CHAPTER 4 RADICAL EXPRESSIONS 4.1 the Nth Root Of

CHAPTER 4 RADICAL EXPRESSIONS

th 4.1 The n Root of a Real Number

A real number a is called the nth root of a real number b if an = b . Thus, for example: 2 is a square root of 4 since 22 = 4 . −2 is also a square root of 4 since ()−22 = 4 .

2 is a cube root of 8 since 23 = 8 . Note that 8 has no other real cube roots.

−2 is a cube root of −8 since ()−23 = − 8 . Note that −8 has no other real cube roots.

2 is a fourth root of 16 since 24 = 16 . −2 is also a fourth root of 16 since ()−24 = 16 .

2 is a fifth root of 32 since 25 = 32 . Note that 32 has no other real fifth roots.

−2 is a fifth root of −32 since ()−25 = − 32 . Note that −32 has no other real fifth roots.

Note: 1. A real number has two nth roots when n is even and 2. only one nth root when n is odd . 3. The odd nth root of a positive number is positive and 4. the odd nth root of a negative number is negative.

We use the notation n b to denote the principal nth root of a real number b. The above notation is called a radical expression , b is called the radicand , n is the index of the radical, and the symbol is called the radical sign . The index n is omitted when the index is 2 .

The principal root is chosen to be the positive root for the case when the index is even ; it is chosen to be the unique root when n is odd . For example, 42=3 82 =4 162 = 5 322 =

3−=−82 5 −=− 322

Note: n b is undefined in the set of real numbers when b is negative and n is even.

Thus, we say that −36 is undefined over the reals or is not a real number . 62

4.2 Simplifying Radical Expressions

There are three conditions for a radical expression to be in simplest form : 1. There are no perfect n power factors under a radical of index n. 2. There can be no fractions under a radical sign OR there can be no radicals in the denominator. 3. The index should be the smallest possible.

In this section we will explain how to accomplish the first condition. We will use the properties nab= n a ⋅ n b for n ≥ 2 an a n = for n ≥ 2 b n b where we assume that a and b are such that the expressions do not become undefined.

Let us first consider numerical radicands.

Examples 1. 98 Solution: 98 is not in simplest form there is a perfect square factor 49 under the radical sign: 98=49 ⋅ 2 OR 7 2 ⋅ 2 =⋅492 =⋅ 7 2 2 =72 = 7 2

2. 3 16 Solution: 3 16 is not in simplest form since there is a perfect cube factor 8 under the radical sign: 316= 3 8 ⋅ 2 OR3 2 3 ⋅ 2 =⋅3 832 =⋅3 2 3 3 2 =232 = 2 ⋅ 3 2

3. 4 80 Solution: 4 80 is not in simplest form since there is a perfect 4 th power factor 16 under the radical sign: 480= 4 16 ⋅ 5 OR4 2 4 ⋅ 5 =⋅416 4 5 =⋅4 2 4 4 5 =24 5 = 2 4 5

Next, let us consider variable radicands . Let us also first consider the case when the index n is odd . We observed in Section 4.1 that odd nth roots follow the sign of the radicand, thus

n xxnn = for odd and any real number x .

For example, 3 xx3=, 5 yy 5 = , 7 mm 7 = , and so on.

Next, let us consider the case when the index n is even . We defined in Section 4.1 the principal nth root to be the positive root when n is even. Consider the following: 22 = 2 and also () − 22 = 2

4 24 = 2 and also4 () − 24 = 2

From the above we conclude that for any real number x and even index n, n xn = x if x is positive, and

n xn = − xif x is negative which can be abbreviated to n xn = x for any real number x .

Thus, for example: x2 = x 4 w4 = w 6 c6 = c

If we want to remove the cumbersome absolute value notation we need to make the assumption that all variables represent positive real numbers . Thus, with this assumption we will have x2 = x 4 w4 = w 6 c6 = c

Let’s now simplify radicals with variable radicands. We will assume that the variables can be any real number .

Examples 4. Simplify: x 4 Solution: x4= x 2 = x 2

5. Simplify: y 6

Solution: y6= y 3

12 6. Simplify: a

12 6 6 Solution: a= a = a

7. Simplify: n22 Solution: n22= n 11

8. Simplify: 3 x 4 Solution: 3x5 = 3x3 ⋅= x 3 x 3 ⋅3 xx = x 3

9. Simplify: 3 m23 Solution: 33m23=m21 ⋅= m 2 33 m 21 ⋅ m 2 = m 7 3 m 2

10. Simplify: 4 p8

Solution: 4 p8= p 2 = p 2

11. Simplify: 4 w 36 Solution: 4 w36= w 9

12. Simplify: 5 n12 Solution: 55n12=n10 ⋅= n 2 55 n 10 ⋅ nn 2 = n 2 5 2

13. Simplify: 6 x18 Solution: 6 x18= x 3

Now let us consider a combination of numbers and variable factors for radicands. We will continue to assume that the variables can represent any real number.

