Intermediate Algebra 2e — Original English

Use the Complex Number System

Evaluate the Square Root of a Negative Number

Whenever we have a situation where we have a square root of a negative number we say there is no real number that equals that square root. For example, to simplify −1, we are looking for a real number x so that x2 = –1. Since all real numbers squared are positive numbers, there is no real number that equals –1 when squared.

Mathematicians have often expanded their numbers systems as needed. They added 0 to the counting numbers to get the whole numbers. When they needed negative balances, they added negative numbers to get the integers. When they needed the idea of parts of a whole they added fractions and got the rational numbers. Adding the irrational numbers allowed numbers like 5. All of these together gave us the real numbers and so far in your study of mathematics, that has been sufficient.

But now we will expand the real numbers to include the square roots of negative numbers. We start by defining the imaginary unit i as the number whose square is –1.

We will use the imaginary unit to simplify the square roots of negative numbers.

We will use this definition in the next example. Be careful that it is clear that the i is not under the radical. Sometimes you will see this written as −b=ib to emphasize the i is not under the radical. But the −b=bi is considered standard form.

Write each expression in terms of i and simplify if possible:

ⓐ −25 ⓑ −7 ⓒ −12.

Solution
ⓐ
Step-by-step simplification of the square root of -25.
−25
Use the definition of the square root of negative numbers. 25i
Simplify. 5i
ⓑ
Steps to simplify the square root of a negative number, showing the use of the imaginary unit 'i'.
−7
Use the definition of the square root of negative numbers. 7i
Simplify. Be careful that it is clear that i is not under the radical sign.
ⓒ
Step-by-step simplification of sqrt(-12) into its complex number form, 2 sqrt(3)i.
−12
Use the definition of the square root of negative numbers. 12i
Simplify 12. 23i

Now that we are familiar with the imaginary number i, we can expand our concept of the number system to include imaginary numbers. The complex number system includes the real numbers and the imaginary numbers. A complex number is of the form a + bi, where a, b are real numbers. We call a the real part and b the imaginary part.

A complex number is in standard form when written as a+bi, where a and b are real numbers.

If b=0, then a+bi becomes a+0·i=a, and is a real number.

If b≠0, then a+bi is an imaginary number.

If a=0, then a+bi becomes 0+bi=bi, and is called a pure imaginary number.

We summarize this here.

a+bi
b=0 a+0·ia Real number
b≠0 a+bi Imaginary number
a=0 0+bibi Pure imaginary number

The standard form of a complex number is a+bi, so this explains why the preferred form is −b=bi when b>0.

The diagram helps us visualize the complex number system. It is made up of both the real numbers and the imaginary numbers.

The table has four rows and three columns. The first row is a header and the second column entry a plus b i. In the second row is b equals zero, a plus 0 i, and “Real number”. The third row contains b is not equal to 0, a plus b i, and “Imaginary number”. The fourth row contains a = 0, 0 plus b i, and “Pure imaginary number”.

Add or Subtract Complex Numbers

We are now ready to perform the operations of addition, subtraction, multiplication and division on the complex numbers—just as we did with the real numbers.

Adding and subtracting complex numbers is much like adding or subtracting like terms. We add or subtract the real parts and then add or subtract the imaginary parts. Our final result should be in standard form.

Add: −12+−27.

Solution
Step-by-step solution for adding square roots of negative numbers, demonstrating simplification into complex numbers.
−12+−27
Use the definition of the square root of negative numbers. 12i+27i
Simplify the square roots. 23i+33i
Add. 53i

Remember to add both the real parts and the imaginary parts in this next example.

Simplify: ⓐ (4−3i)+(5+6i) ⓑ (2−5i)−(5−2i).

Solution
ⓐ
Step-by-step example of adding complex numbers, illustrating the use of the associative property to simplify the expression.
(4−3i)+(5+6i)
Use the Associative Property to put the real
parts and the imaginary parts together.
(4+5)+(−3i+6i)
Simplify. 9+3i
ⓑ
A step-by-step guide demonstrating the subtraction and simplification of two complex numbers.
(2−5i)−(5−2i)
Distribute. 2−5i−5+2i
Use the Associative Property to put the real
parts and the imaginary parts together.
2−5−5i+2i
Simplify. −3−3i

Multiply Complex Numbers

Multiplying complex numbers is also much like multiplying expressions with coefficients and variables. There is only one special case we need to consider. We will look at that after we practice in the next two examples.

