2: 3 x 3 Systems of Equations
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11.2 • 3 ϫ 3 Systems of Equations 457
1. There is one ordered triple that satisfies all three equations. The three planes have a
common point of intersection as indicated in Figure 11.9.
Figure 11.9
2. There are infinitely many ordered triples in the solution set, all of which are coordinates of points on a line common to the planes. This can happen when three planes
have a common line of intersection, as in Figure 11.10(a), or when two of the planes
coincide, and the third plane intersects them, as in Figure 11.10(b).
(a)
(b)
Figure 11.10
3. There are infinitely many ordered triples in the solution set, all of which are coordinates of points on a plane. This happens when the three planes coincide, as illustrated in Figure 11.11.
Figure 11.11
4. The solution set is empty; it is л. This can happen in various ways, as you can see in
Figure 11.12. Notice that in each situation there are no points common to all three planes.
(a) Three parallel planes
(b) Two planes coincide
and the third one is
parallel to the
coinciding planes.
(c) Two planes are
parallel and the third
intersects them in
parallel lines.
(d) No two planes are
parallel, but two of
them intersect in a
line that is parallel
to the third plane.
Figure 11.12
458
Chapter 11 • Additional Topics
Now that you know what possibilities exist, we can consider finding the solution sets for
some systems. Our approach will be the eliminationbyaddition method, whereby systems
are replaced with equivalent systems until we get a system that allows us to easily determine
the solution set. We will start with an example that we can solve without changing to another equivalent system.
Classroom Example
Solve the system:
2x ϩ 3y ϩ z ϭ 4
4y ϩ 3z ϭ 5
Q
P
3z ϭ 9
EXAMPLE 1
Solve the system:
4x ϩ 2y Ϫ z ϭ 5
3y ϩ z ϭ 1
Q
P
2z ϭ 10
(1)
(2)
(3)
Solution
From equation (3), we can find the value of z.
2z ϭ 10
zϭ5
Now we can substitute 5 for z in equation (2).
3y ϩ z ϭ
3y ϩ 5 ϭ
3y ϭ
yϭ
1
1
6
2
Finally, we can substitute Ϫ2 for y and 5 for z in equation (1).
4x ϩ 2y Ϫ z ϭ 5
4x ϩ 21 22 Ϫ 5 ϭ 5
4x Ϫ 9 ϭ 5
4x ϭ 4
xϭ1
The solution set is 511, 2, 526.
Notice the format of the equations in the system in Example 1. The first equation contains all
three variables, the second equation has only two variables, and the third equation has only
one variable. This allowed us to solve the third equation and then use “back substitution” to
find the values of the other variables. Let’s consider another example in which we replace one
equation to make an equivalent system.
Classroom Example
Solve the system:
x Ϫ 2y ϩ 3z ϭ 5
y Ϫ 5z ϭ 3
Q
P
3y Ϫ 7z ϭ 1
EXAMPLE 2
Solve the system:
2x ϩ 4y Ϫ 5z ϭ 8
y ϩ 4z ϭ 7
Q
P
5y ϩ 3z ϭ 1
(1)
(2)
(3)
Solution
In order to achieve the same format as in Example 1, we will need to eliminate the term with
the y variable in equation (3). Using the concept of elimination from Section 8.4, we can
replace equation (3) with an equivalent equation we form by multiplying equation (2) by Ϫ5
and then adding that result to equation (3). The equivalent system is
P
2x ϩ 4y Ϫ
yϩ
5z ϭ 8
4z ϭ
7
Q
17z ϭ  34
(4)
(5)
(6)
11.2 • 3 ϫ 3 Systems of Equations
459
From equation (6) we can find the value of z.
17z ϭ  34
zϭ2
Now we can substitute 2 for z in equation (5).
y ϩ 4z ϭ 7
y ϩ 4122 ϭ 7
y ϭ 1
Finally, we can substitute Ϫ1 for y and 2 for z in equation (4).
2x ϩ 4y Ϫ 5z ϭ 8
2x ϩ 41 12 Ϫ 5122 ϭ 8
2x Ϫ 4 Ϫ 10 ϭ 8
2x Ϫ 14 ϭ 8
2x ϭ 6
xϭ3
The solution set is 513, 1, 226.
