Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.1: A Review of the Properties of Exponents
The first two sections in this chapter provide a review of several rules on how to simplify when working with
exponents. The rules are as follows:
Multiplication of like bases: bmbn = bm+n
Division of like bases:
𝒃𝒎
𝒃𝒏
= 𝒃𝒎−𝒏 (this rule applies as long as “b” does not equal zero)
Power rule of exponents: (𝒃𝒎 )𝒏 = 𝒃𝒎𝒏
Power of a product rule: (ab)m = ambm
𝒂 𝒎
𝒂𝒎
Power of a quotient rule: (𝒃) = 𝒃𝒎 (this rule applies as long as “b” does not equal zero)
Definition of 𝒂−𝒏 :
𝟏
𝒂−𝒏 = 𝒂𝒏 (this rule applies as long as “a” does not equal to zero)
___________________________________________________________________
Example 1: Simplify the expressions: This should help with #1-42 in the section.
a) x2x4
b) z2z3z3
c)
d)
𝒂𝟐
𝒂𝟓
e) (𝒙𝟐 )
g)
𝒂𝟓 𝒃
𝒂𝟐 𝒃𝟐
h) (𝟓𝒙𝒚𝟑 )(𝟐𝟐 𝒙𝟑 𝒚𝟓 )
𝒙𝟓
𝒙𝟐
𝟑
𝟑
f) (𝟑𝒙𝟐 )
𝟑
𝒙𝟐 𝒚
i) (
𝟐𝒛
𝟒
𝟑)
a) x2x4
This problem is supposed to help you remember the multiplication of like bases rule. I will first solve the
problem without the rule. Then I will show you how the rule could have been used to solve the problem.
I could rewrite the problem without the exponents, if I didn’t remember the multiplication of like bases rule.
x2x4 = (xx)(xxxx) ( I put each factor in its own parenthesis to make them stand out)
x2x4 = xxxxxx (now I dropped the parenthesis, as they really were not needed)
x2x4 = x6 (now I wrote the answer using an exponent, rather than repeated multiplication.
67
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.1: A Review of the Properties of Exponents
The other way to solve this problem is using the multiplication of like bases rule.
x2x4 = x2+4 = x6 (The rule tells me that if I am multiplying two terms with the same base, that the
multiplication can be done by adding the exponent, and not changing the base.
Answer: x6
b) z2z3z3
I could do this problem by writing each factor without their exponents. It is a bit more work this way.
z2z3z3 =(zz)(zzz)(zzz)
z2z3z3 = zzzzzzzz
z2z3z3 = z8
However, the multiplication with like bases rule could also be used. The rule doesn’t state that it can be
used to multiply three terms with the same base, but it can.
z2z3z3 = z2+3+3 = z8
Answer: z8
c)
𝑥5
𝑥2
I can solve this problem without the division with like bases rule. By writing out each term without the
exponents, and canceling two of the x’s as follows:
𝑥5
𝑥2
=
𝑥5
𝑥2
= 𝑥3
𝑥𝑥𝑥𝑥𝑥
𝑥𝑥
(𝑛𝑜𝑤 𝐼 𝑐𝑎𝑛 𝑐𝑎𝑛𝑐𝑒𝑙 𝑡ℎ𝑒 2 𝑥 ′ 𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟, 𝑤𝑖𝑡ℎ 2 𝑜𝑓 𝑡ℎ𝑒𝑥 ′ 𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟. )
It would have been easier to solve this problem using the division with like bases rule. The rule tells me that
I can solve this problem by subtracting the exponents.
𝑥5
𝑥2
= 𝑥 5−2 = 𝑥 2
Answer: x2
68
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.1: A Review of the Properties of Exponents
d)
𝑎2
𝑎5
I can solve this problem two ways. I could rewrite the problem without the exponents and cancel two of the
a’s..
𝑎2
𝑎5
=
𝑎2
𝑎5
= 𝑎𝑎𝑎 = 𝑎3 ( 𝐼 𝑐𝑎𝑛𝑐𝑒𝑙𝑒𝑑 𝑡ℎ𝑒 𝑒𝑛𝑡𝑖𝑟𝑒 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 𝑎𝑛𝑑 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑙𝑒𝑎𝑣𝑒 𝑎 1 𝑖𝑛 𝑡ℎ𝑒 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟. )
𝑎𝑎
𝑎𝑎𝑎𝑎𝑎
1
(𝑐𝑎𝑛𝑐𝑒𝑙 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑎′𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 𝑤𝑖𝑡ℎ 2 𝑜𝑓 𝑡ℎ𝑒 𝑎′ 𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟.
1
I will now solve this problem using the division with like bases rule. I will get an answer with a negative
exponent., I need to remember how to make a negative exponent positive. This is covered in detail in
section 4.2. I will use the rule to make a negative exponent positive when I solve this below.
𝑎2
𝑎5
1
= 𝑎2−5 = 𝑎−3 = 𝑎3 (I made the a-3 into my final answer using a rule from section 4.2)
Answer:
𝟏
𝒂𝟑
e) (𝑥 2 )3
I will only solve this problem using the power rule of exponents. I could have solved the problem without
this rule, much like I solved the first few problems in this section without an exponent rule.
(x2)3 = x2*3 = x6 (The power rule allows me to multiply the exponents to get my answer.)
Answer: x6
f) (3𝑥2 )
3
I can still use the power rule to solve this problem. The power rule can be applied when there is an
exponent outside of a parenthesis that DOES NOT HAVE AN ADDITION OR SUBTRACTION IN IT. I will need to
write the 3 with an exponent to use the power rule.
(3x2)3 = (31x2)3 (The exponent of the three is not usually written, but it helps to have it written when using
the power rule.)
= 31*2x2*3 (I applied the power rule to the 3 and the x by multiplying the exponents.)
= 32x6
Answer: 9x6
69
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.1: A Review of the Properties of Exponents
g)
𝑎5 𝑏
𝑎2 𝑏2
I will solve this problem using the division with like bases rule.
𝑎5 𝑏
𝑎2 𝑏2
1
= 𝑎5−2 𝑏1−2 = 𝑎3 𝑏 −1 = 𝑎3 ∙ 𝑏1 =
positive before writing my answer.)
𝑎3
𝑏
(My answer had a negative exponent, and I needed to make it
I could have avoided getting a negative exponent, by solving the problem a little differently. Basically, I will
subtract the exponents of the b’s (2-1) and leave a positive exponent with the b in the denominator,
because that is where the larger exponent of the b exists.
This is another way to solve the problem that avoids negative exponents.
𝑎5 𝑏
𝑎2 𝑏2
𝑎 5−2
= 𝑏2−1 =
𝑎3
𝑏
𝒂𝟑
𝒃
Answer:
h) (5𝑥𝑦 3 )(22 𝑥 3 𝑦 5 )3
First I need to simplify the second parenthesis, by using the power rule to get rid of the exponent of 3
outside of the second parenthesis.
(5𝑥𝑦 3 )(22 𝑥 3 𝑦 5 )3 = (5𝑥𝑦 3 )(22∗3 𝑥 3∗3 𝑦 5∗3 ) = (5𝑥𝑦 3 )(26 𝑥 9 𝑦15 ) = (5𝑥𝑦 3 )(64𝑥 19 𝑦15 )
Now I will multiply the 5 and 64 and add the exponents on the variables to get my answer.
= 5*64x1+19y3+15
Answer: 320x20y18
𝒙𝟐 𝒚
𝟒
i) ( 𝟐𝒛𝟑 )
First I will use the power rule to get the exponent of 4 outside of the parenthesis.
𝑥2 𝑦
4
𝑥 2∗4 𝑦 1∗4
𝑥 8𝑦4
(2𝑥 3 ) = 21∗4 𝑥 3∗4 = 24 𝑥 12 (𝑁𝑜𝑡𝑖𝑐𝑒 ℎ𝑜𝑤 𝐼 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑒𝑑 𝑡ℎ𝑒 𝑦 𝑎𝑛𝑑 𝑡ℎ𝑒 2.
𝐸𝑎𝑐ℎ ℎ𝑎𝑠 𝑎𝑛 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 1 𝑡ℎ𝑎𝑡 𝑤𝑎𝑠 𝑛𝑜𝑡 𝑤𝑟𝑖𝑡𝑡𝑒𝑛.
Answer:
𝒙𝟖 𝒚𝟒
𝟏𝟔𝒙𝟏𝟐
70
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.2: A Review of the Properties of exponents (the power of 0 and negative exponents)
This section focuses more heavily on negative exponents. It also introduces the concept of the
exponent of 0.
Definition of a0:
a0 = 1 (this rule applies as long as “a” does not equal zero)
I will attempt to help you understand the 0 exponent rule with this problem.
23
Simplify: 23
You should be able to simplify this by replacing the 23 with 8, and reduce the fraction that remains.
23
8
=8=1
Now I will simplify the problem using the rules of exponents. The answer I get must be equivalent
to 1. Since that is the answer I got by simplifying the problem in the last step. Using the rules of
exponents can ‘t give me a different answer.