Multiply: 2i(7−5i).

Solution
Step-by-step simplification of the complex number expression 2i(7-5i) into standard form (10 + 14i).
2i(7−5i)
Distribute. 14i−10i2
Simplify i2. 14i−10(−1)
Multiply. 14i+10
Write in standard form. 10+14i

In the next example, we multiply the binomials using the Distributive Property or FOIL.

Multiply: (3+2i)(4−3i).

Solution
A step-by-step guide to multiplying complex numbers using the FOIL method, detailing each calculation stage from the initial product to the simplified result.
(3+2i)(4−3i)
Use FOIL. 12−9i+8i−6i2
Simplify i2 and combine like terms. 12−i−6(−1)
Multiply. 12−i+6
Combine the real parts. 18−i

In the next example, we could use FOIL or the Product of Binomial Squares Pattern.

Multiply: (3+2i)2

Solution
A mathematical expression showing the fraction (a + b) / (3 + 2i) squared. The numerator 'a + b' is displayed in red text.
Use the Product of Binomial Squares Pattern, (a+b)2=a2+2ab+b2. An algebraic identity a^2 + 2ab + b^2 is shown with a specific example: 3^2 + 2 * 3 * 2i + (2i)^2, demonstrating the expansion of (a+b)^2 where a=3 and b=2i.
Simplify. The mathematical expression 9 + 12i + 4i^2 is shown.
Simplify i2. A mathematical expression showing the sum of 9, 12i, and 4 multiplied by -1, which is 9 + 12i + 4(-1).
Simplify. The image shows the complex number expression 5 + 12i in black text against a white background.

Since the square root of a negative number is not a real number, when we have the square roots of two negative numbers, we cannot use the Product Property for Radicals. In order to multiply square roots of negative numbers we should first write them as complex numbers, using −b=bi. This is one place students tend to make errors, so be careful when you see multiplying with a negative square root.

Multiply: −36·−4.

Solution

To multiply square roots of negative numbers, we first write them as complex numbers.

Step-by-step multiplication of square roots of negative numbers, demonstrating conversion to complex numbers and simplification.
−36·−4
Write as complex numbers using −b=bi. 36i·4i
Simplify. 6i·2i
Multiply. 12i2
Simplify i2 and multiply. −12

In the next example, each binomial has a square root of a negative number. Before multiplying, each square root of a negative number must be written as a complex number.

Multiply: (3−−12)(5+−27).

Solution

To multiply square roots of negative numbers, we first write them as complex numbers.

Detailed steps demonstrating the multiplication of two complex numbers, simplifying to their final form.
(3−−12)(5+−27)
Write as complex numbers using −b=bi. (3−23i)(5+33i)
Use FOIL. 15+93i−103i−6·3i2
Combine like terms and simplify i2. 15−3i−6·(−3)
Multiply and combine like terms. 33−3i

We first looked at conjugate pairs when we studied polynomials. We said that a pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference is called a conjugate pair and is of the form (a−b),(a+b).

A complex conjugate pair is very similar. For a complex number of the form a+bi, its conjugate is a−bi. Notice they have the same first term and the same last term, but one is a sum and one is a difference.

We will multiply a complex conjugate pair in the next example.

Multiply: (3−2i)(3+2i).

Solution
Step-by-step multiplication and simplification of complex conjugates (3 - 2i)(3 + 2i) to a real number result.
(3−2i)(3+2i)
Use FOIL. 9+6i−6i−4i2
Combine like terms and simplify i2. 9−4(−1)
Multiply and combine like terms. 13

From our study of polynomials, we know the product of conjugates is always of the form (a−b)(a+b)=a2−b2. The result is called a difference of squares. We can multiply a complex conjugate pair using this pattern.

The last example we used FOIL. Now we will use the Product of Conjugates Pattern.

The quantity a minus b in parentheses times the quantity a plus b in parentheses is written above the expression showing the product of 3 minus 2 i in parentheses and 3 plus 2 i in parentheses. In the next line a squared minus b squared is written above the expression 3 squared minus the quantity 2 i in parentheses squared. Simplifying we get 9 minus 4 i squared. This is equal to 9 minus 4 times negative 1. The final result is 13.

Notice this is the same result we found in Example 9.

When we multiply complex conjugates, the product of the last terms will always have an i2 which simplifies to −1.

