305
Radicals and Radical Functions
So far we have discussed polynomial and rational expressions and functions. In this chapter, we study algebraic expressions that contain radicals. For example, 3 + β2,
βπ₯π₯3 β 1, or 1β5π₯π₯β1
. Such expressions are called radical expressions. Familiarity with
radical expressions is essential when solving a wide variety of problems. For instance, in algebra, some polynomial or rational equations have radical solutions that need to be simplified. In geometry, due to the frequent use of the Pythagorean equation, ππ2 + ππ2 =ππ2, the exact distances are often radical expressions. In sciences, many formulas involve radicals.
We begin the study of radical expressions with defining radicals of various degrees and discussing their properties. Then, we show how to simplify radicals and radical expressions, and introduce operations on radical expressions. Finally, we study the methods of solving radical equations. In addition, similarly as in earlier chapters where we looked at the related polynomial and rational functions, we will also define and look at properties of radical functions.
RD.1 Radical Expressions, Functions, and Graphs
Roots and Radicals
The operation of taking a square root of a number is the reverse operation of squaring a number. For example, a square root of 25 is 5 because raising 5 to the second power gives us 25.
Note: Observe that raising β5 to the second power also gives us 25. So, the square root of 25 could have two answers, 5 or β5. To avoid this duality, we choose the nonnegative value, called the principal square root, for the value of a square root of a number.
The operation of taking a square root is denoted by the symbol β . So, we have
β25 = 5, β0 = 0, β1 = 1, β9 = 3, etc.
What about ββ4 = ? Is there a number such that when it is squared, it gives us β4 ?
Since the square of any real number is nonnegative, the square root of a negative number is not a real number. So, when working in the set of real numbers, we can conclude that
οΏ½ππππππππππππππππ = ππππππππππππππππ, β0 = 0, and οΏ½ππππππππππππππππ = π«π«π«π«π«π«
The operation of taking a cube root of a number is the reverse operation of cubing a number. For example, a cube root of 8 is 2 because raising 2 to the third power gives us 8.
This operation is denoted by the symbol β 3 . So, we have
β83 = 2, β03 = 0, β13 = 1, β273 = 3, etc.
does not
exist
306
Note: Observe that ββ83 exists and is equal to β2. This is because (β2)3 = β8. Generally, a cube root can be applied to any real number and the sign of the resulting value is the same as the sign of the original number.
Thus, we have
οΏ½ππππππππππππππππππ = ππππππππππππππππ, β03 = 0, and οΏ½ππππππππππππππππππ = ππππππππππππππππ
The square or cube roots are special cases of n-th degree radicals.
Definition 1.1 The n-th degree radical of a number ππ is a number ππ such that ππππ = ππ.
Notation:
βππππ = ππ β ππππ = ππ
For example, β164 = 2 because 24 = 16, ββ325 = β2 because (β2)5 = 32, β0.0273 = 0.3 because (0.3)3 = 0.027.
Notice: A square root is a second degree radical, customarily denoted by β rather than β 2 .
Evaluating Radicals
Evaluate each radical, if possible.
a. β0.64 b. β1253 c. ββ164 d. οΏ½β 132
5
a. Since 0.64 = (0.8)2, then β0.64 = 0.8.
Advice: To become fluent in evaluating square roots, it is helpful to be familiar with the following perfect square numbers:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, β¦, 400, β¦, 625, β¦
b. β1253 = 5 as 53 = 125
Solution
take half of the decimal places
radical
radicand
degree (index, order)
radical symbol
307
Advice: To become fluent in evaluating cube roots, it is helpful to be familiar with the following cubic numbers:
1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, β¦
c. ββ164 is not a real number as there is no real number which raised to the 4-th power becomes negative.
d. οΏ½β 132
5 = β12 as οΏ½β 1
2οΏ½5
= β 15
25= β 1
32
Note: Observe that ββ15
β325 = β12
, so οΏ½β 132
5= ββ15
β325 .
Generally, to take a radical of a quotient, οΏ½ππππ
ππ , it is the same as to take the quotient
of radicals, βππππ
βππππ .
Evaluating Radical Expressions
Evaluate each radical expression.
a. β β121 b. β ββ643 c. οΏ½(β3)44 d. οΏ½(β6)33
a. β β121 = β11 b. β ββ643 = β(β4) = 4 c. οΏ½(β3)44 = β814 = 3
Note: If ππ is even, then βππππππ = οΏ½ ππ, ππππ ππ β₯ 0βππ, ππππ ππ < 0 = |ππ|.
For example, β72 = 7 and οΏ½(β7)2 = 7.
d. οΏ½(β6)33 = ββ2163 = β6
Note: If ππ is odd, then βππππππ = ππ. For example, β533 = 5 but οΏ½(β5)33 = β5.
Solution
the result has the same sign
the result is positive
308
An even degree radical is nonnegative, so we must use the absolute value operator.
An odd degree radical assumes the sign of the radicand, so we do not
apply the absolute value operator.
Summary of Properties of ππ-th Degree Radicals
If ππ is EVEN, then
οΏ½ππππππππππππππππππ = ππππππππππππππππ, οΏ½ππππππππππππππππππ = π«π«π«π«π«π«, and βππππππ = |ππ| If ππ is ODD, then
οΏ½ππππππππππππππππππ = ππππππππππππππππ, οΏ½ππππππππππππππππππ = ππππππππππππππππ, and βππππππ = ππ
For any natural ππ β₯ 0, βππππ = ππ and βππππ = ππ.
Simplifying Radical Expressions Using Absolute Value Where Appropriate Simplify each radical, assuming that all variables represent any real number.
a. οΏ½9π₯π₯2π¦π¦4 b. οΏ½β27π¦π¦33 c. βππ204 d. βοΏ½(ππ β 1)44
a. οΏ½9π₯π₯2π¦π¦4 = οΏ½(3π₯π₯π¦π¦2)2 = |3π₯π₯π¦π¦2| = ππ|ππ|ππππ
Recall: As discussed in section L6, the absolute value operator has the following properties:
Note: |π¦π¦2| = π¦π¦2 as π¦π¦2 is already nonnegative.
b. οΏ½β27π¦π¦33 = οΏ½(β3π¦π¦)33 = βππππ
c. βππ204 = οΏ½(ππ5)44 = |ππ5| = |ππ|ππ
Note: To simplify an expression with an absolute value, we keep the absolute value operator as close as possible to the variable(s).
d. βοΏ½(ππ β 1)44 = β|ππβ ππ|
Solution
|ππππ| = |ππ||ππ|
οΏ½πππποΏ½ =
|ππ||ππ|
309
Radical Functions
Since each nonnegative real number π₯π₯ has exactly one principal square root, we can define the square root function, ππ(ππ) = βππ. The domain π·π·ππ of this function is the set of nonnegative real numbers, [ππ,β), and so is its range (as indicated in Figure 1).
To graph the square root function, we create a table of values. The easiest π₯π₯-values for calculation of the corresponding π¦π¦-values are the perfect square numbers. However, sometimes we want to use additional π₯π₯-values that are not perfect squares. Since a square root of such a number, for example β2, β3, β6, etc., is an irrational number, we approximate these values using a calculator.
For example, to approximate β6, we use the sequence of keying: β 6 πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ or
6 ^ ( 1 / 2 ) πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ . This is because a square root operator works the
same way as the exponent of 12.
Note: When graphing an even degree radical function, it is essential that we find its domain first. The end-point of the domain indicates the starting point of the graph, often called the vertex.
For example, since the domain of ππ(π₯π₯) = βπ₯π₯ is [0,β), the graph starts from the point οΏ½0,ππ(0)οΏ½ = (0,0), as in Figure 1.
Since the cube root can be evaluated for any real number, the domain π·π·ππ of the related cube root function, ππ(ππ) = βππππ , is the set of all real numbers, β. The range can be observed in the graph (see Figure 2) or by inspecting the expression βπ₯π₯3 . It is also β. To graph the cube root function, we create a table of values. The easiest π₯π₯-values for calculation of the corresponding π¦π¦-values are the perfect cube numbers. As before, sometimes we might need to estimate additional π₯π₯-values. For example, to approximate β63 , we use the sequence of keying: β 3 6 πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ or 6 ^ ( 1 / 3 ) πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ .
ππ ππ ππ 0 ππππ 1
2
ππ 1 ππ 2 ππ β6 β 2.4
ππ ππ βππ β2 βππ ββ63 β β1.8 βππ β1 β ππ
ππ β 1
2
ππ 0 ππππ 1
2
ππ 1 ππ β63 β 1.8 ππ 2
Figure 1
ππ(ππ) = βππ
π₯π₯
1
1 domain
rang
e
ππ(ππ) = βππππ
π₯π₯
1
1 domain
rang
e
Figure 2
310
or
or
Finding a Calculator Approximations of Roots
Use a calculator to approximate the given root up to three decimal places. a. β3 b. β53 c. β1005 a. β3 β 1.732 5 ^ ( 1 / 3 ) πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ
b. β53 β 1.710 πππππΈπΈππ 4 5 πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ
c. β1005 β2.512 100 ^ ( 1 / 5 ) πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ
5 πππππΈπΈππ 5 100 πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ
Finding the Best Integer Approximation of a Square Root
Without the use of a calculator, determine the best integer approximation of the given root. a. β68 b. β140
a. Observe that 68 lies between the following two consecutive perfect square numbers,
64 and 81. Also, 68 lies closer to 64 than to 81. Therefore, β68 β β64 = 8. b. 140 lies between the following two consecutive perfect square numbers, 121 and 144.
In addition, 140 is closer to 144 than to 121. Therefore, β140 β β144 = 12.
Finding the Domain of a Radical Function
Find the domain of each of the following functions. a. ππ(π₯π₯) = β2π₯π₯ + 3 b. ππ(π₯π₯) = 2 β β1 β π₯π₯
a. When finding domain π·π·ππ of function ππ(π₯π₯) = β2π₯π₯ + 3, we need to protect the radicand 2π₯π₯ + 3 from becoming negative. So, an π₯π₯-value belongs to the domain π·π·ππ if it satisfies the condition
2π₯π₯ + 3 β₯ 0.
This happens for π₯π₯ β₯ β32. Therefore, π«π«ππ = οΏ½β ππ
ππ,β).
b. To find the domain π·π·ππ of function ππ(π₯π₯) = 2 β β1 β π₯π₯, we solve the condition
1 β π₯π₯ β₯ 0 1 β₯ π₯π₯
Thus, π«π«ππ = (ββ,ππ].
Solution
Solution
on a graphing calculator
on a graphing calculator
Solution
/ β3, Γ· 2
/ +π₯π₯
The domain of an even degree radical is the solution set of
the inequality ππππππππππππππππ β₯ ππ
The domain of an odd degree radical
is β.
311
Graphing Radical Functions
For each function, find its domain, graph it, and find its range. Then, observe what transformation(s) of a basic root function result(s) in the obtained graph.
a. ππ(π₯π₯) = ββπ₯π₯ + 3 b. ππ(π₯π₯) = βπ₯π₯3 β 2
a. The domain π·π·ππ is the solution set of the inequality π₯π₯ + 3 β₯ 0, which is equivalent to
π₯π₯ β₯ β3. Hance, π«π«ππ = [βππ,β). The projection of the graph onto the π¦π¦-axis indicates the range of this function, which is (ββ,ππ].
The graph of ππ(π₯π₯) = ββπ₯π₯ + 3 has the same shape as
the graph of the basic square root function ππ(π₯π₯) = βπ₯π₯, except that it is flipped over the π₯π₯-axis and moved to the left by three units. These transformations are illustrated in Figure 3.
b. The domain and range of any odd degree radical are both the set of all real numbers.
So, π«π«ππ = β and ππππππππππππ = β.
The graph of ππ(π₯π₯) = βπ₯π₯3 β 2 has the same shape
as the graph of the basic cube root function ππ(π₯π₯) =βπ₯π₯3 , except that it is moved down by two units. This transformation is illustrated in Figure 4.