23
23
= 23−3 = 20
Since the two methods produce equivalent answers it must be the case that:
20 = 1
In general any number other than 0 raised to the 0 power equals 1.
23
Example 1: Simplify the expression. This should help with #1-12.
a) 30
b) -30
c) 2x0
a) 30 = 1 This is a direct application of the 0 power rule.
Answer: 1
b) -30
If I write 1 for the answer, I will have made a mistake.
I need to think of the problem as: -1*30 (The negative in front of the 3 means to multiply the 3 by 1, now I will simplify by simplifying the exponent before the multiplication.)
-30 = -1*30 = -1*1 = -1
Answer: -1
c) 2x0
I have to think of this problem as 2*x0, and simplify the exponent before doing the multiplication.
2x0 = 2*x0 = 2*1 = 2 ( I used the fact that x0=1)
Answer: 2
Example 2: Simplify the expressions. Write the answer with positive exponents only. This
should help with #13-38.
𝟑
a) 𝟐−𝟑
b) 𝒙−𝟒
c) 𝟓𝒙−𝟐 𝒚𝟓
𝟓 −𝟑
d) (𝒙𝟐 )
g)
e) 𝒙−𝟔 𝒙𝟐
f)
𝒚−𝟓
𝒚𝟔
𝟑𝟐𝒙𝟐 𝒚−𝟔
𝟏𝟎𝒙𝒚𝟑
71
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.2: A Review of the Properties of exponents (the power of 0 and negative exponents)
1
1
a) 2−3 = 23 = 8 (This is a direct application of the a-n rule)
Answer:
𝟏
𝟖
3
b) 𝑥 −4
I can solve this problem without shortcuts as follows:
First rewrite the problem with a division sign, instead of a fraction bar.
3
= 3 ÷ 𝑥 −4
𝑥 −4
Now I can use the a-n rule and make the exponent positive.
1
= 3 ÷ 𝑥4
Now I can change the division to a multiplication.
= 3*x4
=3x4
The answer is 3x4. I could have got the answer without much algebra by moving the x-4 up to the
numerator of the fraction and change its exponent from negative to positive.
SHORTCTU TO MAKE A NEGATIVE EXPONENT POSITIVE: IF A PROBLEM DOES
NOT HAVE ANY ADDITION OR SUBTRACTION- I can make a negative exponent positive
by moving to the opposite side of the fraction.
Answer: 3x4
c) 5𝑥 −2 𝑦 5
I can use the rules of exponents to solve this, or I can get to the answer with a shortcut. I will first
solve this using the rules of exponents.
1
The x-2 can be rewritten as 𝑥 2 , the 5 and the y5 do not have negative exponents and do not need to be
rewritten.
1
5𝑥 −2 𝑦 5 = 5 ∗ 𝑥 2 ∗ 𝑦 5 (now I will write the 5 and the y5 as fractions to make it easier to follow
along)
5
1
= 1 ∗ 𝑥2 ∗
the answer)
𝑦5
1
(now I will multiply across the numerators, and across the denominators to get
𝟓𝒚𝟓
Answer: 𝟐 ( I could have got the same answer by creating a fraction, moving the x term to
𝒙
the denominator and changing its exponent from negative to positive.)
𝟓 −𝟑
d) (𝒙𝟐 )
Most of these problems can be done with some sort of shortcut. The shortcut to this problem is to
make the exponent of (-3) positive by taking the reciprocal of the fraction. I will use the shortcut
without showing the long way to do the problem. Feel free to ask me why the shortcut works if you
are curious.
72
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.2: A Review of the Properties of exponents (the power of 0 and negative exponents)
5 −3
𝑥2
3
(𝑥 2 ) = (51 ) (I made the exponent positive by taking the reciprocal of the fraction. I also wrote
the 5 with an exponent to make the next step easier to follow.)
𝑥 2∗3
𝑥6
𝑥6
=51∗3 = 53 = 125 (I used the power of a quotient rule on this step.)
Answer:
𝒙𝟔
𝟏𝟐𝟓
e) 𝑥 −6 𝑥 2 = 𝑥 −6+2 = 𝑥 −4 (I used the product with like bases rule and added the exponents.)
1
= 𝑥 4 (Now I used the rule to make the negative exponent positive.)
Answer:
f)
𝑦 −5
𝑦6
1
𝑥4
= 𝑦 −5−6 = 𝑦 −11 (I used the quotient with like bases rule and subtracted the exponents.)
𝟏
= 𝒚𝟏𝟏 (Now I used the rule to make the negative exponent positive.)
Answer:
g)
𝟏
𝒚𝟏𝟏
32𝑥 2 𝑦 −6
=
10𝑥𝑦 3
16𝑥 2−1 𝑦 −6−3
5
(I simplified 32/10, and subtracted the exponents of the variables.)
16𝑥𝑦 −11
=
5
16 𝑥
= 5𝑦 11 (I used a shortcut and made the exponent of the y positive by moving it to the
denominator)
𝟏𝟔𝒙
Answer: 𝟓𝒚𝟏𝟏
73
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.3: Definition of nth Root
The reverse operation of squaring a number is to find the numbers square root. For example, finding the
square root of 25 is equivalent to asking what number squared is 25. There are two numbers that satisfy
this. The numbers are 5 and (-5) because (5)2 = 25 and (-5)2 = 25.
In general all positive numbers have two square roots, one positive and one negative. However, we are
usually more concerned with the positive square root than the negative.
Let “a” represent some positive real number:
 √𝒂 is the positive square root of “a”. The positive square root of a number is called the
principal square root of “a”. This is the positive number that when squared gives “a”.
 −√𝒂 is the negative square root of “a”. This is the negative number that when square gives
“a”.
The reverse operation of cubing a number is to find the numbers cubed root. For example, finding the
cubed root of 8 is equivalent to asking what number cubed is 8. Two is only one number that satisfies this,
because (2)3 = 8. (-2) does not satisfy this because (-2)3 = -8, not 8.
Let “a” represent any real number:
 𝟑√𝒂 is the cubed root of “a”. This is the number that when cubed gives “a”. The number will
be positive if “a” is positive, and negative if “a” is negative. There is no principal cubed root,
because there will only be one number that can be cubed to equal the number under the
radical.
The reverse operation of raising a number to the fourth power is to find the numbers fourth root. For
example finding the fourth root of 81 is equivalent to asking what number raised to the fourth power is
equal to 81. There are two numbers that satisfy this. The numbers are 3 and -3, since (3)4 = 81, and (3)4=81.
In general all positive numbers have two fourth roots, one positive and one negative. However, we are
usually more concerned with the positive fourth root than the negative.
Let “a” represent some positive real number:
 𝟒√𝒂 is the positive fourth root of “a”. This is often called the principal fourth root of “a”. This
is the positive number that when raised to the fourth power gives “a”.
 − 𝟒√𝒂 is the negative fourth root of “a”. This is the negative number that when raised to the
fourth power gives “a”.
74
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.3: Definition of nth Root
I could go on for a long time discussing individual roots of a number “a”.
In general:



𝑾𝒉𝒆𝒏 n𝒊𝒔 𝒂𝒏 𝒆𝒗𝒆𝒏 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒊𝒏𝒕𝒆𝒈𝒆𝒓, 𝒂𝒏𝒅
𝒂 > 0 𝑤𝑒 𝑠𝑎𝑦 𝒏√𝒂 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒓𝒊𝒏𝒄𝒊𝒑𝒂𝒍 𝒐𝒓 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒏𝒕𝒉 𝒓𝒐𝒐𝒕 𝒐𝒇 "𝒂".
𝒏
𝑾𝒉𝒆𝒏 "𝒏" 𝒊𝒔 𝒂𝒏 𝒐𝒅𝒅 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒊𝒏𝒕𝒆𝒈𝒆𝒓, 𝒘𝒆 𝒔𝒂𝒚 √𝒂 𝒊𝒔 𝒕𝒉𝒆 𝒏𝒕𝒉 𝒓𝒐𝒐𝒕 𝒐𝒇 "𝒂".
The “n” is called the index of the radical, and the “a” is called the radicand.
___________________________________________________________________
Example 1: Evaluate the roots. Identify those that are not real numbers. This should help with #1-30.
𝟑
a) √𝟖𝟏
b) −√𝟖𝟏
c) √−𝟖𝟏
d) √𝟔𝟒
𝟑
e) √−𝟔𝟒
𝟑
f) − √𝟔𝟒
𝟏𝟔
g) √ 𝟗
h) 𝟑√𝟏𝟔
a) √81
This is asking for the positive number that when squared is 81. The answer is 9, because (9)2 = 81.
Answer: 9
b) −√81
This is asking for the negative number that when squared is 81. The answer is (-9) because (-9)2 = 81
Answer: -9
c) √−81
This is asking for the positive number that when squared gives -81. Any real number squared gives a
positive number. There is no real number that when squared gives a -81.