(a−bi)(a+bi)a2−(bi)2a2−b2i2a2−b2(−1)a2+b2

This leads us to the Product of Complex Conjugates Pattern: (a−bi)(a+bi)=a2+b2

Multiply using the Product of Complex Conjugates Pattern: (8−2i)(8+2i).

Solution
Mathematical expressions illustrating the difference of squares formula, (a - b)(a + b), and its application with complex numbers, (8 - 2i)(8 + 2i).
Use the Product of Complex Conjugates Pattern,
(a−bi)(a+bi)=a2+b2.
A mathematical image displaying a general algebraic expression 'a^2 + b^2' in red text, followed by a specific numerical instance '8^2 + 2^2' in black text below it.
Simplify the squares. The image displays the mathematical expression '64 + 4' in black text against a white background.
Add. The number 68 is prominently displayed in black text against a stark white background.

Divide Complex Numbers

Dividing complex numbers is much like rationalizing a denominator. We want our result to be in standard form with no imaginary numbers in the denominator.

How to Divide Complex Numbers

Divide: 4+3i3−4i.

Solution
Step 1 is to write both the numerator and denominator in standard form. For this example they are both in standard form. Step 2 is to multiply the numerator and denominator by the complex conjugate of the denominator. The complex conjugate of 3 minus 4 i is 3 plus 4 i. The resulting expression is the quantity 4 plus 3 i in parentheses times the quantity 3 plus 4 i in parentheses divided by the product of 3 minus 4 i in parentheses and the quantity 3 plus 4 i in parentheses. Step 3 is to simplify and write the result in standard form. Use the pattern the quantity a plus b i in parentheses equals a squared plus b squared in the denominator. The expression for this example then becomes the quantity 12 plus 16 i plus 9 i plus 12 i squared in parentheses divided by the sum of 9 and 16. Combining like terms we get the quantity 12 plus 25 i minus 12 in parentheses divided by 25. Simplifying we get 25 i divided by 25. Write the result in standard form. The result is i.

We summarize the steps here.

Divide, writing the answer in standard form: −35+2i.

Solution
Step-by-step process for dividing complex numbers by rationalizing the denominator, showing actions and corresponding mathematical expressions.
−35+2i
Multiply the numerator and denominator by the
complex conjugate of the denominator.
−3(5−2i)(5+2i)(5−2i)
Multiply in the numerator and use the Product of
Complex Conjugates Pattern in the denominator.
−15+6i52+22
Simplify. −15+6i29
Write in standard form. −1529+629i

Be careful as you find the conjugate of the denominator.

Divide: 5+3i4i.

Solution
This table illustrates the step-by-step process for dividing complex numbers, transforming a complex fraction into its standard a + bi form.
5+3i4i
Write the denominator in standard form. 5+3i0+4i
Multiply the numerator and denominator by
the complex conjugate of the denominator.
(5+3i)(0−4i)(0+4i)(0−4i)
Simplify. (5+3i)(−4i)(4i)(−4i)
Multiply. −20i−12i2−16i2
Simplify the i2. −20i+1216
Rewrite in standard form. 1216−2016i
Simplify the fractions. 34−54i

Simplify Powers of i

The powers of i make an interesting pattern that will help us simplify higher powers of i. Let’s evaluate the powers of i to see the pattern.

i1i2i3i4i−1i2·ii2·i2−1·i(−1)(−1) −i1 i5i6i7i8 i4·ii4·i2i4·i3i4·i4 1·i1·i21·i31·1 ii2i31 −1−i

We summarize this now.

i1=ii5=i i2=−1i6=−1 i3=−ii7=−i i4=1i8=1

If we continued, the pattern would keep repeating in blocks of four. We can use this pattern to help us simplify powers of i. Since i4 = 1, we rewrite each power, in, as a product using i4 to a power and another power of i.

We rewrite it in the form in=(i4)q·ir, where the exponent, q, is the quotient of n divided by 4 and the exponent, r, is the remainder from this division. For example, to simplify i57, we divide 57 by 4 and we get 14 with a remainder of 1. In other words, 57=4·14+1. So we write i57=(14)14·i1 and then simplify from there.

A mathematical problem demonstrating the simplification of i to the power of 57, using long division (57 / 4 = 14 remainder 1) to determine that i^57 equals i.

Simplify: i86.

