Chapter 7
Radicals, Radical Functions,
and Rational Exponents
§ 7.1
Radical Expressions and Functions
Radicals
In this section, we introduce a new category of expressions and functions
that contain roots.
For example, the reverse operation of squaring a number is finding the
square root of the number.
The symbol
that we use to denote the principal square root
is called a radical sign. The number under the radical sign is called the
radicand. Together we refer to the radical sign and its radicand as a
radical expression.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 7.1
Radical Expressions
EXAMPLE
Index
of the
Radical
n
a
Radical
Sign
Radical
Expression
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 7.1
Radicand
Radical Expressions
Definition of the Principal Square Root
If a is a nonnegative real number, the nonnegative
number b such that b2  a , denoted by b  a , is the
principal square root of a.
P 487
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 7.1
Radical Expressions
EXAMPLE
Evaluate: (a) 16 (b) 144 25 (c) 144  25.
SOLUTION
(a) 16 
(b) 144 25  169  13
(c) 144  25  12  5  17
The principal square root of a
negative number, -16, is not a
real number.
Simplify the radicand. The
principal square root of 169 is
13.
Take the principal square root
of 144, 12, and of 25, 5, and
then add to get 17.
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 7.1
Radical Expressions
Check Point 1 on page 487
Evaluate:
64
8
- 49
7
16
25
0.0081
Principal Square Root means the
answer is nonnegative
-
Denotes the negative
square root of a number
4
5
0.09
9  16  25  5
Is a grouping symbol
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 7.1
9  16  3  4  7
Radical Expressions
Square Root Function
Because each nonnegative real number, x, has precisely one
principal square root, x , there is a square root function defined
by
f ( x)  x
The domain of this function is [0, ).
Bottom of P 487
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 7.1
Radical Expressions
See Figure 7.1 on page 488
x
f ( x) 
x
(x,y)
0 0
(0,0)
0
f (0) 
1
f (1) 
4
f ( 4) 
4 2
(4,2)
9
f (9) 
9 3
(9,3)
16
f (16) 
16  4
(16,4)
1 1
(1,1)
P 488
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 7.1
Radical Functions
EXAMPLE
For the function, find the indicated function value:
 1
g x    2 x  1; g 4, g 1, g   , g  1.
 2
SOLUTION
g 4    24   1
  9  3
g 1   21  1
  3  1.73
Substitute 4 for x in gx   2x 1.
Simplify the radicand and take the
square root of 9.
Substitute 1 for x in gx   2x 1.
Simplify the radicand and take the
square root of 3.
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 7.1
Radical Functions
CONTINUED
 1
 1
g      2    1
 2
 2
 0 0
g  1   2 1  1
  1 
Substitute -1/2 for x in
gx   2x 1.
Simplify the radicand and take the
square root.
Substitute -1 for x in gx   2x 1.
Simplify the radicand. The principal
square root of a negative number is
not a real number.
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 7.1
Radical Functions
Check Point 2
For the function, find the indicated function value:
f x  12x  20; gx   9  3x
SOLUTION
f 3 
12  3  20
g  5   9  3  (5)
4
 24  4.90
P 488
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 7.1
Radical Functions - Domain
We have seen that the domain of a function f is the largest set of
real numbers for which the value of f(x) is defined.
Because only nonnegative numbers have real square roots, the
domain of a square root function is the set of real numbers for
which the radicand is nonnegative.
In other words, we only use “allowable” x in the domain of the function. Not
allowed for x is any value of x that would cause a negative number under a
square root.
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 7.1
Radical Functions - Domain
EXAMPLE
Find the domain of f x  3x 15.
SOLUTION
The domain is the set of real numbers, x, for which the radicand,
3x – 15, is nonnegative. We set the radicand greater than or
equal to 0 and solve the resulting inequality.
3 x  15  0
3x  15
x5
The domain of f is x | x  5 or 5,  .
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 7.1
Radical Functions
Check Point 3
Find the Domain of
f x  9x  27
SOLUTION
domain f x  : x | x  3or [3, )
P 489
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 7.1
Radical Functions in Application
EXAMPLE
Police use the function f x  20x to estimate the speed of a
car, f (x), in miles per hour, based on the length, x, in feet, of its
skid marks upon sudden braking on a dry asphalt road. Use the
function to solve the following problem.
A motorist is involved in an accident. A police officer measures
the car’s skid marks to be 45 feet long. Estimate the speed at
which the motorist was traveling before braking. If the posted
speed limit is 35 miles per hour and the motorist tells the officer
she was not speeding, should the officer believe her? Explain.
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 7.1
Radical Functions in Application
CONTINUED
SOLUTION
f x  20x
Use the given function.
f x  2045
Substitute 45 for x.
f x  900
Simplify the radicand.
f x   30
Take the square root.
The model indicates that the motorist was traveling at 30 miles
per hour at the time of the sudden braking. Since the posted
speed limit was 35 miles per hour, the officer should believe that
she was not speeding.
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 7.1
Radical Expressions
Simplifying a 2 T
For any real number a,
a2  a .
2
In words, the principal square root of a is the absolute
value of a.
The principal root is the positive root.