Answer: Not a real number.
3
d) √64
This is asking for the number that when cubed gives 64. The answer is 4, because (4)3 = 64.
Answer: 4
3
e) − √64
3
I can think of this as −1 ∗ √64. I will find the cubed root of 64, then multiply that by (-1)
3
3
− √64 = −1 ∗ √64 = −1 ∗ 4 = −4
Answer: -4
75
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.3: Definition of nth Root
16
f) √ 9
This is asking for the positive number, that when squared is 16/9. This is not hard. I can find the square root
of a fraction, by finding the square root of the numerator and the denominator.
Answer:
𝟒
𝟑
g) 3√16
First I will find the positive square root of 16, then I will multiply the result by 3.
3√16 = 3 ∗ 4 = 12
Answer: 12
__________________________________________________________________
You will be asked to find the domain of functions that contain radicals in this section. The process for
finding the domain of a radical function depends on whether the index is even or odd. If the index is even,
the function will be a real number only when the radicand is greater than or equal to zero. If the index is
odd, the function will always yield a real number.
Determining the domain of a radical function:
 If “n” is even then the domain of 𝒇(𝒙) = 𝒏√𝒙 can be found be setting the radicand (expression
under the radical) greater than or equal to 0.
 If “n” is odd the domain of 𝒇(𝒙) = 𝒏√𝒙 will be all real numbers.
___________________________________________________________________
Example 2: Find the domain of the following functions. Write your answer in interval notation. This
should help with #43-46.
𝟒
𝟓
a) 𝒇(𝒙) = √𝒙 + 𝟐
b) 𝒉(𝒙) = √𝒙 + 𝟐
4
a) Find the domain of 𝑓(𝑥) = √𝑥 + 2
The index is even, so I will find the domain by setting the radicand greater than or equal to 0.
𝑥+2≥0
𝑥 ≥ −2 This is the domain. I need to write this as an interval.
Answer: domain [−𝟐, ∞)
5
b) Find the domain of ℎ(𝑥) = √𝑥 + 2
The index is odd, so the domain is all real numbers. There is no algebra needed to find the domain.
Answer: Domain (−∞, ∞)
76
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.3: Definition of nth Root
___________________________________________________________________
Finding the nth root of a variable expression is similar to finding the nth root of a number. For principal
roots with an even index care must be taken to make sure the answer is not negative.
For example: √𝑥 2 is asking what positive number squared is x2. I could be wrong if I wrote x for an answer,
even though x2 = x2. I don’t know if x represents a positive or a negative number. The best answer I can give
is:
√𝑥 2 = |𝑥|
When the index is odd, I do not have to worry about absolute values. Only even indexes require positive
answers. The following box summarizes the last few comments.
𝒏
Definition of √𝒂𝒏
 𝒏√𝒂𝒏 = |𝒂|, If n is even and a positive integer
 𝒏√𝒂𝒏 = 𝒂 ,If n is a odd and a positive integer
___________________________________________________________________
Example 3: Simplify the radical expressions. Use absolute values when necessary.
𝟑
a) √𝒙𝟐
b) √𝒃𝟑
a) √𝑥 2 = |𝑥| (I need the absolute value in my answer because the index is even.)
Answer: |𝒙|
3
b) √𝑏 3 = 𝑏 (I don’t need the absolute value in my answer because the index is odd.)
Answer: b
___________________________________________________________________
For the rest of the section, we will assume that any variable in a problem represents a positive number.
This allows us to write answers without absolute values.
___________________________________________________________________
Example 4: Simplify the expressions. Assume all variables represent positive real numbers, so no
absolute values will be needed for any of the answers. This should help with #59-74.
𝟑
a) √𝟏𝟔𝒙𝟐
b) √𝒂𝟏𝟎
c) √𝒃𝟏𝟐
𝟒
d) √𝒛𝟏𝟐
𝟖𝟏
e) √ 𝒙𝟔
𝟑
𝟑
𝟖𝒙
f) √𝟐𝟕𝒛𝟔
a) √16𝑥 2
This is asking what squared equals 16x2. I will solve this by taking the square root of the 16 and the x2.
√16𝑥 2 = 4𝑥, I can check my work by squaring my answer.
(4x)2 = (4x)(4x) = 16x2
Answer: 4x
77
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.3: Definition of nth Root
b) √𝑎10
This is asking the question what squared equals a10. The easiest way to get this answer is to divide the
exponent by the index. The index in this problem is a 2, because this a square root problem.
√𝑎10 = 𝑎10⁄2 = 𝑎5
I can check my work by squaring my answer. If I square my answer I should get a10.
(a5)2 = a5*2 = a10 , I have confirmed my answer is correct.
Answer: a5
3
c) √𝑏12
This is asking the question what cubed equals b12. The easiest way to get this answer is to divide the
exponent by the index of 3.
3
√𝑏12 = 𝑏 12⁄3 = 𝑏 4
I can check my work by cubing my answer. If I cube my answer I should get b12.
(b4)3=b4*3 = b12, I have confirmed my answer is correct.
Answer: b4
4
d) √𝑧12
This is asking the question what to the fourth power equals z12. The easiest way to get this answer is to
divide the exponent by the index of 4.
4 12
√𝑧 = 𝑧 12⁄4 = 𝑧 3
I can check my work by raising my answer to the fourth power. If I raise m y answer to the fourth power I
should get z12.
(z3)4= z3*4 = z12, I have confirmed my answer is correct.
Answer: z3
81
e) √𝑥 6
I need to find the square root of the numerator and the denominator to solve this. The square root of 81 is
9. I can find the square root of x6 by dividing the exponent by 2.
Answer:
3
𝟗
𝒙𝟑
8𝑥 3
f) √27𝑧6
I need to find the cubed root of the numerator and denominator to solve this.
The cubed root of 8 is 2, because (2)3 = 8. I will divide the exponent of the x by three to find its cube root.
The cubed root of 27 is 3, because (3)3 = 27. I will divide the exponent of the z by 3 to find its cubed root.
Answer:
𝟐𝒙
𝟑𝒛𝟐
___________________________________________________________________
78
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.4: Rational Exponents
We simplified radical expression that contained variables in section 4.3 by dividing the exponent of the
variable by the index. The next rule formally states that you can simplify radical expressions by dividing
exponents. It also states that you can make a problem that has a fraction exponent into and expression that
contains a radical.
Radical expressions are equivalent to expressions with fractional exponents. The following rule
describes this relationship.
𝒏
√𝒂 = 𝒂𝟏⁄𝒏
This rule applies, provided n is a positive integer greater than 1 and “a” is a real number.
___________________________________________________________________
Example 1: Evaluate the expression. This will give you practice using the rule, and prepare you for
problems later in the section.
a) 𝟖𝟏𝟏⁄𝟐
b) 𝟖𝟏𝟏⁄𝟒
c) −𝟖𝟏𝟏⁄𝟐
a) 811⁄2
The rule tells me that I can rewrite this problem in radical form.
2
811⁄2 = √81 = √81 = 9 (The rule asks me to write the denominator of the fraction exponent as the
index. However, I don’t really need to write an index when it is a 2. An index of 2 is assumed when no index
is written.
Answer: 9
b) 811⁄4
I will use the rule in the above box to write this is radical form.
4
4
811⁄4 = √81 = 3 (Remember √81 is asking what positive number times itself 4 time s is 81.)
Answer: 3
c) −811⁄2
The negatives can get confusing. It might help to think of the negative sign as multiplication by negative 1.
−811⁄2 = −1 ∗ 811⁄2 ( I will now write this in radical form.)
= −1 ∗ √81
= −1 ∗ 9
Answer: -9
_____________________________________________________________________
79
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.4: Rational Exponents
We now need to work with problems that have more fractional exponents that do not have a 1 in the
numerator.
84⁄3 is an example of such a problem. We can rewrite this problem two ways using the rules of exponents
we studied earlier in the chapter. I will simplify the expression after I rewrite it.
4
4

84⁄3 = (81⁄3 ) = (√8) = 24 = 16

84⁄3 = (84 )1⁄3 = √84 = √4096 = 16
3
3
3
I was able to rewrite each problem so that I could use the rule in the above box and write the problem as a
radical. I should be able to write any problem with a fractional exponent in radical form, and I should be
able to write any problem in radical form with a fractional exponent. The following box tells me how this
can be done.
This rule explains how expressions with fractional exponents that have a numerator other than one
are related to radical expressions. Notice there are two ways to write an expression with a fractional
exponent as a radical expression.
Definition of 𝒂𝒎⁄𝒏
𝒎
𝒎
𝒂𝒎⁄𝒏 = (𝒂𝟏⁄𝒏 ) = ( 𝒏√𝒂) and
𝒏
𝒂𝒎⁄𝒏 = (𝒂𝒎 )𝟏⁄𝒏 = √𝒂𝒎
(provided n is a positive integer greater than 1, and m is an integer not equal to 0)
The fractional exponent of m/n does two things. The numerator raises the base to the mth power and the
denominator takes the nth root.