ππ ππ βππ 0 βππ β1 ππ β2 ππ β3
ππ ππ βππ β4 βππ β3 ππ β2 ππ β1 ππ 0
Solution
ππ(ππ) = ββππ + ππ
π₯π₯ β1 β3 domain
rang
e
start from the endpoint of the
domain
Figure 3
ππ(π₯π₯)
π₯π₯
1
1
Figure 4
ππ(ππ) = βππππ β ππ
π₯π₯ β2
1 domain
rang
e
ππ(ππ) = βππππ β ππ
π₯π₯
1
β1
312
Radicals in Application Problems
Some application problems require evaluation of formulas that involve radicals. For example, the formula ππ = βππ2 + ππ2 allows for finding the hypotenuse in a right angle triangle (see section RD3), Heronβs formula ππ = οΏ½π π (π π β ππ)(π π β ππ)(π π β ππ) allows for finding the area of any triangle given the lengths of its sides (see section T5), the formula
πΈπΈ = 2πποΏ½ππ3
πΊπΊπΊπΊ allows for finding the time needed for a planet to make a complete orbit
around the Sun, and so on.
Using a Radical Formula in an Application Problem
The time π‘π‘, in seconds, for one complete swing of a simple pendulum is given by the formula
π‘π‘ = 2πποΏ½ πΏπΏππ
,
where πΏπΏ is the length of the pendulum in feet, and ππ represents the acceleration due to gravity, which is about 32 ft per sec2. Find the time of a complete swing of a 2-ft pendulum to the nearest tenth of a second.
Since πΏπΏ = 2 ft and ππ = 32 ft/sec2, then
π‘π‘ = 2πποΏ½ 232
= 2πποΏ½ 116
= 2ππ β14
=ππ2β 1.6
So, the approximate time of a complete swing of a 2-ft pendulum is ππ.ππ seconds.
RD.1 Exercises
Vocabulary Check Complete each blank with one of the suggested words, or the most appropriate term or
phrase from the given list: even, index, irrational, nonnegative, odd, principal, radicand, square root.
1. The symbol β stands for the ____________ square root.
2. For any real numbers ππ β₯ 0 and ππ, if ππ2 = ππ, then ππ is the _________ ________ of ππ.
3. The expression under the radical sign is called the ___________.
4. The square root of a negative number ___________ππππ /ππππ ππππππ
a real number.
5. The square root of a number that is not a perfect square is a(n) ______________ number.
Solution
313
6. The cube root of ππ is written βππ3 , where 3 is called the __________ of the radical.
7. The domain of an even degree radical function is the set of all values that would make the radicand ________________.
8. The domain and range of an _____ degree radical functions is the set of all real numbers.
9. If ππ is ________ then βππππππ = |ππ|.
Concept Check Evaluate each radical, if possible.
10. β49 11. ββ81 12. ββ400 13. β0.09
14. β0.0016 15. οΏ½ 64225
16. β643 17. ββ1253
18. β0.0083 19. βββ10003 20. οΏ½ 10.000027
3 21. β164
22. β0.000325 23. ββ17 24. ββ2568 25. βοΏ½ 164
6
Concept Check
26. Decide whether the expression βββππ is positive, negative, 0, or not a real number, given that
a. ππ < 0 b. ππ > 0 c. ππ = 0
27. Assuming that ππ is odd, decide whether the expression ββππππ is positive, negative, or 0, given that
a. ππ < 0 b. ππ > 0 c. ππ = 0
Simplify each radical. Assume that letters can represent any real number.
28. β152 29. οΏ½(β15)2 30. βπ₯π₯2 31. οΏ½(βπ₯π₯)2
32. β81π₯π₯2 33. οΏ½(β12π¦π¦)2 34. οΏ½(ππ + 3)2 35. οΏ½(2 β π₯π₯)2π¦π¦4
36. βπ₯π₯2 β 4π₯π₯ + 4 37. οΏ½9π¦π¦2 + 30π¦π¦ + 25 38. οΏ½(β5)33 39. βπ₯π₯33
40. ββ125ππ33 41. βοΏ½0.008(π₯π₯ β 1)33 42. οΏ½(5π₯π₯)44 43. οΏ½(β10)88
44. οΏ½(π¦π¦ β 3)55 45. οΏ½(ππ + ππ)20172017 46. οΏ½(2ππ β ππ)20182018 47. βπ₯π₯186
48. οΏ½(ππ + 1)124 49. οΏ½(βππ)205 50. οΏ½(βππ)357 51. οΏ½π₯π₯4(βπ¦π¦)84
Find a decimal approximation for each radical. Round the answer to three decimal places.
52. β350 53. ββ0.859 54. β53 55. β35
314
Concept Check Without the use of a calculator, give the best integer approximation of each square root.
56. β67 57. β95 58. β115 59. β87
Questions in Exercises 60 and 61 refer to the accompanying rectangle. Answer these questions without the use of a calculator.
60. Give the best integer estimation of the area of the rectangle.
61. Give the best integer estimation of the perimeter of the rectangle.
Solve each problem. Do not use any calculator.
62. A rectangular yard has a length of β189 m and a width of β48 m. Choose the best estimate of its dimensions. Then estimate the perimeter.
A. 13 m by 7 m B. 14 m by 7 m C. 14 m by 8 m D. 15 m by 7 m
63. If the sides of a triangle are β65 cm, β34 cm, and β27 cm, which one of the following is the best estimate of its perimeter?
A. 20 cm B. 26 cm C. 19 cm D. 24 cm
Graph each function and give its domain and range. Then, discuss the transformations of a basic root function needed to obtain the graph of the given function.
64. ππ(π₯π₯) = βπ₯π₯ + 1 65. ππ(π₯π₯) = βπ₯π₯ + 1 66. β(π₯π₯) = ββπ₯π₯
67. ππ(π₯π₯) = βπ₯π₯ β 3 68. ππ(π₯π₯) = βπ₯π₯ β 3 69. β(π₯π₯) = 2 β βπ₯π₯
70. ππ(π₯π₯) = βπ₯π₯ β 23 71. ππ(π₯π₯) = βπ₯π₯3 + 2 72. β(π₯π₯) = ββπ₯π₯3 + 2
Graph each function and give its domain and range.
73. ππ(π₯π₯) = 2 + βπ₯π₯ β 1 74. ππ(π₯π₯) = 2βπ₯π₯ 75. β(π₯π₯) = ββπ₯π₯ + 3
76. ππ(π₯π₯) = β3π₯π₯ + 9 77. ππ(π₯π₯) = β3π₯π₯ β 6 78. β(π₯π₯) = ββ2π₯π₯ β 4
79. ππ(π₯π₯) = β12 β 3π₯π₯ 80. ππ(π₯π₯) = β8 β 4π₯π₯ 81. β(π₯π₯) = β2ββπ₯π₯
Analytic Skills Graph the three given functions on the same grid and discuss the relationship between them.
82. ππ(π₯π₯) = 2π₯π₯ + 1; ππ(π₯π₯) = β2π₯π₯ + 1; β(π₯π₯) = β2π₯π₯ + 13
83. ππ(π₯π₯) = βπ₯π₯ + 2; ππ(π₯π₯) = ββπ₯π₯ + 2; β(π₯π₯) = ββπ₯π₯ + 23
84. ππ(π₯π₯) = 12π₯π₯ + 1; ππ(π₯π₯) = οΏ½1
2π₯π₯ + 1; β(π₯π₯) = οΏ½1
2π₯π₯ + 1
3
β96
β27
315
Solve each problem.
85. The distance π·π·, in miles, to the horizon from an observerβs point of view over water or βflatβ earth is given by π·π· = β2ππ, where ππ is the height of the point of view, in feet. If a man whose eyes are 6 ft above ground level is standing at the top of a hill 105 ft above βflatβ earth, approximately how far to the horizon he will be able to see? Round the answer to the nearest mile.
86. The threshold body weight πΈπΈ for a person is the weight above which the risk of death increases greatly. The threshold weight in pounds for men aged 40β49 is related to their height β, in inches, by the formula β =12.3βπΈπΈ3 . What height corresponds to a threshold weight of 216 lb for a 43-year-old man? Round your answer to the nearest inch.
87. The time needed for a planet to make a complete orbit around the Sun is given by the formula πΈπΈ = 2πποΏ½ππ3
πΊπΊπΊπΊ,
where ππ is the average distance of the planet from the Sun, πΊπΊ is the universal gravitational constant, and ππ is the mass of the Sun. To the nearest day, find the orbital period of Mercury, knowing that its average distance from the Sun is 5.791 β 107 km, the mass of the Sun is 1.989 β 1030 kg, and πΊπΊ = 6.67408 β10β11 m3/(kgβs2).
88. The time it takes for an object to fall a certain distance is given by the equation π‘π‘ = οΏ½2ππππ
, where π‘π‘ is the time
in seconds, ππ is the distance in feet, and ππ is the acceleration due to gravity. If an astronaut above the moonβs surface drops an object, how far will it have fallen in 3 seconds? The acceleration on the moonβs surface is 5.5 feet per second per second.
Half of the perimeter (semiperimeter) of a triangle with sides a, b, and c is ππ = ππ
ππ(ππ + ππ + ππ). The area of such
a triangle is given by the Heronβs Formula: π¨π¨ = οΏ½ππ(ππ β ππ)(ππ β ππ)(ππ β ππ) . Find the area of a triangular piece of land with the following sides. 89. ππ = 3 m,ππ = 4 m, ππ = 5 m
90. ππ = 80 m,ππ = 80 m, ππ = 140 m
316
RD.2 Rational Exponents In sections P2 and RT1, we reviewed properties of powers with natural and integral exponents. All of these properties hold for real exponents as well. In this section, we give meaning to
expressions with rational exponents, such as ππ12, 8
13, or (2π₯π₯)0.54, and use the rational exponent
notation as an alternative way to write and simplify radical expressions.
Rational Exponents
Observe that β9 = 3 = 32β12 = 9
12. Similarly, β83 = 2 = 23β
13 = 8
13. This suggests the
following generalization:
For any real number ππ and a natural number ππ > 1, we have
βππππ = ππππππ.
Notice: The denominator of the rational exponent is the index of the radical.
Caution! If ππ < 0 and ππ is an even natural number, then ππ1ππ is not a real number.
Converting Radical Notation to Rational Exponent Notation
Convert each radical to a power with a rational exponent and simplify, if possible. Assume that all variables represent positive real numbers.
a. β165 b. β27π₯π₯33 c. οΏ½ 4ππ6
a. β166 = 16
16 = (24)
16 = 2
46 = ππ
ππππ
Observation: Expressing numbers as powers of prime numbers often allows for further simplification.
b. β27π₯π₯33 = (27π₯π₯3)
13 = 27
13 β (π₯π₯3)
13 = (33)
13 β π₯π₯ = ππππ
Note: The above example can also be done as follows:
β27π₯π₯33 = β33π₯π₯33 = (33π₯π₯3)13 = ππππ
c. οΏ½ 9ππ6
= οΏ½ 9ππ6οΏ½12 = οΏ½32οΏ½
12
(ππ6)12
= ππππππ
, as ππ > 0.
Solution
distribution of exponents
change into a power of a prime number
3
317
Observation: βππ4 = ππ42 = ππ2.
Generally, for any real number ππ β 0, natural number ππ > 1, and integral number ππ, we have
βππππππ = (πππΊπΊ)1ππ = ππ
ππππ
Rational exponents are introduced in such a way that they automatically agree with the rules of exponents, as listed in section RT1. Furthermore, the rules of exponents hold not only for rational but also for real exponents.
Observe that following the rules of exponents and the commutativity of multiplication, we have
βππππππ = (πππΊπΊ)1ππ = οΏ½ππ
1πποΏ½
ππ
= οΏ½βππππ οΏ½πΊπΊ
,
provided that βππππ exists.
Converting Rational Exponent Notation to the Radical Notation
Convert each power with a rational exponent to a radical and simplify, if possible.
a. 534 b. (β27)
13 c. 3π₯π₯β
25
a. 534 = β534 = βππππππππ
b. (β27)13 = ββ273 = βππ
c. 3π₯π₯β
25 = 3
π₯π₯25
= πποΏ½ππππππ
Observation: If ππππππ is a real number, then
ππβ ππππ = ππ
ππππππ
,
provided that ππ β 0.