Solution
This table illustrates the step-by-step simplification of the imaginary unit i raised to the power of 86.
i86
Divide 86 by 4 and rewrite i86 in the
in=(i4)q·ir form.
(14)21·i2
A long division calculation of 86 divided by 4. The steps show that 4 goes into 8 two times, resulting in 8. Bringing down the 6, 4 goes into 6 one time, resulting in 4, with a remainder of 2. The quotient is 21 with a remainder of 2.
Simplify. (1)21·(−1)
Simplify. –1

Key Concepts

  • Square Root of a Negative Number
    • If b is a positive real number, then −b=bi
      a+bi
      b=0 a+0·ia Real number
      b≠0 a+bi Imaginary number
      a=0 0+bibi Pure imaginary number
    • A complex number is in standard form when written as a + bi, where a, b are real numbers.
      The diagram has a rectangle with the labels “Complex Numbers” and a plus b i. A second rectangle has the labels “Real Numbers”, a plus b i, b = 0. A third rectangle has the labels “Imaginary Numbers”, a plus b i, b not equal to 0. Arrows go from the Real Numbers rectangle and Imaginary Numbers rectangle and point toward the Complex Numbers rectangle.
  • Product of Complex Conjugates
    • If a, b are real numbers, then
      (a−bi)(a+bi)=a2+b2
  • How to Divide Complex Numbers
    1. Write both the numerator and denominator in standard form.
    2. Multiply the numerator and denominator by the complex conjugate of the denominator.
    3. Simplify and write the result in standard form.

Section Exercises

Practice Makes Perfect

Evaluate the Square Root of a Negative Number

In the following exercises, write each expression in terms of i and simplify if possible.

ⓐ −16 ⓑ −11
ⓒ −8

Solution

ⓐ 4i ⓑ 11i ⓒ 22i

ⓐ −121 ⓑ −1 ⓒ −20

ⓐ −100 ⓑ −13 ⓒ −45

Solution

ⓐ 10i ⓑ 13i ⓒ 35i

ⓐ −49 ⓑ −15 ⓒ −75

Add or Subtract Complex Numbers In the following exercises, add or subtract.

−75+−48

Solution

93i

−12+−75

−50+−18

Solution

82i

−72+−8

(1+3i)+(7+4i)

Solution

8+7i

(6+2i)+(3−4i)

(8−i)+(6+3i)

Solution

14+2i

(7−4i)+(−2−6i)

(1−4i)−(3−6i)

Solution

−2+2i

(8−4i)−(3+7i)

(6+i)−(−2−4i)

Solution

8+5i

(−2+5i)−(−5+6i)

(5−−36)+(2−−49)

Solution

7−13i

(−3+−64)+(5−−16)

(−7−−50)−(−32−−18)

Solution

25−22i

(−5+−27)−(−4−−48)

Multiply Complex Numbers

In the following exercises, multiply.

4i(5−3i)

Solution

12+20i

2i(−3+4i)

−6i(−3−2i)

Solution

−12+18i

−i(6+5i)

(4+3i)(−5+6i)

Solution

−38++9i

(−2−5i)(−4+3i)

(−3+3i)(−2−7i)

Solution

27+15i

(−6−2i)(−3−5i)

In the following exercises, multiply using the Product of Binomial Squares Pattern.

(3+4i)2

Solution

−7+24i

(−1+5i)2

(−2−3i)2

Solution

−5+12i

(−6−5i)2

In the following exercises, multiply.

−25·−36

Solution

−30

−4·−16

−9·−100

Solution

−30

−64·−9

(−2−−27)(4−−48)

Solution

−44−4i3

(5−−12)(−3+−75)

(2+−8)(−4+−18)

Solution

−20−22i

(5+−18)(−2−−50)

(2−i)(2+i)

Solution

5

(4−5i)(4+5i)

(7−2i)(7+2i)

Solution

53

(−3−8i)(−3+8i)

In the following exercises, multiply using the Product of Complex Conjugates Pattern.

(7−i)(7+i)

Solution

50

(6−5i)(6+5i)

(9−2i)(9+2i)

Solution

85

(−3−4i)(−3+4i)

Divide Complex Numbers

In the following exercises, divide.

3+4i4−3i

Solution

i

5−2i2+5i

2+i3−4i

Solution

225+1125i

3−2i6+i

32−3i

Solution

613+913i

24−5i

−43−2i

Solution

−1213−813i

−13+2i

1+4i3i

Solution

43−13i

4+3i7i

−2−3i4i

Solution

−34+12i

−3−5i2i

Simplify Powers of i

In the following exercises, simplify.

i41

Solution

i

i39

i66

Solution

−1

i48

i128

Solution

1

i162

i137

Solution

i

i255

Writing Exercises

Explain the relationship between real numbers and complex numbers.