14. Simplify: 32 x4 y 6

Solution: 32x4 y 6 =16xy46 ⋅= 2 16xy 46 ⋅= 2 4xy 23 242 = x2 y 3

15. Simplify: 128 a10 b 16 c 2 Solution: 128a10 b 16 c 2 =64abc10162 ⋅= 2 64abc 10162 ⋅= 2 8abc 58 282 = a5 c b 8

16. Simplify: 3 54 n5 p 10

Solution: 3354np510=27np39 ⋅= 2 np 2 33 27np 39 ⋅ 2 np 2 = 3np 3 3 2 np 2

17. Simplify: 3 375 x32 y 46 z 21 Solution: 3 375xyz32 46 21=3 125xyz304521 ⋅= 3 xy 2 3 125xyz 304521 ⋅=3 3 xy2 5xyz 10157 3 3 xy 2

18. Simplify: 4 162 a8 m 12 e 26 Solution: 4162ame81226= 481ame81224 ⋅= 2 e 2 44 81ame 81224 ⋅ 2 e 2 =3 a2 m 3 e 6 42e2 = 3 amm 263 4 2 e 2

19. Simplify: 4 16 y30 n 34 Solution: 4 16yn3034=416yn2832 ⋅= 2 yn 22 4 16yn 2832 ⋅=4 2 yn 22 2yn 78 4 222 ynynyn 22 = 7822 4

Next, we have examples showing what to do when there are factors outside the radical expression. 20. Simplify: 3ab25 64 a 6 b 13 Solution: 3643ab261325 ab= ab 5 32a5 b 10 ⋅= 23 ab 32 ab5 32a 5 b 10 ⋅ 5 2 ab 3 =3ab2 ⋅2ab2 ⋅5 262 ab 324 = a b 5 ab 3

21. Simplify: −4w26 64 y 18 w 36

Solution: −4wyww218366 64 =− 4 26 64y18 w 36 =−⋅ 4 w 2 2 y 3 w 6 =− 8 wy 83

Finally, let us look at examples involving rational radical expressions. For these we will assume that the variables represent nonzero real numbers. 63 a16 b 8 22. Simplify: 25 c22 63ab16 8 9a168 b⋅ 7 3 a 84 b 3 7 ab8 4 Solution: = =7 = 25c 22 25c22 5 c 11 5 c11

40 m7 23. Simplify: 3 27 n15 40m7 8m6⋅ 5 m 2m 2 Solution: 3= 3 = 3 5m 27 n15 27n15 3n 5

2 48 x18 y 42 24. Simplify: 4 xy4 z z 4 Solution:

248xy18 42 216xy16 40⋅ 32 xy 2 2 2xy4 10 4 xy3 6 4= 4 =⋅43xy22 = 4 3 xy 22 xyzz4 4 xyz 4 z4 xyz4 z zz

4.3 Adding or Subtracting Radical Expressions

We can add or subtract only like radicals , i.e., those radical expressions that have the same index and the same radicand .

Examples Perform the operations. 1. 23+3 3 + 73 − 43 3 3 3 33 3 Solution: 23++ 373432373 − = + +− 34393 =− 33

2. xy− yx4 −4 xy + 4 yx 4 Solution: xyyx−−444 xy + yxxy 4 =− 4 xyyx −+ 44 433 yx =− xy + yx 4

Recall that sometimes the radicals need to be simplified first. In the following assume that the variables represent positive real numbers .

3. 8y3 − 4 y 18 y Solution: 8418yyy3 − =4y2 ⋅− 24 yyy 9 ⋅= 2 2y 24 yyy −⋅ 3 2 =2yy 2 − 12 yy 2 =− 10 yy 2

4. 6483a5+ a 3 6 a 2 Solution: 648333aaa52+ 66 =8a3 ⋅+ 6 aaa 22 3 66 =⋅ 2a 33 6 aaa 22 + 6 =12aaaa3 62 + 3 6 2 = 136 aa 3 2

5. 3484 xy5 − 6 x4 243 xy

Solution: 3484 x5 y− 6243 x4 xy = 34 16x4 ⋅−⋅ 3 xy 6 x4 81 3 xy

=⋅32x4 3xy −⋅ 6 x 3 44 3 xy = 63 x xy − 183 x 4 xy =− 123 x 4 xy

6. 5−32ab711 − 4 ab 5 ab 26

5 Solution: 5−324ab711 − abab 5 26 =−5 ()2ab5 10 ⋅− ab 2 4 abab5 2 b 5

5 =−5 ()2ab510 ⋅−5ab24 ab 55 b 5 ⋅=− ab 2 24 ab 22 5 ab − ab 22 5 ab =− 6 ab 22 5 ab

4.4 Multiplying Radical Expressions to multiply radicals with the same index n ≥ 2 we use the following: na⋅ n b = n ab .