The next few problems are designed to get you used to working with the rule.
_____________________________________________________________________
Example 2: Write the expression in radical notation. Do not simplify. This should help with #1-6.
a) 𝟓𝟒⁄𝟑
b) (𝟓𝒙)𝟑⁄𝟓
𝟐
𝟕⁄𝟐
c) (𝒂𝒃𝟐)
a) 54⁄3
I am using the rule precisely how it is written in the box. The numerator of the exponent works like the m,
and the denominator of the exponent works like the n.
𝟑
𝟑
𝟒
Answer: √𝟓𝟒 𝒐𝒓 (√𝟓)
b) (5𝑥)3⁄5
I am using the rule precisely how it is written in the box. The numerator of the exponent works like the m,
and the denominator of the exponent works like the n.
𝟓
𝟓
𝟑
Answer: √(𝟓𝒙)𝟑 𝒐𝒓 (√𝟓𝒙)
80
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.4: Rational Exponents
2 7⁄2
)
𝑎𝑏2
c) (
I am using the rule precisely how it is written in the box. The numerator of the exponent works like the m,
and the denominator of the exponent works like the n.
Answer:
𝟕
√( 𝟐 𝟐 )
𝒂𝒃
𝟕
𝒐𝒓
𝟐
(√𝒂𝒃𝟐)
Example 3: Write the expression using rational exponents rather than radical notation. This should help
with #7-12.
𝟑
𝟒
a) √𝒙𝟐
b) 𝟓√𝒚
c) √𝟓𝒙𝟐
3
a) √𝑥 2
I am just using the rule again. The index becomes the denominator in the exponent.
Answer: 𝒙𝟐⁄𝟑
b) 5√𝑦
The 5 will stay separate from the y, and should not have a fractional exponent, or be in a parenthesis. It
might help if I rewrite the problem so that the y has an exponent, and so that the index is written.
2
5√𝑦 = 5 ∗ √𝑦1 (The fraction exponent attached to the y will have a 1 in the numerator and 2 in the
denominator.)
Answer: 𝟓𝒚𝟏⁄𝟐
4
c) √5𝑥 2
I am taking the fourth root of the entire quantity 5x2. I will place the 5x2 in a parenthesis to help make this
easier to follow.
4
4
√5𝑥 2 = √(5𝑥2 )1 (The problem is now ready to be written with a fractional exponent.
𝟏⁄𝟒
Answer: (𝟓𝒙𝟐 )
The next group of problems will teach you how to use multiple rules presented so far in the chapter to solve
problems.
81
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.4: Rational Exponents
____________________________________________________________________
Example 4: Write the expression using positive exponents and radical notation, then simplify.
a) 𝟐𝟓𝟑⁄𝟐
b) −𝟐𝟓𝟑⁄𝟐
c) 𝟐𝟓−𝟑⁄𝟐
d)
𝟒 −𝟓⁄𝟐
𝟒
e) (𝟗)
𝟖−𝟐⁄𝟑
a) 253⁄2
First I will write this in radical form.
3
2
253⁄2 = (√25)
= 53
Answer: 125
(To simplify this I will first find the square root of 25, then cube it.
b) −253⁄2
First I will write the coefficient as a multiplication problem by -1.
−253⁄2 = −1 ∗ 253⁄2 (Now I will write the problem as a radical.)
2
3
= −1 ∗ ( √25) (The quantity in the parenthesis is the same as in part a.)
= -1*125
Answer: -125
c) 25−3⁄2
First, I will write the problem with a positive exponent. Remember I need to create a fraction to do this.
1
25−3⁄2 = 253⁄2 (I already know that the denominator equals 125. I computed it in part a.)
Answer:
𝟏
𝟏𝟐𝟓
4
d) 8−2⁄3
First I will make the negative exponent positive. I will use a shortcut to do this. Ask me in class if you want to
see the long way to make the exponent positive.
4
8−2⁄3
= 4 ∗ 82⁄3 (Now I will write the problem with a radical.)
3
2
= 4 ∗ (√8) (I will now find the cubed root of 8.)
= 4 ∗ 22
=4*4
Answer: 16
82
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.4: Rational Exponents
16 −5⁄2
e) ( 9 )
I will first make the exponent positive by taking the reciprocal of the fraction.
4 −5⁄2
(9)
9 5⁄2
= (4)
(I now will write the problem in radical form.)
5
9
= (√ ) (Next I will find the square root of 9/4.)
4
3 5
= (2) (The last step is to raise both the 3 and the 2 to the fifth power.)
Answer:
𝟐𝟒𝟑
𝟑𝟐
Example 5: Simplify the expression using the properties of rational exponents. Write final answer using
only positive exponents.
a) 𝒙−𝟐⁄𝟑 𝒙𝟏𝟏⁄𝟑
d)
b)
𝒚𝟑⁄𝟓
𝒚−𝟒⁄𝟓
𝟒
c) (𝟐𝒙−𝟏⁄𝟐 )
𝟏𝟔𝒙𝟐
𝟏𝟐𝒙𝟏⁄𝟐
a) 𝑥 −2⁄3 𝑥 11⁄3
I just need to add the exponents to do this multiplication problem.
𝑥 −2⁄3 𝑥 11⁄3 = 𝑥 −2⁄3+11⁄3 = 𝑥 9⁄3
Answer: x3
b)
𝑦 3⁄5
𝑦 −4⁄5
I just need to subtract the exponents to do this division problem.
𝑦 3⁄5
𝑦 −4⁄5
=𝑦
3⁄ −(−4⁄ )
5
5
=𝑦
3⁄ +4⁄
5
5
Answer: 𝒚𝟕⁄𝟓
83
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.4: Rational Exponents
4
c) (2𝑥 −1⁄2 )
I will first write this problem with positive exponents. I will also write the 2 with an exponent of 1 to help
make things easier to follow.
21
4
4
(2𝑥 −1⁄2 ) = (𝑥 1⁄2 ) (I will now multiply the exponents by 4 to clear the parenthesis.)
=
Answer:
d)
𝟏𝟔
𝒙𝟐
21∗4
𝑥 1⁄2∗4
16𝑥 2
12𝑥 1⁄2
I will first reduce the fraction 16/12 by 4, and subtract the exponents of the x’s.
4𝑥 2−1⁄2
16𝑥 2
12𝑥 1⁄2
=
3
denominator)
=
Answer:
( I will now simplify the exponent. I will write the 2 as 4/2 so that I have a common
4𝑥 4⁄2−3⁄2
3
𝟒𝒙𝟑⁄𝟐
𝟑
84
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.5: Properties of Radicals
This section involves simplifying radical expressions. The box below gives a formal definition as to what it
means for a radical expression to be reduces.
Simplified Form of a Radical Expression:
One way to decide of a radical expression is simplified is to first rewrite it as a product of prime factors. A
radical expression where the radicand is written as a product of prime factors is considered in a simplified
form provided each condition is met.
 The exponent of each factor is less than the index.
 There is no radical in the denominator.
We will need to use a couple of properties of radicals that have not been formally stated to simplify radical
expressions. The box below states those properties.
Multiplication and Division Properties of Radicals:
𝒏
𝒏
a. √𝒂𝒃 = 𝒏√𝒂 ∗ √𝒃
𝒏
𝒂
𝒃. √𝒃 =
𝒏
√𝒂
√𝒃
𝒏
𝒏
This theory assumes that a, b, 𝒏√𝒂 𝒂𝒏𝒅 √𝒃 are all real numbers.
This means that this rule does not apply if the index is even and a or b represent negative numbers.
Problems #1-8 in the section will get you used to working with the multiplication rule in the bottom box.
You should notice the statement at the very top of the section. It states that all variables represent positive
real numbers. This statement allows me to use the multiplication and division rules in
_____________________________________________________________________
Example 1: Use the multiplication property of radicals to multiply the expressions. Then simplify the
result.
𝟑
a) √𝟐𝒙𝟐 𝒚𝟓 ∗ 𝟑√𝟒𝒙𝒚
b) √𝒙 + 𝟒 ∗ √𝒙 + 𝟒
3
a) √2𝑥2 𝑦 5 ∗ 3√4𝑥𝑦
3
I will multiply contents of the radicals to get: √8𝑥 3 𝑦 6
To find my answer I know that the cubed root of 8 is 2, and I will divide the exponents of the variables by 3.
Answer: 2xy2
85
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.5: Properties of Radicals
b) √𝑥 + 4 ∗ √𝑥 + 4
I will multiply the contents of the radicals to get: √(𝑥 + 4)2
To find my answer I will divide the exponent by 2.
Answer: x + 4
_____________________________________________________________________
Example 2: Use the division property of radicals to divide the expression. Then simplify the result. This
should help with #9-16.