Now let’s consider some examples in which we replace more than one equation to make
an equivalent system.
Classroom Example
Solve the system:
2x Ϫ y ϩ z ϭ 1
x ϩ 3y Ϫ 2z ϭ 6
P 4x Ϫ y ϩ 3z ϭ 7 Q
EXAMPLE 3
Solve the system:
x ϩ 2y Ϫ 3z ϭ 1
3x Ϫ y ϩ 2z ϭ 13
Q
P
2x ϩ 3y Ϫ 5z ϭ 4
(1)
(2)
(3)
Solution
We start by picking a pair of equations to form a new equation by eliminating a variable. We
will use equations (1) and (2) to form a new equation while eliminating the x variable. We can
replace equation (2) with an equation formed by multiplying equation (1) by Ϫ3 and adding
the result to equation (2). The equivalent system is
x ϩ 2y Ϫ 3z ϭ 1
Ϫ7y ϩ 11z ϭ 10
P
Q
2x ϩ 3y Ϫ 5z ϭ 4
(4)
(5)
(6)
Now we take equation (4) and equation (6) and eliminate the same variable, x. We can replace
equation (6) with a new equation formed by multiplying equation (4) by Ϫ2 and adding the
result to equation (6). The equivalent system is
P
xϩ
2y Ϫ 3z ϭ  1
7y ϩ 11z ϭ  10
Q
y ϩ z ϭ  2
(7)
(8)
(9)
Now we take equations (8) and (9) and form a new equation by eliminating a variable. Either
y or z can be eliminated. For this example we will eliminate y. We can replace equation (8)
with a new equation formed by multiplying equation (9) by Ϫ7 and adding the result to equation (8). The equivalent system is
x ϩ 2y Ϫ 3z ϭ 1
4z ϭ 4
P
Q
y ϩ z ϭ  2
(10)
(11)
(12)
460
Chapter 11 • Additional Topics
From equation (11), we can find the value of z.
4z ϭ 4
zϭ1
Now we substitute 1 for z in equation (12) and determine the value of y.
y ϩ z ϭ 2
y ϩ 1 ϭ 2
y ϭ 3
yϭ3
Finally, we can substitute 3 for y and 1 for z in equation (10).
x ϩ 2y Ϫ 3z ϭ
x ϩ 2132 Ϫ 3112 ϭ
xϩ6Ϫ3ϭ
xϩ3ϭ
xϭ
1
1
1
1
4
The solution set is 51 4, 3, 126.
Classroom Example
Solve the system:
3x ϩ 2y ϩ z ϭ 5
2x Ϫ 4y ϩ 2z ϭ 18
P 5x Ϫ 6y Ϫ 3z ϭ 13 Q
EXAMPLE 4
Solve the system:
2x ϩ 3y Ϫ z ϭ 8
5x ϩ 2y Ϫ 3z ϭ 21
Q
P
3x Ϫ 4y ϩ 2z ϭ 5
(1)
(2)
(3)
Solution
Studying the coefficients in the system indicates that eliminating the z terms from equations (2) and (3) would be easy to do. We can replace equation (2) with an equation
formed by multiplying equation (1) by Ϫ3 and adding the result to equation (2). The
equivalent system is
2x ϩ 3y Ϫ z ϭ 8
x Ϫ 7y
ϭ 3
Q
P
3x Ϫ 4y ϩ 2z ϭ 5
(4)
(5)
(6)
2x ϩ 3y Ϫ z ϭ 8
x Ϫ 7y
ϭ 3
Q
P
7x ϩ 2y
ϭ 21
(7)
(8)
(9)
2x ϩ 3y Ϫ z ϭ 8
x Ϫ 7y
ϭ 3
P
Q
 47y
ϭ 0
(10)
(11)
(12)
Now we replace equation (6) with an equation formed by multiplying equation (4) by 2 and
adding the result to equation (6). The equivalent system is
Now we can eliminate the x term from equation (9). We replace equation (9) with an equation formed by multiplying equation (8) by 7 and adding the result to equation (9). The equivalent system is
From equation (12), we can determine the value of y.