Examples 1. Multiply: 12⋅ 30 Solution: 12⋅ 30 =⋅= 1230 360 =36 ⋅= 10 6 10

Sometimes when the numbers are large it is better to use prime factorization rather than carry out the multiplication.

2. Multiply: 320⋅ 3 54 Solution: 3 3 3 3 3 3 3 3 20⋅ 54 = 2054 ⋅=3 2252333 ⋅⋅⋅⋅⋅⋅= 23 ⋅⋅=⋅ 5 23 565 = 20 54

Let us review multiplication involving variable radicands. We will assume all variables represent positive real numbers.

3. Multiply: xy5⋅ xy 6 Solution: xyxy5⋅ 6 = xyxy 5 ⋅= 6 xy 67 =xy66 ⋅= y xy 33 y

Let us now look at the case where we have both numerical and variable factors. 4. Multiply: 34ab4⋅ 3 − 2 ab 3 2 Solution: 342ab4⋅− 3 ab 32 =3 42 ab 4 ⋅−() ab 32 =−3 8 ab 46 =− 3 8ab3 6 ⋅=− a 2ab 2 3 a

There could be factors sitting outside the radicals. 5. Multiply: (28y4 xy5) ⋅( 34 x 4 xy 2 3 ) Solution: (28y44 xy5) ⋅( 34 x xy 23) =⋅⋅ 238 yx 4 xyxy 523 ⋅ 4 = 632 xy 4 xy 74

=6xy4 16x4 y 4 ⋅=⋅ 26 x3 xy 2xy4 212 x 3223 = xyx 4 2

One or more of the factors could consist of two or more terms. 6. Multiply: 2342( + 76 ) Solution: 2 3( 4 2+ 7 6) =⋅+⋅=+ 2 3 4 2 2 3 7 6 8 6 14 18 =8614 +9 ⋅= 2 86 +⋅ 14 3 286 =+ 422

7. Multiply: (3 5+ 8)( − 2 10 − 3 2 ) Solution: (35+ 8)( − 210 − 32 ) =3 5 ⋅−() 2 10 + 3 5 ⋅−() 3 2 +⋅− 8() 2 10 +⋅− 8() 3 2

=−6 50 − 9 10 − 2 80 − 3 16 =−625 ⋅− 2 910 − 2 16 ⋅ 5 −⋅ 34 =−⋅625 − 910 −⋅ 25 4 − 12 =−30 2 − 9 10 − 8 5 − 12

2 8. Multiply: (63+ 25 ) Solution: One way to do this is to rewrite the problem as (63+ 25)( 63 + 25 ) and perform the multiplication as in Example 7. Another way is to apply the special product formula ()ab+2 =+ a22 abb + 2 which we will do: 2 2 2 (6325+=+) ( 63) 26325( )( ) + ( 25 ) =369 + 2415 + 425 =⋅+363 2415 +⋅ 45 =108 + 2415 + 20 =128 + 24 15

9. Multiply: (48− 56)( 48 + 56 ) Solution: We can multiply as in Example 7 but then again we notice that we can also use the special product formula (abab−)( +=−) a2 b 2 as follows:

2 2 (48− 56)( 48 + 56) =( 48) −( 56) =⋅−⋅=−=− 168 256 128150 22

3 10. Multiply: ( 2+ 6 ) Solution: Using the special product formula ()a+=+ b3 a33 a 2 b + 3 ab 23 + b , we get 332 23 ( 26+=+) ( 232) ( ) ( 6326) +( )( ) + ( 6 ) =23 +⋅⋅ 326 +⋅ 326 + 6 3 =⋅+22 266 + 182 +⋅ 6 2 6 =+22 66 + 182 + 6 6 =20 2 + 12 6

There could be variables in the product. We again assume that all variables represent positive real numbers. 11. Multiply: (xx22+ y)( 3 xyy − 2 ) Solution: We apply FOIL. (xx22+ y)( 3 xyy − 2 ) =xx23 ⋅− xxyx 222 ⋅+⋅− y y 32 x yy ⋅ 2 y

=x6 x2 − xy 4 xy + 23 xy − 22 y y 2 =−x2 62 xy xy + 23 xy − 22 y 2

The index could be larger than 2. 12. Multiply: 43 62( 3 4+ 39 3 ) Solution: 43 62( 3 4+ 39 3 ) =⋅423 64 ⋅+⋅ 43 3 69 ⋅ =833 ⋅⋅⋅+222 122 3 ⋅⋅⋅ 333 OR833 ⋅+ 23 1223 ⋅ 3 3 =⋅823 3 + 12 ⋅ 3 3 2 =163 3 + 363 2

13. Multiply: 3−2x2 y( 3 4 xy 2 − 3 − 32 x 45 y ) Solution: 333−2xy2( 4 xy 2 −− 32 xy 45) =− 3 8 xy 33 − 3 64 xy 66 =−− 24 xyxy 22

14. Multiply: ( 3 7−33 5)( 49 + 3 35 + 3 25 ) Solution: Note that the problem looks like the special product (aba−)( 2 ++ abb 2) =− a 33 b . 2 2  3 7−3333 549 ++=− 35 253 7 3 5 3 7 +⋅+ 3 75 33 5  ()() ()() () 

3 3 =()3 7 −()3 5 =−= 752

4.5 Dividing Radical Expressions; Rationalizing Denominators

To divide radical expressions with the same index n ≥ 2 , we have the following rule: n a a =n ,b ≠ 0 n b b

This means that we simply divide the radicands.

