Chapter 5 Radical Functions and Equations

 

5.1 Roots and Radicals

Learning Objectives

  1. Identify and evaluate square and cube roots.
  2. Determine the domain of functions involving square and cube roots.
  3. Evaluate nth roots.
  4. Simplify radicals using the product and quotient rules for radicals.

Square and Cube Roots

Recall that a square rootA number that when multiplied by itself yields the original number. of a number is a number that when multiplied by itself yields the original number. For example, 5 is a square root of 25, because 52=25. Since (5)2=25, we can say that −5 is a square root of 25 as well. Every positive real number has two square roots, one positive and one negative. For this reason, we use the radical sign to denote the principal (nonnegative) square rootThe positive square root of a positive real number, denoted with the symbol . and a negative sign in front of the radical to denote the negative square root.

25=5Positivesquarerootof2525=5Negativesquarerootof25

Zero is the only real number with one square root.

0=0because02=0

Example 1

Evaluate.

  1. 121
  2. 81

Solution:

  1. 121=112=11
  2. 81=92=9

If the radicandThe expression A within a radical sign, An., the number inside the radical sign, can be factored as the square of another number, then the square root of the number is apparent. In this case, we have the following property:

a2=a       if       a0

Or more generally,

a2=|a|     if     a

The absolute value is important because a may be a negative number and the radical sign denotes the principal square root. For example,

(8)2=|8|=8

Make use of the absolute value to ensure a positive result.

Example 2

Simplify: (x2)2.

Solution:

Here the variable expression x2 could be negative, zero, or positive. Since the sign depends on the unknown quantity x, we must ensure that we obtain the principal square root by making use of the absolute value.

(x2)2=|x2|

Answer: |x2|

The importance of the use of the absolute value in the previous example is apparent when we evaluate using values that make the radicand negative. For example, when x=1,

(x2)2=|x2|=|12|=|1|=1

Next, consider the square root of a negative number. To determine the square root of −25, you must find a number that when squared results in −25:

25=?    or     (?)2=25

However, any real number squared always results in a positive number. The square root of a negative number is currently left undefined. For now, we will state that 25 is not a real number. Therefore, the square root functionThe function defined by f(x)=x. given by f(x)=x is not defined to be a real number if the x-values are negative. The smallest value in the domain is zero. For example, f(0)=0=0 and f(4)=4=2. Recall the graph of the square root function.

The domain and range both consist of real numbers greater than or equal to zero: [0,). To determine the domain of a function involving a square root we look at the radicand and find the values that produce nonnegative results.

Example 3

Determine the domain of the function defined by f(x)=2x+3.

Solution:

Here the radicand is 2x+3. This expression must be zero or positive. In other words,

2x+30

Solve for x.

2x+302x3x32

Answer: Domain: [32,)

A cube rootA number that when used as a factor with itself three times yields the original number, denoted with the symbol 3. of a number is a number that when multiplied by itself three times yields the original number. Furthermore, we denote a cube root using the symbol 3, where 3 is called the indexThe positive integer n in the notation n that is used to indicate an nth root.. For example,

643=4,     because   43=64

The product of three equal factors will be positive if the factor is positive and negative if the factor is negative. For this reason, any real number will have only one real cube root. Hence the technicalities associated with the principal root do not apply. For example,

643=4,     because   (4)3=64

In general, given any real number a, we have the following property:

a33=a     if      a

When simplifying cube roots, look for factors that are perfect cubes.

Example 4

Evaluate.

  1. 83
  2. 03
  3. 1273
  4. 13
  5. 1253

Solution:

  1. 83=233=2
  2. 03=033=0
  3. 1273=(13)33=13
  4. 13=(1)33=1
  5. 1253=(5)33=5

It may be the case that the radicand is not a perfect square or cube. If an integer is not a perfect power of the index, then its root will be irrational. For example, 23 is an irrational number that can be approximated on most calculators using the root button x. Depending on the calculator, we typically type in the index prior to pushing the button and then the radicand as follows:

3yx2=

Therefore, we have

231.260,     because     1.260^32

Since cube roots can be negative, zero, or positive we do not make use of any absolute values.

Example 5

Simplify: (y7)33.

Solution:

The cube root of a quantity cubed is that quantity.

(y7)33=y7

Answer: y7

Try this! Evaluate: 10003.

Answer: −10

Next, consider the cube root functionThe function defined by f(x)=x3.:

f(x)=x3     Cuberootfunction.

Since the cube root could be either negative or positive, we conclude that the domain consists of all real numbers. Sketch the graph by plotting points. Choose some positive and negative values for x, as well as zero, and then calculate the corresponding y-values.

xf(x)=x3OrderedPairs82f(8)=83=2(8,2)11f(1)=13=1(1,1)00f(0)=03=0(0,0)11f(1)=13=1(1,1)82f(8)=83=2(8,2)

Plot the points and sketch the graph of the cube root function.

The graph passes the vertical line test and is indeed a function. In addition, the range consists of all real numbers.

Example 6

Given g(x)=x+13+2, find g(9), g(2), g(1), and g(0). Sketch the graph of g.

Solution:

Replace x with the given values.

xg(x)g(x)=x+13+2OrderedPairs90g(9)=9+13+2=83+2=2+2=0(9,0)21g(2)=2+13+2=13+2=1+2=1(2,1)12g(1)=1+13+2=03+2=0+2=2(1,2)03g(0)=0+13+2=13+2=1+2=3(0,3)

We can also sketch the graph using the following translations:

y=x3Basiccuberootfunctiony=x+13Horizontalshiftleft1unity=x+13+2Verticalshiftup2units

Answer:

nth Roots

For any integer n2, we define an nth rootA number that when raised to the nth power (n2) yields the original number. of a positive real number as a number that when raised to the nth power yields the original number. Given any nonnegative real number a, we have the following property:

ann=a,     if       a0

Here n is called the index and an is called the radicand. Furthermore, we can refer to the entire expression An as a radicalUsed when referring to an expression of the form An.. When the index is an integer greater than or equal to 4, we say “fourth root,” “fifth root,” and so on. The nth root of any number is apparent if we can write the radicand with an exponent equal to the index.

Example 7

Simplify.

  1. 814
  2. 325
  3. 17
  4. 1164

Solution:

  1. 814=344=3
  2. 325=255=2
  3. 17=177=1
  4. 1164=(12)44=12

Note: If the index is n=2, then the radical indicates a square root and it is customary to write the radical without the index; a2=a.

We have already taken care to define the principal square root of a real number. At this point, we extend this idea to nth roots when n is even. For example, 3 is a fourth root of 81, because 34=81. And since (3)4=81, we can say that −3 is a fourth root of 81 as well. Hence we use the radical sign n to denote the principal (nonnegative) nth rootThe positive nth root when n is even. when n is even. In this case, for any real number a, we use the following property:

ann=|a|         Whenniseven

For example,

814=344=|3|=3  814=(3)44=|3|=3

The negative nth root, when n is even, will be denoted using a negative sign in front of the radical n.

814=344=3

We have seen that the square root of a negative number is not real because any real number that is squared will result in a positive number. In fact, a similar problem arises for any even index:

814=?     or     (?)4=81

We can see that a fourth root of −81 is not a real number because the fourth power of any real number is always positive.

4814646}Theseradicalsarenotrealnumbers.

You are encouraged to try all of these on a calculator. What does it say?

Example 8

Simplify.

  1. (10)44
  2. 1044
  3. (2y+1)66

Solution:

Since the indices are even, use absolute values to ensure nonnegative results.

  1. (10)44=|10|=10
  2. 1044=10,0004 is not a real number.
  3. (2y+1)66=|2y+1|

When the index n is odd, the same problems do not occur. The product of an odd number of positive factors is positive and the product of an odd number of negative factors is negative. Hence when the index n is odd, there is only one real nth root for any real number a. And we have the following property:

ann=a         Whennisodd

Example 9

Simplify.

  1. (10)55
  2. 325
  3. (2y+1)77

Solution:

Since the indices are odd, the absolute value is not used.

  1. (10)55=10
  2. 325=(2)55=2
  3. (2y+1)77=2y+1

In summary, for any real number a we have,

ann=|a|Whennisevenann=aWhennisodd

When n is odd, the nth root is positive or negative depending on the sign of the radicand.

273=333=3273=(3)33=3

When n is even, the nth root is positive or not real depending on the sign of the radicand.

164=244=2164=(2)44=|2|=2164Notarealnumber

Try this! Simplify: 8325.

Answer: 16

Simplifying Radicals

It will not always be the case that the radicand is a perfect power of the given index. If it is not, then we use the product rule for radicalsGiven real numbers An and Bn, ABn=AnBn. and the quotient rule for radicalsGiven real numbers An and Bn, ABn=AnBn where B0. to simplify them. Given real numbers An and Bn,

Product Rule for Radicals:

ABn=AnBn

Quotient Rule for Radicals:

ABn=AnBn

A radical is simplifiedA radical where the radicand does not consist of any factors that can be written as perfect powers of the index. if it does not contain any factors that can be written as perfect powers of the index.

Example 10

Simplify: 150.

Solution:

Here 150 can be written as 2352.

150=2352Applytheproductruleforradicals.=2352Simplify.=65=56

We can verify our answer on a calculator:

15012.25     and     5612.25

Also, it is worth noting that

12.252150

Answer: 56

Note: 56 is the exact answer and 12.25 is an approximate answer. We present exact answers unless told otherwise.

Example 11

Simplify: 1603.

Solution:

Use the prime factorization of 160 to find the largest perfect cube factor:

160=255=23225

Replace the radicand with this factorization and then apply the product rule for radicals.

1603=232253Applytheproductruleforradicals.=2332253Simplify.=2203

We can verify our answer on a calculator.

16035.43    and     22035.43

Answer: 2203

Example 12

Simplify: 3205.

Solution:

Here we note that the index is odd and the radicand is negative; hence the result will be negative. We can factor the radicand as follows:

320=13210=(1)5(2)510

Then simplify:

3205=(1)5(2)5105Applytheproductruleforradicals.=(1)55(2)55105Simplify.=12105=2105

Answer: 2105

Example 13

Simplify: 8643.

Solution:

In this case, consider the equivalent fraction with 8=(2)3 in the numerator and 64=43 in the denominator and then simplify.

8643=8643Applythequotientruleforradicals.=(2)33433Simplify.=24=12

Answer: 12

Try this! Simplify: 80814

Answer: 2543

Key Takeaways

  • To simplify a square root, look for the largest perfect square factor of the radicand and then apply the product or quotient rule for radicals.
  • To simplify a cube root, look for the largest perfect cube factor of the radicand and then apply the product or quotient rule for radicals.
  • When working with nth roots, n determines the definition that applies. We use ann=a when n is odd and ann=|a| when n is even.
  • To simplify nth roots, look for the factors that have a power that is equal to the index n and then apply the product or quotient rule for radicals. Typically, the process is streamlined if you work with the prime factorization of the radicand.

Topic Exercises

    Part A: Square and Cube Roots

      Simplify.

    1. 36

    2. 100

    3. 49
    4. 164
    5. 16

    6. 1

    7. (5)2

    8. (1)2

    9. 4

    10. 52

    11. (3)2

    12. (4)2

    13. x2

    14. (x)2

    15. (x5)2

    16. (2x1)2

    17. 643

    18. 2163

    19. 2163

    20. 643

    21. 83

    22. 13

    23. (2)33

    24. (7)33

    25. 183
    26. 8273
    27. (y)33

    28. y33

    29. (y8)33

    30. (2x3)33

      Determine the domain of the given function.

    1. g(x)=x+5

    2. g(x)=x2

    3. f(x)=5x+1

    4. f(x)=3x+4

    5. g(x)=x+1

    6. g(x)=x3

    7. h(x)=5x

    8. h(x)=23x

    9. g(x)=x+43

    10. g(x)=x33

      Evaluate given the function definition.

    1. Given f(x)=x1, find f(1), f(2), and f(5)

    2. Given f(x)=x+5, find f(5), f(1), and f(20)

    3. Given f(x)=x+3, find f(0), f(1), and f(16)

    4. Given f(x)=x5, find f(0), f(1), and f(25)

    5. Given g(x)=x3, find g(1), g(0), and g(1)

    6. Given g(x)=x32, find g(1), g(0), and g(8)

    7. Given g(x)=x+73, find g(15), g(7), and g(20)

    8. Given g(x)=x13+2, find g(0), g(2), and g(9)

      Sketch the graph of the given function and give its domain and range.

    1. f(x)=x+9

    2. f(x)=x3

    3. f(x)=x1+2

    4. f(x)=x+1+3

    5. g(x)=x13

    6. g(x)=x+13

    7. g(x)=x34

    8. g(x)=x3+5

    9. g(x)=x+231

    10. g(x)=x23+3

    11. f(x)=x3

    12. f(x)=x13

    Part B: nth Roots

      Simplify.

    1. 644

    2. 164

    3. 6254

    4. 14

    5. 2564

    6. 10,0004

    7. 2435

    8. 100,0005

    9. 1325
    10. 12435
    11. 164

    12. 16

    13. 325

    14. 15

    15. 1

    16. 164

    17. 6273

    18. 583

    19. 21,0003

    20. 72435

    21. 6164

    22. 12646

    23. 32516
    24. 6169
    25. 5271253
    26. 732755
    27. 58273
    28. 8625164
    29. 2100,0005

    30. 21287

    Part C: Simplifying Radicals

      Simplify.

    1. 96

    2. 500

    3. 480

    4. 450

    5. 320

    6. 216

    7. 5112

    8. 10135

    9. 2240

    10. 3162

    11. 15049
    12. 2009
    13. 675121
    14. 19281
    15. 543

    16. 243

    17. 483

    18. 813

    19. 403

    20. 1203

    21. 1623

    22. 5003

    23. 541253
    24. 403433
    25. 5483

    26. 21083

    27. 8964

    28. 71624

    29. 1605

    30. 4865

    31. 2242435
    32. 5325
    33. 1325
    34. 1646

      Simplify. Give the exact answer and the approximate answer rounded to the nearest hundredth.

    1. 60

    2. 600

    3. 9649
    4. 19225
    5. 2403

    6. 3203

    7. 2881253
    8. 62583
    9. 4864

    10. 2885

      Rewrite the following as a radical expression with coefficient 1.