47y ϭ 0
yϭ0
11.2 • 3 ϫ 3 Systems of Equations
461
Now we can substitute 0 for y in equation (11) and find the value of x.
x Ϫ 7y ϭ  3
x Ϫ 7102 ϭ  3
x ϭ  3
xϭ3
Finally, we can substitute 3 for x and 0 for y in equation (10).
2x ϩ 3y Ϫ z ϭ 8
2132 ϩ 3102 Ϫ z ϭ 8
6Ϫzϭ8
z ϭ 2
z ϭ 2
The solution set is 513, 0, 226.
Classroom Example
Solve the system:
2x Ϫ y ϩ 3z ϭ 5
x ϩ 4y ϩ 3z ϭ 3
P 6x Ϫ 3y ϩ 9z ϭ 11 Q
EXAMPLE 5
Solve the system:
x ϩ 3y Ϫ 2z ϭ 3
3x Ϫ 4y Ϫ z ϭ 4
P
Q
2x ϩ 6y Ϫ 4z ϭ 9
(1)
(2)
(3)
Solution
Studying the coefficients indicates that it would be easy to eliminate the x terms from equations (2) and (3). We can replace equation (2) with an equation formed by multiplying equation (1) by Ϫ3 and adding the result to equation (2). Likewise, we can replace equation (3)
with an equation formed by multiplying equation (1) by Ϫ2 and adding the result to equation
(3). The equivalent system is
xϩ
3y Ϫ 2z ϭ 3
13y ϩ 5z ϭ  5
P
Q
0ϩ
0ϩ0 ϭ 3
(4)
(5)
(6)
The false statement 0 ϭ 3 in equation (6) indicates that the system is inconsistent, and therefore
the solution set is л. (If you were to graph this system, equations (1) and (3) would produce
parallel planes, which is the situation depicted in Figure 11.12(c).)
Classroom Example
Solve the system:
x ϩ 2y ϩ z ϭ 4
2x ϩ y ϩ z ϭ 5
P 5x Ϫ 2y ϩ z ϭ 8 Q
EXAMPLE 6
Solve the system:
xϩyϩ zϭ 6
3x ϩ y Ϫ z ϭ 2
P
Q
5x ϩ y Ϫ 3z ϭ 2
(1)
(2)
(3)
Solution
Studying the coefficients indicates that it would be easy to eliminate the y terms from
equations (2) and (3). We can replace equation (2) with an equation formed by multiplying
equation (1) by Ϫ1 and adding the result to equation (2). Likewise, we can replace equation
(3) with an equation formed by multiplying equation (1) by Ϫ1 and adding the result to equation (3). The equivalent system is
xϩyϩ zϭ 6
2x
Ϫ 2z ϭ 4
P
Q
4x
Ϫ 4z ϭ 8
(4)
(5)
(6)
462
Chapter 11 • Additional Topics
Now we replace equation (6) with an equation formed by multiplying equation (5) by Ϫ2 and
adding the result to equation (6). The equivalent system is
xϩyϩ zϭ 6
2x
Ϫ 2z ϭ 4
P
Q
0ϩ0 ϭ 0
(7)
(8)
(9)
The true numerical statement 0 ϩ 0 ϭ 0 in equation (9) indicates that the system has infinitely
many solutions.
Now we will use the techniques we have presented to solve a geometric problem.
Classroom Example
In a certain triangle, the measure
of Є A is 5° more than four times
the measure of ЄB. The difference
in the measures of Є A and ЄC is
10° less than the measure of ЄB.
Find the measures of all three
angles.
EXAMPLE 7
In a certain triangle, the measure of ЄA is 5˚ more than twice the measure of ЄB. The sum of
the measures of ЄB and ЄC is 10° more than the measure of ЄA. Find the measures of all
three angles.
Solution
We can solve this problem by setting up a system of three linear equations in three variables.