Examples 98 1. Divide: 2 98 98 Solution: = =49 = 7 2 2

3 40 2. Divide: 3 5 3 40 40 Solution: =3 =3 8 = 2 3 5 5

214 64 3. Divide: 354 4 214 64 21 64 7 ⋅ 3 3 6 Solution: =4 = 4 16= ⋅ 2 = 354 4 35 4 7 ⋅ 5 5 5

4x23 48 y 5 4. Divide: 8xy3 6 y Solution:

23 5 2 5 4x 48 y 4xyx 48 4 x3 x =3 =3 8y =3 8y ⋅= y ⋅ 2y 3y= x 3 y 8xy3 6 y 862xy y y 2 y 2y

Multiplication and division can be combined: 33ab2⋅ 3 45 ab 3 6 5. 3 5ab 2 Solution:

32 3 36 2 36 4 8 3ab⋅ 45 ab 3 ab ⋅ 45 ab 3335 ⋅⋅⋅ a b 3 336 2 3 3 =2 = =3ab = 3 ab 3 5ab 2 5ab 5ab 2

There could be two or more terms in the dividend: 4 8x3 y+ 30 50 xy 3 6. 2 2 xy Solution: 48xy3+ 3050 xy 33 48 xy 3050 xy 3 = + =+2 4x2 15 25 y 2 22xy 22 xy 22 xy =⋅22x + 155 ⋅ yx = 4 + 75 y

Note that radicals are NOT ALLOWED IN THE DENOMINATOR if a radical expression is to be considered in simplest form. Let us next consider how the radicals can be removed from the denominator.

Rationalizing the Denominator

Rationalizing the denominator means making the denominator a rational number or removing radicals like 2, 5,3 3,... which are irrational numbers.

Let us consider the case where there is only one radical and only one term in the divisor. If the radical is of index n, the idea is to multiply and divide the expression by a radical that will make the radicand a perfect n power.

Examples Rationalize the denominator. 6 7. Simplify: 3 6 6 3 6363 Solution: =⋅= = = 2 3 3 3 3 32 3

There could be variable factors as well (assume these are positive real numbers): 8 8. Simplify: 2x 8 8 2x 82 x 82 x 42 x Solution: =⋅= = = 2x 2 x 2 x 4x 2 2x x

The index could be larger than 2: 7 9. Simplify: 3 5 7 73 52 7253 725 3 Solution: =⋅ = = 3 3 5 5 3 52 3 53 5

6 10. Simplify: 4 3 6 64 33 6274 627 4 Solution: =⋅= = = 24 27 4 4 3 3 4 33 4 34 3 72

3a 11. Simplify: 5 4a3

3a 3 a5 232 aaaaa 32 5 32 32 5 32 3 Solution: =⋅= = = 5 8a2 54a3 5 2 23 a5 23a 2 5 2 55 a 2a 2

Having a fraction or rational expression under a radical sign is the same as having a radical in the denominator: 4 12. Simplify: 11 4 4 4 11 411⋅ 211 Solution: ==⋅= = 1111 11 11 11 11 4 411 411⋅ 211 OR = ⋅= = 11 11 11 11 11

5 13. Simplify: 3 2xy 2

53 5 3 5 322xy 2 3 20 xy 2 3 = = ⋅ = Solution: 2 2xy32xy2 3 2 xy 2 3 2 22 xy 2 xy

Now let us consider the case where there are two or more terms in the denominator. For the case of index 2 radicals, we can use the fact that (abab−)( +=−) a2 b 2 to remove the radicals from the denominator. 4 14. Simplify: 3− 7

4 4 37+ 4( 3+ 7 ) Solution: = ⋅ = 2 2 3− 7 3 − 73 + 7 ()3− () 7 43( + 7) 4 ( 3+ 7 ) = = = −3 − 7 3− 7 − 4

10 15. Simplify: 23+ 5 2

10 10 2352− 10( 2 3− 5 2 ) Solution: = ⋅ = 2 2 23+ 52 23 + 5223 − 52 ()23− () 52 1023( − 52) 1023( − 52) 1023( − 52 ) = = = 43252⋅−⋅ 1250 − − 38