    1. 215

    2. 37

    3. 510

    4. 103

    5. 273

    6. 363

    7. 254

    8. 324

    9. Each side of a square has a length that is equal to the square root of the square’s area. If the area of a square is 72 square units, find the length of each of its sides.

    10. Each edge of a cube has a length that is equal to the cube root of the cube’s volume. If the volume of a cube is 375 cubic units, find the length of each of its edges.

    11. The current I measured in amperes is given by the formula I=PR where P is the power usage measured in watts and R is the resistance measured in ohms. If a 100 watt light bulb has 160 ohms of resistance, find the current needed. (Round to the nearest hundredth of an ampere.)

    12. The time in seconds an object is in free fall is given by the formula t=s4 where s represents the distance in feet the object has fallen. How long will it take an object to fall to the ground from the top of an 8-foot stepladder? (Round to the nearest tenth of a second.)

    Part D: Discussion Board

    1. Explain why there are two real square roots for any positive real number and one real cube root for any real number.

    2. What is the square root of 1 and what is the cube root of 1? Explain why.

    3. Explain why 1 is not a real number and why 13 is a real number.

    4. Research and discuss the methods used for calculating square roots before the common use of electronic calculators.

Answers

  1. 6

  2. 23

  3. −4

  4. 5

  5. Not a real number

  6. −3

  7. |x|

  8. |x5|

  9. 4

  10. −6

  11. −2

  12. 2

  13. 12

  14. y

  15. y8

  16. [5,)

  17. [15,)
  18. (,1]

  19. (,5]

  20. (,)

  21. f(1)=0; f(2)=1; f(5)=2

  22. f(0)=3; f(1)=4; f(16)=7

  23. g(1)=1; g(0)=0; g(1)=1

  24. g(15)=2; g(7)=0; g(20)=3

  25. Domain: [9,); range: [0,)

  26. Domain: [1,); range: [2,)

  27. Domain: ; range:

  28. Domain: ; range:

  29. Domain: ; range:

  30. Domain: ; range:

  1. 4

  2. 5

  3. 4

  4. 3

  5. 12

  6. −2

  7. −2

  8. Not a real number

  9. 18

  10. −20

  11. Not a real number

  12. 154

  13. 3

  14. 103

  15. 20

  1. 46

  2. 430

  3. 85

  4. 207

  5. 815

  6. 567
  7. 15311
  8. 323

  9. 263

  10. 253

  11. 363

  12. 3235
  13. 1063

  14. 1664

  15. 255

  16. 2753
  17. 12

  18. 215; 7.75

  19. 467; 1.40

  20. 2303; 6.21

  21. 23635; 1.32

  22. 364; 4.70

  23. 60

  24. 250

  25. 563

  26. 804

  27. 62 units

  28. Answer: 0.79 ampere

  1. Answer may vary

  2. Answer may vary

5.2 Simplifying Radical Expressions

Learning Objectives

  1. Simplify radical expressions using the product and quotient rule for radicals.
  2. Use formulas involving radicals.

Simplifying Radical Expressions

An algebraic expression that contains radicals is called a radical expressionAn algebraic expression that contains radicals.. We use the product and quotient rules to simplify them.

Example 1

Simplify: 27x33.

Solution:

Use the fact that ann=a when n is odd.

27x33=33x33Applytheproductruleforradicals.=333x33Simplify.=3x=3x

Answer: 3x

Example 2

Simplify: 16y44.

Solution:

Use the fact that ann=|a| when n is even.

16y44=24y44Applytheproductruleforradicals.=244y44Simplify.=2|y|=2|y|

Since y is a variable, it may represent a negative number. Thus we need to ensure that the result is positive by including the absolute value.

Answer: 2|y|

Important Note

Typically, at this point in algebra we note that all variables are assumed to be positive. If this is the case, then y in the previous example is positive and the absolute value operator is not needed. The example can be simplified as follows.

16y44=24y44=244y44=2y

In this section, we will assume that all variables are positive. This allows us to focus on calculating nth roots without the technicalities associated with the principal nth root problem. For this reason, we will use the following property for the rest of the section,

ann=a,ifa0         nthroot

When simplifying radical expressions, look for factors with powers that match the index.

Example 3

Simplify: 12x6y3.

Solution:

Begin by determining the square factors of 12, x6, and y3.

12=223x6=(x3)2y3=y2y}Squarefactors

Make these substitutions, and then apply the product rule for radicals and simplify.

12x6y3=223(x3)2y2yApplytheproductruleforradicals.=22(x3)2y23ySimplify.=2x3y3y=2x3y3y

Answer: 2x3y3y

Example 4

Simplify: 18a5b8.

Solution:

Begin by determining the square factors of 18, a5, and b8.

18=232a5=a2a2a=(a2)2ab8=b4b4=(b4)2}Squarefactors

Make these substitutions, apply the product and quotient rules for radicals, and then simplify.

18a5b8=232(a2)2a(b4)2Applytheproductandquotientruleforradicals.=32(a2)22a(b4)2Simplify.=3a22ab4

Answer: 3a22ab4

Example 5

Simplify: 80x5y73.

Solution:

Begin by determining the cubic factors of 80, x5, and y7.

80=245=2325x5=x3x2y7=y6y=(y2)3y}Cubicfactors

Make these substitutions, and then apply the product rule for radicals and simplify.

80x5y73=2325x3x2(y2)3y3=233x33(y2)3325x2y3=2xy210x2y3=2xy210x2y3

Answer: 2xy210x2y3

Example 6

Simplify 9x6y3z93.

Solution:

The coefficient 9=32, and thus does not have any perfect cube factors. It will be left as the only remaining radicand because all of the other factors are cubes, as illustrated below:

x6=(x2)3y3=(y)3z9=(z3)3}Cubicfactors

Replace the variables with these equivalents, apply the product and quotient rules for radicals, and then simplify.

9x6y3z93=9(x2)3y3(z3)33=93(x2)33y33(z3)33=93x2yz3=x293yz3

Answer: x293yz3

Example 7

Simplify: 81a4b54.

Solution:

Determine all factors that can be written as perfect powers of 4. Here, it is important to see that b5=b4b. Hence the factor b will be left inside the radical.

81a4b54=34a4b4b4=344a44b44b4=3abb4=3abb4

Answer: 3abb4

Example 8

Simplify: 32x3y6z55.

Solution:

Notice that the variable factor x cannot be written as a power of 5 and thus will be left inside the radical. In addition, y6=y5y; the factor y will be left inside the radical as well.

32x3y6z55=(2)5x3y5yz55=(2)55y55z55x3y5=2yzx3y5=2yzx3y5

Answer: 2yzx3y5

Tip: To simplify finding an nth root, divide the powers by the index.

a6=a3,     which is  a6÷2=a3b63=b2,     which is  b6÷3=b2c66=c ,       which is   c6÷6=c1

If the index does not divide into the power evenly, then we can use the quotient and remainder to simplify. For example,

a5=a2a,      which is  a5÷2=a2r1b53=bb23,       which is   b5÷3=b1r2c145=c2c45,     which is  c14÷5=c2r4

The quotient is the exponent of the factor outside of the radical, and the remainder is the exponent of the factor left inside the radical.

Try this! Simplify: 162a7b5c43.

Answer: 3a2bc6ab2c3

Formulas Involving Radicals

Formulas often consist of radical expressions. For example, the period of a pendulum, or the time it takes a pendulum to swing from one side to the other and back, depends on its length according to the following formula.

T=2πL32

Here T represents the period in seconds and L represents the length in feet of the pendulum.

Example 9

If the length of a pendulum measures 112 feet, then calculate the period rounded to the nearest tenth of a second.

Solution:

Substitute 112=32 for L and then simplify.

T=2πL32=2π3232=2π32132Applythequotientruleforradicals.=2π364Simplify.=2π38=π341.36

Answer: The period is approximately 1.36 seconds.

Frequently you need to calculate the distance between two points in a plane. To do this, form a right triangle using the two points as vertices of the triangle and then apply the Pythagorean theorem. Recall that the Pythagorean theorem states that if given any right triangle with legs measuring a and b units, then the square of the measure of the hypotenuse c is equal to the sum of the squares of the legs: a2+b2=c2. In other words, the hypotenuse of any right triangle is equal to the square root of the sum of the squares of its legs.

Example 10

Find the distance between (−5, 3) and (1, 1).

Solution:

Form a right triangle by drawing horizontal and vertical lines though the two points. This creates a right triangle as shown below:

The length of leg b is calculated by finding the distance between the x-values of the given points, and the length of leg a is calculated by finding the distance between the given y-values.

a=31=2 unitsb=1(5)=1+5=6 units

Next, use the Pythagorean theorem to find the length of the hypotenuse.

c=22+62=4+36=40=410=210 units

Answer: The distance between the two points is 210 units.

Generalize this process to produce a formula that can be used to algebraically calculate the distance between any two given points.

Given two points, (x1,y1) and (x2,y2), the distance, d, between them is given by the distance formulaGiven two points (x1,y1) and (x2,y2), calculate the distance d between them using the formula d=(x2x1)2+(y2y1)2., d=(x2x1)2+(y2y1)2.

Example 11

Calculate the distance between (−4, 7) and (2, 1).

Solution:

Use the distance formula with the following points.

(x1,y1)(x2,y2)(4,7)(2,1)

It is a good practice to include the formula in its general form before substituting values for the variables; this improves readability and reduces the probability of making errors.

d=(x2x1)2+(y2y1)2=(2(4))2+(17)2=(2+4)2+(17)2=(6)2+(6)2=36+36=72=362=62

Answer: The distance between the two points is 62 units.

Example 12

Do the three points (2, −1), (3, 2), and (8, −3) form a right triangle?

Solution:

The Pythagorean theorem states that having side lengths that satisfy the property a2+b2=c2 is a necessary and sufficient condition of right triangles. In other words, if you can show that the sum of the squares of the leg lengths of the triangle is equal to the square of the length of the hypotenuse, then the triangle must be a right triangle. First, calculate the length of each side using the distance formula.

Geometry

Calculation

Points: (2, −1) and (8, −3)

a=(82)2+[3(1)]2=(6)2+(3+1)2=36+(2)2=36+4=40=210

Points: (2, −1) and (3, 2)

b=(32)2+[2(1)]2=(1)2+(2+1)2=1+(3)2=1+9=10

Points: (3, 2) and (8, −3)

c=(83)2+(32)2=(5)2+(5)2=25+25=50=52

Now we check to see if a2+b2=c2.

a2+b2=c2(210)2+(10)2=(52)24(10)2+(10)2=25(2)2410+10=25250=50

Answer: Yes, the three points form a right triangle.

Try this! The speed of a vehicle before the brakes were applied can be estimated by the length of the skid marks left on the road. On wet concrete, the speed v in miles per hour can be estimated by the formula v=23d, where d represents the length of the skid marks in feet. Estimate the speed of a vehicle before applying the brakes if the skid marks left behind measure 27 feet. Round to the nearest mile per hour.

Answer: 18 miles per hour

Key Takeaways

  • To simplify a radical expression, look for factors of the radicand with powers that match the index. If found, they can be simplified by applying the product and quotient rules for radicals, as well as the property ann=a, where a is nonnegative.
  • A radical expression is simplified if its radicand does not contain any factors that can be written as perfect powers of the index.
  • We typically assume that all variable expressions within the radical are nonnegative. This allows us to focus on simplifying radicals without the technical issues associated with the principal nth root. If this assumption is not made, we will ensure a positive result by using absolute values when simplifying radicals with even indices.

Topic Exercises

    Part A: Simplifying Radical Expressions

      Assume that the variable could represent any real number and then simplify.

    1. 9x2

    2. 16y2

    3. 8y33

    4. 125a33

    5. 64x44

    6. 81y44

    7. 36a4

    8. 100a8

    9. 4a6

    10. a10

    11. 18a4b5

    12. 48a5b3

    13. 128x6y86

    14. a6b7c86

    15. (5x4)2

    16. (3x5)4

    17. x26x+9

    18. x210x+25

    19. 4x2+12x+9

    20. 9x2+6x+1

      Simplify. (Assume all variable expressions represent positive numbers.)

    1. 49a2

    2. 64b2

    3. x2y2

    4. 25x2y2z2

    5. 180x3

    6. 150y3

    7. 49a3b2

    8. 4a4b3c

    9. 45x5y3

    10. 50x6y4

    11. 64r2s6t5

    12. 144r8s6t2

    13. (x+1)2

    14. (2x+3)2

    15. 4(3x1)2

    16. 9(2x+3)2

    17. 9x325y2
    18. 4x59y4
    19. m736n4
    20. 147m9n6
    21. 2r2s525t4
    22. 36r5s2t6
    23. 27a33

    24. 125b33

    25. 250x4y33

    26. 162a3b53

    27. 64x3y6z93

    28. 216x12y33

    29. 8x3y43

    30. 27x5y33

    31. a4b5c63

    32. a7b5c33

    33. 8x427y33
    34. x5125y63
    35. 360r5s12t133

    36. 540r3s2t93

    37. 81x44

    38. x4y44

    39. 16x4y84

    40. 81x12y44

    41. a4b5c64

    42. 54a6c84

    43. 128x64

    44. 243y74

    45. 32m10n55
    46. 37m9n105
    47. 34x2

    48. 79y2

    49. 5x4x2y

    50. 3y16x3y2

    51. 12aba5b3

    52. 6a2b9a7b2

    53. 2x8x63

    54. 5x227x33

    55. 2ab8a4b53

    56. 5a2b27a3b33

      Rewrite the following as a radical expression with coefficient 1.

    1. 3x6x

    2. 5y5y

    3. ab10a

    4. 2ab2a

    5. m2nmn

    6. 2m2n33n

    7. 2x3x3

    8. 3yy23

    9. 2y24y4

    10. x2y9xy25

    Part B: Formulas Involving Radicals

      The period T in seconds of a pendulum is given by the formula T=2πL32 where L represents the length in feet of the pendulum. Calculate the period, given each of the following lengths. Give the exact value and the approximate value rounded to the nearest tenth of a second.