We let
x ϭ measure of ЄA
y ϭ measure of ЄB
z ϭ measure of ЄC
Knowing that the sum of the measures of the angles in a triangle is 180° gives us the equation x ϩ y ϩ z ϭ 180. The information “the measure of ЄA is 5° more than twice the measure
of ЄB” gives us the equation x ϭ 2y ϩ 5 or an equivalent form x Ϫ 2y ϭ 5. The information
“the sum of the measures of ЄB and ЄC is 10° more than the measure of ЄA” gives us the
equation y ϩ z ϭ x ϩ 10 or an equivalent form x Ϫ y Ϫ z ϭ Ϫ10. Putting the three equations
together, we get the system of equations
x ϩ y ϩ z ϭ 180
x Ϫ 2y
ϭ 5
Q
P
x Ϫ y Ϫ z ϭ10
(1)
(2)
(3)
To solve the system, we first replace equation (3) with an equation formed by adding equation (1) and equation (3). The equivalent system is
x ϩ y ϩ z ϭ 180
x Ϫ 2y
ϭ 5
P
Q
2x
ϭ 170
From equation (6), we can determine that x ϭ 85.
Now we can substitute 85 for x in equation (5) and find the value of y.
x Ϫ 2y ϭ 5
85 Ϫ 2y ϭ 5
2y ϭ 80
y ϭ 40
(4)
(5)
(6)
11.2 • 3 ϫ 3 Systems of Equations
463
Finally, we can substitute 40 for y and 85 for x in equation (4) and find the value of z.
x ϩ y ϩ z ϭ 180
85 ϩ 40 ϩ z ϭ 180
125 ϩ z ϭ 180
z ϭ 55
The measures of the angles are ЄA ϭ 85°, ЄB ϭ 40°, and ЄC ϭ 55°.
Concept Quiz 11.2
For Problems 1–10, answer true or false.
1.
2.
3.
4.
The graph of a linear equation in three variables is a line.
A system of three linear equations in three variables produces three planes when graphed.
Three planes can be related by intersecting in exactly two points.
One way three planes can be related is if two of the planes are parallel, and the third
plane intersects them in parallel lines.
5. A system of three linear equations in three variables always has an infinite number of
solutions.
6. A system of three linear equations in three variables can have one ordered triple as a
solution.
2x Ϫ y ϩ 3z ϭ 4
7. The solution set of the system
y Ϫ z ϭ 12 is {(5, 15, 3)}.
Q
P
2z ϭ 6
8. The solution set of the system
xϪyϩ zϭ4
x Ϫ y Ϫ z ϭ 6 is {(3, 1, 2)}.
Q
P
3y Ϫ 2z ϭ 9
x ϩ y Ϫ 2z ϭ 8
9. The solution set of the system 2x Ϫ y ϩ z ϭ 7 is {(1, 1, 4)}.
Q
P
3x ϩ y ϩ z ϭ 9
x ϩ 2y Ϫ z ϭ
0
10. The solution set of the system 3x Ϫ y Ϫ 2z ϭ 20 is {(0, 4, 8)}.
Q
P
2x ϩ y ϩ z ϭ 12
Problem Set 11.2
For Problems 1–16, solve each system of equations.
(Objective 1)
1.
2.
3.
3x ϩ 6y ϩ  2z ϭ  6
3x ϩ 6y ϩ  5z ϭ 4
P
3x ϩ 6y ϩ  4z ϭ  8 Q
2x Ϫ 3y ϩ 3z ϭ  20
2x Ϫ 2y Ϫ 5z ϭ 98
P
2x Ϫ 3y ϩ 3z ϭ  12 Q
x ϩ 2y Ϫ z ϭ 11
x ϩ 2 y ϩ 2z ϭ 11
P x ϩ 2y Ϫ z ϭ 12 Q
4.
5.
6.