523()− 5 2 103− 252 − 103 + 252 = = or −19 − 19 19

There could also be two terms in the numerator: 73

5 3+ 6 16. Simplify: 2 5− 9 53+ 6 53 + 625 + 9 5325 ⋅+⋅+⋅+⋅ 539 625 69 Solution: = ⋅ = 2 25− 9 25 − 925 + 9 ()2 5− 9 2 10 15+++ 45 3 12 5 54 10 15 +++ 45 3 12 5 54 = = 4⋅ 5 − 81 20 − 81

10 15+ 45 3 + 12 5 + 54 = −61

There could be variables involved: 2x− 3 x 17. Simplify: 3x+ 2 x 23x− xx 23 − xx 32 − x Solution: = ⋅ 32x+ xx 32 + xx 32 − x 64x2 −−+ xx 9 xx 6613 x x2 − xx + 6 x =2 = 2 ()3x2 − () 2 x 9x− 4 x

There could be radicals of index 3: 1 18. Simplify: 32+ 3 7 Solution: As in the case for index 2 radicals we can use a special product formula, namely (aba+)( 2 −+ abb 2) =+ a 33 b , which we actually learned as a factoring formula. 1 13 22−⋅+3 277 3 2 33 414 −+3 49 33 414 −+ 3 49 =⋅ = = 32+ 3 7 3 2 + 3 7 322−3 27 ⋅ + 3 7 2 2+ 7 9

3 x 19. Simplify: 3 x2 +23 x + 4 2 Solution: The denominator looks like ( 3x) + 3 x ⋅+2 2 2 = a 2 + ab + b 2 so if we multiply the numerator and denominator by a− b =3 x − 2 and apply the formula (aba−)( 2 ++ abb 233) =− a b , we will get

33 33 3x 3 x − 2 xx( −2) xx( − 2 ) ⋅ = = . 3 3 3 x2 +23 x + 4 x − 2 ()3 x − 23 x − 8

4.6 Rational Exponents

In this section we will learn how to manipulate expressions involving rational or fractional exponents.

1 We define an where a is a real number and n is an integer with n ≥ 2 as

1 an = n a .

The following will show how to use the above definition to simplify expressions 1 that look like an .

Examples 1 1. Evaluate: 42 1 Solution: 42 = 4 = 2

1 2 ⋅2 2 Note that ()41 2 = 42 = 4 and ( 4) = 4 , thus the definition makes sense.

1 2. Evaluate: ()−8 3

1 3 Solution: ()−83 = −=− 8 2

1 16  4 3. Evaluate:   81 

1 16  4 164 16 2 Solution:   =4 = = 81  814 81 3

Now let us consider the case when the numerator in the rational exponent is not 1. Let us look at how we can handle this case:

m 1 1 m⋅ aan= n =() amn = n a m OR

m 1 1 m ⋅m   m n n n n aa= = a  = () a  

Let us look at how these will work in the following examples.

2 4. Evaluate: 83 2 2 2 2 Solution: 83 =3 82 =3 644 = OR 83 =()3 8 =() 24 =

Note that the second way seems “better” in the sense that we deal with smaller numbers. 5 5. Evaluate: 16 4 5 5 Solution: 164 =()4 16 = 25 = 32

5 1  3 6. Evaluate:   27 

5 5 5 13  1   11 Solution: =3  =  = 27 27   3 243

4 7. Evaluate: ()−32 5

4 4 5 4 Solution: ()−325 =−() 32 =−() 2 = 16

3 8. Evaluate: −16 4 3 3 Solution: −164 =−()4 16 =−=− 23 8

5 9. Evaluate: 22 Solution: For this problem it would make more sense to use the first way of interpreting the rational exponent. 5 22 = 25 = 2222 4 ⋅= 2 .

The exponent could be negative: 3 − 10. Evaluate: 25 2 3 − 1 1 11 Solution: 25 2 = = == 3 3 53 125 25 2 ()25

2 − 11. Evaluate: ()−8 3

2 − 1 1 11 3 Solution: ()−=8 2 = 2 = 2 = 3 −2 4 ()−8 3 ()−8 ()

3 − 12. Evaluate: −81 4 3 − 1 1 11 −4 =− =− =− =− Solution: 81 3 3 3 4 3 27 81 4 ()81 76

5 − 8  3 13. Evaluate:   27 

5 5 − + 5 5 83 27 3  27   3 243 Solution: = =3  ==  27 8 8   2 32

3 − 14. Evaluate: 2 2 3 − 2 11 1 1 Solution: 2 ==3 = = 3 2 ⋅ 2 2 22 2 2 2

4 − 1  3 15. Evaluate: −  3  4 − 4 1  3 + 4 3 Solution: −  =−()33 =−3() 3 =− 3 ()() 3 ⋅−=−−= 333333 3 3 

Let us next consider expressions involving variable factors. Again let us assume all the variables represent positive real numbers. 1 16. Simplify: ()27 x3 y 6 3

1 Solution: ()27x36 y3 =3 27 x 36 y = 3 xy 2

3 17. Simplify: ()16 a8 b 16 4

3 3 3 Solution: ()16ab8164 =( 4 16 ab 816) =() 2 ab 24 == 2 3612 ab 8 ab 612

2 27 m12 n 9  3 18. Simplify: −  8p6  2 2 2 27mn1293  27 mn 129   3 mn 43 9 mn 86 Solution: − =−3  =−  = 6 6  24 8p 8 p   24 pp

4.7 Multiplying or Dividing Radicals with Different Indices

Now that we know about rational exponents we can talk about multiplying or dividing radicals whose indices are not the same.