P 490
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 7.1
Radical Expressions
EXAMPLE
Simplify each expression: (a) 81x 4
(b)
x 2  14x  49.
SOLUTION
The principal square root of an expression squared is the
absolute value of that expression. In both exercises, it will first
be necessary to express the radicand as an expression that is
squared.
(a) To simplify 81x 4, first write 81x 4 as an expression that is
2
squared: 81x 4  9 x 2  . Then simplify.
81x 
4
9x 
2 2
 9 x 2 or 9 x 2
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 7.1
Radical Expressions
CONTINUED
(b) To simplify x2 14x  49, first write x 2  14x  49 as an
expression that is squared: x2  14x  49  x  72. Then simplify.
x 2  14x  49 
x  72
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 7.1
 x7
Radical Functions
Check Point 5
Simplify each expression:
a.
(7) 2
b.
( x  8) 2
c.
d.
10
49 x
x2  6x  9
7
 x 8
 7x 5
 x 3
P 490
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 7.1
Radical Expressions
Definition of the Cube Root of a Number
The cube root of a real number a is written
3
3
.a
a  b meansthat b3  a.
P 491
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 7.1
Radical Functions
EXAMPLE
For the function, find the indicated function value:
gx  3 2x 1; g13, g 0, g  63.
SOLUTION
g 13  3 213  1
 3 27  3
g 0    20   1
3
  1  1
Substitute 13 for x in gx  3 2x 1.
Simplify the radicand and take the
cube root of 27.
Substitute 0 for x in gx  3 2x 1.
Simplify the radicand and take the
cube root of 1.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 7.1
Radical Functions
CONTINUED
g  63  3 2 63  1
 3  125   5  5
Substitute -63 for x in gx  3 2x 1.
Simplify the radicand and take the
cube root of -125 and then simplify.
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 7.1
Radical Expressions
See Figure 7.2 on page 491
x
f ( x)  3 x
(x,y)
-8
f ( 8)  3  8  2
(-8,-2)
-1
f ( 1)  3  1  1
(-1,-1)
0
f (0)  3 0  0
(0,0)
1
f (1)  3 1  1
(1,1)
8
f (8)  3 8  2
(8,2)
P 491
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 7.1
Radical Functions
Check Point 6
For the function, find the indicated function value:
a. f  x   3 x  6
f 33  3 33  6  3 27
3
b. g  x   3 2 x  2
g  5 
3
2  (5)  2  3  8
 2
P 492
Blitzer, Intermediate Algebra, 5e – Slide #26 Section 7.1
Radical Expressions
Simplifying
3
a3 T
For any real number a,
3
a3  a.
In words, the cube root of any expression is that expression
cubed.
Blitzer, Intermediate Algebra, 5e – Slide #27 Section 7.1
Radical Expressions
EXAMPLE
Simplify: 3 125x3 .
SOLUTION
Begin by expressing the radicand as an expression that is cubed:
3
125x3   5x . Then simplify.
3
 125 x 3  3  5 x   5 x
3
We can check our answer by cubing -5x:
 5x3   53 x3  125x3
By obtaining the original radicand, we know that our
simplification is correct.
Blitzer, Intermediate Algebra, 5e – Slide #28 Section 7.1
Radical Functions
Check Point 7
Simplify:
3
 27x 3
 3x
P 492
Blitzer, Intermediate Algebra, 5e – Slide #29 Section 7.1
Radical Expressions
EXAMPLE
Find the indicated root, or state that the expression is not a real
number:
(a) 5 1 (b) 8 1.
SOLUTION
(a) 5 1  1 because 15  11111  1. An odd root
of a negative real number is always negative.
(b) 8 1 is not a real number because the index, 8, is even and the
radicand, -1, is negative. No real number can be raised to the
eighth power to give a negative result such as -1. Real numbers to
even powers can only result in nonnegative numbers.
Blitzer, Intermediate Algebra, 5e – Slide #30 Section 7.1
Radical Expressions
Simplifying
n
an T
For any real number a,
1) If n is even,
n
an  a .
2) If n is odd,
n
an  a .
Blitzer, Intermediate Algebra, 5e – Slide #31 Section 7.1
Radical Expressions
EXAMPLE
Simplify: (a)
4
x  54
(b) 5  32x  2 .
5
SOLUTION
Each expression involves the nth root of a radicand raised to the
nth power. Thus, each radical expression can be simplified.
Absolute value bars are necessary in part (a) because the index,
n, is even.
(a) 4 x  5  x  5
4
(b) 5  32 x  2   5  2   x  2 
5
5
5
n
a n  a if n is even.
n
a n  a if n is odd.
 5  2 x  2 
5
 2 x  2 
Blitzer, Intermediate Algebra, 5e – Slide #32 Section 7.1
Radical Functions
Check Point 8
Find the indicated root, or state not real:
a. 4 16
b. - 4 16
c.
4
d.
 16
5
-1
2
 2
not real
 1
P 493
Blitzer, Intermediate Algebra, 5e – Slide #33 Section 7.1
Radical Functions
Check Point 9
Simply:
a.
4
x  64
 x6
b.
5
3x - 25
 3x  2
c.
6
(8) 6
8
P 494
Blitzer, Intermediate Algebra, 5e – Slide #34 Section 7.1
DONE