𝟑
𝟒
a)
√𝟗𝟔
b)
𝟒
√𝟔
√𝟓𝟒𝒚𝟏𝟏
𝟑
√𝟐𝒚𝟐
4
a)
√96
4
√6
4
I will divide the contents of the radicals to get: √16
This is asking what number times itself 4 times is 16. The answer is 2.
Answer: 2
3
b)
√54𝑦 11
3
√2𝑦 2
3
I will divide the contents of the radicals to get: √27𝑦 9
I know that the cubed root of 27 is 3. I will divide the exponent of the y by three and write my answer.
Answer: 3y3
_____________________________________________________________________
Example 3: Simplify the radicals.
𝟑
a) √𝟏𝟐𝟎
b) √𝟏𝟓𝟎
d) √𝟒𝟖𝟔
e) √𝒚𝟏𝟎
g) √𝟏𝟖𝒙𝟑 𝒚𝟐
h) √𝟓𝟒𝒙𝒚𝟔 𝒛𝟕
𝟑
𝟒
c) √𝟐𝟖𝟖
𝟒
f) √𝒛𝟏𝟏
𝟑
86
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.5: Properties of Radicals
3
a) √120
I will first make a factor tree to find the prime factorization of 120.
3
3
√120 = √23 ∗ 3 ∗ 5
Now I will use the multiplication property of radicals and separate the factors whose exponents are multiple
of 3 from the other factors.
3
3
= √23 ∗ √3 ∗ 5
To get the answer I will divide the exponent in the left radical by three, and I will multiply the numbers in the
right radical.
𝟑
Answer: 𝟐 √𝟏𝟓
b) √150
I will first make a factor tree to find the prime factorization of 150.
√150 = √2 ∗ 3 ∗ 52
Now I will use the multiplication property of radicals and separate the factors whose exponents are multiple
of 2 from the other factors.
= √52 ∗ √2 ∗ 3
To get the answer I will divide the exponent in the left radical by 2, and I will multiply the numbers in the
right radical.
Answer: 𝟓√𝟔
4
c) √288
I will first make a factor tree to find the prime factorization of 288.
4
4
√288 = √25 ∗ 32
The exponent on the 2 is not a multiple of 4, but it is bigger than 4. I need to rewrite the 25 as 24 * 21. To
make it so I can divide the exponent of the two by 4.
4
= √24 ∗ 2 ∗ 32
87
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.5: Properties of Radicals
Now I will use the multiplication property of radicals and separate the factors whose exponents are multiple
of 4 from the other factors.
4
4
= √24 ∗ √2 ∗ 32
To get the answer I will divide the exponent in the left radical by 4, and I will multiply the numbers in the
right radical.
𝟒
Answer: 𝟐 ∗ √𝟏𝟖
d) √486
I will first make a factor tree to find the prime factorization of 486.
√486 = √2 ∗ 35
The exponent on the 3 is not a multiple of 2, but it is bigger than 2. I need to rewrite the 35 as 34 * 31. To
make it so I can divide the exponent of the 3 by 2.
= √34 ∗ √2 ∗ 3
To get the answer I will divide the exponent in the left radical by 2, and I will multiply the numbers in the
right radical.
= 32 ∗ √6
Answer: 𝟗√𝟔
3
e) √𝑦10
The exponent of the y is not a multiple of 3, and it is bigger than 3. I need to rewrite the y10 as
𝑦 9 ∗ 𝑦1 to meet the requirement of having an exponent that is a multiple of the index and one that is less
than the index.
3
3
√𝑦10 = √𝑦 9 ∗ 3√𝑦
To get the answer I will divide the exponent in the left radical by 3.
Answer: 𝒚𝟑 ∗ 𝟑√𝒚
88
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.5: Properties of Radicals
4
f) √𝑧11
The exponent of the z is not a multiple of 4, and it is bigger than 4. I need to rewrite the z11 as
𝑧 8 ∗ 𝑧 3 to meet the requirement of having an exponent that is a multiple of the index and one that is less
than the index.
4
4
4
√𝑧11 = √𝑧 8 ∗ √𝑧 3
To get the answer I will divide the exponent in the left radical by 4.
𝟒
Answer: 𝒛𝟐 ∗ √𝒛𝟑
g) √18𝑥 3 𝑦 2
I will work the 18, the x3 and the y2 separately.
Since 18 = 32 * 2, I will put a 32 in the left radical, and a 2 in the right radical.
Since x3 = x2 * x, I will put a x2 in the left radical, and a x in the right radical.
Since the exponent of the y is a multiple of 2, I will place the y2 in the left radical, and no y in the right
radical.
√18𝑥 3 𝑦 2 = √32 𝑥2 𝑦 2 ∗ √2𝑥
Now I will divide the exponents in the left radical by 2 and write my answer.
Answer: 𝟑𝒙𝒚√𝟐𝒙
3
h) √54𝑥𝑦 6 𝑧 7
I will work with the 54, x, y6 and z7 separately.
Since 54 = 2*33 I will place a 33 in the left radical, and a 2 in the right radical.
Since the exponent of the x is not a multiple of 3, and it is less than 3 the entire x will go in the right radical.
I can put the y6 in the left radical since its exponent is a multiple of 6.
Since z7 = z6 * z1, I will place a z6 in the left radical, and a z1 in the right radical.
3
3
3
√54𝑥𝑦 6 𝑧 7 = √33 𝑦 6 𝑧 6 ∗ √2𝑥𝑧
Now I will divide the exponents in the left radical by 3 and write my answer.
𝟑
Answer: 𝟑𝒚𝟐 𝒛𝟐 ∗ √𝟐𝒙𝒛
89
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.6: Addition and Subtraction of Radicals
This is a section on adding and subtracting radical expressions. “Like radicals” can be added or subtracted.
To be “like radicals” two radicals have to have the same index and same radicand.
These two radicals are “like radicals” and could be added or subtracted.
3
3
√3𝑥 𝑎𝑛𝑑 5√3𝑥
These two radicals are not “like radicals” and could not be added or subtracted.
3
4
√𝑦 𝑎𝑛𝑑 √𝑦 (These are not like, because they have a different index.
_____________________________________________________________________
Example 1: Add or subtract the radical expressions if possible. This will help with #1-12 in the section.
a) 𝟐√𝟕 + 𝟗√𝟕
b)
𝟏
𝟕
𝒙√𝟓 − 𝟐 𝒙√𝟓
𝟐
a) 2√7 + 9√7
These are like radicals and can be added. To add like radicals just add the coefficients and leave the rest of
the problem unchanged.
Answer: 𝟏𝟏√𝟕
b)
1
7
𝑥√5 − 2 𝑥√5
2
These are like radicals and can be subtracted. To subtract like radicals just subtract the coefficients and
1
7
leave the rest of the problem unchanged. I need to subtract the 2 𝑥 − 2 𝑥. I have to subtract the
numerators of the fractions to get my answer.
1
𝑥√5
2
7
6
− 2 𝑥√5 = − 2 𝑥√5
Answer: −𝟑𝒙√𝟓
____________________________________________________________________
Example 2: Add or subtract the radical expressions if possible. This will help with #13-28 in the section.
a) 𝟓√𝟏𝟖 + 𝟕√𝟓𝟎
𝟑
𝟑
b) 𝟐 ∗ √𝟏𝟔𝒙𝒚𝟓 − 𝟑𝒚 ∗ √𝟓𝟒𝒙𝒚𝟐
90
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.6: Addition and Subtraction of Radicals
a) 5√18 + 7√50
These radicals are not like and can’t be added as they are written. However, both radicals can be reduced. I
will be able to add the radicals after they are reduced. I will follow the steps taught in section 4.5 to reduce
the radicals.
5√18 + 7√50 = 5√32 ∗ 2 + 7√2 ∗ 52 = 5√32 √2 + 7√52 √2 = 45√2 + 35√2
Answer: 𝟖𝟎√𝟐
3
3
b) 2 ∗ √16𝑥𝑦 5 − 3𝑦 ∗ √54𝑥𝑦 2
These radicals are not like and can’t be subtracted as they are written. However, both radicals can be
reduced. I will be able to subtract the radicals after they are reduced. I will follow the steps taught in
section 4.5 to reduce the radicals.
3
3
3
3
2 ∗ √16𝑥𝑦 5 − 3𝑦 ∗ √54𝑥𝑦 2 = 2 ∗ √23 ∗ 2 ∗ 𝑥 ∗ 𝑦 3 𝑦 2 − 3𝑦 ∗ √2 ∗ 33 𝑥𝑦 2
3
3
3
3
= 2 ∗ √23 𝑦 3 ∗ √2𝑥𝑦 2 − 3𝑦 ∗ √33 ∗ √2𝑥𝑦 2
3
3
= 2 ∗ 2𝑦 ∗ √2𝑥𝑦 2 − 3𝑦 ∗ 3 ∗ √2𝑥𝑦 2
3
3
= 4𝑦 ∗ √2𝑥𝑦 2 − 9𝑦 ∗ √2𝑥𝑦 2
𝟑
Answer: −𝟓𝒚 ∗ √𝟐𝒙𝒚𝟐
91
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.7: Multiplication of Radicals.