The idea is to write each radical using rational exponents. The goal is to first make the indices the same. Since the index corresponds to the denominator of the rational exponent, the idea is similar to finding the LCD for the exponents and then converting each exponent to its equivalent fraction.

Examples 1. Multiply: 2⋅ 3 2 1 1 1⋅3 1 ⋅ 2 32 Solution: 222222⋅=⋅=3 2 3 2⋅3 ⋅ 3 ⋅ 2 =⋅= 22 66 6 223 ⋅ 6 2 = 6 2 5 = 6 32

2. Multiply: x⋅4 x3 ⋅ 6 x 5 1 3 5 1⋅6 3 ⋅ 3 5 ⋅ 2 6 9 10 Solution: xx⋅4 3 ⋅6 xxxxxxx 5 =⋅⋅=2 4 6 2⋅6 ⋅⋅ 4 ⋅ 3 6 ⋅ 2 =⋅⋅ xxx12 12 12 6910+ + 25 =x12 == x 12 12 x25 = 12 xxxx 24 ⋅= 2 12

2 3. Divide: 3 2 1 1 13− 2 1 2 2 2 − Solution: ==223 = 2 6 == 2 6 6 2 3 2 1 23

3 y 4. Divide: 4 y

1 1 1 4− 3 1 3 y y 3 − Solution: ==y3 4 = y12 == y 12 12 y 4 y 1 y 4

2x⋅ 3 2 x 5. Simplify: 4 2x Solution: 1 1 3 2 3 1116+ 4 − 3 7 2x⋅ 2 x ()2x() 2 x + − 7 12 7 = =()2x234 =() 2 x 12 ===() 2 xx 12 12 () 2 128 x 4 2x 1 ()2x 4

4.8 Reducing the Index

The notion of rational exponents we can explain why an answer like 6 x2 is NOT considered to be in simplest form. To see why not, we have 2 1 6 x2 = x6 = x 3 = 3 x and thus the simplest form of 6 x2 is 3 x , where we assume x is positive.

Another way to arrive at the same answer is to use the following:

m na= mn a which one can easily prove using rational exponents. Thus, 6 x2=3 x 2 = 3 x . On the other hand, 6x3= 3 x 3 = x . 78

Examples 24. Simplify: 6 8 11 3 1 Solution: 6 88==6() 23 6 === 22 6 2 2 OR 68= 3 8 = 2

25. Simplify: 12 x 4 4 1 Solution: 12 x4= x12 = x3 = 3 x OR 12x4=3 4 x 4 = 3 x

26. Simplify: 15 32 a10 b 5 1 1 1 2 1 Solution: 15 32ab105=() 32 ab 10515 =() 2 5105 ab 15 == 23 ab 3 3 3 2 ab 2

OR 1532ab105=3 5 32 ab 105 = 3 2 ab 2

4.9 Complex Numbers

In our previous work whenever we encountered an expression like −3, − 16,... we concluded that the expression is undefined (over the reals) or is not a real number. When we were solving equations in elementary algebra, whenever we reached a step, say, x = − 9 , we concluded that the equation has no real solution. In this section we will enlarge the set of numbers that we are considering to include numbers such as these numbers.

We define a complex number to be any number of the form a+ bi , where a and b are real numbers and i = − 1 . a is called the real part of the complex 1 3 number and b is called the imaginary part . −+3 2,i − i , − 0.3 ii − 4.5 are 2 5 examples of complex numbers. All the real numbers are complex numbers – the imaginary part is 0 for a real number. When the real part is 0, the complex number is a pure imaginary number . Two complex numbers a+ bi and c+ di are said to be equal if and only if the real parts are equal and the imaginary parts are equal, i.e.

a+=+⇔= bi c di a c and b = d .

Adding or Subtracting Complex Numbers

To add or subtract two complex numbers a+ bi and c + di , we simply add/subtract the real part of one to/from the real part of the other, the imaginary part of one to/from the imaginary part of the other, i.e. (abi+) ++( cdi) =+++( ac) ( bdi) (abi+) −+( cdi) =−+−( ac) ( bdi)

Examples 1. Add: 30− 28i and − 42 + 15 i Solution: (30− 28i) +−+( 42 15 i) =− 30 28 ii −+ 42 15 =30 − 42 − 28i + 15 i =−− 12 13 i

2. Subtract 25− 16i from − 13 + 12 i . Solution: (−+13 12i) −( 25 − 16 i) =−+ 13 12 i −+ 25 16 i =−13 − 25 + 12i + 16 i =− 38 + 28 i

3. Simplify: 6− − 4 + − 95 − Solution: 6−−+−−=− 4 95641 ⋅−+() 915 ⋅−−() =−6 4 ⋅−+ 1 9 ⋅−−=− 1562351i + i −=+ i

Multiplying Complex Numbers

To multiply two complex numbers a+ bi and c+ di we simply apply the distributive property and remember that i = − 1 so that i 2 = − 1.

