Download Simplifying Radicals

Product Rule for Radicals
If n a and n b are real numbers and
n is a natural number, then
Simplifying Radicals
Objective: To simplify radical
expressions using the product
and quotient rules.
n
a
( b) =
n
n
ab .
In other words, the product of two
radicals is the radical of the product.
Product Rule Practice
( 3)
23 ( t )
7x ( 2 y )
3 y ( 6 yz )
5a ( 4ab )
SOLUTIONS
( 3 ) = 30
23 ( t ) = 23t
7 x ( 2 y ) = 14 xy
3 y ( 6 yz ) = 18 y z
5a ( 4ab ) = cannot multiply
a ) 10
a) 10
b)
b)
c)
3
d)
4
e)
3
2
6
4
5
Quotient Rule for Radicals
If
n
a and n b are real numbers and
c)
3
d)
4
e)
3
6
n
In other words, the quotient of two
radicals is the radical of the quotient.
4
4
3
5
Quotient Rule Practice
f)
16
49
g) 3
216
27
b ≠ 0, and n is a natural number, then
n
a
a
= n .
b
b
2
3
h) 3 −
8
125
i) − 4
625
16
1
SOLUTIONS
16 4
=
49 7
f)
g)
3
h)
3
8
2
=−
125
5
i) − 4
625
5
=−
16
2
Rules for Simplifying Radicals
2.
3.
4.
z For
both the product and
quotient properties, you
must have the same index!
216 6
= =2
27 3
−
1.
Special Note
No factor under the radical has a higher
power than the root.
No fractions under radical.
N radical
No
di l iin denominator.
d
i t
No common factors between root and
exponents in radical.
Practice Simplifying Radicals
Removing Perfect-Square
Factors
z
Re-write the radicand as a product of
perfect-square factors times any
remaining factors.
For example:
75 = 25
( 3) = 5
3
SOLUTIONS
i ) 48
i ) 48 = 16i3 = 4 3
j ) − 24
j ) − 24 = − 4i6 = −2 6
k ) − 5 300
k ) − 5 300 = −5 100
00i3 = −5(10)
5( 0) 3 = −50 3
l ) 18
m)
3
24
l ) 18 = 9i2 = 3 2
m)
3
24 = 3 8i3 = 2 3 3
2
Practice Simplifying Radicals
SOLUTIONS
n) −150
o)
3
n) −150 = not real
−250
o) 3 −250 = 3 −125i2 = −5 3 2
p ) − 4 1250
p ) − 4 1250 = − 4 625i2 = −5 4 2
q ) 5 128
q) 5 128 = 5 32i4 = 2 5 4
Removing Variable Factors
z
Practice with Variables
Divide variable exponents by the root and
leave any remainder under the radical.
For example:
11
2
1
2
x = x =x x =x
11
5
r ) 18m 2
s) 256 z12
5
x
t ) − 200 p13
SOLUTIONS
r ) 18m 2 = 9i2im 2 = 3m 2
s)
256 z12 = 16 z 6
t ) − 200 p13 = − 100i2i p12 i p = −10 p 6 2 p
Practice with Variables
u ) − 3 216 y15 x 6 z 3
v) 23k 9 p14
w) − 4 32k 5 m10
3
SOLUTIONS
u ) − 3 216 y15 x 6 z 3 = −6 y 5 x 2 z
v) 23k 9 p14 = 23k 8 ik i p14 = k 4 p 7 23k
w) − 4 32k 5 m10 = − 4 16i2ik 4 ik im8 im 2 = −2km 2 4 2m 2 k
4