3x ϩ y Ϫ 2z ϭ  0
3x ϩ y Ϫ 3z ϭ 5
P 2x Ϫ y Ϫ 5z ϭ  1 Q
4x ϩ 3y Ϫ 2z ϭ 29
2x ϩ 3 y2z Ϫ ϭ 27
P 3x Ϫ 2y Ϫ 2z ϭ 21 Q
2x Ϫ 5y ϩ 3z ϭ 43
3x Ϫ 5 y ϩ 3z ϭ 25
P 2x ϩ 3y ϩ 3z ϭ  5 Q
Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
464
Chapter 11 • Additional Topics
2x ϩ 2y Ϫ 3z ϭ  11
7. 2x Ϫ 3 y ϩ 2z ϭ 13
P 4x ϩ 3y ϩ 2z ϭ  1 6 Q
18. One binder, 2 reams of paper, and 5 spiral notebooks
cost $14.82. Three binders, 1 ream of paper, and 4 spiral notebooks cost $14.32. Two binders, 3 reams of
paper, and 3 spiral notebooks cost $19.82. Find the cost
for each item.
8.
19. In a certain triangle, the measure of ЄA is five times the
measure of ЄB. The sum of the measures of ЄB and ЄC
is 60˚ less than the measure of ЄA. Find the measure of
each angle.
9.
10.
11.
12.
13.
14.
15.
16.
2x Ϫ 2y ϩ 2z ϭ 10
3x ϩ 2 y ϩ 2z ϭ 15
P 2x ϩ 3y Ϫ 3z ϭ 19 Q
4x Ϫ 3y ϩ 3z ϭ 14
2x ϩ 3 y Ϫ 3z ϭ 16
P 3x Ϫ 4y ϩ 2z ϭ 19 Q
5x ϩ 2y ϩ 2z ϭ  13
2x Ϫ 3y ϩ 2z ϭ 15
P 2x Ϫ 2y Ϫ 3z ϭ  10 Q
2x ϩ 2y ϩ 4z ϭ  15
5x Ϫ 2y ϩ 2 z ϭ 10
P 3x ϩ 3y Ϫ 2z ϭ  14 Q
4x Ϫ 2y ϩ 2z ϭ  1
3x ϩ 2y Ϫ 2 z ϭ 11
P 2x ϩ 3y ϩ 4z ϭ  5 Q
3x ϩ 3y Ϫ 4z ϭ 11
3x Ϫ 3 y ϩ 2z ϭ 15
P 2x ϩ 5y Ϫ 2z ϭ 18 Q
4x Ϫ 2y ϩ 3z ϭ 13
4x ϩ 2 y Ϫ 2z ϭ 8
P 2x ϩ 3y ϩ 2z ϭ  3 Q
3x ϩ 3y Ϫ 2z ϭ  3
2x Ϫ 3y ϩ 4z ϭ 2
P
Q
4x ϩ 3y ϩ 4z ϭ  6
2x ϩ 2y ϩ 3z ϭ 11
2x Ϫ 2y ϩ 4z ϭ 1 0
P 3x ϩ 2y ϩ 2z ϭ 11 Q
For Problems 17– 26, solve each problem by setting up
and solving a system of three linear equations in three
variables. (Objective 2)
17. Brooks has 20 coins consisting of quarters, dimes, and
nickels worth $3.40. The sum of the number of dimes
and nickels is equal to the number of quarters. How
many coins of each kind are there?
20. Shannon purchased a skirt, blouse, and sweater for $72.
The cost of the skirt and sweater was $2 more than six
times the cost of the blouse. The skirt cost twice the sum
of the costs of the blouse and sweater. Find the cost of
each item.
21. The wages for a crew consisting of a plumber, an
apprentice, and a laborer are $80 an hour. The plumber
earns $20 an hour more than the sum of the wages of the
apprentice and the laborer. The plumber earns five times
as much as the laborer. Find the hourly wage of each.
22. Martha has 24 bills consisting of $1, $5, and $20 bills
worth $150. The number of $20 bills is twice the number of $5 bills. How many bills of each kind are there?
23. Two pounds of peaches, 1 pound of cherries, and 3
pounds of pears cost $5.64. One pound of peaches, 2
pounds of cherries, and 2 pounds of pears cost $4.65.
Two pounds of peaches, 4 pounds of cherries, and 1
pound of pears cost $7.23. Find the price per pound for
each item.
24. In a certain triangle, the measure of ЄC is 40° more than
the sum of the measures of ЄA and ЄB. The measure of
ЄA is 20° less than the measure of ∠ B. Find the measure
of each angle.
