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Radicals and nth roots
Solving equations with exponents
and radicals
But first, some Domain and range review.
Find the domain
and range of the
blue graph
𝐷: −∞ < 𝑥 < ∞
R: −4 ≤ 𝑦 < ∞
Find the domain and range
of the pink graph
𝐷: −∞ < 𝑥 < ∞
R: y= 1.2
Finding Domain and Range
State the domain and range of the function in the red
SOLUTION
𝐷: 2 ≤ 𝑥 < ∞
R: −∞ < 𝑦 ≤ 1
Find the domain and range of the graph
Domain: x 0, Range: y  0
Domain and range: all real numbers
Objectives/Assignment
• Change between radical and exponent
notation
• Evaluate nth roots of real numbers using both
radical notation and rational exponent
notation.
• Use nth roots to solve equations containing
radicals and exponents other than 1 or 2.
Ex. 5: Using nth Roots in “Real Life”
• The total mass M (in kilograms) of a spacecraft that
can be propelled by a magnetic sail is, in theory,
given by:
0.015m
M
4
3
fd
2
where m is the mass (in kilograms) of the magnetic
sail, f is the drag force (in newtons) of the
spacecraft, and d is the distance (in astronomical
units) to the sun. Find the total mass of a
spacecraft that can be sent to Mars using m = 5,000
kg, f = 4.52 N, and d = 1.52 AU.
Solution
The spacecraft can have a total mass of about 47,500
kilograms. (For comparison, the liftoff weight for a space
shuttle is usually about 2,040,000 kilograms.
Ex. 6: Solving an Equation Using an nth Root
• NAUTICAL SCIENCE. The Olympias is a
reconstruction of a trireme, a type of Greek
galley ship used over 2,000 years ago. The power
P (in kilowatts) needed to propel the Olympias at
a desired speed, s (in knots) can be modeled by
this equation:
P = 0.0289s3
A volunteer crew of the Olympias was able to
generate a maximum power of about 10.5
kilowatts. What was their greatest speed?
SOLUTION
• The greatest speed attained by the Olympias was
approximately 7 knots (about 8 miles per hour).
Goal
1
Solve equations that contain square roots
Solve the equation. Check for extraneous solutions.
Simple Radical
check your solutions!!
Ex.1)
Key Step:
x 3
 x   3
2
2
To raise each side of the
equation to the same
power.
x 9
6.6 Solving Radical Equations
One Radical
Ex.3)
2(46)  8  4  6
2x  8  4  6
4 4

2 x  8 2 10

100  4  6
2 x  8  10 
2 x  8  100
8
2
10  4  6
8
2x  92
Don’t forget to
check your
solutions!!
2
2
x  46
6.6 Solving Radical Equations
Radicals with an Extraneous Solution
Ex.5)
x  3  4x
x  3
2

 4x 
9  3  4(9)
2
x  6 x  9  4x
2
 4x
 4x
x  10 x  9  0
2
( x  9)( x  1)  0
Don’t forget to
check your
solutions!!
x 9
1  3  4(1)
x 1
6.6 Solving Radical Equations
Two Radicals
Ex.4)
12  2 x  2 x  0
2 x 2 x
 12  2 x   2 x 
12  2 x 2 2 x
12  2x  4x
 2x
2
Steps for two
12
 2(2equations
) 2 2  0
radical
• Set radical equal
to 8
radical
2 2  0
• Square both
sides
• Solve for x
2 2 2 2  0
 2x
12  6x
Don’t forget to
check your
solutions!!
6
6
2 x
6.6 Solving Radical Equations
Reflection on the Section
Without solving, explain why
2 x  4  8
has no solution.
6.6 Solving Radical Equations
Radical notation
Radical
Index
Number
n
a a
Radicand
1
n
n>1
The index number becomes the
denominator of the exponent.
More on Radicals
n
a
• If n is odd a can be any number
• If n is even, then a must be a positive
number or zero, otherwise there is no
real solution.
Example: Radical form to Exponential
Form
Change to exponential form.
3
x
2
x
or
2
3
 
 x
1
2
3
or
 x
2

1
3
Example: Exponential to Radical Form
Change to radical form.
x
2
3
 x
3
2
or
 x
3
2
The denominator of the
exponent becomes the index
number of the radical.
Using Rational Exponent Notation
Rewrite the expression using RADICAL notation.
Ex)
Ex)
24 3
1
28
1
4
3
4
24
28
6.1 nth Roots and Rational Exponents
Ex. 2 Evaluating Expressions with
Rational Exponents
A. 9
3
2
 ( 9 )  3  27
3
3
Using radical notation
Using rational exponent
notation.
9  (9 )  3  27
1
1 1
B. 32 2 5  1 