    1. 8 feet

    2. 32 feet

    3. 12 foot

    4. 18 foot

      The time t in seconds an object is in free fall is given by the formula t=s4 where s represents the distance in feet the object has fallen. Calculate the time it takes an object to fall, given each of the following distances. Give the exact value and the approximate value rounded to the nearest tenth of a second.

    1. 48 feet

    2. 80 feet

    3. 192 feet

    4. 288 feet

    5. The speed of a vehicle before the brakes were applied can be estimated by the length of the skid marks left on the road. On dry pavement, the speed v in miles per hour can be estimated by the formula v=26d, where d represents the length of the skid marks in feet. Estimate the speed of a vehicle before applying the brakes on dry pavement if the skid marks left behind measure 27 feet. Round to the nearest mile per hour.

    6. The radius r of a sphere can be calculated using the formula r=6π2V32π, where V represents the sphere’s volume. What is the radius of a sphere if the volume is 36π cubic centimeters?

      Given the function find the y-intercept

    1. f(x)=x+12

    2. f(x)=x+83

    3. f(x)=x83

    4. f(x)=x+273

    5. f(x)=x+163

    6. f(x)=x+331

      Use the distance formula to calculate the distance between the given two points.

    1. (5, −7) and (3, −8)

    2. (−9, 7) and (−8, 4)

    3. (−3, −4) and (3, −6)

    4. (−5, −2) and (1, −6)

    5. (−1, 1) and (−4, 10)

    6. (8, −3) and (2, −12)

    7. (0, −6) and (−3, 0)

    8. (0, 0) and (8, −4)

    9. (12,12) and (1,32)

    10. (13,2) and (53,23)

      Determine whether or not the three points form a right triangle. Use the Pythagorean theorem to justify your answer.

    1. (2,−1), (−1,2), and (6,3)

    2. (−5,2), (−1, −2), and (−2,5)

    3. (−5,0), (0,3), and (6,−1)

    4. (−4,−1), (−2,5), and (7,2)

    5. (1,−2), (2,3), and (−3,4)

    6. (−2,1), (−1,−1), and (1,3)

    7. (−4,0), (−2,−10), and (3,−9)

    8. (0,0), (2,4), and (−2,6)

    Part D: Discussion Board

    1. Give a value for x such that x2x. Explain why it is important to assume that the variables represent nonnegative numbers.

    2. Research and discuss the accomplishments of Christoph Rudolff. What is he credited for?

    3. What is a surd, and where does the word come from?

    4. Research ways in which police investigators can determine the speed of a vehicle after an accident has occurred. Share your findings on the discussion board.

Answers

  1. 3|x|

  2. 2y

  3. 2|x|

  4. 6a2

  5. 2|a3|
  6. 3a2b22b

  7. 2|xy|2y26

  8. |5x4|

  9. |x3|

  10. |2x+3|

  11. 7a

  12. xy

  13. 6x5x

  14. 7aba

  15. 3x2y5xy

  16. 8rs3t2t

  17. x+1

  18. 2(3x1)

  19. 3xx5y
  20. m3m6n2
  21. rs22s5t2
  22. 3a

  23. 5xy2x3

  24. 4xy2z3

  25. 2xyy3

  26. abc2ab23

  27. 2xx33y
  28. 2rs4t445r2t3

  29. 3x

  30. 2xy2

  31. abcbc24

  32. 2x8x24

  33. 2m2n
  34. 6x

  35. 10x2y

  36. 12a3b2ab

  37. 4x3

  38. 4a2b2ab23

  39. 54x3

  40. 10a3b2

  41. m5n3

  42. 24x43

  43. 64y94

  1. π seconds; 3.1 seconds

  2. π4 seconds; 0.8 seconds

  3. 3 seconds; 1.7 seconds

  4. 23 seconds; 3.5 seconds

  5. 25 miles per hour

  6. (0,23)

  7. (0,2)

  8. (0,223)
  9. 5 units

  10. 210 units

  11. 310 units

  12. 35 units

  13. 52 units

  14. Right triangle

  15. Not a right triangle

  16. Right triangle

  17. Right triangle

  1. Answer may vary

  2. Answer may vary

5.3 Adding and Subtracting Radical Expressions

Learning Objectives

  1. Add and subtract like radicals.
  2. Simplify radical expressions involving like radicals.

Adding and Subtracting Like Radicals

Adding and subtracting radical expressions is similar to adding and subtracting like terms. Radicals are considered to be like radicalsRadicals that share the same index and radicand., or similar radicalsTerm used when referring to like radicals., when they share the same index and radicand. For example, the terms 26 and 56 contain like radicals and can be added using the distributive property as follows:

26+56=(2+5)6=76

Typically, we do not show the step involving the distributive property and simply write,

26+56=76

When adding terms with like radicals, add only the coefficients; the radical part remains the same.

Example 1

Add: 753+353.

Solution:

The terms are like radicals; therefore, add the coefficients.

753+353=1053

Answer: 1053

Subtraction is performed in a similar manner.

Example 2

Subtract: 410510.

Solution:

410510=(45)10=110=10

Answer: 10

If the radicand and the index are not exactly the same, then the radicals are not similar and we cannot combine them.

Example 3

Simplify: 105+629572.

Solution:

105+629572=10595+6272=52

We cannot simplify any further because 5 and 2 are not like radicals; the radicands are not the same.

Answer: 52

Caution: It is important to point out that 5252. We can verify this by calculating the value of each side with a calculator.

520.8252=31.73

In general, note that an±bna±bn.

Example 4

Simplify: 5103+310103210.

Solution:

5103+310103210=5103103+310210=4103+10

We cannot simplify any further, because 103 and 10 are not like radicals; the indices are not the same.

Answer: 4103+10

Adding and Subtracting Radical Expressions

Often, we will have to simplify before we can identify the like radicals within the terms.

Example 5

Subtract: 3218+50.

Solution:

At first glance, the radicals do not appear to be similar. However, after simplifying completely, we will see that we can combine them.

3218+50=16292+252=4232+52=62

Answer: 62

Example 6

Simplify: 1083+243323813.

Solution:

Begin by looking for perfect cube factors of each radicand.

1083+243323813=2743+8338432733Simplify.=343+233243333Combinelikeradicals.=4333

Answer: 4333

Try this! Simplify: 20+2735212.

Answer: 53

Next, we work with radical expressions involving variables. In this section, assume all radicands containing variable expressions are nonnegative.

Example 7

Simplify: 95x32x3+105x3.

Solution:

Combine like radicals.

95x32x3+105x3=95x3+105x32x3=5x32x3

We cannot combine any further because the remaining radical expressions do not share the same radicand; they are not like radicals. Note: 5x32x35x2x3.

Answer: 5x32x3

We will often find the need to subtract a radical expression with multiple terms. If this is the case, remember to apply the distributive property before combining like terms.

Example 8

Simplify: (5x4y)(4x7y).

Solution:

(5x4y)(4x7y)=5x4y4x+7yDistribute.=5x4x4y+7y=x+3y

Answer: x+3y

Until we simplify, it is often unclear which terms involving radicals are similar. The general steps for simplifying radical expressions are outlined in the following example.

Example 9

Simplify: 53x43+24x33(x24x3+43x33).

Solution:

Step 1: Simplify the radical expression. In this case, distribute and then simplify each term that involves a radical.

53x43+24x33(x24x3+43x33)=53x43+24x33x24x343x33=53xx33+83x33x83x343x33=5x3x3+2x332x3x34x33

Step2: Combine all like radicals. Remember to add only the coefficients; the variable parts remain the same.

=5x3x3+2x332x3x34x33=3x3x32x33

Answer: 3x3x32x33

Example 10

Simplify: 2a125a2ba280b+420a4b.

Solution:

2a125a2ba280b+420a4b=2a255a2ba2165b+445(a2)2bFactor.=2a5a5ba245b+42a25bSimplify.=10a25b4a25b+8a25bCombineliketerms.=14a25b

Answer: 14a25b

Try this! 2x6y3+xy33(y27x32x2x3y3)

Answer: 3x22y32yx3

Tip

Take careful note of the differences between products and sums within a radical. Assume both x and y are nonnegative.

ProductsSumsx2y2=xyx3y33=xyx2+y2x+yx3+y33x+y

The property abn=anbn says that we can simplify radicals when the operation in the radicand is multiplication. There is no corresponding property for addition.

Example 11

Calculate the perimeter of the triangle formed by the points (2,1), (3,6), and (2,1).

Solution:

The formula for the perimeter of a triangle is P=a+b+c where a, b, and c represent the lengths of each side. Plotting the points we have,

Use the distance formula to calculate the length of each side.

a=[3(2)]2+[6(1)]2=(3+2)2+(6+1)2=(1)2+(7)2=1+49=50=52b=[2(2)]2+[1(1)]2=(2+2)2+(1+1)2=(4)2+(2)2=16+4=20=25

Similarly we can calculate the distance between (−3, 6) and (2,1) and find that c=52 units. Therefore, we can calculate the perimeter as follows:

P=a+b+c=52+25+52=102+25

Answer: 102+25 units

Key Takeaways

  • Add and subtract terms that contain like radicals just as you do like terms. If the index and radicand are exactly the same, then the radicals are similar and can be combined. This involves adding or subtracting only the coefficients; the radical part remains the same.
  • Simplify each radical completely before combining like terms.

Topic Exercises

    Part A: Adding and Subtracting Like Radicals

      Simplify

    1. 10353

    2. 15686

    3. 93+53

    4. 126+36

    5. 457525

    6. 310810210

    7. 646+26

    8. 5101510210

    9. 1376257+52

    10. 10131215+5131815

    11. 65(4335)

    12. 122(66+2)

    13. (25310)(10+35)

    14. (83+615)(315)

    15. 463353+663

    16. 103+51034103

    17. (793433)(93333)

    18. (853+253)(253+6253)

      Simplify. (Assume all radicands containing variable expressions are positive.)

    1. 2x42x

    2. 53y63y

    3. 9x+7x

    4. 8y+4y

    5. 7xy3xy+xy

    6. 10y2x12y2x2y2x

    7. 2ab5a+6ab10a

    8. 3xy+6y4xy7y

    9. 5xy(3xy7xy)

    10. 8ab(2ab4ab)

    11. (32x3x)(2x73x)

    12. (y42y)(y52y)

    13. 5x312x3

    14. 2y33y3

    15. a3b5+4a3b5a3b5

    16. 8ab4+3ab42ab4

    17. 62a42a3+72a2a3

    18. 43a5+3a393a5+3a3

    19. (4xy4xy3)(24xy4xy3)

    20. (56y65y)(26y6+3y)

    21. 2x23x3(x23x3x3x3)

    22. 5y36y(6y4y36y)

    Part B: Adding and Subtracting Radical Expressions

      Simplify.

    1. 7512

    2. 2454

    3. 32+278

    4. 20+4845

    5. 2827+6312

    6. 90+244054

    7. 4580+2455

    8. 108+48753

    9. 42(2772)

    10. 35(2050)

    11. 163543

    12. 813243

    13. 1353+40353

    14. 108332343

    15. 227212

    16. 350432

    17. 324321848

    18. 6216224296

    19. 218375298+448

    20. 24512+220108

    21. (2363396)(712254)

    22. (2288+3360)(272740)

    23. 3543+525034163

    24. 416232384337503

      Simplify. (Assume all radicands containing variable expressions are positive.)

    1. 81b+4b

    2. 100a+a

    3. 9a2b36a2b

    4. 50a218a2

    5. 49x9y+x4y

    6. 9x+64y25xy

    7. 78x(316y218x)

    8. 264y(332y81y)

    9. 29m2n5m9n+m2n

    10. 418n2m2n8m+n2m

    11. 4x2y9xy216x2y+y2x

    12. 32x2y2+12x2y18x2y227x2y

    13. (9x2y16y)(49x2y4y)

    14. (72x2y218x2y)(50x2y2+x2y)

    15. 12m4nm75m2n+227m4n

    16. 5n27mn2+212mn4n3mn2

    17. 227a3ba48aba144a3b

    18. 298a4b2a162a2b+a200b

    19. 125a327a3

    20. 1000a2364a23

    21. 2x54x3216x43+52x43

    22. x54x33250x63+x223

    23. 16y24+81y24

    24. 32y45y45

    25. 32a34162a34+52a34

    26. 80a4b4+5a4b4a5b4

    27. 27x33+8x3125x33

    28. 24x3128x381x3

    29. 27x4y38xy33+x64xy3yx3

    30. 125xy33+8x3y3216xy33+10xy3

    31. (162x4y3250x4y23)(2x4y23384x4y3)
    32. (32x2y65243x6y25)(x2y65xxy25)

      Calculate the perimeters of the triangles formed by the following sets of vertices.

    1. {(−4, −5), (−4, 3), (2, 3)}

    2. {(−1, 1), (3, 1), (3, −2)}

    3. {(−3, 1), (−3, 5), (1, 5)}

    4. {(−3, −1), (−3, 7), (1, −1)}

    5. {(0,0), (2,4), (−2,6)}

    6. {(−5,−2), (−3,0), (1,−6)}

    7. A square garden that is 10 feet on each side is to be fenced in. In addition, the space is to be partitioned in half using a fence along its diagonal. How much fencing is needed to do this? (Round to the nearest tenth of a foot.)

    8. A garden in the shape of a square has an area of 150 square feet. How much fencing is needed to fence it in? (Hint: The length of each side of a square is equal to the square root of the area. Round to the nearest tenth of a foot.)

    Part C: Discussion Board

    1. Choose values for x and y and use a calculator to show that x+yx+y.

    2. Choose values for x and y and use a calculator to show that x2+y2x+y.

Answers

  1. 53

  2. 143

  3. 55

  4. 6

  5. 872

  6. 9543

  7. 5410

  8. 1063353

  9. 69333

  10. 32x

  11. 16x

  12. 5xy

  13. 8ab15a

  14. 9xy

  15. 22x+63x

  16. 7x3

  17. 4a3b5

  18. 132a52a3

  19. 4xy4

  20. x23x3+x3x3

  1. 33

  2. 22+33

  3. 5753

  4. 55

  5. 10233

  6. 23

  7. 453

  8. 23

  9. 23362

  10. 82+3

  11. 8366

  12. 2623

  13. 11b

  14. 3ab

  15. 8x5y

  16. 202x12y

  17. 8mn

  18. 2xy2yx

  19. 4xy

  20. 3m23n

  21. 2a3ab12a2ab

  22. 2a3

  23. 7x2x3

  24. 5y24

  25. 42a34

  26. 2x+2x3

  27. 7xxy33yx3

  28. 7x6xy36x2xy23

  29. 24 units

  30. 8+42 units

  31. 45+210 units

  32. 54.1 feet

  1. Answer may vary

5.4 Multiplying and Dividing Radical Expressions

Learning Objectives

  1. Multiply radical expressions.
  2. Divide radical expressions.
  3. Rationalize the denominator.

Multiplying Radical Expressions

When multiplying radical expressions with the same index, we use the product rule for radicals. Given real numbers An and Bn,

AnBn=ABn

Example 1

Multiply: 12363.