25. Mike bought a motorcycle helmet, jacket, and gloves for
$650. The jacket costs $100 more than the helmet. The
cost of the helmet and gloves together was $50 less than
the cost of the jacket. How much did each item cost?
26. A catering group that has a chef, a salad maker, and a
server costs the customer $70 per hour. The salad maker
costs $5 per hour more than the server. The chef costs the
same as the salad maker and the server cost together.
Find the cost per hour of each.
Thoughts Into Words
27. Give a stepbystep description of how to solve this system of equations.
2x ϩ 3y Ϫ 6z ϭ  13
2x ϩ 4y ϩ 3z ϭ 1 2
P
2x ϩ 3y ϩ 6z ϭ  12 Q
28. Describe how you would solve this system of equations.
2x ϩ 3y Ϫ z ϭ 7
2x ϩ 2y Ϫ z ϭ 4
P 3x Ϫ 4y Ϫ z ϭ 2 Q
11.3 • Fractional Exponents
465
Further Investigations
For Problems 29 – 32, solve each system of equations, indicating whether the solution is л or contains infinitely many
solutions.
29.
30.
2x ϩ 2y Ϫ 3z ϭ 1
2x Ϫ 2 y ϩ 2z ϭ 3
P 3x ϩ 2y Ϫ 2z ϭ 4 Q
31.
32.
2x ϩ 3y ϩ 2z ϭ  7
2x Ϫ 4y ϩ 2z ϭ 3
P
Q
3x Ϫ 3y ϩ 3z ϭ  5
Answers to the Concept Quiz
1. False
2. True
3. False
4. True
11.3
5. False
3x Ϫ 3y ϩ 2z ϭ  1
2x ϩ 3y ϩ 2 z ϭ  8
P 8x ϩ 2y ϩ 5z ϭ  4 Q
2x ϩ 2y Ϫ 2z ϭ  16
3x Ϫ 2y ϩ 4z ϭ 10
P 5x ϩ 3y ϩ 2z ϭ  12 Q
6. True
7. True
8. False
9. False
10. True
Fractional Exponents
OBJECTIVES
1
Review the concepts of square root and cube root
2
Use the definition of rational exponents to simplify numerical expressions
3
Use fractional exponents to simplify algebraic expressions
At the beginning of Chapter 9, we defined and discussed the concept of square root. For your
convenience and for the sake of continuity, we will repeat some of that material at this time.
To square a number means to raise it to the second power—that is, to use the number as a
factor twice.
32 ϭ 3 и 3 ϭ 9
72 ϭ 7 и 7 ϭ 49
1 322 ϭ 1 321  32 ϭ 9
1 722 ϭ 1 721  72 ϭ 49
A square root of a number is one of its two equal factors. Thus, 3 is a square root of 9
because 3 и 3 ϭ 9. Likewise, Ϫ3 is also a square root of 9 because (Ϫ3)(Ϫ3) ϭ 9. The concept of square root is defined as follows.
Definition 11.1
a is a square root of b if a2 ϭ b.
The following generalizations are a direct consequence of Definition 11.1:
1. Every positive real number has two square roots; one is positive and the other is negative. They are opposites of each other.
2. Negative real numbers have no real number square roots. (This follows from
Definition 11.1, because any nonzero real number is positive when squared.)
3. The square root of 0 is 0.
Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
466
Chapter 11 • Additional Topics
We use the symbol 1 , called a radical sign, to indicate the nonnegative square root.
Consider the next examples.
249 ϭ 7
249 ϭ 7
20 ϭ 0
24
24
249 indicates the nonnegative or principal square root of 49
 249 indicates the negative square root of 49
Zero has only one square root
2  4 is not a real number
 2 4 is not a real number
To cube a number means to raise it to the third power—that is, to use the number as a
factor three times.