Solution
Caution: A negative exponent indicates a reciprocal not a negative number!
Also, the exponent refers to π₯π₯ only, so 3 remains in the numerator.
Notice that β2713 = ββ273 = β3, so (β27)
13 = β27
13.
However, (β9)12 β β9
12, as (β9)
12 is not a real
number while β912 = ββ9 = β3.
318
Caution! Make sure to distinguish between a negative exponent and a negative result. Negative exponent leads to a reciprocal of the base. The result can be either
positive or negative, depending on the sign of the base. For example,
8β13 = 1
813
= 12, but (β8)β
13 = 1
(β8)13
= 1β2
= β12 and β8β
13 = β 1
813
= β12 .
Applying Rules of Exponents When Working with Rational Exponents
Simplify each expression. Write your answer with only positive exponents. Assume that all variables represent positive real numbers.
a. ππ34 β 2ππβ
53 b. 4
13
453 c. οΏ½π₯π₯
38 β π¦π¦
52οΏ½
43
a. ππ34 β 2ππβ
23 = 2ππ
34+οΏ½β
23οΏ½ = 2ππ
912β
812 = ππππ
ππππππ
b. 413
453
= 413β
53 = 4β43 = ππ
ππππππ
c. οΏ½π₯π₯38 β π¦π¦
52οΏ½
43 = π₯π₯
38β43 β π¦π¦
52β43 = ππ
ππππππ
ππππππ
Evaluating Powers with Rational Exponents
Evaluate each power.
a. 64β13 b. οΏ½β 8
125οΏ½23
a. 64β13 = (26)β
13 = 2β2 = 1
22= ππ
ππ
b. οΏ½β 8125
οΏ½23 = οΏ½οΏ½β 2
5οΏ½3οΏ½23
= οΏ½β25οΏ½2
= ππππππ
Observe that if ππ in βππππππ is a multiple of ππ, that is if ππ = ππππ for some integer ππ, then
βππππππππ = ππππππππ = ππππ
Simplifying Radical Expressions by Converting to Rational Exponents
Simplify. Assume that all variables represent positive real numbers. Leave your answer in simplified single radical form.
Solution It is helpful to change the base into a
power of prime number, if possible.
Solution
2
319
a. β3205 b. βπ₯π₯ β βπ₯π₯34 c. οΏ½2β23
a. β3205 = (320)
15 = 34 = ππππ
b. βπ₯π₯ β βπ₯π₯34 = π₯π₯
12 β π₯π₯
34 = π₯π₯
1β22β2 + 34 = π₯π₯
54 = π₯π₯ β π₯π₯
14 = ππβππππ
c. οΏ½2β23
= οΏ½2 β 212οΏ½
13 = οΏ½21+
12οΏ½
13 = οΏ½2
32οΏ½
13 = 2
12 = βππ
Another solution:
οΏ½2β23
= 213 β οΏ½2
12οΏ½
13 = 2
13 β 2
16 = 2
1β23β2+
16 = 2
12 = βππ
RT.2 Exercises
Concept Check Match each expression from Column I with the equivalent expression from Column II.
1. Column I Column II 2. Column I Column II
a. 2512 A. 1
5 a. (β32)
25 A. 2
b. 25β12 B. 5 b. β27
23 B. 1
4
c. β2532 C. β125 c. 32
15 C. β8
d. β25β12 D. not a real number d. 32β
25 D. β9
e. (β25)12 E. 1
125 e. β4
32 E. not a real number
f. 25β32 F. β1
5 f. (β4)
32 F. 4
Concept Check Write the base as a power of a prime number to evaluate each expression, if possible.
3. 3215 4. 27
43 5. β49
32 6. 16
34
7. β100β 12 8. 125β 13 9. οΏ½6481οΏ½34 10. οΏ½ 8
27οΏ½β 23
Solution
divide at the exponential level
add exponents as 54=1+14
t
This bracket is essential!
320
11. (β36)12 12. (β64)
13 13. οΏ½β1
8οΏ½β 13 14. (β625)β 14
Concept Check Rewrite with rational exponents and simplify, if possible. Assume that all variables represent
positive real numbers.
15. β5 16. β63 17. βπ₯π₯6 18. οΏ½π¦π¦25
19. β64π₯π₯63 20. οΏ½16π₯π₯2π¦π¦33 21. οΏ½25π₯π₯5
22. οΏ½16ππ6
4
Concept Check Rewrite without rational exponents, and simplify, if possible. Assume that all variables represent
positive real numbers.
23. 452 24. 8
34 25. π₯π₯
35 26. ππ
73
27. (β3)23 28. (β2)
35 29. 2π₯π₯β
12 30. π₯π₯
13π¦π¦β
12
Concept Check Use the laws of exponents to simplify. Write the answers with positive exponents. Assume that all variables represent positive real numbers.
31. 334 β 3
18 32. π₯π₯
23 β π₯π₯β 14 33. 2
58
2β 18 34. ππ
13
ππ23
35. οΏ½5158 οΏ½
23 36. οΏ½π¦π¦
23οΏ½
β 37 37. οΏ½π₯π₯38 β π¦π¦
52οΏ½
43 38. οΏ½ππβ 23 β ππ
58οΏ½β4
39. οΏ½π¦π¦β 32
π₯π₯β53οΏ½
13
40. οΏ½ππβ 23
ππβ56οΏ½
34
41. π₯π₯23 β 5π₯π₯β
25 42. π₯π₯
25 β οΏ½4π₯π₯β
45οΏ½
β 14
Use rational exponents to simplify. Write the answer in radical notation if appropriate. Assume that all variables represent positive real numbers.
43. βπ₯π₯26 44. οΏ½βππππ3 οΏ½15
45. οΏ½π¦π¦β186 46. οΏ½π₯π₯4π¦π¦β6
47. β816 48. β12814 49. οΏ½8π¦π¦63 50. οΏ½81ππ64
51. οΏ½(4π₯π₯3π¦π¦)23 52. οΏ½64(π₯π₯ + 1)105 53. οΏ½16π₯π₯4π¦π¦24 54. β32ππ10ππ155
Use rational exponents to rewrite in a single radical expression in a simplified form. Assume that all variables represent positive real numbers.
55. β53 β β5 56. β23 β β34 57. βππ β β3ππ3 58. βπ₯π₯3 β β2π₯π₯5
59. βπ₯π₯56 β βπ₯π₯23 60. βπ₯π₯π₯π₯3 β βπ₯π₯ 61. οΏ½π₯π₯5
βπ₯π₯8 62.
οΏ½ππ53
βππ3
321
63. β8π₯π₯3
βπ₯π₯34 64. οΏ½βππ3 65. οΏ½οΏ½π₯π₯π¦π¦34
66. οΏ½οΏ½(3π₯π₯)23
67. οΏ½οΏ½βπ₯π₯43 68. οΏ½3β33
69. οΏ½π₯π₯βπ₯π₯4 70. οΏ½2βπ₯π₯
3 Discussion Point
71. Suppose someone claims that βππππ + ππππππ must equal ππ + ππ, since, when ππ = 1 and ππ = 0, the two expressions are equal: βππππ + ππππππ = β1ππ + 0ππππ = 1 = 1 + 9 = ππ + ππ. Explain why this is faulty reasoning.
Analytic Skills Solve each problem.
72. One octave on a piano contains 12 keys (including both the black and white keys). The frequency of each
successive key increases by a factor of 2112. For example, middle C is two keys below the first D above it.
Therefore, the frequency of this D is
2112 β 2
112 = 2
16 β 1.12
times the frequency of the middle C.
a. If two tones are one octave apart, how do their frequencies compare?
b. The A tone below middle C has a frequency of 220 cycles per second. Middle C is 3 keys above this A note. Estimate the frequency of middle C.
73. According to one model, an animalβs heart rate varies according to its weight. The formula
πΈπΈ(π€π€) = 885π€π€β12 gives an estimate for the average number πΈπΈ of beats per minute for an animal that weighs π€π€ pounds. Use the formula to estimate the heart rate for a horse that weighs 800 pounds.
74. Meteorologists can determine the duration of a storm by using the function defined by
πΈπΈ(π·π·) = 0.07π·π·32, where π·π· is the diameter of the storm in miles and πΈπΈ is the time in hours.
Find the duration of a storm with a diameter of 16 mi. Round your answer to the nearest tenth of an hour.
one octave
C C D A
322
RD.3 Simplifying Radical Expressions and the Distance Formula
In the previous section, we simplified some radical expressions by replacing radical signs with rational exponents, applying the rules of exponents, and then converting the resulting expressions back into radical notation. In this section, we broaden the above method of simplifying radicals by examining products and quotients of radicals with the same indexes, as well as explore the possibilities of decreasing the index of a radical. In the second part of this section, we will apply the skills of simplifying radicals in problems involving the Pythagorean Theorem. In particular, we will develop the distance formula and apply it to calculate distances between two given points in a plane.
Multiplication, Division, and Simplification of Radicals
Suppose we wish to multiply radicals with the same indexes. This can be done by converting each radical to a rational exponent and then using properties of exponents as follows:
βππππ β βππππ = ππ1ππ β ππ
1ππ = (ππππ)
1ππ = βππππππ
This shows that the product of same index radicals is the radical of the product of their radicands.
Similarly, the quotient of same index radicals is the radical of the quotient of their radicands, as we have
βππππ
βππππ =ππ1ππ
ππ1ππ
= οΏ½πππποΏ½1ππ
= οΏ½ππππ
ππ
So, β2 β β8 = β2 β 8 = β16 = ππ. Similarly, β163
β23 = οΏ½162
3= β83 = ππ.
Attention! There is no such rule for addition or subtraction of terms. For instance, βππ + ππ β βππ Β± βππ,
and generally βππ Β± ππππ β βππππ Β± βππππ .
Here is a counterexample: β2ππ = β1 + 1ππ β β1ππ + β1ππ = 1 + 1 = 2
Multiplying and Dividing Radicals of the Same Indexes
Perform the indicated operations and simplify, if possible. Assume that all variables are positive.
a. β10 β β15 b. β2π₯π₯3 οΏ½6π₯π₯π¦π¦
c. β10π₯π₯β5
d. β32π₯π₯34
β2π₯π₯4
PRODUCT RULE
QUOTIENT RULE
323
a. β10 β β15 = β10 β 15 = β2 β 5 β 5 β 3 = β5 β 5 β 2 β 3 = β25 β β6 = ππβππ
b. β2π₯π₯3 οΏ½6π₯π₯π¦π¦ = οΏ½2 β 2 β 3π₯π₯4π¦π¦ = β4π₯π₯4 β οΏ½3π¦π¦ = πππππποΏ½ππππ
c. β10π₯π₯β5
= οΏ½10π₯π₯5
= βππππ
d. β32π₯π₯34
β2π₯π₯4 = οΏ½32π₯π₯3
2π₯π₯
4= β16π₯π₯24 = β164 β βπ₯π₯24 = ππβππ
Caution! Remember to indicate the index of the radical for indexes higher than two.
The product and quotient rules are essential when simplifying radicals.
To simplify a radical means to:
1. Make sure that all power factors of the radicand have exponents smaller than the index of the radical.
For example, οΏ½24π₯π₯8π¦π¦3 = β23π₯π₯63 β οΏ½2π₯π₯2π¦π¦3 = 2π₯π₯οΏ½2π₯π₯2π¦π¦3 .
2. Leave the radicand with no fractions.
For example, οΏ½2π₯π₯25
= β2π₯π₯β25
= β2π₯π₯5
.
3. Rationalize any denominator. (Make sure that denominators are free from radicals.)
For example, οΏ½4π₯π₯
= β4βπ₯π₯
= 2ββπ₯π₯βπ₯π₯ββπ₯π₯
= 2βπ₯π₯π₯π₯
, providing that π₯π₯ > 0.
4. Reduce the power of the radicand with the index of the radical, if possible.
For example, βπ₯π₯24 = π₯π₯24 = π₯π₯
12 = βπ₯π₯.