Solution:

Apply the product rule for radicals, and then simplify.

12363=1263Multiplytheradicands.=723Simplify.=23323=2323=293

Answer: 293

Often, there will be coefficients in front of the radicals.

Example 2

Multiply: 3652

Solution:

Using the product rule for radicals and the fact that multiplication is commutative, we can multiply the coefficients and the radicands as follows.

3652=3562Multiplicationiscommutative.=1512Multiplythecoefficientsandtheradicands.=1543Simplify.=1523=303

Typically, the first step involving the application of the commutative property is not shown.

Answer: 303

Example 3

Multiply: 34y23516y3.

Solution:

34y23516y3=1564y33Multiplythecoefficientsandthenmultiplytheradicands.=1543y33Simplify.=154y=60y

Answer: 60y

Use the distributive property when multiplying rational expressions with more than one term.

Example 4

Multiply: 52x(3x2x).

Solution:

Apply the distributive property and multiply each term by 52x.

52x(3x2x)=52x3x52x2xDistribute.=152x254x2Simplify.=15x252x=15x210x

Answer: 15x210x

Example 5

Multiply: 6x2y3(9x2y2354xy3).

Solution:

Apply the distributive property, and then simplify the result.

6x2y3(9x2y2354xy3)=6x2y39x2y236x2y354xy3=54x4y33524x3y23=272xx3y33583x3y23=3xy2x352x3y23=3xy2x310x3y23

Answer: 3xy2x310x3y23

The process for multiplying radical expressions with multiple terms is the same process used when multiplying polynomials. Apply the distributive property, simplify each radical, and then combine like terms.

Example 6

Multiply: (x5y)2.

Solution:

(x5y)2=(x5y)(x5y)

Begin by applying the distributive property.

=xx+x(5y)+(5y)x+(5y)(5y)=x25xy5xy+25y2=x10xy+25y

Answer: x10xy+25y

The binomials (a+b) and (ab) are called conjugatesThe factors (a+b) and (ab) are conjugates.. When multiplying conjugate binomials the middle terms are opposites and their sum is zero.

Example 7

Multiply: (10+3)(103).

Solution:

Apply the distributive property, and then combine like terms.

(10+3)(103)=1010+10(3)+310+3(3)Distribute.=10030+309Simplify.=1030+303oppositesaddto0=103=7

Answer: 7

It is important to note that when multiplying conjugate radical expressions, we obtain a rational expression. This is true in general

(x+y)(xy)=x2xy+xyy2=xy

Alternatively, using the formula for the difference of squares we have,

(a+b)(ab)=a2b2Differenceofsquares.(x+y)(xy)=(x)2(y)2=xy

Try this! Multiply: (32y)(3+2y). (Assume y is positive.)

Answer: 94y

Dividing Radical Expressions

To divide radical expressions with the same index, we use the quotient rule for radicals. Given real numbers An and Bn,

AnBn=ABn

Example 8

Divide: 96363.

Solution:

In this case, we can see that 6 and 96 have common factors. If we apply the quotient rule for radicals and write it as a single cube root, we will be able to reduce the fractional radicand.

96363=9663Applythequotientruleforradicalsandreducetheradicand.=163Simplify.=823=223

Answer: 223

Example 9

Divide: 50x6y48x3y.

Solution:

Write as a single square root and cancel common factors before simplifying.

50x6y48x3y=50x6y48x3yApplythequotientruleforradicalsandcancel.=25x3y34Simplify.=25x3y34=5xyxy2

Answer: 5xyxy2

Rationalizing the Denominator

When the denominator (divisor) of a radical expression contains a radical, it is a common practice to find an equivalent expression where the denominator is a rational number. Finding such an equivalent expression is called rationalizing the denominatorThe process of determining an equivalent radical expression with a rational denominator..

RadicalexpressionRationaldenominator12=22

To do this, multiply the fraction by a special form of 1 so that the radicand in the denominator can be written with a power that matches the index. After doing this, simplify and eliminate the radical in the denominator. For example:

12=1222=24=22

Remember, to obtain an equivalent expression, you must multiply the numerator and denominator by the exact same nonzero factor.

Example 10

Rationalize the denominator: 25x.

Solution:

The goal is to find an equivalent expression without a radical in the denominator. The radicand in the denominator determines the factors that you need to use to rationalize it. In this example, multiply by 1 in the form 5x5x.

25x=25x5x5xMultiplyby5x5x.=10x25x2Simplify.=10x5x

Answer: 10x5x

Sometimes, we will find the need to reduce, or cancel, after rationalizing the denominator.

Example 11

Rationalize the denominator: 3a26ab.

Solution:

In this example, we will multiply by 1 in the form 6ab6ab.

3a26ab=3a26ab6ab6ab=3a12ab36a2b2Simplify.=3a43ab6ab=6a3ab6abCancel.=3abb

Notice that b does not cancel in this example. Do not cancel factors inside a radical with those that are outside.

Answer: 3abb

Try this! Rationalize the denominator: 9x2y.

Answer: 32xy2y

Up to this point, we have seen that multiplying a numerator and a denominator by a square root with the exact same radicand results in a rational denominator. In general, this is true only when the denominator contains a square root. However, this is not the case for a cube root. For example, 1x3x3x3=x3x23 Note that multiplying by the same factor in the denominator does not rationalize it. In this case, if we multiply by 1 in the form of x23x23, then we can write the radicand in the denominator as a power of 3. Simplifying the result then yields a rationalized denominator.

1x3=1x3x23x23=x23x33=x23x

Therefore, to rationalize the denominator of a radical expression with one radical term in the denominator, begin by factoring the radicand of the denominator. The factors of this radicand and the index determine what we should multiply by. Multiply the numerator and denominator by the nth root of factors that produce nth powers of all the factors in the radicand of the denominator.

Example 12

Rationalize the denominator: 23253.

Solution:

The radical in the denominator is equivalent to 523. To rationalize the denominator, we need: 533. To obtain this, we need one more factor of 5. Therefore, multiply by 1 in the form of 5353.

23253=235235353Multiplybythecuberootof  factorsthatresultinpowersof3.=103533Simplify.=1035

Answer: 1035

Example 13

Rationalize the denominator: 27a2b23.

Solution:

In this example, we will multiply by 1 in the form 22b322b3.

27a2b23=33a32b23Applythequotientruleforradicals.=3a32b2322b322b3Multiplybythecuberootoffactorsthatresultinpowersof3.=322ab323b33Simplify.=34ab32b

Answer: 34ab32b

Example 14

Rationalize the denominator: 2x554x3y5.

Solution:

In this example, we will multiply by 1 in the form 23x2y4523x2y45.

2x554x3y5=2x5522x3y523x2y4523x2y45Multiplybythefifthrootoffactorsthatresultinpowersof5.=2x523x2y4525x5y55Simplify.=2x40x2y452xy=40x2y45y

Answer: 40x2y45y

When two terms involving square roots appear in the denominator, we can rationalize it using a very special technique. This technique involves multiplying the numerator and the denominator of the fraction by the conjugate of the denominator. Recall that multiplying a radical expression by its conjugate produces a rational number.

Example 15

Rationalize the denominator: 153.

Solution:

In this example, the conjugate of the denominator is 5+3. Therefore, multiply by 1 in the form (5+3)(5+3).

153=1(53)(5+3)(5+3)Multiplynumeratoranddenominatorbytheconjugateofthedenominator.=5+325+15159Simplify.=5+353=5+32

Answer: 5+32

Notice that the terms involving the square root in the denominator are eliminated by multiplying by the conjugate. We can use the property (a+b)(ab)=ab to expedite the process of multiplying the expressions in the denominator.

Example 16

Rationalize the denominator: 102+6.

Solution:

Multiply by 1 in the form 2626.

102+6=(10)(2+6)(26)(26)Multiplybytheconjugateofthedenominator.=206026Simplify.=454154=252154=2(515)4=5152=5152=5+152

Answer: 1552

Example 17

Rationalize the denominator: xyx+y.

Solution:

In this example, we will multiply by 1 in the form xyxy.

xyx+y=(xy)(x+y)(xy)(xy)Multiplybytheconjugateofthedenominator.=x2xyxy+y2xySimplify.=x2xy+yxy

Answer: x2xy+yxy

Try this! Rationalize the denominator: 2353

Answer: 53+311

Key Takeaways

  • To multiply two single-term radical expressions, multiply the coefficients and multiply the radicands. If possible, simplify the result.
  • Apply the distributive property when multiplying a radical expression with multiple terms. Then simplify and combine all like radicals.
  • Multiplying a two-term radical expression involving square roots by its conjugate results in a rational expression.
  • It is common practice to write radical expressions without radicals in the denominator. The process of finding such an equivalent expression is called rationalizing the denominator.
  • If an expression has one term in the denominator involving a radical, then rationalize it by multiplying the numerator and denominator by the nth root of factors of the radicand so that their powers equal the index.
  • If a radical expression has two terms in the denominator involving square roots, then rationalize it by multiplying the numerator and denominator by the conjugate of the denominator.

Topic Exercises

    Part A: Multiplying Radical Expressions

      Multiply. (Assume all variables represent non-negative real numbers.)

    1. 37

    2. 25

    3. 612

    4. 1015

    5. 26

    6. 515

    7. 77

    8. 1212

    9. 25710

    10. 31526

    11. (25)2

    12. (62)2

    13. 2x2x

    14. 5y5y

    15. 3a12

    16. 3a2a

    17. 42x36x

    18. 510y22y

    19. 3393

    20. 43163

    21. 153253

    22. 1003503

    23. 43103

    24. 18363

    25. (593)(263)

    26. (243)(343)

    27. (223)3

    28. (343)3

    29. 3a239a3

    30. 7b349b23

    31. 6x234x23

    32. 12y39y23

    33. 20x2y310x2y23

    34. 63xy312x4y23

    35. 5(35)

    36. 2(32)

    37. 37(273)

    38. 25(6310)

    39. 6(32)

    40. 15(5+3)

    41. x(x+xy)

    42. y(xy+y)

    43. 2ab(14a210b)

    44. 6ab(52a3b)

    45. 63(93203)

    46. 123(363+143)

    47. (25)(3+7)

    48. (3+2)(57)

    49. (234)(36+1)

    50. (526)(723)

    51. (53)2

    52. (72)2

    53. (23+2)(232)

    54. (2+37)(237)

    55. (a2b)2

    56. (ab+1)2

    57. What is the perimeter and area of a rectangle with length measuring 53 centimeters and width measuring 32 centimeters?

    58. What is the perimeter and area of a rectangle with length measuring 26 centimeters and width measuring 3 centimeters?

    59. If the base of a triangle measures 62 meters and the height measures 32 meters, then calculate the area.

    60. If the base of a triangle measures 63 meters and the height measures 36 meters, then calculate the area.

    Part B: Dividing Radical Expressions

      Divide. (Assume all variables represent positive real numbers.)

    1. 753
    2. 36010
    3. 7275
    4. 9098
    5. 90x52x
    6. 96y33y
    7. 162x7y52xy
    8. 363x4y93xy
    9. 16a5b232a2b23
    10. 192a2b732a2b23

    Part C: Rationalizing the Denominator

      Rationalize the denominator. (Assume all variables represent positive real numbers.)

    1. 15
    2. 16
    3. 23
    4. 37
    5. 5210
    6. 356
    7. 353
    8. 622
    9. 17x
    10. 13y
    11. a5ab
    12. 3b223ab
    13. 2363
    14. 1473
    15. 14x3
    16. 13y23
    17. 9x239xy23
    18. 5y2x35x2y3
    19. 3a23a2b23
    20. 25n325m2n3
    21. 327x2y5
    22. 216xy25
    23. ab9a3b5
    24. abcab2c35
    25. 3x8y2z5
    26. 4xy29x3yz45
    27. 3103
    28. 262
    29. 15+3
    30. 172
    31. 33+6
    32. 55+15
    33. 10535
    34. 22432
    35. 3+535
    36. 10210+2
    37. 233243+2
    38. 65+2252
    39. xyx+y
    40. xyxy
    41. x+yxy
    42. xyx+y
    43. aba+b
    44. ab+2ab2
    45. x52x
    46. 1xy
    47. x+2y2xy
    48. 3xyx+3y
    49. 2x+12x+11
    50. x+11x+1
    51. x+1+x1x+1x1
    52. 2x+32x32x+3+2x3
    53. The radius of the base of a right circular cone is given by r=3Vπh where V represents the volume of the cone and h represents its height. Find the radius of a right circular cone with volume 50 cubic centimeters and height 4 centimeters. Give the exact answer and the approximate answer rounded to the nearest hundredth.

    54. The radius of a sphere is given by r=3V4π3 where V represents the volume of the sphere. Find the radius of a sphere with volume 135 square centimeters. Give the exact answer and the approximate answer rounded to the nearest hundredth.