We have already multiplied a few radicals in section 4.5. This multiplication problems in this section will
require more work to simplify after the multiplication is complete than the problems in section 4.5, but the
idea is the same.
_____________________________________________________________________
Example 1: Multiply the radicals.
𝟑
𝟑
a) √𝟒 ∗ √𝟏𝟖
b) (𝟐√𝟏𝟒)(𝟑√𝟐𝟏)
d) 𝟐√𝟑(𝟓√𝟑 − 𝟒√𝟔)
3
𝟒
𝟒
c) √𝟖𝒙𝒚𝟑 ∗ √𝟔𝒙𝟑 𝒚𝟒
e) (𝟐 − 𝟑√𝒙)(𝟒 + 𝟓√𝒙)
3
a) √4 ∗ √18
I can multiply the 4 and the 18 without changing the problem because the indexes are the same.
3
3
3
√4 ∗ √18 = √72 (Now I will reduce the answer by finding the prime factorization of 72.)
3
3
3
= √23 ∗ 32 = √23 ∗ √32 (Now I will divide the exponent in the left radical, and multiply out the
right radical.)
𝟑
Answer: 𝟐 √𝟗
b) (2√14)(3√21)
First I will multiply the 2 with the 3, and the 14 with the 21.
(2√14)(3√21) = 6√294 (Next I will find the prime factorization of 294 and reduce the radical.)
= 6√2 ∗ 3 ∗ 72 = 6√72 ∗ √2 ∗ 3
= 6 ∗ 7√6
Answer: 𝟒𝟐√𝟔
4
4
c) √8𝑥𝑦 3 ∗ √6𝑥 3 𝑦 4
First I will multiply the 8 and the 6, then add the exponents of the x’s and y’s.
4
4
4
√8𝑥𝑦 3 ∗ √6𝑥 3 𝑦 4 = √48𝑥 4 𝑦 7 (Next I will get the radical ready to be simplified.)
4
4
= √24 𝑥 4 𝑦 4 ∗ √3𝑦 3
𝟑
Answer: 𝟐𝒙𝒚 ∗ √𝟑𝒚𝟑
92
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.7: Multiplication of Radicals.
d) 2√3(5√3 − 4√6)
First I will clear the parenthesis by using the distributive property.
2√3(5√3 − 4√6) = (2√3)(5√3) − (2√3)(4√6) (Next I will complete the multiplication that I started.)
= 10√9 − 8√18
= 10 ∗ 3 − 8√32 ∗ √2
= 30 − 8 ∗ 3√2
Answer: 𝟑𝟎 − 𝟐𝟒√𝟐
e) (2 − 3√𝑥)(4 + 5√𝑥)
First I will FOIL to clear the parenthesis.
(2 − 3√𝑥)(4 + 5√𝑥) = 2 ∗ 4 + 2 ∗ 5√𝑥 − 3√𝑥 ∗ 4 − 3√𝑥 ∗ 5√𝑥
= 8 + 10√𝑥 − 12√𝑥 − 15√𝑥2 (Next I will combine the inners and outers, and
simplify the −15√𝑥2
Answer: 𝟖 − 𝟐√𝒙 − 𝟏𝟓𝒙
_____________________________________________________________________
Example 2: Multiply the special products. This should help with #27-36.
a) (𝟑 − 𝟒√𝒙)(𝟑 + 𝟒√𝒙)
𝟐
b) (𝟑 − 𝟒√𝒙)
a) (3 − 4√𝑥)(3 + 4√𝑥)
This is just a FOIL problem. The contents of the parenthesis are identical, except for the signs. The inners
and outers will cancel. To save time, I shouldn’t multiply the inners and outers. I will show the inners and
outers for this example, but it is not necessary.
(3 − 4√𝑥)(3 + 4√𝑥) = 3 ∗ 3 + (3)(4√𝑥) − (4√𝑥)(3) − (4√𝑥)(4√𝑥)
= 9 + 12√𝑥 − 12√𝑥 − 16√𝑥2 (Next, I combine the inners and outers, and reduce
the last term.)
Answer: 𝟗 − 𝟏𝟔𝒙
93
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.7: Multiplication of Radicals.
2
b) (3 − 4√𝑥)
This is also a FOIL problem. The inners and outers do not cancel. I will show you a shortcut to simplify this
problem in class.
First I will write out the problem so that it is more obviously a FOIL problem.
2
(3 − 4√𝑥) = (3 − 4√𝑥)(3 − 4√𝑥) (Next I will FOIL.)
= 9 − 12√𝑥 − 12√𝑥 + 16√𝑥2 (Next I will combine the inners and outers, and simplify the
last term.)
Answer: 𝟗 − 𝟐𝟒√𝒙 + 𝟏𝟔𝒙
94
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.8: Rationalization
A rule in section 4.5 stated that one of the conditions for a radical expression to be in its most simplified
form is that the denominator can’t contain a radical.
Rationalization is the process of simplifying a radical expression by getting the radical out of the
denominator.
It is necessary to multiply a rational expression by a carefully selected version of 1 to rationalize any radical
expression that contains a fraction.
___________________________________________________________________
Example 1: Rationalize the denominator. This will help with the square root problems in #1-22.
a)
a)
𝟓
b)
√𝟔
𝟖
√𝟏𝟎
c)
𝒙
√𝟑𝒙
5
√6
I can multiply by
√6
√6
to rationalize the denominator. In fact if the denominator of a fraction contains a
monomial that is a square root I can rationalize the fraction by multiplying by “the denominator over the
denominator.”
5
√6
=
5
√6
∗
√6 √6
=
5√6
√36
Answer:
b)
(Next I will multiply the numerators and denominators, then simplify.)
𝟓√𝟔
𝟔
𝒐𝒓
𝟓
√𝟔
𝟔
8
√10
This is a fraction that has a monomial square root in the denominator. I just need to multiply by the
denominator over the denominator.
8
√10
=
8
√10
=
8√10
√100
=
8√10
10
Answer:
∗
√10
√10
(I need to reduce the 8/10 that is not part of the radical.)
𝟒√𝟏𝟎
𝟓
𝟒
𝒐𝒓 𝟓 √𝟏𝟎
95
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.8: Rationalization
c)
𝑥
√3𝑥
This is a fraction that has a monomial square root in the denominator. I just need to multiply by the
denominator over the denominator.
𝑥
√3𝑥
𝑥
√3𝑥
=
√3𝑥
√3𝑥
𝑥 √3𝑥
=
=
∗
√ 9𝑥 2
𝑥√3𝑥
3𝑥
(Next I can cancel the x’s that are not under the square root.)
𝟏
√𝟑𝒙
𝟑
Answer:
𝒐𝒓 𝟑 √𝟑𝒙
____________________________________________________________________
Example 2: Rationalize the denominator. This will help with the cube and fourth root problems in #1-22.
𝟓
a)
𝟒
a)
4
b)
√ 𝒙𝟓
𝟐
𝟑
√𝟑
5
√𝑥 5
I have to be more careful when solving problems that are not square roots. In fact IT WOULD BE WRONG
𝟒
TO MULTIPLY BY
√𝑿𝟓
𝟒
√𝑿𝟓
. It takes thought to determine what to multiply by in non square root problems. I
4
√𝑥 3
need to multiply by something so that my exponent becomes a multiple of 4. I will multiply by 4
5
4
√𝑥 5
=
√𝑥 3
4
√x3
5
∗4
4
√x3
√𝑥 5
(Next I will multiply the denominator by adding exponents.)
4
=
5∗ √𝑥 3
4
√𝑥 8
(Next I will divide the exponent in the denominator by 4 to get rid of the radical.)
𝟒
Answer:
𝟓∗ √𝒙𝟑
𝒙𝟐
96
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.8: Rationalization
b)
2
3
√3
I have to be more careful when solving problems that are not square roots. In fact IT WOULD BE WRONG
𝟑
√𝟑
TO MULTIPLY BY
𝟑
. It takes thought to determine what to multiply by in non square root problems. I
√𝟑
3
√32
need to multiply by something so that my exponent becomes a multiple of 3. I will multiply by 3
√32
2
3
√3
=
.
3
√32
2
3
√3
∗3
1
(Now I will add the exponents to complete the multiplication in the denominator.)
√32
3
=
2∗ √32
3
√33
(Next I will divide the exponent in the denominator by 3 to clear the radical, and square
the 3 under the radical in the numerator.)
𝟑
Answer:
𝟐∗ √𝟗
𝟑
𝒐𝒓
𝟐𝟑
√𝟗
𝟑
_____________________________________________________________________
Example 3: Rationalize the denominators by multiplying by the conjugate. This should help with #23-34.
a)
√𝟔
𝟑+√𝟔
a)
√6
3+√6
b)
𝟓
√𝒙−𝟑
When I denominator of a fraction contains a binomial with a square root, I have to multiply by the conjugate
3−√6
to
√6
of the denominator to rationalize the denominator. In this problem I will multiply by 3−
rationalize the
denominator.