Simplifying Radicals
Simplify each radical. Assume that all variables are positive.
a. οΏ½96π₯π₯7π¦π¦155 b. οΏ½ ππ12
16ππ44
c. οΏ½25π₯π₯2
8π₯π₯3 d. β27ππ156
Solution
2
product rule prime factorization commutativity of multiplication
product rule
use commutativity of multiplication to isolate perfect
square factors
Here the multiplication sign is assumed, even if it
is not indicated.
quotient rule
16 2
Recall that
βπ₯π₯24 = π₯π₯24 = π₯π₯
12 = βπ₯π₯.
2
324
a. οΏ½96π₯π₯7π¦π¦155 = οΏ½25 β 3π₯π₯7π¦π¦155 = ππππππππβππππππππ
Generally, to simplify βπ₯π₯ππππ , we perform the division
ππ Γ· ππ = πππππππ‘π‘πππππππ‘π‘ ππ + ππππππππππππππππππ ππ,
and then pull the ππ-th power of π₯π₯ out of the radical, leaving the ππ-th power of π₯π₯ under the radical. So, we obtain βππππππ = ππππ βππππππ
b. οΏ½ ππ12
16ππ44
= βππ124
β24ππ44 = ππππ
ππππ
c. οΏ½25π₯π₯2
8π₯π₯3= οΏ½ 25
23π₯π₯= β25
β23π₯π₯= 5
2β2π₯π₯β β2π₯π₯β2π₯π₯
= 5β2π₯π₯2β2π₯π₯
= ππβππππππππ
d. β27ππ156 = β33ππ156 = ππ2β33ππ36 = ππ2 β οΏ½(3ππ)36 = ππππβππππ
Simplifying Expressions Involving Multiplication, Division, or Composition of Radicals with Different Indexes
Simplify each expression. Leave your answer in simplified single radical form. Assume that all variables are positive.
a. οΏ½π₯π₯π¦π¦5 β οΏ½π₯π₯4π¦π¦3 b. βππ2ππ34
βππππ3 c. οΏ½π₯π₯2β2π₯π₯3
a. οΏ½π₯π₯π¦π¦5 β οΏ½π₯π₯2π¦π¦3 = π₯π₯12π¦π¦
52 β π₯π₯
23π¦π¦
13 = π₯π₯
1β32β3 + 2β23β2π¦π¦
5β32β3 + 1β23β2 = π₯π₯
76 π¦π¦
176 = (π₯π₯7π¦π¦17)
16
= οΏ½π₯π₯7 π¦π¦176 = ππππππ οΏ½ππππππππ
b. βππ2ππ34
βππππ3 = ππ24 ππ
34
ππ13ππ
13
= ππ1β32β3 β 1β23β2ππ
3β34β3 β 1β43β4 = ππ
1β26β2 ππ
512 = (ππ2 ππ5 )
112 = βππππ ππππ ππππ
c. οΏ½π₯π₯2 β2π₯π₯3
= π₯π₯23 β οΏ½(2π₯π₯)
12οΏ½
13 = π₯π₯
23 β 2
16 β π₯π₯
16 = π₯π₯
2β23β2 + 16 β 2
16 = 2
16π₯π₯
56 = (2π₯π₯5)
16 = βππππππππ
Solution
οΏ½π₯π₯75 = π₯π₯οΏ½π₯π₯25
Solution
οΏ½π¦π¦155 = π¦π¦3
1
2
2
If radicals are of different indexes, convert them to
exponential form.
Bring the exponents to the LCD in order to leave the answer as a single radical.
325
Pythagorean Theorem and Distance Formula
One of the most famous theorems in mathematics is the Pythagorean Theorem.
Pythagorean Suppose angle πΆπΆ in a triangle πππ΄π΄πΆπΆ is a 90Β° angle. Theorem Then the sum of the squares of the lengths of the two legs, ππ and ππ, equals to the square
of the length of the hypotenuse ππ:
ππππ + ππππ = ππππ
Using The Pythagorean Equation For the first two triangles, find the exact length π₯π₯ of the unknown side. For triangle c., express length π₯π₯ in terms of the unknown ππ. a. b. c.
a. The length of the hypotenuse of the given right triangle is equal to π₯π₯. So, the
Pythagorean equation takes the form π₯π₯2 = 42 + 52.
To solve it for π₯π₯, we take a square root of each side of the equation. This gives us
οΏ½π₯π₯2 = οΏ½42 + 52 π₯π₯ = β16 + 25 π₯π₯ = βππππ
b. Since 10 is the length of the hypotenuse, we form the Pythagorean equation
102 = π₯π₯2 + β242
.
To solve it for π₯π₯, we isolate the π₯π₯2 term and then apply the square root operator to both sides of the equation. So, we have
102 β β242
= π₯π₯2
100β 24 = π₯π₯2
π₯π₯2 = 76
π₯π₯ = β76 = β4 β 19 = ππβππππ
c. The length of the hypotenuse is βππ, so we form the Pythagorean equation as below.
οΏ½βπποΏ½2
= 12 + π₯π₯2
Solution
Caution: Generally, οΏ½π₯π₯2 = |π₯π₯|
However, the length of a side of a triangle is
positive. So, we can write οΏ½π₯π₯2 = π₯π₯
ππ πΆπΆ
π΄π΄
ππ
ππ
ππ
4
9
ππ β24
10
ππ 1
βππ ππ
Customary, we simplify each
root, if possible.
326
To solve this equation for π₯π₯, we isolate the π₯π₯2 term and then apply the square root operator to both sides of the equation. So, we obtain
ππ2 = 1 + π₯π₯2 ππ2 β 1 = π₯π₯2 π₯π₯ = οΏ½ππππ β ππ
Note: Since the hypotenuse of length βππ must be longer than the leg of length 1, then ππ > 1. This means that ππ2 β 1 > 0, and therefore βππ2 β 1 is a positive real number.
The Pythagorean Theorem allows us to find the distance between any two given points in a plane. Suppose ππ(π₯π₯1,π¦π¦1) and π΄π΄(π₯π₯2,π¦π¦2) are two points in a coordinate plane. Then |π₯π₯2 β π₯π₯1| represents the horizontal distance between ππ and π΄π΄ and |π¦π¦2 β π¦π¦1| represents the vertical distance between ππ and π΄π΄, as shown in Figure 1. Notice that by applying the absolute value operator to each difference of the coordinates we guarantee that the resulting horizontal and vertical distance is indeed a nonnegative number. Applying the Pythagorean Theorem to the right triangle shown in Figure 1, we form the equation
ππ2 = |π₯π₯2 β π₯π₯1|2 + |π¦π¦2 β π¦π¦1|2,
where d is the distance between ππ and π΄π΄.
Notice that |π₯π₯2 β π₯π₯1|2 = (π₯π₯2 β π₯π₯1)2 as a perfect square automatically makes the expression nonnegative. Similarly, |π¦π¦2 β π¦π¦1|2 = (π¦π¦2 β π¦π¦1)2. So, the Pythagorean equation takes the form
ππ2 = (π₯π₯2 β π₯π₯1)2 + (π¦π¦2 β π¦π¦1)2 After solving this equation for ππ, we obtain the distance formula:
ππ = οΏ½(ππππ β ππππ)ππ + (ππππ β ππππ)ππ
Note: Observe that due to squaring the difference of the corresponding coordinates, the distance between two points is the same regardless of which point is chosen as first, (π₯π₯1,π¦π¦1), and second, (π₯π₯2,π¦π¦2).
Finding the Distance Between Two Points
Find the exact distance between the points (β2,4) and (5, 3). Let (β2,4) = (π₯π₯1,π¦π¦1) and (5, 3) = (π₯π₯2,π¦π¦2). To find the distance ππ between the two points, we follow the distance formula:
Solution
1
βππ οΏ½ππ2 β 1
ππ(π₯π₯1, π¦π¦1)
π΄π΄(π₯π₯2, π¦π¦2)
π₯π₯2 π₯π₯1
π¦π¦2
π¦π¦1 |π₯π₯2 β π₯π₯1|
| π¦π¦2βπ¦π¦ 1
|
ππ
π₯π₯
π¦π¦
Figure 1
327
ππ = οΏ½οΏ½5 β (β2)οΏ½2 + (3 β 4)2 = οΏ½72 + (β1)2 = β49 + 1 = β50 = ππβππ
So, the points (β2,4) and (5, 3) are 5β2 units apart.
RD.3 Exercises
Vocabulary Check Complete each blank with the most appropriate term or phrase from the given list: distance,
index, lower, product, Pythagorean, right, simplified.
1. The product of the same __________ radicals is the radical of the ___________ of the radicands.
2. The radicand of a simplified radical contains only powers with ________ exponents than the index of the radical.
3. The radical βπ₯π₯46 can still be ______________.
4. The _______________ between two points (π₯π₯1,π¦π¦1) and (π₯π₯2,π¦π¦2) on a coordinate plane is equal to οΏ½(π₯π₯2 β π₯π₯1)2 + (π¦π¦2 β π¦π¦1)2.
5. The _________________ equation applies to __________ triangles only. Concept Check Multiply and simplify, if possible. Assume that all variables are positive.
6. β5 β β5 7. β18 β β2 8. β6 β β3 9. β15 β β6
10. β45 β β60 11. β24 β β75 12. β3π₯π₯3 β β6π₯π₯5 13. οΏ½5π¦π¦7 β β15ππ3
14. οΏ½12π₯π₯3π¦π¦ οΏ½8π₯π₯4π¦π¦2 15. β30ππ3ππ4 β18ππ2ππ5 16. β4π₯π₯23 β2π₯π₯43 17. β20ππ34 β4ππ54
Concept Check Divide and simplify, if possible. Assume that all variables are positive.
18. β90β5
19. β48β6
20. β42ππβ7ππ
21. β30π₯π₯3
β10π₯π₯
22. β52ππππ3
β13ππ 23. οΏ½56π₯π₯π¦π¦3
β8π₯π₯ 24. οΏ½128π₯π₯
2π¦π¦2β2
25. β48ππ3ππ2β3
26. β804
β54 27. β1083
β43 28. οΏ½96ππ5ππ23
οΏ½12ππ2ππ3 29.
οΏ½48π₯π₯9π¦π¦134
οΏ½3π₯π₯π¦π¦54
Concept Check Simplify each expression. Assume that all variables are positive.
30. οΏ½144π₯π₯4π¦π¦9 31. βοΏ½81ππ8ππ5 32. ββ125ππ6ππ9ππ123 33. οΏ½50π₯π₯3π¦π¦4
34. οΏ½ 116ππ8ππ204
35. βοΏ½β 127π₯π₯
2π¦π¦73 36. β7ππ7ππ6 37. οΏ½75ππ3ππ4
328
38. οΏ½64π₯π₯12π¦π¦155 39. οΏ½ππ14ππ7ππ235 40. βοΏ½162ππ15ππ104 41. βοΏ½32π₯π₯5π¦π¦104
42. οΏ½1649
43. οΏ½ 27125
3 44. οΏ½121
π¦π¦2 45. οΏ½64
π₯π₯4
46. οΏ½81ππ5
64
3 47. οΏ½36π₯π₯5
π¦π¦6 48. οΏ½16π₯π₯12
π¦π¦4π§π§164 49. οΏ½32π¦π¦8
π₯π₯105
50. β364 51. β276 52. β οΏ½π₯π₯2510 53. βπ₯π₯4412
54. βοΏ½ 1π₯π₯3π¦π¦ 55. οΏ½64π₯π₯15
π¦π¦4π§π§53
56. οΏ½ π₯π₯13
π¦π¦6π§π§126
57. οΏ½ππ9ππ24
ππ186
Discussion Point
58. When asked to simplify the radical βπ₯π₯3 + π₯π₯2, a student wrote βπ₯π₯3 + π₯π₯2 = π₯π₯βπ₯π₯ + π₯π₯ = π₯π₯οΏ½βπ₯π₯ + 1οΏ½. Is this correct? If yes, explain the studentβs reasoning. If not, discuss how such radical could be simplified.
Perform operations. Leave the answer in simplified single radical form. Assume that all variables are positive.