    Part D: Discussion

    1. Research and discuss some of the reasons why it is a common practice to rationalize the denominator.

    2. Explain in your own words how to rationalize the denominator.

Answers

  1. 21

  2. 62

  3. 23

  4. 7

  5. 702

  6. 20

  7. 2x

  8. 6a

  9. 24x3

  10. 3

  11. 533

  12. 253

  13. 3023

  14. 16

  15. 3a

  16. 2x3x3

  17. 2xy25x3

  18. 355

  19. 42321

  20. 3223

  21. x+xy

  22. 2a7b4b5a

  23. 3232153

  24. 6+141535

  25. 182+231264

  26. 8215

  27. 10

  28. a22ab+2b

  29. Perimeter: (103+62) centimeters; area: 156 square centimeters

  30. 18 square meters

  1. 5

  2. 265
  3. 3x25

  4. 9x3y2

  5. 2a

  1. 55
  2. 63
  3. 104
  4. 3153
  5. 7x7x
  6. ab5b
  7. 633
  8. 2x232x
  9. 36x2y3y
  10. 9ab32b
  11. 9x3y45xy
  12. 27a2b453
  13. 12xy3z452yz
  14. 310+9

  15. 532
  16. 1+2

  17. 5352
  18. 415

  19. 157623
  20. xy

  21. x2+2xy+yx2y
  22. a2ab+bab
  23. 5x+2x254x
  24. x2+3xy+y22xy
  25. 2x+1+2x+12x
  26. x+x21

  27. 56π2π centimeters; 3.45 centimeters

  1. Answer may vary

5.5 Rational Exponents

Learning Objectives

  1. Write expressions with rational exponents in radical form.
  2. Write radical expressions with rational exponents.
  3. Perform operations and simplify expressions with rational exponents.
  4. Perform operations on radicals with different indices.

Rational Exponents

So far, exponents have been limited to integers. In this section, we will define what rational (or fractional) exponents mean and how to work with them. All of the rules for exponents developed up to this point apply. In particular, recall the product rule for exponents. Given any rational numbers m and n, we have xmxn=xm+n For example, if we have an exponent of 1/2, then the product rule for exponents implies the following: 51/251/2=51/2+1/2=51=5 Here 51/2 is one of two equal factors of 5; hence it is a square root of 5, and we can write 51/2=5 Furthermore, we can see that 21/3 is one of three equal factors of 2. 21/321/321/3=21/3+1/3+1/3=23/3=21=2 Therefore, 21/3 is a cube root of 2, and we can write 21/3=23 This is true in general, given any nonzero real number a and integer n2, a1/n=an In other words, the denominator of a fractional exponent determines the index of an nth root.

Example 1

Rewrite as a radical.

  1. 61/2
  2. 61/3

Solution:

  1. 61/2=62=6
  2. 61/3=63

Example 2

Rewrite as a radical and then simplify.

  1. 161/2
  2. 161/4

Solution:

  1. 161/2=16=42=4
  2. 161/4=164=244=2

Example 3

Rewrite as a radical and then simplify.

  1. (64x3)1/3
  2. (32x5y10)1/5

Solution:

a.

(64x3)1/3=64x33=43x33=4x

b.

(32x5y10)1/5=32x5y105=(2)5x5(y2)55=2xy2

Next, consider fractional exponents where the numerator is an integer other than 1. For example, consider the following:

52/352/352/3=52/3+2/3+2/3=56/3=52

This shows that 52/3 is one of three equal factors of 52. In other words, 52/3 is a cube root of 52 and we can write:

52/3=523

In general, given any nonzero real number a where m and n are positive integers (n2),

am/n=amn

An expression with a rational exponentThe fractional exponent m/n that indicates a radical with index n and exponent m: am/n=amn. is equivalent to a radical where the denominator is the index and the numerator is the exponent. Any radical expression can be written with a rational exponent, which we call exponential formAn equivalent expression written using a rational exponent..

RadicalformExponentialformx25=x2/5

Example 4

Rewrite as a radical.

  1. 62/5
  2. 33/4

Solution:

  1. 62/5=625=365
  2. 33/4=334=274

Example 5

Rewrite as a radical and then simplify.

  1. 272/3
  2. (12)5/3

Solution:

We can often avoid very large integers by working with their prime factorization.

a.

272/3=2723=(33)23Replace27with33.=363Simplify.=32=9

b.

(12)5/3=(12)53Replace12with223.=(223)53Applytherulesforexponents.=210353Simplify.=29233323=2332323=24183

Given a radical expression, we might want to find the equivalent in exponential form. Assume all variables are positive.

Example 6

Rewrite using rational exponents: x35.

Solution:

Here the index is 5 and the power is 3. We can write

x35=x3/5

Answer: x3/5

Example 7

Rewrite using rational exponents: y36.

Solution:

Here the index is 6 and the power is 3. We can write

y36=y3/6=y1/2

Answer: y1/2

It is important to note that the following are equivalent.

am/n=amn=(an)m

In other words, it does not matter if we apply the power first or the root first. For example, we can apply the power before the nth root:

272/3=2723=(33)23=363=32=9

Or we can apply the nth root before the power:

272/3=(273)2=(333)2=(3)2=9

The results are the same.

Example 8

Rewrite as a radical and then simplify: (8)2/3.

Solution:

Here the index is 3 and the power is 2. We can write

(8)2/3=(83)2=(2)2=4

Answer: 4

Try this! Rewrite as a radical and then simplify: 1003/2.

Answer: 1,000

Some calculators have a caret button ^ which is used for entering exponents. If so, we can calculate approximations for radicals using it and rational exponents. For example, to calculate 2=21/2=2^(1/2)1.414, we make use of the parenthesis buttons and type

2^(1÷2)=

To calculate 223=22/3=2^(2/3)1.587, we would type

2^(2÷3)=

Operations Using the Rules of Exponents

In this section, we review all of the rules of exponents, which extend to include rational exponents. If given any rational numbers m and n, then we have

Product rule for exponents:

xmxn=xm+n

Quotient rule for exponents:

xmxn=xmn,x0

Power rule for exponents:

(xm)n=xmn

Power rule for a product:

(xy)n=xnyn

Power rule for a quotient:

(xy)n=xnyn,y0

Negative exponents:

xn=1xn

Zero exponent:

x0=1,x0

These rules allow us to perform operations with rational exponents.

Example 9

Simplify: 71/374/9.

Solution:

71/374/9=71/3+4/9Applytheproductrulexmxn=xm+n.=73/9+4/9=77/9

Answer: 77/9

Example 10

Simplify: x3/2x2/3.

Solution:

x3/2x2/3=x3/22/3Applythequotientrulexmxn=xmn.=x9/64/6=x5/6

Answer: x5/6

Example 11

Simplify: (y3/4)2/3.

Solution:

(y3/4)2/3=y(3/4)(2/3)Applythepowerrule(xm)n=xmn.=y6/12Multiplytheexponentsandreduce.=y1/2

Answer: y1/2

Example 12

Simplify: (81a8b12)3/4.

Solution:

(81a8b12)3/4=(34a8b12)3/4Rewrite81as34.=(34)3/4(a8)3/4(b12)3/4Applythepowerruleforaproduct(xy)n=xnyn.=34(3/4)a8(3/4)b12(3/4)Applythepowerruletoeachfactor.=33a6b9Simplify.=27a6b9

Answer: 27a6b9

Example 13

Simplify: (9x4)3/2.

Solution:

(9x4)3/2=1(9x4)3/2Applythedefinitionofnegativeexponentsxn=1xn.=1(32x4)3/2Write9as32andapplytherulesofexponents.=132(3/2)x4(3/2)=133x6=127x6

Answer: 127x6

Try this! Simplify: (125a1/4b6)2/3a1/6.

Answer: 25b4

Radical Expressions with Different Indices

To apply the product or quotient rule for radicals, the indices of the radicals involved must be the same. If the indices are different, then first rewrite the radicals in exponential form and then apply the rules for exponents.

Example 14

Multiply: 223.

Solution:

In this example, the index of each radical factor is different. Hence the product rule for radicals does not apply. Begin by converting the radicals into an equivalent form using rational exponents. Then apply the product rule for exponents.

223=21/221/3Equivalentsusingrationalexponents=21/2+1/3Applytheproductruleforexponents.=25/6=256

Answer: 256

Example 15

Divide: 4325.

Solution:

In this example, the index of the radical in the numerator is different from the index of the radical in the denominator. Hence the quotient rule for radicals does not apply. Begin by converting the radicals into an equivalent form using rational exponents and then apply the quotient rule for exponents.

4325=22325=22/321/5Equivalentsusingrationalexponents=22/31/5Applythequotientruleforexponents.=27/15=2715

Answer: 2715

Example 16

Simplify: 43.

Solution:

Here the radicand of the square root is a cube root. After rewriting this expression using rational exponents, we will see that the power rule for exponents applies.

43=223=(22/3)1/2Equivalentsusingrationalexponents=2(2/3)(1/2)Applythepowerruleforexponents.=21/3=23

Answer: 23

Key Takeaways

  • Any radical expression can be written in exponential form: amn=am/n.
  • Fractional exponents indicate radicals. Use the numerator as the power and the denominator as the index of the radical.
  • All the rules of exponents apply to expressions with rational exponents.
  • If operations are to be applied to radicals with different indices, first rewrite the radicals in exponential form and then apply the rules for exponents.

Topic Exercises

    Part A: Rational Exponents

      Express using rational exponents.

    1. 10

    2. 6

    3. 33

    4. 54

    5. 523

    6. 234

    7. 493

    8. 93

    9. x5

    10. x6

    11. x76

    12. x45

    13. 1x
    14. 1x23

      Express in radical form.

    1. 101/2

    2. 111/3

    3. 72/3

    4. 23/5

    5. x3/4

    6. x5/6

    7. x1/2

    8. x3/4

    9. (1x)1/3
    10. (1x)3/5
    11. (2x+1)2/3

    12. (5x1)1/2

      Write as a radical and then simplify.

    1. 641/2

    2. 491/2

    3. (14)1/2
    4. (49)1/2
    5. 41/2

    6. 91/2

    7. (14)1/2
    8. (116)1/2
    9. 81/3

    10. 1251/3

    11. (127)1/3
    12. (8125)1/3
    13. (27)1/3

    14. (64)1/3

    15. 161/4

    16. 6251/4

    17. 811/4

    18. 161/4

    19. 100,0001/5

    20. (32)1/5

    21. (132)1/5
    22. (1243)1/5
    23. 93/2

    24. 43/2

    25. 85/3

    26. 272/3

    27. 163/2

    28. 322/5

    29. (116)3/4
    30. (181)3/4
    31. (27)2/3

    32. (27)4/3

    33. (32)3/5

    34. (32)4/5

      Use a calculator to approximate an answer rounded to the nearest hundredth.

    1. 21/2

    2. 21/3

    3. 23/4

    4. 32/3

    5. 51/5

    6. 71/7

    7. (9)3/2

    8. 93/2

    9. Explain why (−4)^(3/2) gives an error on a calculator and −4^(3/2) gives an answer of −8.

    10. Marcy received a text message from Mark asking her age. In response, Marcy texted back “125^(2/3) years old.” Help Mark determine Marcy’s age.

    Part B: Operations Using the Rules of Exponents

      Perform the operations and simplify. Leave answers in exponential form.

    1. 53/251/2

    2. 32/337/3

    3. 51/251/3

    4. 21/623/4

    5. y1/4y2/5

    6. x1/2x1/4

    7. 511/352/3
    8. 29/221/2
    9. 2a2/3a1/6
    10. 3b1/2b1/3
    11. (81/2)2/3
    12. (36)2/3
    13. (x2/3)1/2
    14. (y3/4)4/5
    15. (y8)1/2
    16. (y6)2/3
    17. (4x2y4)1/2
    18. (9x6y2)1/2
    19. (2x1/3y2/3)3
    20. (8x3/2y1/2)2
    21. (36x4y2)1/2
    22. (8x3y6z3)1/3
    23. (a3/4a1/2)4/3
    24. (b4/5b1/10)10/3
    25. (4x2/3y4)1/2
    26. (27x3/4y9)1/3
    27. y1/2y2/3y1/6
    28. x2/5x1/2x1/10
    29. xyx1/2y1/3
    30. x5/4yxy2/5
    31. 49a5/7b3/27a3/7b1/4
    32. 16a5/6b5/48a1/2b2/3
    33. (9x2/3y6)3/2x1/2y
    34. (125x3y3/5)2/3xy1/3
    35. (27a1/4b3/2)2/3a1/6b1/2
    36. (25a2/3b4/3)3/2a1/6b1/3
    37. (16x2y1/3z2/3)3/2

    38. (81x8y4/3z4)3/4

    39. (100a2/3b4c3/2)1/2

    40. (125a9b3/4c1)1/3

    Part C: Radical Expressions with Different Indices

      Perform the operations.

    1. 9335

    2. 5255

    3. xx3

    4. yy4

    5. x23x4

    6. x35x3

    7. 100310
    8. 16543
    9. a23a
    10. b45b3
    11. x23x35
    12. x34x23
    13. 165

    14. 93

    15. 253

    16. 553

    17. 73

    18. 33

    Part D: Discussion Board

    1. Who is credited for devising the notation that allows for rational exponents? What are some of his other accomplishments?

    2. When using text, it is best to communicate nth roots using rational exponents. Give an example.

Answers

  1. 101/2

  2. 31/3

  3. 52/3

  4. 72/3

  5. x1/5

  6. x7/6

  7. x1/2

  8. 10

  9. 493

  10. x34

  11. 1x
  12. x3

  13. (2x+1)23

  14. 8

  15. 12

  16. 12

  17. 2

  18. 2

  19. 13

  20. −3

  21. 2

  22. 13

  23. 10

  24. 12

  25. 27

  26. 32

  27. 64

  28. 18

  29. 9

  30. −8

  31. 1.41

  32. 1.68

  33. 1.38

  34. Not a real number

  35. Answer may vary

  1. 25

  2. 55/6

  3. y13/20

  4. 125

  5. 2a1/2

  6. 2

  7. x1/3

  8. 1y4
  9. 2xy2

  10. 8xy2

  11. 16x2y
  12. a1/3

  13. 2x1/3y2
  14. y

  15. x1/2y2/3

  16. 7a2/7b5/4

  17. 27x1/2y8

  18. 9b1/2

  19. y1/264x3z
  20. a1/3b3/410b2
  1. 31315

  2. x56

  3. x1112

  4. 106

  5. a6

  6. x15

  7. 45

  8. 215

  9. 76

  1. Answer may vary

5.6 Solving Radical Equations

Learning Objectives

  1. Solve equations involving square roots.
  2. Solve equations involving cube roots.

Radical Equations

A radical equationAny equation that contains one or more radicals with a variable in the radicand. is any equation that contains one or more radicals with a variable in the radicand. Following are some examples of radical equations, all of which will be solved in this section:

2x1=3

4x2+732=0

x+2x=1

We begin with the squaring property of equalityGiven real numbers a and b, where a=b, then a2=b2.; given real numbers a and b, we have the following:

Ifa=b,thena2=b2.