(a+ bi)( c +=+++ di) ac adi bci bdi2 =++ ac( ad bci) − bd

=()()ac − bd + ad + bci

It is not necessary to memorize the formula; one can simply perform the multiplication. It might be interesting to note some pattern to make the multiplication faster: a+ bi ×c + di ()()ac− bd + ad + bci

Examples 4. Multiply: (25−i)( − 47 + i ) Solution: (2− 5ii)( −+ 4 7) =−+ 8 14 iii + 20 − 352 =−+ 8 34 i + 35 = 27 + 34 i

5. Expand: ()−3 − 6 i 2 Solution: ()()()()()−−36i2 =−−− 3 2 236 ii + 6 2 =+ 93636 ii + 2 =+9 36i − 36 =− 27 + 36 i

6. Multiply: (2−− 53)( −− 6 ) Solution: (2−− 53)( −−=− 6) ( 2i 53)( − i 662635) =− iii − + 2 30 =−62635i − i − 30 =−( 6 30) −( 2635 + ) i Check what happens if you do not do the second step and simply multiplied directly. Can you explain why you got a slightly different answer? 80

Dividing Complex Numbers

Suppose we want to divide the complex number 2+ 3 i by the complex number 1− 2 i . We can write the problem as

2+ 3 i . 1− 2 i

What do we do with the above expression? Note that since i = − 1 if we rewrite the problem it will look like an expression with a radical in the denominator and following what we discussed in a previous section we can remove i = − 1 from the denominator by following a similar procedure as in rationalizing the denominator. More precisely, we would need the following:

The conjugate of a complex number a+ bi is the complex number a− bi . Let us give a few examples:

Complex Number Conjugate

1− 2 i 1+ 2 i

−3 + 5 i −3 − 5 i

12 i −12 i

− − − 24i 9 24i 9

Note that the product of a complex number and its conjugate is a real number: ()()()abiabi+ −=− a2 bi2 =− a 22222 bi =+ a b

To divide a complex number by another complex number, multiply both the dividend and the divisor by the conjugate of the divisor:

a+ bi a + bi c − di = ⋅ c+ di c + di c − di

Examples 2+ 3 i 7. Divide: 1− 2 i Solution: 23+i 2312 + ii + 2436 +++ iii2 276 +−−+ i 47 i −+ 47 i 47 =⋅= = = = =−+ i 121212−−+iii12 − () 2 i 2 14 −+ i 2 14 5 55 −8 + 2 i 8. Divide: 5i −+8 2i −+− 8 2 iiii 5 40 − 102 40 i + 10 10 40 2 8 Solution: = ⋅= = =+=+i i 5i 5 ii− 5 − 25 i 2 25 25 25 5 5 −+82i −+ 82 ii −+ 82 ii2 −− 8228 i OR = ⋅= = =+ i 5i 5 ii 5 i 2 − 555

Powers of i

We have the following first few powers of i: i = − 1 i5 = i i9 = i i 2 = − 1 i 6 = − 1 i10 = − 1 … i3 = − i i7 = − i i11 = − i i 4 = 1 i 8 = 1 i12 = 1

Do you notice any pattern? Let us try the following:

9. Evaluate: i 21 Solution: We note above that i 4n = 1, n a positive integer, i.e. i raised to a positive multiple of 4 always gives the answer 1. Thus, one way of figuring out what i 21 is:

i21= ii 20 ⋅=1 5 ⋅= ii

There is a faster way but let us look at another example first:

10. Evaluate: i 35 Solution: i35= ii 32 ⋅ 3 =⋅1 i 3 =− i

Did you observe that all we really need is the remainder when the exponent is divided by 4 and then we can look at the first column of powers of I given above to find the answer? Let’s try the following:

i46= i 2 = − 1 i96= i 0 = 1

i55= i 3 = − i i201= i 1 = i

4.10 Chapter Review

A real number b is called the nth root of a real number a if an = b . For example, the 5th root of 32 is 2 since 25 = 32 . We use the notation n b to denote the principal nth root of a real number b. The above notation is called a radical expression , b is called the radicand , n is the index of the radical, and the symbol is called the radical sign . The index n is omitted when the index is 2.

The principal root is chosen to be the positive root for the case when the index is even; it is chosen to be the unique root when n is odd. For example, 6 64= 2 and 7 −128 = − 2 .