59. β3 β β43 60. βπ₯π₯ β βπ₯π₯5 61. βπ₯π₯23 β βπ₯π₯4 62. β43 β β85
63. βππ23
βππ 64. βπ₯π₯
βπ₯π₯4 65. οΏ½π₯π₯2π¦π¦34
βπ₯π₯π¦π¦3 66. β16ππ25
β2ππ23
67. οΏ½2βπ₯π₯3 68. οΏ½π₯π₯ β2π₯π₯23 69. οΏ½3 β934
70. οΏ½π₯π₯2 βπ₯π₯343
Concept Check For each right triangle, find length π₯π₯. Simplify the answer if possible. In problems 73 and 74,
expect the length x to be an expression in terms of n. 71. 72. 73.
74. 75. 76.
Concept Check Find the exact distance between each pair of points.
77. (8,13) and (2,5) 78. (β8,3) and (β4,1) 79. (β6,5) and (3,β4)
80. οΏ½57
, 114οΏ½ and οΏ½1
7, 1114οΏ½ 81. οΏ½0,β6οΏ½ and οΏ½β7, 0οΏ½ 82. οΏ½β2,β6οΏ½ and οΏ½2β2,β4β6οΏ½
5
7
ππ
3
ππ ππ
β2
2β2
ππ
2β6
3β8 ππ
2
βππ + 4 ππ
329
83. οΏ½ββ5, 6β3οΏ½ and οΏ½β5,β3οΏ½ 84. (0,0) and (ππ, ππ) 85. (π₯π₯ + β,π¦π¦ + β) and (π₯π₯,π¦π¦) (assume that β > 0)
Analytic Skills Solve each problem.
86. The length of the diagonal of a box is given by the formula π·π· = βππ2 + πΏπΏ2 + ππ2, where ππ, πΏπΏ, and ππ are, respectively, the width, length, and height of the box. Find the length of the diagonal π·π· of a box that is 4 ft long, 2 ft wide, and 3 ft high. Give the exact value, and then round to the nearest tenth of a foot.
87. The screen of a 32-inch television is 27.9-inch wide. To the nearest tenth of an inch, what is the measure of its height? TVs are measured diagonally, so a 32-inch television means that its screen measures diagonally 32 inches.
88. Find all ordered pairs on the π₯π₯-axis of a Cartesian coordinate system that are 5 units from the point (0, 4).
89. Find all ordered pairs on the π¦π¦-axis of a Cartesian coordinate system that are 5 units from the point (3, 0).
90. During the summer heat, a 2-mi bridge expands 2 ft in length. If we assume that the bulge occurs in the middle of the bridge, how high is the bulge? The answer may surprise you. In reality, bridges are built with expansion spaces to avoid such buckling.
πΏπΏ ππ
ππ π·π·
bulge
330
RD.4 Operations on Radical Expressions; Rationalization of Denominators
Unlike operations on fractions or decimals, sums and differences of many radicals cannot be simplified. For instance, we cannot combine β2 and β3, nor simplify expressions such as β23 β 1. These types of radical expressions can only be approximated with the aid of a calculator. However, some radical expressions can be combined (added or subtracted) and simplified. For example, the sum of 2β2 and β2 is 3β2, similarly as 2π₯π₯ + π₯π₯ = 3π₯π₯. In this section, first, we discuss the addition and subtraction of radical expressions. Then, we show how to work with radical expressions involving a combination of the four basic operations. Finally, we examine how to rationalize denominators of radical expressions.
Addition and Subtraction of Radical Expressions
Recall that to perform addition or subtraction of two variable terms we need these terms to be like. This is because the addition and subtraction of terms are performed by factoring out the variable βlikeβ part of the terms as a common factor. For example,
π₯π₯2 + 3π₯π₯2 = (1 + 3)π₯π₯2 = 4π₯π₯2
The same strategy works for addition and subtraction of the same types of radicals or radical terms (terms containing radicals).
Definition 4.1 Radical terms containing radicals with the same index and the same radicands are referred to as like radicals or like radical terms.
For example, β5π₯π₯ and 2β5π₯π₯ are like (the indexes and the radicands are the same) while 5β2 and 2β5 are not like (the radicands are different) and βπ₯π₯ and βπ₯π₯3 are not like radicals (the indexes are different).
To add or subtract like radical expressions we factor out the common radical and any other common factor, if applicable. For example,
4β2 + 3β2 = (4 + 3)β2 = 7β2,
and 4π₯π₯π¦π¦β2 β 3π₯π₯β2 = (4π¦π¦ + 3)π₯π₯β2.
Caution! Unlike radical expressions cannot be combined. For example, we are unable to perform the addition β6 + β3. Such a sum can only be approximated using a calculator.
Notice that unlike radicals may become like if we simplify them first. For example, β200 and β50 are not like, but β200 = 10β2 and β50 = 5β2. Since 10β2 and 5β2 are like radical terms, they can be combined. So, we can perform, for example, the addition:
β200 + β50 = 10β2 + 5β2 = 15β2
2β2
β2 β2
2 2
1 β2
331
Adding and Subtracting Radical Expressions
Perform operations and simplify, if possible. Assume that all variables represent positive real numbers.
a. 5β3 β 8β3 b. 3β25 β 7π₯π₯β25 + 6β25
c. 7β45 + β80 β β12 d. 3οΏ½π¦π¦53 β 7οΏ½π¦π¦23 + οΏ½8π¦π¦83
e. οΏ½ π₯π₯16
+ 2οΏ½π₯π₯3
9 f. β25π₯π₯ β 25 ββ9π₯π₯ β 9
a. To subtract like radicals, we combine their coefficients via factoring.
5β3 β 8β3 = (5 β 8)β3 = βππβππ
b. 3β25 β 7π₯π₯β25 + 6β25 = (3 β 7π₯π₯ + 6)β25 = (9 β 7π₯π₯)β25
Note: Even if not all coefficients are like, factoring the common radical is a useful strategy that allows us to combine like radical expressions.
c. The expression 7β45 + β80 ββ12 consists of unlike radical terms, so they cannot be combined in this form. However, if we simplify the radicals, some of them may become like and then become possible to combine.
7β45 + β80 β β12 = 7β9 β 5 + β16 β 5 β β4 β 3 = 7 β 3β5 + 4β5 β 2β3
= 21β5 + 4β5 β 2β3 = ππππβππ β ππβππ
d. As in the previous example, we simplify each radical expression before attempting to combine them.
3οΏ½π¦π¦53 β 5π¦π¦οΏ½π¦π¦23 + οΏ½32π¦π¦75 = 3π¦π¦οΏ½π¦π¦23 β 5π¦π¦οΏ½π¦π¦23 + 2π¦π¦οΏ½π¦π¦25
= (3π¦π¦ β 5π¦π¦)οΏ½π¦π¦23 + 2π¦π¦οΏ½π¦π¦25 = βπππποΏ½ππππππ + πππποΏ½ππππππ
Note: The last two radical expressions cannot be combined because of different indexes.
e. To perform the addition οΏ½ π₯π₯16
+ 2οΏ½π₯π₯3
9, we may simplify each radical expression first.
Then, we add the expressions by bringing them to the least common denominator and finally, factor the common radical, as shown below.
οΏ½ π₯π₯16
+ 2οΏ½π₯π₯3
9= βπ₯π₯
β16+ 2 βπ₯π₯3
β9= βπ₯π₯
4+ 2 βπ₯π₯3
3= 3βπ₯π₯+2β4βπ₯π₯βπ₯π₯
12= οΏ½ππ+ππππ
πππποΏ½βππ
Solution
Remember to write the index with each radical.
The brackets are essential here.
332
f. In an attempt to simplify radicals in the expression β25π₯π₯2 β 25 ββ9π₯π₯2 β 9, we factor each radicand first. So, we obtain
οΏ½25π₯π₯2 β 25 βοΏ½9π₯π₯2 β 9 = οΏ½25(π₯π₯2 β 1) βοΏ½9(π₯π₯2 β 1) = 5οΏ½π₯π₯2 β 1 βοΏ½π₯π₯2 β 1
= πποΏ½ππππ β ππ
Caution! The root of a sum does not equal the sum of the roots. For example,
β5 = βππ + ππ β βππ + βππ = 1 + 2 = 3
So, radicals such as β25π₯π₯2 β 25 or β9π₯π₯2 β 9 can be simplified only via factoring a perfect square out of their radicals while βπ₯π₯2 β 1 cannot be simplified any further.
Multiplication of Radical Expressions with More than One Term
Similarly as in the case of multiplication of polynomials, multiplication of radical expressions where at least one factor consists of more than one term is performed by applying the distributive property.
Multiplying Radical Expressions with More than One Term
Multiply and then simplify each product. Assume that all variables represent positive real numbers.
a. 2β2οΏ½3β2π₯π₯ β β6οΏ½ b. βπ₯π₯3 οΏ½β3π₯π₯23 β β81π₯π₯23 οΏ½
c. οΏ½2β3 + β2οΏ½οΏ½β3 β 3β2οΏ½ d. οΏ½π₯π₯βπ₯π₯ β οΏ½π¦π¦οΏ½οΏ½π₯π₯βπ₯π₯ + οΏ½π¦π¦οΏ½
e. οΏ½3β2 + 2βπ₯π₯3 οΏ½οΏ½3β2 β 2βπ₯π₯3 οΏ½ f. οΏ½οΏ½5π¦π¦ + π¦π¦οΏ½π¦π¦οΏ½2
a. 5β2οΏ½3β2π₯π₯ β β6οΏ½ = 15β4π₯π₯ β 5β2 β 2 β 3 = 15 β 2βπ₯π₯ β 5 β 2β3 = ππππβππ β ππππβππ
b.
βπ₯π₯3 οΏ½β3π₯π₯23 β β81π₯π₯23 οΏ½ = β3π₯π₯2 β π₯π₯3 β β81π₯π₯2 β π₯π₯3 = π₯π₯β33 β 3π₯π₯β33 = βππππβππππ
c. To multiply two binomial expressions involving radicals we may use the FOIL method. Recall that the acronym FOIL refers to multiplying the First, Outer, Inner, and Last terms of the binomials.
Solution
These are unlike terms. So, they cannot be combined. 5β2 β 3β2π₯π₯ = 5 β 3β2 β 2π₯π₯
distribution
simplification
combining like terms
333
F O I L οΏ½2β3 + β2οΏ½οΏ½β3 β 3β2οΏ½ = 2 β 3 β 6β3 β 2 + β2 β 3 β 3 β 2 = 6 β 6β6 + β6 β 6
= βππβππ
d. To multiply two conjugate binomial expressions we follow the difference of squares formula, (ππ β ππ)(ππ + ππ) = ππ2 β ππ2. So, we obtain
οΏ½π₯π₯βπ₯π₯ β οΏ½π¦π¦οΏ½οΏ½π₯π₯βπ₯π₯ + οΏ½π¦π¦οΏ½ = οΏ½π₯π₯βπ₯π₯οΏ½2β οΏ½οΏ½π¦π¦οΏ½
2= π₯π₯2 β π₯π₯ β π¦π¦ = ππππ β ππ
e. Similarly as in the previous example, we follow the difference of squares formula.