In other words, equality is retained if we square both sides of an equation.

3=3(3)2=(3)29=9

The converse, on the other hand, is not necessarily true,

9=9(3)2=(3)233

This is important because we will use this property to solve radical equations. Consider a very simple radical equation that can be solved by inspection,

x=5

Here we can see that x=25 is a solution. To solve this equation algebraically, make use of the squaring property of equality and the fact that (a)2=a2=a when a is nonnegative. Eliminate the square root by squaring both sides of the equation as follows:

(x)2=(5)2x=25

As a check, we can see that 25=5 as expected. Because the converse of the squaring property of equality is not necessarily true, solutions to the squared equation may not be solutions to the original. Hence squaring both sides of an equation introduces the possibility of extraneous solutionsA properly found solution that does not solve the original equation., which are solutions that do not solve the original equation. For example,

x=5

This equation clearly does not have a real number solution. However, squaring both sides gives us a solution:

(x)2=(5)2x=25

As a check, we can see that 255. For this reason, we must check the answers that result from squaring both sides of an equation.

Example 1

Solve: 3x+1=4.

Solution:

We can eliminate the square root by applying the squaring property of equality.

3x+1=4(3x+1)2=(4)2Squarebothsides.3x+1=16Solve.3x=15x=5

Next, we must check.

3(5)+1=415+1=416=44=4

Answer: The solution is 5.

There is a geometric interpretation to the previous example. Graph the function defined by f(x)=3x+1 and determine where it intersects the graph defined by g(x)=4.

As illustrated, f(x)=g(x) where x=5.

Example 2

Solve: x3=x5.

Solution:

Begin by squaring both sides of the equation.

x3=x5(x3)2=(x5)2Squarebothsides.x3=x210x+25

The resulting quadratic equation can be solved by factoring.

x3=x210x+250=x211x+280=(x4)(x7)x4=0orx7=0x=4x=7

Checking the solutions after squaring both sides of an equation is not optional. Use the original equation when performing the check.

Checkx=4

Checkx=7

x3=x543=451=11=1

x3=x573=754=22=2

After checking, you can see that x=4 is an extraneous solution; it does not solve the original radical equation. Disregard that answer. This leaves x=7 as the only solution.

Answer: The solution is 7.

Geometrically we can see that f(x)=x+3 is equal to g(x)=x5 where x=7.

In the previous two examples, notice that the radical is isolated on one side of the equation. Typically, this is not the case. The steps for solving radical equations involving square roots are outlined in the following example.

Example 3

Solve: 2x1+2=x.

Solution:

Step 1: Isolate the square root. Begin by subtracting 2 from both sides of the equation.

2x1+2=x2x1=x2

Step 2: Square both sides. Squaring both sides eliminates the square root.

(2x1)2=(x2)22x1=x24x+4

Step 3: Solve the resulting equation. Here we are left with a quadratic equation that can be solved by factoring.

2x1=x24x+40=x26x+50=(x1)(x5)x1=0orx5=0x=1x=5

Step 4: Check the solutions in the original equation. Squaring both sides introduces the possibility of extraneous solutions; hence the check is required.

Checkx=1

Checkx=5

2x1+2=x2(1)1+2=11+2=11+2=13=1

2x1+2=x2(5)1+2=59+2=53+2=55=5

After checking, we can see that x=1 is an extraneous solution; it does not solve the original radical equation. This leaves x=5 as the only solution.

Answer: The solution is 5.

Sometimes there is more than one solution to a radical equation.

Example 4

Solve: 22x+5x=4.

Solution:

Begin by isolating the term with the radical.

22x+5x=4Addxtobothsides.22x+5=x+4

Despite the fact that the term on the left side has a coefficient, we still consider it to be isolated. Recall that terms are separated by addition or subtraction operators.

22x+5=x+4(22x+5)2=(x+4)2Squarebothsides.4(2x+5)=x2+8x+16

Solve the resulting quadratic equation.

4(2x+5)=x2+8x+168x+20=x2+8x+160=x240=(x+2)(x2)x+2=0orx2=0x=2x=2

Since we squared both sides, we must check our solutions.

Checkx=2

Checkx=2

22x+5x=422(2)+5(2)=424+5+2=421+2=42+2=44=4

22x+5x=422(2)+5(2)=424+52=4292=462=44=4

After checking, we can see that both are solutions to the original equation.

Answer: The solutions are ±2.

Sometimes both of the possible solutions are extraneous.

Example 5

Solve: 411xx+2=0.

Solution:

Begin by isolating the radical.

411xx+2=0Isolatetheradical.411x=x2(411x)2=(x2)2Squarebothsides.411x=x24x+4Solve.0=x2+7x0=x(x+7)

x=0orx+7=0x=7

Since we squared both sides, we must check our solutions.

Checkx=0

Checkx=7

411xx+2=0411(0)0+2=04+2=02+2=04=0

411xx+2=0411(7)(7)+2=04+77+7+2=081+9=09+9=018=0

Since both possible solutions are extraneous, the equation has no solution.

Answer: No solution, Ø

The squaring property of equality extends to any positive integer power n. Given real numbers a and b, we have the following:

Ifa=b,thenan=bn.

This is often referred to as the power property of equalityGiven any positive integer n and real numbers a and b where a=b, then an=bn.. Use this property, along with the fact that (an)n=ann=a, when a is nonnegative, to solve radical equations with indices greater than 2.

Example 6

Solve: 4x2+732=0.

Solution:

Isolate the radical, and then cube both sides of the equation.

4x2+732=0Isolatetheradical.4x2+73=2(4x2+73)3=(2)3Cubebothsides.4x2+7=8Solve.4x21=0(2x+1)(2x1)=02x+1=0or2x1=02x=12x=1x=12x=12

Check.

Checkx=12

Checkx=12

4x2+732=04(12)2+732=0414+732=01+732=0832=022=00=0

4x2+732=04(12)2+732=0414+732=01+732=0832=022=00=0

Answer: The solutions are ±12.

Try this! x33x+1=3

Answer: The solution is 33.

It may be the case that the equation has more than one term that consists of radical expressions.

Example 7

Solve: 5x3=4x1.

Solution:

Both radicals are considered isolated on separate sides of the equation.

5x3=4x1(5x3)2=(4x1)2Squarebothsides.5x3=4x1Solve.x=2

Check x=2.

5x3=4x15(2)3=4(2)1103=817=7

Answer: The solution is 2.

Example 8

Solve: x2+x143=x+503.

Solution:

Eliminate the radicals by cubing both sides.

x2+x143=x+503(x2+x143)3=(x+503)3Cubebothsides.x2+x14=x+50Solve.x264=0(x+8)(x8)=0x+8=0orx8=0x=8x=8

Check.

Checkx=8

Checkx=8

x2+x143=x+503(8)2+(8)143=(8)+503648143=423423=423

x2+x143=x+503(8)2+(8)143=(8)+50364+8143=583583=583

Answer: The solutions are ±8.

It may not be possible to isolate a radical on both sides of the equation. When this is the case, isolate the radicals, one at a time, and apply the squaring property of equality multiple times until only a polynomial remains.

Example 9

Solve: x+2x=1

Solution:

Begin by isolating one of the radicals. In this case, add x to both sides of the equation.

x+2x=1x+2=x+1

Next, square both sides. Take care to apply the distributive property to the right side.

(x+2)2=(x+1)2x+2=(x+1)(x+1)x+2=x2+x+x+1x+2=x+2x+1

At this point we have one term that contains a radical. Isolate it and square both sides again.

x+2=x+2x+11=2x(1)2=(2x)21=4x14=x

Check to see if x=14 satisfies the original equation x+2x=1.

14+214=19412=13212=122=11=1

Answer: The solution is 14.

Note: Because (A+B)2A2+B2, we cannot simply square each term. For example, it is incorrect to square each term as follows.

(x+2)2(x)2=(1)2Incorrect!

This is a common mistake and leads to an incorrect result. When squaring both sides of an equation with multiple terms, we must take care to apply the distributive property.

Example 10

Solve: 2x+10x+6=1

Solution:

Begin by isolating one of the radicals. In this case, add x+6 to both sides of the equation.

2x+10x+6=12x+10=x+6+1

Next, square both sides. Take care to apply the distributive property to the right side.

(2x+10)2=(x+6+1)22x+10=x+6+2x+6+12x+10=x+7+2x+6

At this point we have one term that contains a radical. Isolate it and square both sides again.

2x+10=x+7+2x+6x+3=2x+6(x+3)2=(2x+6)2x2+6x+9=4(x+6)x2+6x+9=4x+24x2+2x15=0(x3)(x+5)=0x3=0orx+5=0x=3x=5

Check.

Checkx=3

Checkx=5

2x+10x+6=12(3)+103+6=1169=143=11=1

2x+10x+6=12(5)+105+6=101=101=11=1

Answer: The solution is 3.

Try this! Solve: 4x+212x+22=1

Answer: The solution is 7.

Key Takeaways

  • Solve equations involving square roots by first isolating the radical and then squaring both sides. Squaring a square root eliminates the radical, leaving us with an equation that can be solved using the techniques learned earlier in our study of algebra.
  • Squaring both sides of an equation introduces the possibility of extraneous solutions. For this reason, you must check your solutions in the original equation.
  • Solve equations involving nth roots by first isolating the radical and then raise both sides to the nth power. This eliminates the radical and results in an equation that may be solved with techniques you have already mastered.
  • When more than one radical term is present in an equation, isolate them one at a time, and apply the power property of equality multiple times until only a polynomial remains.

Topic Exercises

    Part A: Solving Radical Equations

      Solve

    1. x=7

    2. x=4

    3. x+8=9

    4. x4=5

    5. x+7=4

    6. x+3=1

    7. 5x1=0

    8. 3x2=0

    9. 3x+1=2

    10. 5x4=4

    11. 7x+4+6=11

    12. 3x5+9=14

    13. 2x13=0

    14. 3x+12=0

    15. x+1=x+1

    16. 2x1=2x1

    17. 4x1=2x1

    18. 4x11=2x1

    19. x+8=x4

    20. 25x1=5x+1

    21. x3=3

    22. x3=4

    23. 2x+93=3

    24. 4x113=1

    25. 5x+73+3=1

    26. 3x63+5=2

    27. 42x+23=0

    28. 632x33=0

    29. 3(x+10)5=2

    30. 4x+35+5=4

    31. 8x+11=3x+1

    32. 23x4=2(3x+1)

    33. 2(x+10)=7x15

    34. 5(x4)=x+4

    35. 5x23=4x3

    36. 9(x1)3=3(x+7)3

    37. 3x+13=2(x1)3

    38. 9x3=3(x6)3

    39. 3x55=2x+85

    40. x+35=2x+55

    41. 4x+21=x

    42. 8x+9=x

    43. 4(2x3)=x

    44. 3(4x9)=x

    45. 2x1=x

    46. 32x9=x

    47. 9x+9=x+1

    48. 3x+10=x+4

    49. x1=x3

    50. 2x5=x4

    51. 163x=x6

    52. 73x=x3

    53. 32x+10=x+9

    54. 22x+5=x+4

    55. 3x11=x

    56. 22x+21=x

    57. 10x+415=x

    58. 6(x+3)3=x

    59. 8x24x+1=2x

    60. 18x26x+1=3x

    61. 5x+2=x+8

    62. 42(x+1)=x+7

    63. x225=x

    64. x2+9=x

    65. 3+6x11=x

    66. 2+9x8=x

    67. 4x+25x=7

    68. 8x+73x=10

    69. 24x+33=2x

    70. 26x+33=3x

    71. 2x4=1410x

    72. 3x6=3324x

    73. x2243=1

    74. x2543=3

    75. x2+6x3+1=4

    76. x2+2x3+5=7

    77. 25x210x73=2

    78. 9x212x233=3

    79. 4x2132=0

    80. 4x231=0

    81. x(2x+1)51=0

    82. 3x220x52=0

    83. 2x215x+25=(x+5)(x5)

    84. x24x+4=x(5x)

    85. 2(x2+3x20)3=(x+3)23

    86. 3x2+3x+403=(x5)23

    87. 2x5+2x=5

    88. 4x+132x=3

    89. 8x+1722x=3

    90. 3x62x3=1

    91. 2(x2)x1=1

    92. 2x+5x+3=2

    93. 2(x+1)3x+41=0

    94. 65x+33x1=0

    95. x21=2(x3)

    96. 1411x+79x=1

    97. x+1=32x

    98. 2x+9x+1=2

    99. x1/210=0

    100. x1/26=0

    101. x1/3+2=0

    102. x1/3+4=0

    103. (x1)1/23=0

    104. (x+2)1/26=0

    105. (2x1)1/3+3=0

    106. (3x1)1/32=0

    107. (4x+15)1/22x=0

    108. (3x+2)1/23x=0

    109. (2x+12)1/2x=6

    110. (4x+36)1/2x=9

    111. 2(5x+26)1/2=x+10

    112. 3(x1)1/2=x+1

    113. x1/2+(3x2)1/2=2

    114. (6x+1)1/2(3x)1/2=1

    115. (3x+7)1/2+(x+3)1/22=0

    116. (3x)1/2+(x+1)1/25=0

      Determine the roots of the given functions. Recall that a root is a value in the domain that results in zero. In other words, find x where f(x)=0.

    1. f(x)=x+52

    2. f(x)=2x31

    3. f(x)=2x+28

    4. f(x)=3x76

    5. f(x)=x+13+2

    6. f(x)=2x13+6

      Solve for the indicated variable.

    1. Solve for P: r=P1

    2. Solve for x: y=xh+k

    3. Solve for s: t=2sg

    4. Solve for L: T=2πL32

    5. Solve for R: I=PR

    6. Solve for h: r=3Vπh

    7. Solve for V: r=3V4π3

    8. Solve for c: a=b2π2c3

    9. The square root of 1 less than twice a number is equal to 2 less than the number. Find the number.

    10. The square root of 4 less than twice a number is equal to 6 less than the number. Find the number.

    11. The square root of twice a number is equal to one-half of that number. Find the number.

    12. The square root of twice a number is equal to one-third of that number. Find the number.

    13. The distance d in miles a person can see an object on the horizon is given by the formula d=6h2 where h represents the height in feet of the person’s eyes above sea level. How high must a person’s eyes be to see an object 5 miles away?