Solution

Answers will vary.

Aniket multiplied as follows and he got the wrong answer. What is wrong with his reasoning?

−7·−7497

Why is −64=8i but −643=−4.

Solution

Answers will vary.

Explain how dividing complex numbers is similar to rationalizing a denominator.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

The table has 4 columns and 4 rows. The first row is a header row with the headers “I can…”, “Confidently”, “With some help.”, and “No – I don’t get it!”. The first column contains the phrases “evaluate the square root of a negative number”, “add or subtract complex numbers”, “multiply complex numbers”, “divide complex numbers”, and “simplify powers of i”. The other columns are left blank so the learner can indicate their level of understanding.

ⓑ On a scale of 1−10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Chapter Review Exercises

Simplify Expressions with Roots

Simplify Expressions with Roots

In the following exercises, simplify.

ⓐ 225 ⓑ −16

Solution

ⓐ 15 ⓑ −4

ⓐ −169 ⓑ −8

ⓐ 83 ⓑ 814 ⓒ 2435

Solution

ⓐ 2 ⓑ 3 ⓒ 3

ⓐ −5123 ⓑ −814 ⓒ −15

Estimate and Approximate Roots

In the following exercises, estimate each root between two consecutive whole numbers.

ⓐ 68 ⓑ 843

Solution

ⓐ 8<68<9
ⓑ 4<843<5

In the following exercises, approximate each root and round to two decimal places.

ⓐ 37 ⓑ 843 ⓒ 1254

Simplify Variable Expressions with Roots

In the following exercises, simplify using absolute values as necessary.


ⓐ a33
ⓑ b77

Solution

ⓐ a ⓑ b


ⓐ a14
ⓑ w24


ⓐ m84
ⓑ n205

Solution

ⓐ m2 ⓑ n4


ⓐ 121m20
ⓑ −64a2


ⓐ 216a63
ⓑ 32b205

Solution

ⓐ 6a2 ⓑ 2b4


ⓐ 144x2y2
ⓑ 169w8y10
ⓒ 8a51b63

Simplify Radical Expressions

Use the Product Property to Simplify Radical Expressions

In the following exercises, use the Product Property to simplify radical expressions.

125

Solution

55

675

ⓐ 6253 ⓑ 1286

Solution

ⓐ 553 ⓑ 226

In the following exercises, simplify using absolute value signs as needed.


ⓐ a23
ⓑ b83
ⓒ c138


ⓐ 80s15
ⓑ 96a75
ⓒ 128b76

Solution

ⓐ 4|s7|5s ⓑ 2a3a25
ⓒ 2|b|2b6


ⓐ 96r3s3
ⓑ 80x7y63
ⓒ 80x8y94


ⓐ −325
ⓑ −18

Solution

ⓐ −2 ⓑ not real


ⓐ 8+96
ⓑ 2+402

Use the Quotient Property to Simplify Radical Expressions

In the following exercises, use the Quotient Property to simplify square roots.

ⓐ 7298 ⓑ 24813 ⓒ 6964

Solution

ⓐ 67 ⓑ 23 ⓒ 12

ⓐ y4y8 ⓑ u21u115 ⓒ v30v126

300m564

Solution

5m23m4


ⓐ 28p7q2
ⓑ 81s8t33
ⓒ 64p15q124


ⓐ 27p2q108p4q3
ⓑ 16c5d7250c2d23
ⓒ 2m9n7128m3n6

Solution

ⓐ 12|pq| ⓑ 2cd5d23
ⓒ |mn|2


ⓐ 80q55q
ⓑ −625353
ⓒ 80m745m4

Simplify Rational Exponents

Simplify expressions with a1n

In the following exercises, write as a radical expression.

ⓐ r12 ⓑ s13 ⓒ t14

Solution

ⓐ r ⓑ s3 ⓒ t4

In the following exercises, write with a rational exponent.

ⓐ 21p ⓑ 8q4 ⓒ 436r6

In the following exercises, simplify.


ⓐ 62514
ⓑ 24315
ⓒ 3215

Solution

ⓐ 5 ⓑ 3 ⓒ 2


ⓐ (−1,000)13
ⓑ −1,00013
ⓒ (1,000)−13


ⓐ (−32)15
ⓑ (243)−15
ⓒ −12513

Solution

ⓐ −2 ⓑ 13 ⓒ −5

Simplify Expressions with amn

In the following exercises, write with a rational exponent.