οΏ½3β2 + 2βπ₯π₯3 οΏ½οΏ½3β2 β 2βπ₯π₯3 οΏ½ = οΏ½3β2οΏ½2β οΏ½2βπ₯π₯3 οΏ½
2= 9 β 2 β 4οΏ½π₯π₯23 = ππππ β πποΏ½ππππππ
f. To multiply two identical binomial expressions we follow the perfect square formula, (ππ + ππ)(ππ + ππ) = ππ2 + 2ππππ + ππ2. So, we obtain
οΏ½οΏ½5π¦π¦ + π¦π¦οΏ½π¦π¦οΏ½2
= οΏ½οΏ½5π¦π¦οΏ½2
+ 2οΏ½οΏ½5π¦π¦οΏ½οΏ½π¦π¦οΏ½π¦π¦οΏ½ + οΏ½π¦π¦οΏ½π¦π¦οΏ½2
= 5π¦π¦ + 2π¦π¦οΏ½5π¦π¦2 + π¦π¦2π¦π¦
= ππππ + ππβππππππ + ππππ
Rationalization of Denominators
As mentioned in section RD3, a process of simplifying radicals involves rationalization of any emerging denominators. Similarly, a radical expression is not in its simplest form unless all its denominators are rational. This agreement originated before the days of calculators when computation was a tedious process performed by hand. Nevertheless, even in present time, the agreement of keeping denominators rational does not lose its validity, as we often work with variable radical expressions. For example, the expressions 2
β2 and
β2 are equivalent, as 2β2
=2β2
ββ2β2
=2β2
2= β2
Similarly, π₯π₯βπ₯π₯
is equivalent to βπ₯π₯, as
π₯π₯βπ₯π₯
=π₯π₯βπ₯π₯
ββπ₯π₯βπ₯π₯
=π₯π₯βπ₯π₯π₯π₯
= βπ₯π₯
While one can argue that evaluating 2β2
is as easy as evaluating β2 when using a calculator,
the expression βπ₯π₯ is definitely easier to use than π₯π₯βπ₯π₯
in any further algebraic manipulations.
Definition 4.2 The process of removing radicals from a denominator so that the denominator contains only rational numbers is called rationalization of the denominator.
square each factor
οΏ½βπ₯π₯οΏ½2
= π₯π₯
334
Rationalization of denominators is carried out by multiplying the given fraction by a factor of 1, as shown in the next two examples.
Rationalizing Monomial Denominators Simplify, if possible. Leave the answer with a rational denominator. Assume that all variables represent positive real numbers.
a. β13β5
b. 5β32π₯π₯3 c. οΏ½81π₯π₯5
π¦π¦4
a. Notice that β5 can be converted to a rational number by multiplying it by another β5.
Since the denominator of a fraction cannot be changed without changing the numerator in the same way, we multiply both, the numerator and denominator of β1
3β5 by β5. So,
we obtain β1
3β5ββ5β5
=ββ53 β 5
= ββππππππ
b. First, we may want to simplify the radical in the denominator. So, we have
5β32π₯π₯3 =
5β8 β 4π₯π₯3 =
52β4π₯π₯3
Then, notice that since β4π₯π₯3 = β22π₯π₯3 , it is enough to multiply it by β2π₯π₯23 to nihilate the radical. This is because β22π₯π₯3 β β2π₯π₯23 = β23π₯π₯33 = 2π₯π₯. So, we proceed
5β32π₯π₯3 =
52β4π₯π₯3 β
β2π₯π₯23
β2π₯π₯23 =5β2π₯π₯23
2 β 2π₯π₯=ππβππππππππ
ππππ
Caution: A common mistake in the rationalization of β4π₯π₯3 is the attempt to multiply it by a copy of β4π₯π₯3 . However, β4π₯π₯3 β β4π₯π₯3 = β16π₯π₯23 = 2β3π₯π₯23 is still not rational. This is because we work with a cubic root, not a square root. So, to rationalize β4π₯π₯3 we must look for βfillingβ the radicand to a perfect cube. This is achieved by multiplying 4π₯π₯ by 2π₯π₯2 to get 8π₯π₯3.
c. To simplify οΏ½81π₯π₯5
π¦π¦4
, first, we apply the quotient rule for radicals, then simplify the
radical in the numerator, and finally, rationalize the denominator. So, we have
οΏ½81π₯π₯5
π¦π¦4
=β81π₯π₯54
οΏ½π¦π¦4 =3π₯π₯βπ₯π₯4
οΏ½π¦π¦4 βοΏ½π¦π¦34
οΏ½π¦π¦34 =πππποΏ½ππππππππ
ππ
Solution
335
To rationalize a binomial containing square roots, such as 2 β βπ₯π₯ or β2 β β3, we need to find a way to square each term separately. This can be achieved through multiplying by a conjugate binomial, in order to benefit from the difference of squares formula. In particular, we can rationalize denominators in expressions below as follows:
12 β βπ₯π₯
=1
οΏ½2 β βπ₯π₯οΏ½βοΏ½2 + βπ₯π₯οΏ½οΏ½2 + βπ₯π₯οΏ½
=ππ + βππππ β ππ
or
β2β2 + β3
=β2
οΏ½β2 + β3οΏ½βοΏ½β2 β β3οΏ½οΏ½β2 β β3οΏ½
=2 β β62 β 3
=2 β β6β1
= βππ β ππ
Rationalizing Binomial Denominators
Rationalize each denominator and simplify, if possible. Assume that all variables represent positive real numbers.
a. 1ββ31+β3
b. βπ₯π₯π¦π¦2βπ₯π₯ββπ¦π¦
a. 1ββ31+β3
β οΏ½1ββ3οΏ½οΏ½1ββ3οΏ½
= 1β2β3+31β3
= 4β2β3β2
= β2οΏ½β2+β3οΏ½β2
= βππ β ππ
b. βπ₯π₯π¦π¦2βπ₯π₯ββπ¦π¦
β οΏ½2βπ₯π₯+βπ¦π¦οΏ½οΏ½2βπ₯π₯+βπ¦π¦οΏ½
= 2π₯π₯βπ¦π¦+π¦π¦βπ₯π₯4π₯π₯βπ¦π¦
Some of the challenges in algebraic manipulations involve simplifying quotients with
radical expressions, such as 4β2β3β2
, which appeared in the solution to Example 4a. The key
concept that allows us to simplify such expressions is factoring, as only common factors can be reduced.
Writing Quotients with Radicals in Lowest Terms
Write each quotient in lowest terms.
a. 15β6β56
b. 3π₯π₯+β8π₯π₯2
9π₯π₯
a. To reduce this quotient to the lowest terms we may factor the numerator first,
15β 6β56
=3οΏ½5 β 2β5οΏ½
6=ππ β ππβππ
ππ,
Solution
Solution
Apply the difference of squares formula:
(ππ β ππ)(ππ+ ππ) = ππππ β ππππ
factor
2
336
or alternatively, rewrite the quotient into two fractions and then simplify,
15 β 6β56
=156β
6β56
=ππππβ βππ.
Caution: Here are the common errors to avoid:
15β6β56
= 15β β5 - only common factors can be reduced!
15β6β56
= 9β56
= 3β52
- subtraction is performed after multiplication!
b. To reduce this quotient to the lowest terms, we simplify the radical and factor the numerator first. So,
3π₯π₯ + β8π₯π₯2
6π₯π₯=
3π₯π₯ + 2π₯π₯β26π₯π₯
=π₯π₯οΏ½3 + 2β2οΏ½
6π₯π₯=ππ + ππβππ
ππ
RD.4 Exercises
Vocabulary Check Complete each blank with the most appropriate term or phrase from the given list: coefficients, denominator, distributive, factoring, like, rationalize, simplified.
1. Radicals that have the same index and the same radicand are called __________ radicals.
2. Some unlike radicals can become like radicals after they are _____________.
3. Like radicals can be added by combining their _______________ and then multiplying this sum by the common radical.
4. To multiply radical expressions containing more than one term, we use the _____________ property.
5. To _____________ the denominator means to rewrite an expression that has a radical in its denominator as an equivalent expression that does not have one.
6. To rationalize a denominator with two terms involving square roots, multiply both the numerator and the denominator by the conjugate of the _____________.
7. The key strategy for simplifying quotients such as ππ+ππβππππ
is _____________ the numerator, if possible.
3
This expression
cannot be simplified any further.
337 Concept Check
8. A student claims that 24β 4βπ₯π₯ = 20βπ₯π₯ because for π₯π₯ = 1 both sides of the equation equal to 20. Is this a valid justification? Explain.
9. For most values of ππ and ππ, βππ + βππ β βππ + ππ. Are there any real numbers ππ and ππ such that βππ + βππ =βππ + ππ ? Justify your answer.
Perform operations and simplify, if possible. Assume that all variables represent positive real numbers.
10. 2β3 + 5β3 11. 6βπ₯π₯3 β 4βπ₯π₯3 12. 9π¦π¦β3π₯π₯ + 4π¦π¦β3π₯π₯
13. 12ππβ5ππ β 4ππβ5ππ 14. 5β32 β 3β8 + 2β3 15. β2β48 + 4β75 β β5
16. β163 + 3β543 17. β324 β 3β24 18. β5ππ + 2β45ππ3
19. β24π₯π₯3 β β3π₯π₯43 20. 4βπ₯π₯3 β 2β9π₯π₯ 21. 7β27π₯π₯3 + β3π₯π₯
22. 6β18π₯π₯ β β32π₯π₯ + 2β50π₯π₯ 23. 2β128ππ β β98ππ + 2β72ππ
24. β6π₯π₯43 + β48π₯π₯3 β β6π₯π₯3 25. 9οΏ½27π¦π¦2 β 14οΏ½108π¦π¦2 + 2οΏ½48π¦π¦2
26. 3β98ππ2 β 5β32ππ2 β 3β18ππ2 27. β4π¦π¦οΏ½π₯π₯π¦π¦3 + 7π₯π₯οΏ½π₯π₯3π¦π¦
28. 6ππβππππ5 β 9ππβππ3ππ 29. οΏ½β125ππ93 + πποΏ½β8ππ63
30. 3οΏ½π₯π₯5π¦π¦4 + 2π₯π₯οΏ½π₯π₯π¦π¦4 31. β125ππ5 β 2β125ππ43 32. π₯π₯β16π₯π₯3 + β2 β β2π₯π₯43
33. β9ππ β 9 + βππ β 1 34. β4π₯π₯ + 12 β βπ₯π₯ + 3 35. βπ₯π₯3 β π₯π₯2 β β4π₯π₯ β 4
36. β25π₯π₯ β 25 ββπ₯π₯3 β π₯π₯2 37. 4β33β 2β3
9 38. β27
2β 3β3
4
39. οΏ½49π₯π₯4
+ οΏ½81π₯π₯8
40. 2ππ οΏ½ ππ16
4 β 5ππ οΏ½ ππ81
4 41. β4 οΏ½ 4π¦π¦9
3 + 3 οΏ½ 9π¦π¦12
3
Discussion Point
42. A student simplifies an expression incorrectly:
β8 + β163 ?=
β4 β 2 + β8 β 23
?=
β4 β β2 + β83 β β23
?=
2β2 + 2β23
?=
4β4
?=
8
Explain any errors that the student made. What would you do differently?
338
Concept Check
43. Match each expression from Column I with the equivalent expression in Column II. Assume that A and B represent positive real numbers.
Column I Column II
A. οΏ½ππ + βπ΄π΄οΏ½οΏ½ππ β βπ΄π΄οΏ½ a. ππ β π΄π΄
B. οΏ½βππ + π΄π΄οΏ½οΏ½βππ β π΄π΄οΏ½ b. ππ + 2π΄π΄βππ + π΄π΄2
C. οΏ½βππ + βπ΄π΄οΏ½οΏ½βππ β βπ΄π΄οΏ½ c. ππ β π΄π΄2
D. οΏ½βππ + βπ΄π΄οΏ½2 d. ππ β 2βπππ΄π΄ + π΄π΄
E. οΏ½βππ β βπ΄π΄οΏ½2 e. ππ2 β π΄π΄
F. οΏ½βππ + π΄π΄οΏ½2 f. ππ + 2βπππ΄π΄ + π΄π΄
Multiply, and then simplify each product. Assume that all variables represent positive real numbers.
