CLASSIFYING THE GIVEN RADICAL AS REAL OR IMAGINARY

Identify each as Real (R) or Imaginary (C). Then simplify.

Problem 1 :

√49

Solution :

√49

√49 = √(7 × 7)

= 7

So, √49 is real term.

Problem 2 :

√-49

Solution :

√-49

Square root of a negative number is considered to be an imaginary value.

= √49 × √-1

= √(7 × 7) × √-1

= 7i

So, √-49 is imaginary term.

Problem 3 :

√16

Solution :

√16

√16 = √(4 × 4)

= 4

So, √16 is real term.

Problem 4 :

√-25

Solution :

√-25

Square root of a negative number is considered to be an imaginary value.

= √25 × √-1

= √(5 × 5) × √-1

= 5i

So, √-25 is imaginary term.

Problem 5 :

√-81

Solution :

√-81

Square root of a negative number is considered to be an imaginary value.

= √81 × √-1

= √(9 × 9) × √-1

= 9i

So, √-81 is imaginary term.

Problem 6 :

Which expressions are equivalent to 1 + i? Select all that apply.

a) (4 − i) − (3 − 2i)

b) −2/(1 + i)

c) (2 − i)(i + 1)

d)  i²⁰ − i²¹

Solution :

a) (4 − i) − (3 − 2i)

= 4 - i - 3 + 2i

= 4 - 3 - i + 2i

= 1 + i

b) −2/(1 + i)

Conjugate the denominator is 1 - i

= [−2/(1 + i)] [(1 - i)/(1 - i)]

= -2(1 - i) / (1 + i)(1 - i)

= (-2 + 2i)/(1² - i²)

= (-2 + 2i)/(1 + 1)

= (-2 + 2i)/2

= -2/2 + (2/2)i

= -1 + i

c) (2 − i)(i + 1)

Using distributive property, we get

= 2i + 2 - i² − i

= 2i + 2 - (-1) − i

= 2i - i + 2 + 1

= i + 3

d)  i²⁰ − i²¹

= (i²)¹⁰  − (i²)¹⁰ i

= (-1)¹⁰  − (-1)¹⁰ i

= 1 - i

So, options a, b and d are equal to 1 - i.

Problem 7 :

Write each product as a complex number in standard form.

a. (2 − 3i)³

b. (3 + i)⁴

Solution :

a. (2 − 3i)³

= (2 − 3i)² (2 - 3i)

= [2² - 2(2)(3i) + (3i)²] (2 - 3i)

= [4 - 12i + 9i²](2 - 3i)

= [4 - 12i + 9(-1)](2 - 3i)

= [4 - 12i - 9](2 - 3i)

= (-12i - 5) (2 - 3i)

Using distributive property, we get

= -24i + 36i² - 10 + 15i

= -24i - 36 - 10 + 15i

= -9i - 46

b. (3 + i)⁴

(3 + i)⁴ = [(3 + i)²]²

= [3² + 2(3)i + i²]²

= [9 + 6i - 1]²

= [8 + 6i]²

= 8² + 2(8)(6i) + (6i)²

= 64 + 96i + 36i²

= 64 + 96i - 36

= 28 + 96i

Problem 8 :

find the zeros of the function.

f (x) = 3x² + 6

g(x) = 7x² + 21

h(x) = 2x² + 72

k(x) = −5x² − 125

Solution :

f(x) = 3x² + 6

To solve for x, we apply y = 0

0 = 3x² + 6

3x² = -6

x² = -6/3

x² = -2

x = √-2

g(x) = 7x² + 21

To solve for x, we apply y = 0

0 = 7x² + 21

7x² = -21

x² = -21/7

x² = -3

x = √-3

h(x) = 2x² + 72

To solve for x, we apply y = 0

0 = 2x² + 72

2x² = -72

x² = -72/2

x² = -36

x = √-36

x = -6i and 6i

k(x) = −5x² − 125

−5x² − 125 = 0

−5x² = 125

x² = -125/5

x² = -25

x = √-25

x = -5i and 5i

Problem 9 :

Write each expression as a complex number in standard form.

a. √−9 + √−4 − √16

b. √−16 + √8 + √-36

Solution :

a. √−9 + √−4 − √16

√−9 = 3i

√-4 = 2i

√16 = 4

applying these values, we get

= 3i + 2i - 4

= 5i - 4

b. √−16 + √8 + √-36

√−9 = 3i

√-4 = 2i

√16 = 4

applying these values, we get

= 3i + 2i - 4

= 5i - 4

Problem 10 :

Find the square root of the number.

1. √−36

2. √−64

3. √−18

Solution :

1. √−36

= √(−1 ⋅ 6 ⋅ 6)

= √(i² ⋅ 6 ⋅ 6)

= 6i

2. √−64

= √(−1 ⋅ 8 ⋅ 8)

= √(i² ⋅ 8 ⋅ 8)

= 8i

3. √−18

= √(−1 ⋅ 18)

= √(i² ⋅ 3 ⋅ 3 ⋅ 2)

= 3i√2