takes the planet to
travel around the sun. You can
find the orbital period P (in Earth
years) using the formula P = d 3
where d is the average distance
(in astronomical units,
abbreviated AU) of the planet
from the sun.
a.
Simplify the formula.
b.
Jupiter’s average distance from the sun is shown
in the diagram. What is Jupiter’s orbital period?
11.2 Example 7
SOLUTION
a.
P =
d
3
Write formula.
2
= d d
=
d
2
= d d
b.
Factor using perfect square factor.
d
Product property of radicals
Simplify.
Substitute 5.2 for d in the simplified formula.
P = d d = 5.2 5.2
ANSWER
The orbital period of Jupiter is 5.2 5.2 , or about 11.9,
Earth years.
11.2 Guided Practice
8.
ASTRONOMY
Neptune’s average distance from the sun is about 6 times
Jupiter’s average distance from the sun. Is the orbital
period of Neptune 6 times the orbital period of Jupiter?
Explain.
ANSWER
No. Neptune’s orbital period is
Thus, Neptune’s orbital period is 6 6 times the orbital
period of Jupiter.
11.2 Lesson Quiz
Simplify the expression.
1.
56a2
ANSWER
2.
2a 14
b
64
b
8
ANSWER
3.
5 2 –2 8+
6
ANSWER
2+
6
11.2 Lesson Quiz
4.
A person is standing on a cliff, with an eye level at
60 feet, looking out across the ocean. Using the
formula d = 3h where d (in miles) is the distance a
2
person can see to the horizon and h (in feet) is the
person’s eye level above the water to find the
number of miles the person can see to the horizon.
ANSWER
about 9.5 mi