2
2
2
5
5
2
4
32
( 32 )
OR
OR
3
1
2
32
2
5
2
3
3
1
1 1



1
2
5 2
2
4
(32 )
Evaluate both ways (a) proficient 163/2 and (b) advanced 32–3/5.
SOLUTION
Rational Exponent Form
a. 163/2= (161/2)3= 43 = 64
b.
1
1
=
= 3/5
(321/5)3
32
1 = 1
= 3
8
2
32–3/5
Radical Form
3
3/2
(
)
16 =  16 = 43 = 64
32-3/5
1
1
= 3/5 = 5
32
(  32 )3
= 13 = 1
8
2
Cancelling radicals to solve equations
a.
b.
c.
d.
x2 
x
6
x 
x
11
y11 
y
4
6
r 
8
4
r4 r4  4 r 4  4 r 4
 r r  r 2
Equations with “nth root” radicals
Ex.2)
3
x  6  12
6 6
3
Key Step:
x 6
 x   6
3
3
x
Before raising each side to the
same power, you should isolate
the radical expression on one
side of the equation.
3
 216
Don’t forget
to check it.
6.6 Solving Radical Equations
3
Goal
Ex)
4
Solving Equations by taking the “nth root” of both sides
4
x 4  81
x   81
4
x  3
When the exponent is EVEN you
must use the Plus/Minus
2 x  64
5
Ex)
5
5
x 5  32
x  32
5
x2
When the exponent is ODD you
don’t use the Plus/Minus
6.1 nth Roots and Rational Exponents
Solving Equations
Ex)
4
4
( x  4)  256
4
Take the Square 1st.
x  4   256
4
x  4  4
x4  4
x 8
Very Important
2 answers !
x  4  4
x0
6.1 nth Roots and Rational Exponents
Example:
Solve the equation:
x  7  9993
4
x  7  7  9993  7
4
x  10000
4
4
x  4 10000
x  10
4
Note: index number
is even, therefore,
two answers.
1 x5 = 512
2
SOLUTION
1 x5 = 512
2
x5 = 1024
x =
5
x = 4
1024
Multiply each side by 2.
take 5th root of each side.
Simplify.
( x – 2 )3 = –14
SOLUTION
( x – 2 )3= –14
(x–2)=
3
–14
x =
3
–14 + 2
x =
3
–14 + 2
x = – 0.41
Use a calculator.
( x + 5 )4 = 16
SOLUTION
( x + 5 )4 = 16
4
(x+5) =+
x =+
4
16
16 – 5
take 4th root of each side.
add 5 to each side.
x = 2 – 5 or
x = – 2 – 5
Write solutions separately.
x = –3
x = –7
Use a calculator.
or
Ex. 4 Solving Equations Using nth Roots
A. 2x4 = 162
B. (x – 2)3 = 10
2 x  162
(x – 2)  10
x  81
x - 2  3 10
x   81
x  10  2
x  4.15
4
4
4
x  3
3
3
Goal
2
Ex. 6)
Solve equations that contain Rational exponents.
2x
32
2
32
32 23
x 
x
 250
2
 125 2 3
 125

x  52

13 2
x  125
it
2(25)
32
 250
2(253 )1 2
12
2(15625)
2(125)
 250
x  25
6.6 Solving Radical Equations
Ex: Simplify the Expression.
Assume all variables are positive.
a. 3
27z 9  3 27  3 z 9
5
x5
y10

x

y2
5
5
d.
18rs
2
3
1
4 3
6r t
b. (16g4h2)1/2
= 161/2g4/2h2/2
= 4g2h
c.
 3z
3
3r
3
4
2
3 3
3r s t
x5
y10
1
1
4
2
3
s t3