√6
3+√6
=
(3−√6)
√6
∗
(3+√6) (3−√6)
(Next I will FOIL the denominator, remember the outers and inners should cancel.
I also will multiply out the numerator.)
=
3√6−√36
9−3√6+3√6−√36
=
3√6−6
9−6
=
3(√6−2)
3
(I will cancel the inners and outers in the denominator, and replace √36 with 6.
(I will factor out the GCF of 3 from the numerator.)
(Next I will cancel the 3’s.)
Answer: √𝟔 − 𝟐
97
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.8: Rationalization
b)
5
√𝑥−3
When I denominator of a fraction contains a binomial with a square root, I have to multiply by the conjugate
of the denominator to rationalize the denominator. In this problem I will multiply by
√𝑥+3
to
√𝑥+3
rationalize the
denominator.
5
(√𝑥−3)
∗
(√𝑥+3)
(√𝑥+3)
=
=
Answer:
𝟓(√𝒙+𝟑)
𝒙−𝟗
5√𝑥+15
√𝑥2 +3√𝑥−3√𝑥−9
5(√𝑥+3)
𝑥−9
𝒐𝒓
𝟓√𝒙+𝟏𝟓
𝒙−𝟗
_____________________________________________________________________
98
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.9: Radical Equations
An equation that contains one or more radicals is called a radical equation. We need to remember
a rule from earlier in the chapter to make this section easier. The rule is:
𝑛
𝑛
𝑛
( √𝑥) = 𝑥, 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 √𝑥 𝑖𝑠 𝑎 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟
These are a few examples of how we will use this rule in section 4.10.
__________________________________________________________________
Example 1: Simplify the following. Assume each variable represents a positive real number.
These are to familiarize you with the above rule.
𝟑
𝟑
3
3
a) ( √𝟐𝒙)
b) (√𝟑)
𝟐
𝟒
c) ( √𝒙 + 𝟑)
𝟒
a) ( √2𝑥 )
I will do this problem and the next problem the long way. First I will write without the exponent
and write it with repeated multiplication.
3
3
3
3
3
( √2𝑥) = ( √2𝑥 )( √2𝑥)( √2𝑥) (Next I will multiply the 2x’s under the radicals.)
3
= √(2𝑥)3 (Next I will divide the exponent by the index to get rid of the radical.)
= (2𝑥)3⁄3
Answer: 2x
2
b) (√3)
I will do this problem and the next problem the long way. First I will write without the exponent
and write it with repeated multiplication.
2
(√3) = (√3)(√3) (Next I will multiply the 3’s under the radicals.)
= √9
Answer: 3
4
4
c) ( √𝑥 + 3)
I hope you see that if you raise a radical to the power of the index you can just drop the radical.
This assumes that the original expression represents a real number. I will write this answer down
without performing any algebra.
Answer: x+3
___________________________________________________________________
99
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.9: Radical Equations
The following box contains a list of steps that will help you solve the problems in this section.
Steps to solve a radical equation.
1. Rewrite the equation so that the term with the radical is isolated. If the equation has two
radicals you need to isolate one of them. I usually try to isolate a radical so that is has
positive coefficient.
2. Get rid of the radical that you isolated by raising each side of the equation to the power of
the index of the isolated radical. Note if the side of the equation opposite the isolated radical
has a binomial you will probably need to FOIL, (assuming the radical is a square root).
3. Solve the equation obtained in step 2. If the equation still has a radical repeat steps 1 and
2.
4. Check the potential solutions in the original equation.
*You can generate solutions that do not check in problems that have an even index. It is
imperative that you check answers to equations of problems that contain an even index.
___________________________________________________________________
Example 2: Solve the equation. Be sure to check your answers. If a solution is extraneous, say
so in your solution. This should help with #1-16.
𝟑
a) √𝒃 = 𝟒
b) √𝒙 − 𝟒 = 𝟑
3
a) √𝑏 = 4
The radical is isolated. I can skip to step 2 and raise both sides of the equation to the third power.
3
3
( √𝑏) = (4)3 (Example 1 tells me I can drop the radical on the left side of the equation.)
𝑏 = 64
Check: I will replace the b in the original problem with 4 to check my work.
3
√64 = 4 ( I know the cubed root of 64 is 4. You could make a prime factor tree to simplify the
radical if you do not know this fact.)
4 = 4, The answer checks. b = 64 is a solution.
Answer: b = 64
b) √𝑥 − 4 = 3
The radical is isolated. I can skip to step 2 and raise both sides of the equation to the second
power.
2
(√𝑥 − 4) = (3)2 (Example 1 tells me I can drop the radical on the left side of the equation.)
𝑥−4=9
𝑥 = 13
Check: I will replace the x in the original problem with 13 to check my work.
√13 − 4 = 3
√9 = 3
3 = 3 The answer checks. x = 13 is a solution.
Answer: x = 13
100
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
___________________________________________________________________
Section 4.9: Radical Equations
Example 3: Solve the equation. Be sure to check your answers. If a solution is extraneous, say
so in your solution. This should help with #17-24.
a) 𝒙𝟏⁄𝟑 = 𝟐
b) (𝒙 + 𝟏)𝟏⁄𝟐 = 𝟑
a) 𝑥 1⁄3 = 2
I need to write this problem as a radical to solve it using the steps given in the section.
3
√𝑥1 = 2 (Now I will cube both sides to remove the radical.)
3
3
( √𝑥) = (2)3
x=8
Check:
81⁄3 = 2 (8 to the 1/3 power is the same as the cubed root of 8. The left side of the equation
equals 2.)
2=2
Answer: x = 8
b) (𝑥 + 1)1⁄2 = 3
I need to write this problem as a radical to solve it using the steps given in the section.
2
√(𝑥 + 1)1 = 3 (I will write the left side without the index or the exponent, as neither is
necessary.)
√𝑥 + 1 = 3 (Now I will square both sides to get rid of the radical.)
2
(√𝑥 + 1) = (3)2
x+1 = 9
x=8
Check:
(8 + 1)1⁄2 = 3
91⁄2 = 3 (The ½ power is the same as square root. I will simplify the left side by finding the
square root of 9.)
3 = 3 (The solution checks.)
Answer: x = 8
___________________________________________________________________
Example 4: Solve the equation. Be sure to check your answers. If a solution is extraneous, say
so in your solution. This should help with #25-32.
a) 𝟓 + √𝟐𝒙 = 𝟗
b) √𝟓𝒙 + 𝟔 = 𝒙
a) 5 + √2𝑥 = 9
I need to isolate the radical by subtracting 5 from each side of the equation.
√2𝑥 = 4 (Next, I will square both sides of the equation to get rid of the radical.)
2
(√2𝑥) = (4)2
2𝑥 = 16
Answer: x = 8
101
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.9: Radical Equations
Check (example 4a):
5 + √2 ∗ 8 = 9
5 + √16 = 9
5 + 4 = 9 (The solution checks.)
Answer: x = 8
b) √5𝑥 + 6 = 𝑥
The radical is isolated. I will square both sides to get rid of the radical.
2
(√5𝑥 + 6) = (𝑥)2
5𝑥 + 6 = 𝑥 2 ( I have to set this equal to zero and factor to solve.)
0 = 𝑥 2 − 5𝑥 − 6
0 = (𝑥 + 1)(𝑥 − 6)
𝑥 + 1 = 0 𝑜𝑟 𝑥 − 6 = 0
𝑥 = −1 𝑜𝑟 𝑥 = 6
Check x = -1
Check x = 6
√5(−1) + 6 = −1
√5(6) + 6 = 6
√−5 + 6 = −1
√30 + 6 = 6
√1 = −1
√36 = 6
1 = −1 (The solution does not check and is
extraneous.)
6 = 6 (The solution checks, and x = 6 is a
solution.)
Answer: x = 6, x = -1 is extraneous
___________________________________________________________________
Example 5: Solve the equation. Be sure to check your answers. If a solution is extraneous, say
so in your solution. This should help with #33-46.
a) √𝟐𝒚 − 𝟒 = √𝟑𝒚 − 𝟐𝟎
b) √𝟑𝒛 + 𝟏 − √𝒛 + 𝟒 = 𝟏
a) √2𝑦 − 4 = √3𝑦 − 20
I will square both sides to get rid of the radicals.
2
2
(√2𝑦 − 4) = (√3𝑦 − 20)
(This allows me to drop both the radicals.)
2𝑦 − 4 = 3𝑦 − 20 (I will add 20 to both sides, and subtract 2y from both sides to solve this.)
-2y +20 -2y +20
16 = y
102
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.9: Radical Equations
Check (example 5a):
√2(16) − 4 = √3(16) − 20
√28 = √28 (This is a true statement. I do not need to find the decimal equivalence. y = 16 is a
solution.