23 ϭ 2 и 2 и 2 ϭ 8
43 ϭ 4 и 4 и 4 ϭ 64
8
2 3 2 2 2
a b ϭ и и ϭ
3
3 3 3
27
1 223 ϭ 1 221  221 22 ϭ 8
A cube root of a number is one of its three equal factors. Thus Ϫ2 is a cube root of
Ϫ8 because (Ϫ2)(Ϫ2)(Ϫ2) ϭ Ϫ8. In general, the concept of cube root can be defined as
follows:
Definition 11.2
a is a cube root of b if a3 ϭ b.
The following generalizations are a direct consequence of 11.2:
1. Every positive real number has one positive real number cube root.
2. Every negative real number has one negative real number cube root.
3. The cube root of 0 is 0.
(Technically, every nonzero real number has three cube roots, but only one of them is a real
number. The other two cube roots are imaginary numbers.)
3
The symbol 1 is used to designate the real number cube root. Thus we can write
1
1
ϭ
A 27
3
3
3
28 ϭ 2
3
2 8 ϭ 2
1
1
ϭ A 27
3
3

We can extend the concept of root to fourth roots, fifth roots, sixth roots, and in general,
nth roots. We can make these generalizations: If n is an even positive integer, then the following statements are true.
1. Every positive real number has exactly two real nth roots, one positive and one negative.
n
For example, the real fourth roots of 16 are 2 and Ϫ2. We use the symbol 2
to designate the positive root. Thus we write 216 ϭ 2.
2. Negative real numbers do not have real nth roots. For example, there are no real fourth
roots of Ϫ16.
If n is an odd positive integer greater than 1, then the following statements are true.
1. Every real number has exactly one real nth root, and we designate this root by the
n
symbol 2 .
2. The real nth root of a positive number is positive. For example, the fifth root
of 32 is 2, and we write 232 ϭ 2.
3. The nth root of a negative number is negative. For example, the fifth root of Ϫ32
is Ϫ2, and we write 2 32 ϭ  2.
11.3 • Fractional Exponents
467
n
To complete our terminology, we call the n in the radical 2
b the index of the radical. If
2
n ϭ 2, we commonly write 2b instead of 2b.
Merging of Exponents and Roots
In Section 5.6 we used the basic properties of positive integer exponents to motivate a
definition for the use of zero and negative integers as exponents. Now we can use the properties of integer exponents to motivate definitions for the use of all rational numbers as exponents. This material is commonly referred to as “fractional exponents.”
Consider the following comparisons.
If 1bn2m ϭ bmn is to hold when n equals
From the meaning of root,
we know that
a rational number of the form 1 , where p
p
is a positive integer greater than 1, then
15222 ϭ 52122 ϭ 51 ϭ 5
12522 ϭ 5
1
12 823 ϭ 8
1
18323 ϭ 83132 ϭ 81 ϭ 8
3
1
122124 ϭ 21
1
121424 ϭ 214142 ϭ 211 ϭ 21
1
1
The following definition is motivated by such examples.
Definition 11.3
1
n
n
If b is a real number, n is a positive integer greater than 1, and 2
b exists, then bn ϭ 2
b
1
Definition 11.3 states that bn means the nth root of b. The following examples illustrate this
definition.
1
1
252 ϭ 225 ϭ 5
164 ϭ 216 ϭ 2
1
1
3
3
1
3
36 2
36
6
a b ϭ
ϭ
49
B 49
7
8 ϭ 28 ϭ 2
1
( 27)3 ϭ 2 27 ϭ  3
(32)5 ϭ 2  32 ϭ  2
Now the next definition provides the basis for the use of all rational numbers as exponents.
Definition 11.4
m
If is a rational number, where n is a positive integer greater than 1 and b is a
n
n
real number such that 2b exists, then
m
n
n
b n ϭ 2bm ϭ (2b)m
m
is in reduced form
n
n m
n
Whether we use the form 1
b or (1
b)m for computational purposes depends somewhat on
the magnitude of the problem. We use both forms on two problems to illustrate this point.
2
3
83 ϭ 282
3
ϭ 264
ϭ4
2
3
or
273 ϭ 2272 or
3
ϭ 2729
ϭ9
83 ϭ 12822
ϭ (2)2
ϭ4
2
3
3
273 ϭ 12
2722
ϭ 32
ϭ9
2