Simplifying Radical Expressions

A radical expression to be in simplest form if the following are satisfied (in the examples we will assume all variables represent positive real numbers):

1. There are no perfect n power factors under a radical of index n. 8xy5 2 =4xy4 2 ⋅ 2 x = 2xy 2 2 x

327 ab6 10 = 3 27a69 b ⋅ b = 3a 23 b 3 b

2. There can be no fractions under a radical sign or there can be no radicals in the denominator. 22x 2x 2 x = ⋅= = xxx x2 x

4 43a 2 12a23 12 a 2 3 =3 ⋅= 3 = 9aa 93a 2 27 aa3 3

y y 5y yy 5 yy 5 y 5y 5y = ⋅= = = = 25y 25 y 5 y 2 25 y 2 2⋅ 5 y 10 y 10

x x2 x+ 32 xxxxxx + 32 + 3 = ⋅ = = 2x− 32 x − 32 x + 3 4x2 − 9 4x − 9

1 1 3a2−+3 ab 33 b 22 a −+ 3 ab 3 b 2 = ⋅ = 3a+ 3 b 3 a + 3 b 3a2−3 ab + 3 b 2 a+ b

3. The index should be the smallest possible. 4 4a26 b= 4 a 26 b = 2 ab 3 = b 2 ab

664xyz31521= 3 64 xyz 31521 = 4 xyz 57 = 2 yzxyz 23

Adding or Subtracting Radical Expressions

We can add or subtract only like radicals , i.e., those radical expressions that have the same index and the same radicand .

Examples 1. 5x− 458 xxxxx −+ 655 =−− 8 4665 x + x =−+ 3 xx 25

2. −320b aba3 + 845 ab 3 =− 3 b4a2 ⋅+ 5 aba 8 9b 2 ⋅ 5 ab =−⋅3b2a 5 ab +⋅ 8 a 3b 5 ab =− 65 abab + 245 abab = 185 abab

Multiplying Radical Expressions

A. For radical expressions with the same index n ≥ 2 we use the following:

na⋅ n b = n ab .

Example 34xy2⋅ 3 2 xy 25 = 3 8 xy 37 = 2 xyy 2 3

B. For radical expressions with different indices n ≥ 2 use rational exponents .

11 3 2 3 2 Example xy⋅=3 xy22() xy2() xy3 =() xy 6 () xy 2 6 =⋅6 () xy6 () xy 2

=66xy33 ⋅ xy 24 = 6 xy 57 = yxy 6 5

Apply the distributive property and/or special product formulas when multiplying radicals involving two or more terms.

Example 2 2 (23ab+=) ( 23 a) + 223( abb ⋅+=⋅+) 2 4343 abab +=+ 2 1243 abab + 2

2 One can also perform the expansion above by writing (2 3 a+ b ) as (23ab+)( 23 ab + ) and applying FOIL.

Dividing Radical Expressions

A. For radical expressions with the same index n ≥ 2 use:

n a a =n ,b ≠ 0 n b b

4 80abc717 80 abc 717 16 ab 416 2 ab 4 =4 = 4 = Example 3 5 4 4 5a3 bc 5 5abc c c

B. For radical expressions with different indices n ≥ 2 use rational exponents .

Example 1 4 2 2 3 4x2 ()4x3() 4 x 12 4212248 xxxx 88 111112 1111 = = =12 =12 =⋅= 12 4 3 1 3 39 99 1111 8x 3 3 8x 2 x 22 xx 2 x ()8x4() 8 x 12

Rationalizing Denominators See condition 2 for simplified radicals.

Rational Exponents 1 For a real number a and n an integer with n ≥ 2 , we have: an = n a .

Also, for an integer m, we have:

m m an = n am or ()n a . . 1 3 3 4 5 3 Examples 164 = 16 = 2 ; ()−325 =−() 32 =−() 2 =− 8

7 − 2 11 1 1 1 5 ===7 = = 7 6 ⋅ 53 5 125 5 52 5 5 5

Complex Numbers

We define a complex number to be any number of the form a+ bi , where a and b are real numbers and i = − 1 . We call a the real part of the complex number and we call b the imaginary part . −4 + 8 i is a complex number; the real part is −4 and the imaginary part is 8.

Operations on Complex Numbers

Examples

1. Add: (35−i) +−−( 42 i) =( 34 −) +−−( 52 ii) =−− 17 i 2. Subtract: (−−64i) −( 26 − i) =−− 6426 ii −+ =−+ 82 i 3. Multiply: (42−ii)( −+=−+ 3) 124 iii +− 6 22 =−+ 1210 i +=−+ 2 1010 i

2 2 4. Expand: ()()()()36+=+i 32 236 ii + 6 =++ 936 ii 36 2 =+9 36i − 36 =− 27 + 36 i 46+i 46 + i 32 + i 128 +++ iii 18 12 2 5. Divide: = ⋅ = 32−−+iii 3232 94 − i 2 12+ 26i − 12 26 i = = = 2i 94+ 13 12 6. Evaluate: i49= iii 48 ⋅=() 4 ⋅= ii