    14. The current I measured in amperes is given by the formula I=PR where P is the power usage measured in watts and R is the resistance measured in ohms. If a light bulb requires 1/2 amperes of current and uses 60 watts of power, then what is the resistance through the bulb?

      The period of a pendulum T in seconds is given by the formula T=2πL32 where L represents the length in feet. Calculate the length of a pendulum given the period. Give the exact value and the approximate value rounded to the nearest tenth of a foot.

    1. 1 second

    2. 2 seconds

    3. 12 second

    4. 13 second

      The time t in seconds, an object is in free fall is given by the formula t=s4 where s represents the distance it has fallen, in feet. Calculate the distance an object will fall given the amount of time.

    1. 1 second

    2. 2 seconds

    3. 12 second

    4. 14 second

    Part B: Discussion Board

    1. Discuss reasons why we sometimes obtain extraneous solutions when solving radical equations. Are there ever any conditions where we do not need to check for extraneous solutions? Why or why not?

    2. If an equation has multiple terms, explain why squaring all of them is incorrect. Provide an example.

Answers

  1. 49

  2. 1

  3. Ø

  4. 125

  5. 1

  6. 3

  7. 134

  8. 0

  9. 14

  10. Ø

  11. 27

  12. 9

  13. −3

  14. 6

  15. 23

  16. 2

  17. 7

  18. 2

  19. −3

  20. 13

  21. 7

  22. 2, 6

  23. 2

  24. −1, 8

  25. 5

  26. Ø

  27. −3, 3

  28. 2, 5

  29. −4, 4

  30. 12

  31. 2, 7

  32. Ø

  33. 10

  34. −6, −4

  35. 12,32

  36. Ø

  37. −5, 5

  38. −9, 3

  39. 15

  40. 32,32

  41. −1, 1/2

  42. 5, 10

  43. −7, 7

  44. 92

  45. 1

  46. 10

  47. Ø

  48. 3

  49. −1, 2

  50. 100

  51. −8

  52. 10

  53. −13

  54. 52

  55. −6, −4

  56. −2, 2

  57. 1

  58. −2

  59. −1

  60. 14

  61. −9

  62. P=(r+1)2

  63. s=gt22

  64. R=PI2

  65. V=4πr33

  66. 5

  67. 0, 8

  68. 1623 feet

  69. 8π2 feet; 0.8 feet

  70. 2π2 feet; 0.2 feet

  71. 16 feet

  72. 4 feet

  1. Answer may vary

5.7 Complex Numbers and Their Operations

Learning Objectives

  1. Define the imaginary unit and complex numbers.
  2. Add and subtract complex numbers.
  3. Multiply and divide complex numbers.

Introduction to Complex Numbers

Up to this point the square root of a negative number has been left undefined. For example, we know that 9 is not a real number.

9=?or(?)2=9

There is no real number that when squared results in a negative number. We begin to resolve this issue by defining the imaginary unitDefined as i=1 where i2=1., i, as the square root of −1.

i=1       andi2=1

To express a square root of a negative number in terms of the imaginary unit i, we use the following property where a represents any non-negative real number:

a=1a=1a=ia

With this we can write

9=19=19=i3=3i

If 9=3i, then we would expect that 3i squared will equal −9:

(3i)2=9i2=9(1)=9

In this way any square root of a negative real number can be written in terms of the imaginary unit. Such a number is often called an imaginary numberA square root of any negative real number..

Example 1

Rewrite in terms of the imaginary unit i.

  1. 7
  2. 25
  3. 72

Solution:

  1. 7=17=17=i7
  2. 25=125=125=i5=5i
  3. 72=1362=1362=i62=6i2

Notation Note: When an imaginary number involves a radical, we place i in front of the radical. Consider the following:

6i2=62i

Since multiplication is commutative, these numbers are equivalent. However, in the form 62i, the imaginary unit i is often misinterpreted to be part of the radicand. To avoid this confusion, it is a best practice to place i in front of the radical and use 6i2.

A complex numberA number of the form a+bi, where a and b are real numbers. is any number of the form, a+bi where a and b are real numbers. Here, a is called the real partThe real number a of a complex number a+bi. and b is called the imaginary partThe real number b of a complex number a+bi.. For example, 34i is a complex number with a real part of 3 and an imaginary part of −4. It is important to note that any real number is also a complex number. For example, 5 is a real number; it can be written as 5+0i with a real part of 5 and an imaginary part of 0. Hence, the set of real numbers, denoted , is a subset of the set of complex numbers, denoted .

={a+bi|a,b}

Complex numbers are used in many fields including electronics, engineering, physics, and mathematics. In this textbook we will use them to better understand solutions to equations such as x2+4=0. For this reason, we next explore algebraic operations with them.

Adding and Subtracting Complex Numbers

Adding or subtracting complex numbers is similar to adding and subtracting polynomials with like terms. We add or subtract the real parts and then the imaginary parts.

Example 2

Add: (52i)+(7+3i).

Solution:

Add the real parts and then add the imaginary parts.

(52i)+(7+3i)=52i+7+3i=5+72i+3i=12+i

Answer: 12+i

To subtract complex numbers, we subtract the real parts and subtract the imaginary parts. This is consistent with the use of the distributive property.

Example 3

Subtract: (107i)(9+5i).

Solution:

Distribute the negative sign and then combine like terms.

(107i)(9+5i)=107i95i=1097i5i=112i

Answer: 112i

In general, given real numbers a, b, c and d:

(a+bi)+(c+di)=(a+c)+(b+d)i(a+bi)(c+di)=(ac)+(bd)i

Example 4

Simplify: (5+i)+(23i)(47i).

Solution:

(5+i)+(23i)(47i)=5+i+23i4+7i=3+5i

Answer: 3+5i

In summary, adding and subtracting complex numbers results in a complex number.

Multiplying and Dividing Complex Numbers

Multiplying complex numbers is similar to multiplying polynomials. The distributive property applies. In addition, we make use of the fact that i2=1 to simplify the result into standard form a+bi.

Example 5

Multiply: 6i(23i).

Solution:

We begin by applying the distributive property.

6i(23i)=(6i)2(6i)3iDistribute.=12i+18i2Substitutei2=1.=12i+18(1)Simplify.=12i18=1812i

Answer: 1812i

Example 6

Multiply: (34i)(4+5i).

Solution:

(34i)(4+5i)=34+35i4i44i5iDistribute.=12+15i16i20i2Substitutei2=1.=12+15i16i20(1)=12i+20=32i

Answer: 32i

In general, given real numbers a, b, c and d:

(a+bi)(c+di)=ac+adi+bci+bdi2=ac+adi+bci+bd(1)=ac+(ad+bc)ibd=(acbd)+(ad+bc)i

Try this! Simplify: (32i)2.

Answer: 512i

Given a complex number a+bi, its complex conjugateTwo complex numbers whose real parts are the same and imaginary parts are opposite. If given a+bi, then its complex conjugate is abi. is abi. We next explore the product of complex conjugates.

Example 7

Multiply: (5+2i)(52i).

Solution:

(5+2i)(52i)=5552i+2i52i2i=2510i+10i4i2=254(1)=25+4=29

Answer: 29

In general, the product of complex conjugatesThe real number that results from multiplying complex conjugates: (a+bi)(abi)=a2+b2. follows:

(a+bi)(abi)=a2abi+biab2i2=a2abi+abib2(1)=a2+b2

Note that the result does not involve the imaginary unit; hence, it is real. This leads us to the very useful property

(a+bi)(abi)=a2+b2

To divide complex numbers, we apply the technique used to rationalize the denominator. Multiply the numerator and denominator by the conjugate of the denominator. The result can then be simplified into standard form a+bi.

Example 8

Divide: 123i.

Solution:

In this example, the conjugate of the denominator is 2+3i. Therefore, we will multiply by 1 in the form (2+3i)(2+3i).

123i=1(23i)(2+3i)(2+3i)=(2+3i)22+32=2+3i4+9=2+3i13

To write this complex number in standard form, we make use of the fact that 13 is a common denominator.

2+3i13=213+3i13=213+313i

Answer: 213+313i

Example 9

Divide: 15i4+i.

Solution:

15i4+i=(15i)(4+i)(4i)(4i)=4i20i+5i242+12=421i+5(1)16+1=421i517=121i17=1172117i

Answer: 1172117i

In general, given real numbers a, b, c and d where c and d are not both 0:

(a+bi)(c+di)=(a+bi)(c+di)(cdi)(cdi)=acadi+bcibdi2c2+d2=(ac+bd)+(bcad)ic2+d2=(ac+bdc2+d2)+(bcadc2+d2)i

Example 10

Divide: 83i2i.

Solution:

Here we can think of 2i=0+2i and thus we can see that its conjugate is 2i=02i.

83i2i=(83i)(2i)(2i)(2i)=16i+6i24i2=16i+6(1)4(1)=16i64=616i4=6416i4=324i

Because the denominator is a monomial, we could multiply numerator and denominator by 1 in the form of ii and save some steps reducing in the end.

83i2i=(83i)(2i)ii=8i3i22i2=8i3(1)2(1)=8i+32=8i2+32=4i32

Answer: 324i

Try this! Divide: 3+2i1i.

Answer: 12+52i

When multiplying and dividing complex numbers we must take care to understand that the product and quotient rules for radicals require that both a and b are positive. In other words, if an and bn are both real numbers then we have the following rules.

Product rule for radicals:abn=anbnQuotient rule for radicals:abn=anbn

For example, we can demonstrate that the product rule is true when a and b are both positive as follows:

49=3623=66=6

However, when a and b are both negative the property is not true.

49=?362i3i=66i2=66=6

Here 4 and 9 both are not real numbers and the product rule for radicals fails to produce a true statement. Therefore, to avoid some common errors associated with this technicality, ensure that any complex number is written in terms of the imaginary unit i before performing any operations.

Example 11

Multiply: 615.

Solution:

Begin by writing the radicals in terms of the imaginary unit i.

615=i6i15

Now the radicands are both positive and the product rule for radicals applies.

615=i6i15=i2615=(1)90=(1)910=(1)310=310

Answer: 310

Example 12

Multiply: 10(610).

Solution:

Begin by writing the radicals in terms of the imaginary unit i and then distribute.

10(610)=i10(i610)=i260i100=(1)415i100=(1)215i10=21510i

Answer: 21510i

In summary, multiplying and dividing complex numbers results in a complex number.

Try this! Simplify: (2i2)2(3i5)2.

Answer: 12+6i5

Key Takeaways

  • The imaginary unit i is defined to be the square root of negative one. In other words, i=1 and i2=1.
  • Complex numbers have the form a+bi where a and b are real numbers.
  • The set of real numbers is a subset of the complex numbers.
  • The result of adding, subtracting, multiplying, and dividing complex numbers is a complex number.
  • The product of complex conjugates, a+bi and abi, is a real number. Use this fact to divide complex numbers. Multiply the numerator and denominator of a fraction by the complex conjugate of the denominator and then simplify.
  • Ensure that any complex number is written in terms of the imaginary unit i before performing any operations.

Topic Exercises

    Part A: Introduction to Complex Numbers

      Rewrite in terms of imaginary unit i.

    1. 81

    2. 64

    3. 4

    4. 36

    5. 20

    6. 18

    7. 50

    8. 48

    9. 45

    10. 8

    11. 116

    12. 29

    13. 0.25

    14. 1.44

      Write the complex number in standard form a+bi.

    1. 524

    2. 359

    3. 2+38

    4. 4218

    5. 3246
    6. 2+7510
    7. 63512
    8. 72+824

      Given that i2=1 compute the following powers of i.

    1. i3

    2. i4

    3. i5

    4. i6

    5. i15

    6. i24

    Part B: Adding and Subtracting Complex Numbers

      Perform the operations.

    1. (3+5i)+(74i)

    2. (67i)+(52i)

    3. (83i)+(5+2i)

    4. (10+15i)+(1520i)

    5. (12+34i)+(1618i)

    6. (2516i)+(11032i)

    7. (5+2i)(83i)

    8. (7i)(69i)

    9. (95i)(8+12i)

    10. (11+2i)(137i)

    11. (114+32i)(4734i)

    12. (3813i)(1212i)

    13. (2i)+(3+4i)(65i)

    14. (7+2i)(6i)(34i)

    15. (13i)(112i)(16+16i)

    16. (134i)+(52+i)(1458i)

    17. (53i)(2+7i)(110i)

    18. (611i)+(2+3i)(84i)

    19. 16(31)

    20. 100+(9+7)

    21. (1+1)(11)

    22. (381)(539)

    23. (5225)(3+41)

    24. (121)(349)

    Part C: Multiplying and Dividing Complex Numbers

      Perform the operations.

    1. i(1i)

    2. i(1+i)

    3. 2i(74i)

    4. 6i(12i)

    5. 2i(34i)

    6. 5i(2i)

    7. (2+i)(23i)

    8. (35i)(12i)

    9. (1i)(89i)

    10. (1+5i)(5+2i)

    11. (4+3i)2

    12. (1+2i)2

    13. (25i)2

    14. (5i)2

    15. (1+i)(1i)

    16. (2i)(2+i)

    17. (42i)(4+2i)

    18. (6+5i)(65i)

    19. (12+23i)(1312i)
    20. (2313i)(1232i)
    21. (2i)3

    22. (13i)3

    23. 2(26)

    24. 1(1+8)

    25. 6(106)

    26. 15(310)

    27. (232)(2+32)

    28. (1+5)(15)

    29. (134)(2+9)

    30. (231)(1+216)

    31. (23i2)(3+i2)

    32. (1+i3)(22i3)

    33. 3i
    34. 5i
    35. 15+4i
    36. 134i
    37. 1512i
    38. 295+2i
    39. 20i13i
    40. 10i1+2i
    41. 105i3i
    42. 52i12i
    43. 5+10i3+4i
    44. 24i5+3i
    45. 26+13i23i
    46. 4+2i1+i
    47. 3i2i
    48. 5+2i4i
    49. 1abi
    50. ia+bi
    51. 111+1
    52. 1+919
    53. 618+4
    54. 12227

      Given that in=1in compute the following powers of i.

