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Name ________________________________________ Date __________________ Class__________________
Reteach
LESSON
5-5
Complex Numbers and Roots
An imaginary number is the square root of a negative number.
Use the definition −1 = i to simplify square roots.
Simplify.
−25
( 25 )( −1)
Factor out −1.
( 25 )
Separate roots.
−1
5 −1
Simplify.
5i
Express in terms of i.
− −48
−
( 48 )( −1)
Factor out −1.
−
( 48 )
Separate roots.
−1
− 16 3 −1
Factor the perfect square.
−4 3 −1
Simplify.
−4i 3
Express in terms of i.
Real
Imaginary
Complex numbers are numbers that can be written in the form a + bi.
Write as a + bi
Find 0 − 5i = −5i
The complex conjugate of a + bi is a − bi.
The complex conjugate of 5i is −5i.
Express each number in terms of i.
1.
−72
( 36 )( 2 )( −1)
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4. 5 −54
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2. 4 −45
4
3.
−100
( 9 )( 5 )( −1)
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5. 2 −64
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6. − −98
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________________________
Find each complex conjugate.
7. −9i
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8. 1 + 4i
9. 12 − i
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________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
5-38
Holt Algebra 2
Name ________________________________________ Date __________________ Class__________________
Reteach
LESSON
5-5
Complex Numbers and Roots (continued)
−1 = i to solve quadratic equations with
You can use the square root property and
imaginary solutions.
Solve x2 = −64.
x 2 = ± −64
Take the square root of both sides.
x = ±8i
Express in terms of i.
Check each root: (8i) = 64i = 64(−1) = −64
2
2
Remember:
(
−1
)
2
= i2 = −1
(−8i)2 = 64i 2 = 64(−1) = −64
Solve 5x2 + 80 = 0.
5x2 = −80
2
Subtract 80 from both sides.
x = −16
Divide both sides by 5.
x 2 = ± −16
Take the square root of both sides.
x = ± 4i
Express in terms of i.
Check each root:
5(4i )2 + 80
5(−4i )2 + 80
5(16)i 2 + 80
5(16)i 2 + 80
80(−1) + 80
80(−1) + 80
0
0
Solve each equation.
10. x2 + 18 = 0
x2 = −18
x=±
( 9 )( 2 )( −1)
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2
13. x + 100 = 0
11. 6x2 + 24 = 0
12. x2 + 49 = 0
6x2 = −24
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________________________
2
14. 3x + 108 = 0
15. x2 + 12 = 0
________________________
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________________________
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________________________
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________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
5-39
Holt Algebra 2
12. Imaginary; possible answer: since a is
positive, the parabola opens upward and
the vertex is at the minimum. Since the
function is in vertex form, you can tell that
the vertex is at (1, 5). With a minimum at
5, the function never crosses the x-axis,
so the zeros have to be imaginary.
LESSON 5-5
Practice A
1. 3i; 7i − 1; −2i
2. i
3. −1
4. a
5. bi
6. 2i
7. 9i
8. −9i
9. 8i
13. a. The beginning and end of the flight
when the speed of the rocket is 0
10. 5i
11. 21i
12. 1 − 2i
b. t = −3 ± 5i
13. −5i
14. 2 + 3i
c. No; possible answer: the zeros are
imaginary because the graph never
crosses the x-axis so the function
never equals 0. The speed of the
rocket must be 0 before takeoff and
after landing.
15. a. x =
−25 , so x = 5i and −5i.
b. Possible answer: You could multiply
(x + 5i)(x − 5i) to get the original
expression.
16. a. x =
−16 , so x = 4i and −4i.
Reteach
b. Possible answer: You could multiply
(x + 4i)(x − 4i) to get the original
expression.
Practice B
1. 4i 2
3.
2. 6i 2
1
i
3
6. x = ±4i 3
7. x = ±i 21
8. x = 4, y = 5
1
1
9. x = − , y =
3
2
12. −3 − i
13. −4 − 3i
14. −11i
4. 15i 6
5. 16i
6. −7i 2
7. 9i
8. 1 − 4i
10. x = ±3i 2
x = ±2i
12. x2 = −49
x = ± −49
x = ±7i
13. x2 = −100
x = ± −100
x = ±10i
15. 3 ± i 11
14. x2 = −36
Practice C
1. x = ±2i 14
2. x = ±i 11
3. x = 3 ± i
4. x = 2 ± 2i
11
6. x =
7. x = −6, y = −8
8. x = 2, y = 0.25
10.
x = ± −36
x = ±6i
15. x2 = −12
5
,y=1
3
5. x = 1 ± i 2
9. −25 − i 3
3. 10i
11. x = ± −4
10. x = 1 ± i 3
11. x = −3 ± i 5
2. 12i 2
9. 12 + i
4. x = ±3i 3
5. x = ±i 7
1. 6i 2
x = ± (4)(3)( −1)
x = ±2i 3
12
+ 5i
5
Challenge
11. −2 + 1.5i
1. 2i, −7i
2. −6i, −8i
3. 9i, −12i
4. 5i, 49i
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A58
Holt Algebra 2