Answer: y = -16
b) √3𝑧 + 1 − √𝑧 + 4 = 1
First I need to isolate a radical. I will add √𝑧 + 4 to both sides and isolate the √2𝑧 + 1.
√3𝑧 + 1 − √𝑧 + 4 = 1
+√𝑧 + 4 + √𝑧 + 4
√3𝑧 + 1 = 1 + √𝑧 + 4 (Now I will square both sides. This will get rid of the radical on the right
side.)
2
2
(√3𝑧 + 1) = (1 + √𝑧 + 4) (I have to FOIL the right side.)
3𝑧 + 1 = (1 + √𝑧 + 4)(1 + √𝑧 + 4)
2
3𝑧 + 1 = 1 + √𝑧 + 4 + √𝑧 + 4 + (√𝑧 + 4)
3𝑧 + 1 = 1 + 2√𝑧 + 4 + 𝑧 + 4 (I will combine like terms on the right side.)
3𝑧 + 1 = 𝑧 + 5 + 2√𝑧 + 4 (I need to isolate the radical by subtracting z and 4 from both sides.)
−𝑧 − 5 − 𝑧 − 5
2𝑧 − 4 = 2√𝑧 + 4 (I am going to divide both sides by 2 to make the numbers easier to work with.)
2𝑧−4
2
=
2√𝑧+4
2
𝑧 − 2 = √𝑧 + 4 (I will square both sides to get rid of the radical.)
2
(𝑧 − 2)2 = (√𝑧 + 4) (I have to FOIL the left side.)
𝑧 2 − 2𝑧 − 2𝑧 + 4 = 𝑧 + 4
𝑧 2 − 4𝑧 + 4 = 𝑧 + 4 (I will set this equal to zero, and factor.)
-z -4 -z - 4
𝑧 2 − 5𝑧 = 0
𝑧(𝑧 − 5) = 0
𝑧 = 0 𝑜𝑟 𝑧 − 5 = 0
Potential solutions: z = 0, or z = 5
103
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.9: Radical Equations
Check z = 0
Check z = 5
√3(5) + 1 − √5 + 4 = 1
√3(0) + 1 − √0 + 4 = 1
√16 − √9 = 1
√1 − √4 = 1
4−3=1
1−2=1
This solution checks. z = 5 is a solution.
−1 = 1
This solution does not check and is
extraneous.
Answer: z = 5, z = 0 is extraneous.
___________________________________________________________________
104
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.10: Complex Numbers
Earlier in the chapter we learned that there are no real-valued square roots of a negative number.
For example: √−16 is not a real number because when a real number equals a positive number when
squared, so that no real number squared equals (-16). However, there is a mathematical answer to the
question what is the square root of (-16). The solution is not a real number. The solution to the question
what is the square root of (-16) is an imaginary number. Most of the work we will do in this section will
involve the imaginary number, i.
Definition of the imaginary number i:
𝒊 = √−𝟏
𝒊𝟐 = −𝟏
The imaginary number, i, will help us calculate the square roots of negative numbers.
Definition of the square root of a negative number:
√−𝒙 = 𝒊√𝒙 (This rule applies when x is a greater than or equal to zero, and –x is less than 0.)
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Example 1: Simplify the expressions. This should help with #1-16.
a) √−𝟒𝟗
b) √−𝟐𝟒
c) √−𝟔 ∙ √−𝟖
a) √−49
The rule in above the box states: √−49 = 𝑖√49
Answer: 7i
b) √−24
√−24 = 𝑖√24 = 𝑖√4 ∗ √6
Answer: 𝟐𝒊√𝟔
105
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.10: Complex Numbers
c) √−6 ∙ √−8
We have multiplied square roots before. The rule for multiplying square roots specifically states that the
values under each square root must be positive before the square roots can be multiplied. This means I
need to pull the i’s out before I multiply.
√−6 ∙ √−8 = 𝑖√6 ∙ 𝑖√8 = 𝑖 2 √48 = 𝑖 2 √16 ∙ √3 = −1 ∗ 4 ∗ √3
Answer: −𝟒√𝟑
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Definition of a Complex N umber:
A complex number is a number written in the form a + bi, where a and b are real numbers.
We call “a” the real part of a complex number.
We call b the imaginary part of a complex number.
In this section we will be asked to add, subtract, and multiply complex numbers. For the most part these
operations can be done by treating the i as any variable. The only exception is that i2 must be replaced with
the number (-1).
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Example 2: Perform the indicated operations, write your answer in standard form. This should help with
#17-44.
a) (𝟔 − 𝟑𝒊) − (𝟒 + 𝟓𝒊)
b) (𝟒𝒊)(−𝟓𝒊)
c) (𝟑 − 𝟐𝒊)(𝟓 − 𝟔𝒊)
d) (𝟓 + 𝟒𝒊)(𝟓 − 𝟒𝒊)
a) (6 − 3𝑖) − (4 + 5𝑖)
I will drop the parenthesis, and combine like terms.
(6 − 3𝑖) − (4 + 5𝑖) = 6 − 3𝑖 − 4 − 5𝑖 = (6 − 4) + (−3𝑖 − 5𝑖) = 2 − 8𝑖
Answer: 𝟐 − 𝟖𝒊
106
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.10: Complex Numbers
b) (4𝑖)(−5𝑖)
(4𝑖)(−5𝑖) = −20𝑖 2 = −20(−1) = 20
Answer: 20
c) (3 − 2𝑖)(5 − 6𝑖)
This is a FOIL problem.
(3 − 2𝑖)(5 − 6𝑖) = 15 − 18𝑖 − 10𝑖 + 12𝑖 2
= 15 − 28𝑖 + 12(−1) (I combined the inners and outers, and replaced i2 with (-1).)
= 15 − 28𝑖 − 12
Answer: 𝟑 − 𝟐𝟖𝒊
d) (5 + 4𝑖)(5 − 4𝑖)
This is a special FOIL problem. The factors are conjugate pairs. The inners and outers will cancel.
(5 + 4𝑖)(5 − 4𝑖) = 25 − 20𝑖 + 20𝑖 − 16𝑖 2
= 25 − 16(−1) (I canceled the inners and outers, and replaced i2 with (-1).)
= 25 + 16
Answer: 𝟒𝟏
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Example 3: Perform the division by multiplying by a factor equivalent to 1 that will take the i out of the
denominator. This will help with #45-56.
a)
𝟔
𝟕𝒊
b)
𝟐
𝟔+𝟓𝒊
c)
−𝟑𝒊
𝟐−𝟕𝒊
107
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.10: Complex Numbers
6
a) 7𝑖
There are two kinds of problems within example 3. This problem has a monomial in the denominator. I will
𝑖
multiply by to get the desired answer. All problems between #45-56 with a monomial in the denominator
𝑖
𝑖
can be solved by multiplying by . The next two problems have a binomial in the denominator and they will
𝑖
be solved differently.
6
7𝑖
𝑖
6𝑖
6𝑖
6𝑖
∗ 𝑖 = 7𝑖2 = 7(−1) = −7 (I will move the negative out of the denominator when I write the answer.)
−𝟔𝒊
𝟔
Answer: 𝟕 𝒐𝒓 − 𝟕 𝒊 (The negative can be written in the numerator or as a coefficient. The i can be
written in the numerator or after the fraction.)
b)
2
6+5𝑖
6−5𝑖
This problem has a binomial in the denominator. I will multiply
to get the desired answer. I call this
6−5𝑖
multiplying by conjugate of the denominator, although I am really multiplying by the conjugate over the
conjugate. I can solve all problems between, #45-56, by multiplying by the conjugate of the denominator.
2
6+5𝑖
6−5𝑖
12−10𝑖
∗ 6−5𝑖 = 36−30𝑖+30𝑖−25𝑖2 (I distributed the 2 in the numerator and FOILed the denominator.)
Answer:
=
12−10𝑖
36−25(−1)
=
12−10𝑖
36+25
=
12−10𝑖
61
(I canceled the inners and outers, and changed the i2 to (-1).)
(I will write this as two fractions in my answer.)
𝟏𝟐
𝟏𝟎
− 𝟔𝟏 𝒊
𝟔𝟏
108
Chapter 4: Radicals and Complex Numbers
Examples and Explanations
Section 4.10: Complex Numbers
c)
−3𝑖
2−7𝑖
I need to multiply by the conjugate of the denominator to solve this problem.
−3𝑖
2−7𝑖
−6𝑖−21𝑖 2
2+7𝑖
∗ 2+7𝑖 = 4+14𝑖−14𝑖−49𝑖2 (I distributed the -3i in the numerator, and FOILed the denominator.)
=
=
=
Answer:
−6𝑖−21(−1)
4−49(−1)
(I replaced the i2 with (-1) and canceled the inners and outers.)
−6𝑖+21
49+4
21−6𝑖
53
𝟐𝟏
𝟔
− 𝒊
𝟓𝟑
𝟓𝟑
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109