ⓐ r74
ⓑ (2pq5)3
ⓒ (12m7n)34

In the following exercises, simplify.


ⓐ 2532
ⓑ 9−32
ⓒ (−64)23

Solution

ⓐ 125 ⓑ 127 ⓒ 16


ⓐ −6432
ⓑ −64−32
ⓒ (−64)32

Use the Laws of Exponents to Simplify Expressions with Rational Exponents

In the following exercises, simplify.


ⓐ 652·612
ⓑ (b15)35
ⓒ w27w97

Solution

ⓐ 63 ⓑ b9 ⓒ 1w


ⓐ a34·a−14a−104
ⓑ (27​b23​c−52b−73c12)13

Add, Subtract and Multiply Radical Expressions

Add and Subtract Radical Expressions

In the following exercises, simplify.


ⓐ 72−32
ⓑ 7p3+2p3
ⓒ 5x3−3x3

Solution

ⓐ 42 ⓑ 9p3 ⓒ 2x3


ⓐ 11b−511b+311b
ⓑ 811cd4+511cd4−911cd4


ⓐ 48+27
ⓑ 543+1283
ⓒ 654−32804

Solution

ⓐ 73 ⓑ 723 ⓒ 354


ⓐ 80c7−20c7
ⓑ 2162r104+432r104

375y2+8y48−300y2

Solution

37y3

Multiply Radical Expressions

In the following exercises, simplify.


ⓐ (56)(−12)
ⓑ (−2184)(−94)


ⓐ (32x3)(718x2)
ⓑ (−620a23)(−216a33)

Solution

ⓐ 126x2x ⓑ 48a5a23

Use Polynomial Multiplication to Multiply Radical Expressions

In the following exercises, multiply.


ⓐ 11(8+411)
ⓑ 33(93+183)


ⓐ (3−27)(5−47)
ⓑ (x3−5)(x3−3)

Solution

ⓐ 71−227
ⓑ x23−8x3+15

(27−511)(47+911)


ⓐ (4+11)2
ⓑ (3−25)2

Solution

ⓐ 27+811 ⓑ 29−125

(7+10)(7−10)

(3x3+2)(3x3−2)

Solution

9x23−4

Divide Radical Expressions

Divide Square Roots

In the following exercises, simplify.


ⓐ 4875
ⓑ 813243


ⓐ 320mn−545m−7n3
ⓑ 16x4y−23−54x−2y43

Solution

ⓐ 8m43n4 ⓑ −2x23y2

Rationalize a One Term Denominator

In the following exercises, rationalize the denominator.

ⓐ 83 ⓑ 740 ⓒ 82y

ⓐ 1113 ⓑ 7543 ⓒ 33x23

Solution

ⓐ 121311 ⓑ 2836 ⓒ 9x3x

ⓐ 144 ⓑ 9324 ⓒ 69x34

Rationalize a Two Term Denominator

In the following exercises, simplify.

72−6

Solution

−7(2+6)2

5n−7

x+8x−8

Solution

(x+22)x−82

Solve Radical Equations

Solve Radical Equations

In the following exercises, solve.

4x−3=7

5x+1=−3

Solution

no solution

4x−13=3

u−3+3=u

Solution

u=3,u=4

4x+53−2=−5

(8x+5)13+2=−1

Solution

x=−4

y+4−y+2=0

28r+1−8=2

Solution

r=3

Solve Radical Equations with Two Radicals

In the following exercises, solve.

10+2c=4c+16

2x2+9x−183=x2+3x−23

Solution

x=−8,x=2

r+6=r+8

x+1−x−2=1

Solution

x=3

Use Radicals in Applications

In the following exercises, solve. Round approximations to one decimal place.

Landscaping Reed wants to have a square garden plot in his backyard. He has enough compost to cover an area of 75 square feet. Use the formula s=A to find the length of each side of his garden. Round your answer to the nearest tenth of a foot.

Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was 175 feet. Use the formula s=24d to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Solution

64.8 feet

Use Radicals in Functions

Evaluate a Radical Function

In the following exercises, evaluate each function.

g(x)=6x+1, find
ⓐ g(4)
ⓑ g(8)

G(x)=5x−1, find
ⓐ G(5)
ⓑ G(2)

Solution

ⓐ G(5)=26 ⓑ G(2)=3

h(x)=x2−43, find
ⓐ h(−2)
ⓑ h(6)

For the function
g(x)=4−4x4, find
ⓐ g(1)
ⓑ g(−3)

Solution

ⓐ g(1)=0 ⓑ g(−3)=2

Find the Domain of a Radical Function

In the following exercises, find the domain of the function and write the domain in interval notation.

g(x)=2−3x

F(x)=x+3x−2

Solution

(2,∞)

f(x)=4x2−163

F(x)=10−7x4

Solution

−∞,107

Graph Radical Functions

In the following exercises, ⓐ find the domain of the function ⓑ graph the function ⓒ use the graph to determine the range.

g(x)=x+4

g(x)=2x

Solution

ⓐ domain: [0,∞)
ⓑ
The figure shows a square root function graph on the x y-coordinate plane. The x-axis of the plane runs from 0 to 8. The y-axis runs from 0 to 8. The function has a starting point at (0, 0) and goes through the points (1, 2) and (4, 4).
ⓒ range: [0,∞)

f(x)=x−13

f(x)=x3+3

Solution

ⓐ domain: (−∞,∞)
ⓑ
The figure shows a cube root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis runs from negative 2 to 6. The function has a center point at (0, 3) and goes through the points (negative 1, 2) and (1, 4).
ⓒ range: (−∞,∞)

Use the Complex Number System

Evaluate the Square Root of a Negative Number

In the following exercises, write each expression in terms of i and simplify if possible.


ⓐ −100
ⓑ −13
ⓒ −45

Add or Subtract Complex Numbers

In the following exercises, add or subtract.

−50+−18

Solution

82i

(8−i)+(6+3i)

(6+i)−(−2−4i)

Solution

8+5i

(−7−−50)−(−32−−18)

Multiply Complex Numbers

In the following exercises, multiply.

(−2−5i)(−4+3i)

Solution

23+14i

−6i(−3−2i)

−4·−16

Solution

−8

(5−−12)(−3+−75)

In the following exercises, multiply using the Product of Binomial Squares Pattern.

(−2−3i)2

Solution

−5+12i

In the following exercises, multiply using the Product of Complex Conjugates Pattern.

(9−2i)(9+2i)

Divide Complex Numbers

In the following exercises, divide.

2+i3−4i

Solution

225+1125i

−43−2i

Simplify Powers of i

In the following exercises, simplify.

i48

Solution

1

i255

Practice Test

In the following exercises, simplify using absolute values as necessary.

125x93

Solution

5x3

169x8y6

72x8y43

Solution

2x2y9x2y3

45x3y4180x5y2

In the following exercises, simplify. Assume all variables are positive.

ⓐ 256−14 ⓑ −4932

Solution

ⓐ 14 ⓑ −343

−45

x−14·x54x−34

Solution

x74

(8​x23​y−52x−73y12)13

48x5−75x5

Solution

−x23x

27x2−4x12+108x2

212x5·36x3

Solution

36x42

43(163−63)

(4−33)(5+23)

Solution

2−73

1283543

245xy−445x−4y3

Solution

7x2x3|y3|y

153

32+3

Solution

3(2−3)

−4·−9

−4i(−2−3i)

Solution

−12+8i

4+i3−2i

i172

Solution

1

In the following exercises, solve.

2x+5+8=6

x+5+1=x

Solution

x=4

2x2−6x−233=x2−3x+53

In the following exercise, ⓐ find the domain of the function ⓑ graph the function ⓒ use the graph to determine the range.

g(x)=x+2

Solution

ⓐ domain: [−2,∞)
ⓑ
The figure shows a square root function graph on the x y-coordinate plane. The x-axis of the plane runs from negative 2 to 6. The y-axis runs from 0 to 8. The function has a starting point at (negative 2, 0) and goes through the points (negative 1, 1) and (2, 2).
ⓒ range: [0,∞)

complex conjugate pair
A complex conjugate pair is of the form a + bi, a – bi.
complex number
A complex number is of the form a + bi, where a and b are real numbers. We call a the real part and b the imaginary part.
complex number system
The complex number system is made up of both the real numbers and the imaginary numbers.
imaginary unit
The imaginary unit i is the number whose square is –1. i2 = –1 or i=−1.
standard form
A complex number is in standard form when written as a+bi, where a, b are real numbers.