44. β5οΏ½3 β 2β5οΏ½ 45. β3οΏ½3β3 β β2οΏ½ 46. β2οΏ½5β2 β β10οΏ½
47. β3οΏ½β4β3 + β6οΏ½ 48. β23 οΏ½β43 β 2β323 οΏ½ 49. β33 οΏ½β93 + 2β213 οΏ½
50. οΏ½β3 β β2οΏ½οΏ½β3 + β2οΏ½ 51. οΏ½β5 + β7οΏ½οΏ½β5 β β7οΏ½ 52. οΏ½2β3 + 5οΏ½οΏ½2β3 β 5οΏ½
53. οΏ½6 + 3β2οΏ½οΏ½6 β 3β2οΏ½ 54. οΏ½5 β β5οΏ½2 55. οΏ½β2 + 3οΏ½
2
56. οΏ½βππ + 5βπποΏ½οΏ½βππ β 5βπποΏ½ 57. οΏ½2βπ₯π₯ β 3βπ¦π¦οΏ½οΏ½2βπ₯π₯ + 3βπ¦π¦οΏ½ 58. οΏ½β3 + β6οΏ½2
59. οΏ½β5 β β10οΏ½2 60. οΏ½2β5 + 3β2οΏ½
2 61. οΏ½2β3 β 5β2οΏ½
2
62. οΏ½4β3 β 5οΏ½οΏ½β3 β 2οΏ½ 63. οΏ½4β5 + 3β3οΏ½οΏ½3β5 β 2β3οΏ½ 64. οΏ½οΏ½2π¦π¦3 β 5οΏ½οΏ½οΏ½2π¦π¦3 + 1οΏ½
65. οΏ½βπ₯π₯ + 5 β 3οΏ½οΏ½βπ₯π₯ + 5 + 3οΏ½ 66. οΏ½βπ₯π₯ + 1 β βπ₯π₯οΏ½οΏ½βπ₯π₯ + 1 + βπ₯π₯οΏ½ 67. οΏ½βπ₯π₯ + 2 + βπ₯π₯ β 2οΏ½2
Concept Check
Given ππ(π₯π₯) and ππ(π₯π₯), find (ππ + ππ)(π₯π₯) and (ππππ)(π₯π₯).
68. ππ(π₯π₯) = 5π₯π₯β20π₯π₯ and ππ(π₯π₯) = 3β5π₯π₯3 69. ππ(π₯π₯) = 2π₯π₯β64π₯π₯4 and ππ(π₯π₯) = β3β4π₯π₯54
Rationalize each denominator and simplify, if possible. Assume that all variables represent positive real numbers.
70. β52β2
71. 35β3
72. 12β6
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73. β 15β24
74. β 10β20
75. οΏ½3π₯π₯20
76. οΏ½5π¦π¦32
77. β7ππ3
β3ππ3 78. οΏ½2π¦π¦43
β6π₯π₯43
79. β3ππ43
β5πΊπΊ23 80. πππποΏ½ππ3ππ4 81. 2π₯π₯
β18π₯π₯85
82. 176+β2
83. 43ββ5
84. 2β3β3ββ2
85. 6β33β2ββ3
86. 33β5+2β3
87. β2+β3β3+5β2
88. πΊπΊβ4βπΊπΊ+2
89. 4βπ₯π₯β2βπ¦π¦
90. β3+2βπ₯π₯β3β2βπ₯π₯
91. βπ₯π₯β23βπ₯π₯+βπ¦π¦
92. 2βππβππββππ
93. βπ₯π₯ββπ¦π¦βπ₯π₯+βπ¦π¦
Write each quotient in lowest terms. Assume that all variables represent positive real numbers.
94. 10β20β510
95. 12+6β36
96. 12β9β7218
97. 2π₯π₯+β8π₯π₯2
2π₯π₯ 98. 6ππβοΏ½24ππ3
3ππ 99. 9π₯π₯+β18
15
Discussion Point
100. In a certain problem in trigonometry, a student obtained the answer β3+11ββ3
. The textbook answer to this problem
was β2β β3. Was the studentβs answer equivalent to the textbook answer?
Analytic Skills Solve each problem.
101. The Great Pyramid at Giza has a square base with an area of 52,900 m2. What is the perimeter of its base?
102. The areas of two types of square floor tiles sold at a home improvement store are shown. How much longer is the side of the larger tile? Express the answer as a radical in simplest form and as a decimal to the nearest tenth.
Area = 72 ππππ2
Area = 128 ππππ2
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RD.5 Radical Equations In this section, we discuss techniques for solving radical equations. These are equations containing at least one radical expression with a variable, such as β3π₯π₯ β 2 = π₯π₯, or a variable
expression raised to a fractional exponent, such as (2π₯π₯)13 + 1 = 5.
At the end of this section, we revisit working with formulas involving radicals as well as application problems that can be solved with the use of radical equations.
Radical Equations
Definition 5.1 A radical equation is an equation in which a variable appears in one or more radicands. This includes radicands βhiddenβ under fractional exponents.
For example, since (π₯π₯ β 1)12 = βπ₯π₯ β 1, then the base π₯π₯ β 1 is, in fact, the βhiddenβ
radicand.
Some examples of radical equations are
π₯π₯ = β2π₯π₯, βπ₯π₯ + βπ₯π₯ β 2 = 5, (π₯π₯ β 4)32 = 8, β3 + π₯π₯3 = 5
Note that π₯π₯ = β2 is not a radical equation since there is no variable under the radical sign.
The process of solving radical equations involves clearing radicals by raising both sides of an equation to an appropriate power. This method is based on the following property of equality.
Power Rule: For any odd natural number ππ, the equation ππ = ππ is equivalent to the equation ππππ = ππππ.
For any even natural number ππ, if an equation ππ = ππ is true, then ππππ = ππππ is true.
When rephrased, the power rule for odd powers states that the solution sets to both equations, ππ = ππ and ππππ = ππππ, are exactly the same.
However, the power rule for even powers states that the solutions to the original equation ππ = ππ are among the solutions to the βpowerβ equation ππππ = ππππ.
Unfortunately, the reverse implication does not hold for even numbers ππ. We cannot conclude that ππ = ππ from the fact that ππππ = ππππ is true. For instance, 32 = (β3)2 is true but 3 β β3. This means that not all solutions of the equation ππππ = ππππ are in fact true solutions to the original equation ππ = ππ. Solutions that do not satisfy the original equation are called extraneous solutions or extraneous roots. Such solutions must be rejected.
For example, to solve β2 β π₯π₯ = π₯π₯, we may square both sides of the equation to obtain the quadratic equation
2 β π₯π₯ = π₯π₯2.
Then, we solve it via factoring and the zero-product property:
π₯π₯2 + π₯π₯ β 2 = 0
(π₯π₯ + 2)(π₯π₯ β 1) = 0
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So, the possible solutions are π₯π₯ = β2 and π₯π₯ = 1. Notice that π₯π₯ = 1 satisfies the original equation, as β2 β 1 = 1 is true. However, π₯π₯ = β2 does not satisfy the original equation as its left side equals to οΏ½2 β (β2) = β4 = 2, while the right side equals to β2. Thus, π₯π₯ = β2 is the extraneous root and as such, it does not belong to the solution set of the original equation. So, the solution set of the original equation is {1}.
Caution: When the power rule for even powers is used to solve an equation, every solution of the βpowerβ equation must be checked in the original equation.
Solving Equations with One Radical Solve each equation.
a. β3π₯π₯ + 4 = 4 b. β2π₯π₯ β 5 + 4 = 0
c. 2βπ₯π₯ + 1 = π₯π₯ β 7 d. βπ₯π₯ β 83 + 2 = 0
a. Since the radical in β3π₯π₯ + 4 = 4 is isolated on one side of the equation, squaring both
sides of the equation allows for clearing (reversing) the square root. Then, by solving the resulting polynomial equation, one can find the possible solution(s) to the original equation.
οΏ½β3π₯π₯ + 4οΏ½2
= (4)2
3π₯π₯ + 4 = 16
3π₯π₯ = 12
π₯π₯ = 4
To check if 4 is a true solution, it is enough to check whether or not π₯π₯ = 4 satisfies the original equation.
β3 β 4 + 4 ?= 4
β16 ?= 4
4 = 4
Since π₯π₯ = 4 satisfies the original equation, the solution set is {ππ}.
b. To solve β2π₯π₯ β 5 + 4 = 0, it is useful to isolate the radical on one side of the equation. So, consider the equation
β2π₯π₯ β 5 = β4
Notice that the left side of the above equation is nonnegative for any π₯π₯-value while the right side is constantly negative. Thus, such an equation cannot be satisfied by any π₯π₯-value. Therefore, this equation has no solution.
Solution
οΏ½βπποΏ½ππ
= οΏ½πππππποΏ½
ππ= ππ
true
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c. Squaring both sides of the equation gives us
οΏ½2βπ₯π₯ + 1οΏ½2
= (π₯π₯ β 7)2
4(π₯π₯ + 1) = π₯π₯2 β 14π₯π₯ + 49
4π₯π₯ + 4 = π₯π₯2 β 14π₯π₯ + 49
π₯π₯2 β 18π₯π₯ + 45 = 0
(π₯π₯ β 3)(π₯π₯ β 15) = 0
So, the possible solutions are π₯π₯ = 3 or π₯π₯ = 15. We check each of them by substituting them into the original equation.
If π₯π₯ = 3, then If π₯π₯ = 15, then
2β3 + 1?
= 3 β 7 2β15 + 1?
= 15β 7
2β4?
= β 4 2β16?
= 8
4 β β4 8 = 8
Since only 15 satisfies the original equation, the solution set is {15}.
d. To solve βπ₯π₯ β 83 + 2 = 0, we first isolate the radical by subtracting 2 from both sides
of the equation. βπ₯π₯ β 83 = β2
Then, to clear the cube root, we raise both sides of the equation to the third power.
οΏ½βπ₯π₯ β 83 οΏ½3
= (β2)3 So, we obtain
π₯π₯ β 8 = β8
π₯π₯ = 0
Since we applied the power rule for odd powers, the obtained solution is the true solution. So the solution set is {0}.
Observation: When using the power rule for odd powers checking the obtained solutions against the original equation is not necessary. This is because there is no risk of obtaining extraneous roots when applying the power rule for odd powers.
To solve radical equations with more than one radical term, we might need to apply the power rule repeatedly until all radicals are cleared. In an efficient solution, each application of the power rule should cause clearing of at least one radical term. For that reason, it is a good idea to isolate a single radical term on one side of the equation before each application of the power rule. For example, to solve the equation
the bracket is essential here
apply the perfect square formula
(ππ β ππ)2 = ππππ β ππππππ + ππππ
true false So π₯π₯ = 3 is the extraneous root.
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βπ₯π₯ β 3 + βπ₯π₯ + 5 = 4,
we isolate one of the radicals before squaring both sides of the equation. So, we have
οΏ½βπ₯π₯ β 3οΏ½2
= οΏ½4 β βπ₯π₯ + 5οΏ½2
π₯π₯ β 3 = 16οΏ½ππ2β 8βπ₯π₯ + 5οΏ½οΏ½οΏ½οΏ½οΏ½
2ππππ+ π₯π₯ + 5οΏ½οΏ½οΏ½
ππ2
Then, we isolate the remaining radical term and simplify, if possible. This gives us
8βπ₯π₯ + 5 = 24
βπ₯π₯ + 5 = 3
Squaring both sides of the last equation gives us
π₯π₯ + 5 = 9
π₯π₯ = 4
The reader is encouraged to check that π₯π₯ = ππ is the true solution to the original equation.
A general strategy for solving radical equations, including those with two radical terms, is as follows. Summary of Solving a Radical Equation
1. Isolate one of the radical terms. Make sure that one radical term is alone on one side of the equation.
2. Apply an appropriate power rule. Raise each side of the equation to a power that is the same as the index of the isolated radical.
3. Solve the resulting equation. If it still contains a radical, repeat steps 1 and 2.
4. Check all proposed solutions in the original equation.
5. State the solution set to the original equation.
Solving Equations Containing Two Radical Terms
Solve each equation.
a. β3π₯π₯ + 1 ββπ₯π₯ + 4 = 1 b. β4π₯π₯ β 53 = 2βπ₯π₯ + 13 a. We start solving the equation β3π₯π₯ + 1 β βπ₯π₯ + 4 = 1 by isolating one radical on one
side of the equation. This can be done by adding βπ₯π₯ + 4 to both sides of the equation. So, we have
β3π₯π₯ + 1 = 1 + βπ₯π₯ + 4
Solution
Remember that the
perfect square formula consists of three terms.