    1. i1

    2. i2

    3. i3

    4. i4

      Perform the operations and simplify.

    1. 2i(2i)i(34i)

    2. i(5i)3i(16i)

    3. 53(1i)2

    4. 2(12i)2+3i

    5. (1i)22(1i)+2

    6. (1+i)22(1+i)+2

    7. (2i2)2+5

    8. (3i5)2(i3)2

    9. (2i)2(2+i)2

    10. (i3+1)2(4i2)2

    11. (11+i)2
    12. (11+i)3
    13. (abi)2(a+bi)2

    14. (a2+ai+1)(a2ai+1)

    15. Show that both 2i and 2i satisfy x2+4=0.

    16. Show that both i and i satisfy x2+1=0.

    17. Show that both 32i and 3+2i satisfy x26x+13=0.

    18. Show that both 5i and 5+i satisfy x210x+26=0.

    19. Show that 3, 2i, and 2i are all solutions to x33x2+4x12=0.

    20. Show that −2, 1i, and 1+i are all solutions to x32x+4=0.

    Part D: Discussion Board.

    1. Research and discuss the history of the imaginary unit and complex numbers.

    2. How would you define i0 and why?

    3. Research what it means to calculate the absolute value of a complex number |a+bi|. Illustrate your finding with an example.

    4. Explore the powers of i. Look for a pattern and share your findings.

Answers

  1. 9i

  2. 2i

  3. 2i5

  4. 5i2

  5. 3i5

  6. i4

  7. 0.5i

  8. 54i

  9. 2+6i2

  10. 1263i
  11. 51274i
  12. i

  13. i

  14. i

  1. 10+i

  2. 3i

  3. 23+58i

  4. 3+5i

  5. 1717i

  6. 12+94i
  7. 1+8i

  8. 5623i
  9. 2

  10. 3+5i

  11. 2i

  12. 814i

  1. 1+i

  2. 8+14i

  3. 86i

  4. 74i

  5. 117i

  6. 7+24i

  7. 2120i

  8. 2

  9. 20

  10. 12136i
  11. 211i

  12. 22i3

  13. 6+2i15

  14. 22

  15. 209i

  16. 127i2

  17. 3i

  18. 541441i
  19. 3+6i

  20. 6+2i

  21. 7212i
  22. 11525i
  23. 1+8i

  24. 1232i
  25. aa2+b2+ba2+b2i
  26. i

  27. 6113311i
  28. i

  29. i

  30. 2+i

  31. 5+6i

  32. 0

  33. −3

  34. 4i2

  35. i2

  36. 4abi

  37. Proof

  38. Proof

  39. Proof

  1. Answer may vary

  2. Answer may vary

5.8 Review Exercises and Sample Exam

Review Exercises

    Roots and Radicals

      Simplify.

    1. 121

    2. (7)2

    3. (xy)2

    4. (6x7)2

    5. 1253

    6. 273

    7. (xy)33

    8. (6x+1)33

    9. Given f(x)=x+10, find f(1) and f(6).

    10. Given g(x)=x53, find g(4) and g(13).

    11. Determine the domain of the function defined by g(x)=5x+2.

    12. Determine the domain of the function defined by g(x)=3x13.

      Simplify.

    1. 2503

    2. 41203

    3. 31083

    4. 101325
    5. 681164
    6. 1286

    7. 1925

    8. 3420

    Simplifying Radical Expressions

      Simplify.

    1. 20x4y3

    2. 454x6y3

    3. x214x+49

    4. (x8)4

      Simplify. (Assume all variable expressions are nonzero.)

    1. 100x2y4

    2. 36a6b2

    3. 8a2b4
    4. 72x4yz6
    5. 10x150x7y4

    6. 5n225m10n6

    7. 48x6y3z23

    8. 270a10b8c33

    9. a3b564c63
    10. a2632b5c105
    11. The period T in seconds of a pendulum is given by the formula T=2πL32 where L represents the length in feet of the pendulum. Calculate the period of a pendulum that is 212 feet long. Give the exact answer and the approximate answer to the nearest hundredth of a second.

    12. The time in seconds an object is in free fall is given by the formula t=s4 where s represents the distance in feet the object has fallen. How long does it take an object to fall 28 feet? Give the exact answer and the approximate answer to the nearest tenth of a second.

    13. Find the distance between (−5, 6) and (−3,−4).

    14. Find the distance between (23,12) and (1,34).

      Determine whether or not the three points form a right triangle. Use the Pythagorean theorem to justify your answer.

    1. (−4,5), (−3,−1), and (3,0)

    2. (−1,−1), (1,3), and (−6,1)

    Adding and Subtracting Radical Expressions

      Simplify. Assume all radicands containing variables are nonnegative.

    1. 72+52

    2. 815215

    3. 143+525362

    4. 22ab5ab+7ab2ab

    5. 7x(3x+2y)

    6. (8yx7xy)(5xy12yx)

    7. (35+26)+(8536)

    8. (433123)(5332123)

    9. (210x+3y)(1+210x6y)

    10. (3aab23+6a2b3)+(9aab2312a2b3)
    11. 45+122075

    12. 2432+54232

    13. 23x2+45xx27+20x

    14. 56a2b+8a2b2224a2ba18b2

    15. 5y4x2y(x16y329x2y3)

    16. (2b9a2c3a16b2c)(64a2b2c9ba2c)

    17. 216x3125xy38x3

    18. 128x332x543+32x33

    19. 8x3y32x8y3+27x3y3+xy3

    20. 27a3b338ab33+a64b3ba3

    21. Calculate the perimeter of the triangle formed by the following set of vertices: {(3,2),(1,1),(1,2)}.

    22. Calculate the perimeter of the triangle formed by the following set of vertices: {(0,4),(2,0),(3,0)}.

    Multiplying and Dividing Radical Expressions

      Multiply.

    1. 615

    2. (42)2

    3. 2(210)

    4. (56)2

    5. (53)(5+3)

    6. (26+3)(253)

    7. (a5b)2

    8. 3xy(x2y)

    9. 3a2318a3

    10. 49a2b37a2b23

      Divide. Assume all variables represent nonzero numbers and rationalize the denominator where appropriate.

    1. 729
    2. 104864
    3. 55
    4. 152
    5. 326
    6. 2+510
    7. 183x
    8. 23x6xy
    9. 13x23
    10. 5ab25a2b3
    11. 5xz249x2y2z3
    12. 18x4y2z5
    13. 9x2y81xy2z35
    14. 27ab315a4bc25
    15. 153
    16. 32+1
    17. 36210
    18. xyxy
    19. 262+6
    20. a+bab
    21. The base of a triangle measures 26 units and the height measures 315 units. Find the area of the triangle.

    22. If each side of a square measures 5+210 units, find the area of the square.

    Rational Exponents

      Express in radical form.

    1. 111/2

    2. 22/3

    3. x3/5

    4. a4/5

      Write as a radical and then simplify.

    1. 161/2

    2. 721/2

    3. 82/3

    4. 321/3

    5. (19)3/2
    6. (1216)1/3

      Perform the operations and simplify. Leave answers in exponential form.

    1. 61/263/2

    2. 31/331/2

    3. 65/263/2
    4. 43/441/4
    5. (64x6y2)1/2

    6. (27x12y6)1/3

    7. (a4/3a1/2)2/5
    8. (16x4/3y2)1/2
    9. 56x3/4y3/214x1/2y2/3
    10. (4a4b2/3c4/3)1/22a2b1/6c2/3
    11. (9x4/3y1/3)3/2

    12. (16x4/5y1/2z2/3)3/4

      Perform the operations with mixed indices.

    1. yy25

    2. y3y35

    3. y23y5

    4. y23

    Solving Radical Equations

      Solve.

    1. 2x+3=13

    2. 3x2=4

    3. x5+4=8

    4. 5x+3+7=2

    5. 4x3=2x+15

    6. 8x15=x

    7. x1=13x

    8. 4x3=2x3

    9. x+5=5x

    10. x+3=3x1

    11. 2(x+1)x+2=1

    12. 6x+x2=2

    13. 3x2+x1=1

    14. 9x=x+161

    15. 4x33=2

    16. x83=1

    17. x(3x+10)3=2

    18. 2x2x3+4=5

    19. 3(x+4)(x+1)3=5x+373

    20. 3x29x+243=(x+2)23

    21. y1/23=0

    22. y1/3+3=0

    23. (x5)1/22=0

    24. (2x1)1/35=0

    25. (x1)1/2=x1/21

    26. (x2)1/2(x6)1/2=2

    27. (x+4)1/2(3x)1/2=2

    28. (5x+6)1/2=3(x+3)1/2

    29. Solve for g: t=2sg.

    30. Solve for x: y=x+432.

    31. The period in seconds of a pendulum is given by the formula T=2πL32 where L represents the length in feet of the pendulum. Find the length of a pendulum that has a period of 112 seconds. Find the exact answer and the approximate answer rounded off to the nearest tenth of a foot.

    32. The outer radius of a spherical shell is given by the formula r=3V4π3+2 where V represents the inner volume in cubic centimeters. If the outer radius measures 8 centimeters, find the inner volume of the sphere.

    33. The speed of a vehicle before the brakes are applied can be estimated by the length of the skid marks left on the road. On dry pavement, the speed v in miles per hour can be estimated by the formula v=26d, where d represents the length of the skid marks in feet. Estimate the length of a skid mark if the vehicle is traveling 30 miles per hour before the brakes are applied.

    34. Find the real root of the function defined by f(x)=x33+2.

    Complex Numbers and Their Operations

      Write the complex number in standard form a+bi.

    1. 516

    2. 256

    3. 3+810
    4. 1246

      Perform the operations.

    1. (612i)+(4+7i)

    2. (3+2i)(64i)

    3. (12i)(3432i)
    4. (5815i)+(3223i)
    5. (52i)(67i)+(44i)

    6. (103i)+(20+5i)(3015i)

    7. 4i(23i)

    8. (2+3i)(52i)

    9. (4+i)2

    10. (83i)2

    11. (3+2i)(32i)

    12. (1+5i)(15i)

    13. 2+9i2i
    14. i12i
    15. 4+5i2i
    16. 32i3+2i
    17. 105(23i)2

    18. (23i)2(23i)+4

    19. (11i)2
    20. (1+2i3i)2
    21. 8(34)

    22. (118)(32)

    23. (510)2

    24. (12)2(1+2)2

    25. Show that both 5i and 5i satisfy x2+25=0.

    26. Show that both 12i and 1+2i satisfy x22x+5=0.

Answers

  1. −11

  2. |xy|

  3. 5

  4. xy

  5. f(1)=3; f(6)=4

  6. [25,)
  7. 523

  8. 943

  9. −9

  10. 265

  1. 2x2|y|5y

  2. |x7|

  3. 10xy2

  4. 2a2b2
  5. 50x4y26x

  6. 2x2y6z23

  7. abb234c2
  8. π54 seconds; 1.76 seconds

  9. 226 units

  10. Right triangle

  1. 122

  2. 932

  3. 4x2y

  4. 1156

  5. 1310x+9y

  6. 533

  7. x3+55x

  8. 12xyy

  9. 4x35xy3

  10. 2xy3

  11. 4+213 units

  1. 310

  2. 225

  3. 22

  4. a10ab+25b

  5. 3a23

  6. 22

  7. 5

  8. 64
  9. 63xx
  10. 9x33x
  11. 35x2yz37xy
  12. 3xy3x4y3z25z
  13. 5+32
  14. 6+15

  15. 2+3

  16. 910 square units

  1. 11

  2. x35

  3. 4

  4. 4

  5. 1/27

  6. 36

  7. 6

  8. 8x3y

  9. a1/3

  10. 4x1/4y5/6

  11. x227y1/2
  12. y910

  13. y715

  1. 25

  2. 21

  3. 9

  4. 4

  5. 4

  6. 7

  7. 1

  8. 114

  9. 4,23

  10. 5,53

  11. 9

  12. 9

  13. 1

  14. 12

  15. g=2st2

  16. 18π2 feet; 1.8 feet

  17. 37.5 feet

  1. 54i

  2. 310+25i
  3. 105i

  4. 14+12i

  5. 3+i

  6. 12+8i

  7. 15+8i

  8. 13

  9. 92i

  10. 35+145i

  11. 35+60i

  12. 12i

  13. 42+2i6

  14. 15+102

  15. Answer may vary

Sample Exam

      Simplify. (Assume all variables are positive.)

    1. 5x121x2y4

    2. 2xy264x6y93

    3. Calculate the distance between (5,3) and (2,6).

    4. The time in seconds an object is in free fall is given by the formula t=s4 where s represents the distance in feet that the object has fallen. If a stone is dropped into a 36-foot pit, how long will it take to hit the bottom of the pit?

      Perform the operations and simplify. (Assume all variables are positive and rationalize the denominator where appropriate.)

    1. 150xy2218x3+y24x+x128x

    2. 316x3y23(2x250y2354x3y23)
    3. 22(236)

    4. (105)2

    5. 62+3
    6. 2x2xy
    7. 18xy2z45
    8. Simplify: 813/4.

    9. Express in radical form: x3/5.

      Simplify. Assume all variables are nonzero and leave answers in exponential form.

    1. (81x4y2)1/2

    2. (25a4/3b8)3/2a1/2b

      Solve.

    1. x5=1

    2. 5x23+6=4

    3. 52x+52x=11

    4. 43x+2=x

    5. 2x+5x+3=2

    6. The time in seconds an object is in free fall is given by the formula t=s4 where s represents the distance in feet that the object has fallen. If a stone is dropped into a pit and it takes 4 seconds to reach the bottom, how deep is the pit?

    7. The width in inches of a container is given by the formula w=4V32+1 where V represents the inside volume in cubic inches of the container. What is the inside volume of the container if the width is 6 inches?

      Perform the operations and write the answer in standard form.

    1. 3(63)

    2. 4+3i2i

    3. 63(23i)2

Answers

  1. 55x2y2

  2. 310 units

  3. 7y6x+2x2x

  4. 4123

  5. 23+32

  6. 4x4y3z52xyz
  7. 1x35
  8. 125a3/2b11

  9. 65

  10. Ø

  11. 256 feet

  12. 3+3i2

  13. 21+36i