/ ( )2
/ Γ· 8
/ β5
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which after squaring give us
οΏ½β3π₯π₯ + 1οΏ½2
= οΏ½1 + βπ₯π₯ + 4οΏ½2
3π₯π₯ + 1 = 1 + 2βπ₯π₯ + 4 + π₯π₯ + 4
2π₯π₯ β 4 = 2βπ₯π₯ + 4
π₯π₯ β 2 = βπ₯π₯ + 4
To clear the remaining radical, we square both sides of the above equation again.
(π₯π₯ β 2)2 = οΏ½βπ₯π₯ + 4οΏ½2
π₯π₯2 β 4π₯π₯ + 4 = π₯π₯ + 4
π₯π₯2 β 5π₯π₯ = 0
The resulting polynomial equation can be solved by factoring and applying the zero-product property. Thus,
π₯π₯(π₯π₯ β 5) = 0.
So, the possible roots are π₯π₯ = 0 or π₯π₯ = 5.
We check each of them by substituting to the original equation.
If π₯π₯ = 0, then If π₯π₯ = 5, then
β3 β 0 + 1 β β0 + 4?
= 1 β3 β 5 + 1 ββ5 + 4?
= 1
β1 β β4?
= 1 β16 β β9?
= 1
1 β 2?
= 1 4 β 3?
= 1
β1 β 1 1 = 1 Only 5 satisfies the original equation. So, the solution set is {ππ}.
b. To solve the equation β4π₯π₯ β 53 = 2βπ₯π₯ + 13 , we would like to clear the cubic roots. This
can be done by cubing both of its sides, as shown below.
οΏ½β4π₯π₯ β 53 οΏ½3
= οΏ½2βπ₯π₯ + 13 οΏ½3
4π₯π₯ β 5 = 23(π₯π₯ + 1)
4π₯π₯ β 5 = 8π₯π₯ + 8
β13 = 4π₯π₯
π₯π₯ = β ππππππ
Since we applied the power rule for cubes, the obtained root is the true solution of the original equation.
/ Γ· 2
true false
the bracket is essential here
/ β4π₯π₯,β8
/ Γ· (β13)
Since π₯π₯ = 0 is the extraneous root, it
does not belong to the solution set.
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Formulas Containing Radicals
Many formulas involve radicals. For example, the period πΈπΈ, in seconds, of a pendulum of length πΏπΏ, in feet, is given by the formula
πΈπΈ = 2πποΏ½πΏπΏ
32
Sometimes, we might need to solve a radical formula for a specified variable. In addition to all the strategies for solving formulas for a variable, discussed in sections L2, F4, and RT6, we may need to apply the power rule to clear the radical(s) in the formula.
Solving Radical Formulas for a Specified Variable
Solve each formula for the indicated variable.
a. πΈπΈ = 12πποΏ½
ππππ for ππ b. ππ = οΏ½π΄π΄
π·π·3
β 1 for π·π·
a. Since ππ appears in the radicand, to solve πΈπΈ = 12πποΏ½
ππππ for ππ, we may want to clear the
radical by squaring both sides of the equation. So, we have
πΈπΈ2 = οΏ½1
2πποΏ½πππποΏ½
2
πΈπΈ2 =1
(2ππ)2 βππππ
πππ π πππ«π«ππππ = ππ
Note: We could also first multiply by 2ππ and then square both sides of the equation.
b. First, observe the position of π·π· in the equation ππ = οΏ½π΄π΄π·π·
3β 1. It appears in the
denominator of the radical. Therefore, to solve for π·π·, we may plan to isolate the cube root first, cube both sides of the equation to clear the radical, and finally bring π·π· to the numerator. So, we have
ππ = οΏ½πππ·π·
3β 1
(ππ + 1)3 = οΏ½οΏ½πππ·π·
3οΏ½
3
Solution
/ β 4ππ2ππ
/ +1
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(ππ + 1)3 =πππ·π·
π·π· =π¨π¨
(ππ + ππ)ππ
Radicals in Applications
Many application problems in sciences, engineering, or finances translate into radical equations.
Finding the Velocity of a Skydiver
When skydivers initially fall from an airplane, their velocity π£π£ in kilometers per hour after free falling ππ meters can be approximated by π£π£ = 15.9βππ. Approximately how far, in meters, do skydivers need to fall to attain 100 kph? We may substitute π£π£ = 100 into the equation π£π£ = 15.9βππ and solve it for ππ. Thus,
100 = 15.9βππ
6.3 β βππ
ππππ β ππ
So, skydivers fall at 100 kph approximately after 40 meters of free falling.
RD.5 Exercises
Vocabulary Check Complete each blank with the most appropriate term or phrase from the given list: even, extraneous, original, power, radical, solution.
1. A __________ equation contains at least one radical with a variable.
2. Radical equations are solved by applying an appropriate ________ rule to both sides of the equation.
3. The roots obtained in the process of solving a radical equation that do not satisfy the original equation are called _____________ roots. They do not belong to the ____________ set of the original equation.
4. If a power rule for _______ powers is applied, every possible solution must be checked in the ___________ equation.
Solution
/ β ππ, Γ· (ππ+ 1)3
/ Γ· 15.9
/ π π ππππππππππ πππππ‘π‘β π π πππππππ π
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Concept Check True or false.
5. β2π₯π₯ = π₯π₯2 β β5 is a radical equation.
6. When raising each side of a radical equation to a power, the resulting equation is equivalent to the original equation.
7. β3π₯π₯ + 9 = π₯π₯ cannot have negative solutions.
8. β9 is a solution to the equation βπ₯π₯ = β3. Solve each equation.
9. β7π₯π₯ β 3 = 6 10. οΏ½5π¦π¦ + 2 = 7 11. β6π₯π₯ + 1 = 3 12. β2ππ β 4 = 6
13. βπ₯π₯ + 2 = β6 14. οΏ½π¦π¦ β 3 = β2 15. βπ₯π₯3 = β3 16. βππ3 = β1
17. οΏ½π¦π¦ β 34 = 2 18. βππ + 14 = 3 19. 5 = 1βππ
20. 1
βπ¦π¦= 3
21. β3ππ + 1 β 4 = 0 22. β5π₯π₯ β 4 β 9 = 0 23. 4 βοΏ½π¦π¦ β 2 = 0
24. 9 β β4ππ + 1 = 0 25. π₯π₯ β 7 = βπ₯π₯ β 5 26. π₯π₯ + 2 = β2π₯π₯ + 7
27. 2βπ₯π₯ + 1 β 1 = π₯π₯ 28. 3βπ₯π₯ β 1 β 1 = π₯π₯ 29. π¦π¦ β 4 = οΏ½4 β π¦π¦
30. π₯π₯ + 3 = β9 β π₯π₯ 31. π₯π₯ = βπ₯π₯2 + 4π₯π₯ β 20 32. π₯π₯ = βπ₯π₯2 + 3π₯π₯ + 9 Concept Check
33. When solving the equation β3π₯π₯ + 4 = 8 β π₯π₯, a student wrote the following for the first step.
3π₯π₯ + 4 = 64 + π₯π₯2
Is this correct? Justify your answer.
34. When solving the equation β5π₯π₯ + 6 β βπ₯π₯ + 3 = 3, a student wrote the following for the first step.
(5π₯π₯ + 6) + (π₯π₯ + 3) = 9
Is this correct? Justify your answer. Solve each equation.
35. β5π₯π₯ + 1 = β2π₯π₯ + 7 36. οΏ½5π¦π¦ β 3 = οΏ½2π¦π¦ + 3 37. οΏ½ππ + 53 = οΏ½2ππ β 43
38. βπ₯π₯2 + 5π₯π₯ + 13 = βπ₯π₯2 + 4π₯π₯3 39. 2βπ₯π₯ β 3 = β7π₯π₯ + 15 40. β6π₯π₯ β 11 = 3βπ₯π₯ β 7
41. 3β2π‘π‘ + 3 β βπ‘π‘ + 10 = 0 42. 2οΏ½π¦π¦ β 1 βοΏ½3π¦π¦ β 1 = 0 43. βπ₯π₯ β 9 + βπ₯π₯ = 1
44. οΏ½π¦π¦ β 5 + οΏ½π¦π¦ = 5 45. β3ππ + βππ β 2 = 4 46. βπ₯π₯ + 5 β 2 = βπ₯π₯ β 1
47. β14 β ππ = βππ + 3 + 3 48. οΏ½ππ + 15βοΏ½2ππ + 7 = 1 49. β4ππ + 1 β βππ β 2 = 3
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50. 4 β βππ + 6 = βππ β 2 51. βπ₯π₯ β 5 + 1 = ββπ₯π₯ + 3
52. β3π₯π₯ β 5 + β2π₯π₯ + 3 + 1 = 0 53. β2ππβ 3 + 2 β βππ + 7 = 0
54. βπ₯π₯ + 2 + β3π₯π₯ + 4 = 2 55. β6π₯π₯ + 7 β β3π₯π₯ + 3 = 1
56. β4π₯π₯ + 7 β 4 = β4π₯π₯ β 1 57. οΏ½5π¦π¦ + 4 β 3 = οΏ½2π¦π¦ β 2
58. οΏ½2βπ₯π₯ + 11 = β4π₯π₯ + 2 59. οΏ½1 + β24 + 10π₯π₯ = β3π₯π₯ + 5
60. (2π₯π₯ β 9)12 = 2 + (π₯π₯ β 8)
12 61. (3ππ + 7)
12 = 1 + (ππ + 2)
12
62. (π₯π₯ + 1)12 β (π₯π₯ β 6)
12 = 1 63. οΏ½(π₯π₯2 β 9)
12 = 2
64. οΏ½βπ₯π₯ + 4 = βπ₯π₯ β 2 65. βππ2 + 30ππ = ππ + β5ππ
Discussion Point
66. Can the expression οΏ½7 + 4β3 βοΏ½7 β 4β3 be evaluated without the use of a calculator?
Solve each formula for the indicated variable.
67. ππ = οΏ½πΏπΏπΆπΆ for πΏπΏ 68. ππ = οΏ½2πΎπΎ
πΊπΊ for πΎπΎ 69. ππ = οΏ½2πΎπΎ
πΊπΊ for ππ
70. ππ = οΏ½πΊπΊπΊπΊπΉπΉ
for ππ 71. ππ = οΏ½πΊπΊπΊπΊπΉπΉ
for πΉπΉ 72. ππ = βπΏπΏ2 + πΈπΈ2 for πΈπΈ
73. πΉπΉ = 12ππβπΏπΏπΆπΆ
for πΆπΆ 74. πΈπΈ = 12πποΏ½
ππππ for ππ 75. πΈπΈ = 1
2πποΏ½ππππ for ππ
Analytic Skills Solve each problem.
76. According to Einsteinβs theory of relativity, time passes more quickly for bodies that travel very close to the
speed of light. The aging rate compared to the time spent on earth is given by the formula ππ =οΏ½ππππβππππ
οΏ½ππππ, where
ππ is the speed of light, and π£π£ is the speed of the traveling body. For example, the aging rate of 0.5 means that one year for the person travelling at the speed π£π£ corresponds to two years spent on earth.
a. Find the aging rate for a person traveling at 90% of the speed of light. b. Find the elapsed time on earth for one year of travelling time at 90% of the
speed of light.
77. Before determining the dosage of a drug for a patient, doctors will sometimes calculate the patientβs Body Surface Area (or BSA). One way to determine a personβs BSA, in square meters, is to use the formula
π©π©π©π©π¨π¨ = οΏ½ ππππππππππππ
, where w is the weight, in pounds, and h is the height, in centimeters, of the patient. Jason
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weighs 160 pounds and has a BSA of about 2β2 m2. How tall is he? Round the answer to the nearest centimeter.
78. The distance, ππ, to the horizon for an object β miles above the Earthβs surface is given by the equation ππ = β8000β + β2. How many miles above the Earthβs surface is a satellite if the distance to the horizon is 900 miles?
79. To calculate the minimum speed S, in miles per hour, that a car was traveling before skidding to a stop, traffic accident investigators use the formula ππ = οΏ½30πππΏπΏ, where ππ is the drag factor of the road surface and πΏπΏ is the length of a skid mark, in feet. Calculate the length of the skid marks for a car traveling at a speed of 30 mph that skids to a stop on a road surface with a drag factor of 0.5.