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Ch. 4.6 : I Can define and use imaginary
and complex numbers and solve quadratic
equations with complex roots
Do Now:
Find the zeros of each function.
1. f(x) = x2 – 18x + 16
Success Criteria:
 I can use complex numbers
 I can apply the conjugates
Today’s Agenda
 Check HW
 Notes
 Assignment
Ch. 4.6 : I Can define and use imaginary
and complex numbers and solve quadratic
equations with complex roots
Do Now:
Find the zeros of each function.
1. 3x2 + 96 = 0
Success Criteria:
 I can use complex numbers
 I can apply the conjugates
2. x2 + 8x +20 = 0
Today’s Agenda
 Notes
 Assignment
You can see in the graph of f(x) = x2 + 1 below
that f has no real zeros. If you solve the
corresponding equation 0 = x2 + 1, you find
that x =
,which has no real solutions.
However, you can find solutions if you
define the square root of negative
numbers, which is why imaginary
numbers were invented. The
imaginary unit i is defined
as
. You can use the imaginary
unit to write the square root of
any negative number.
Example 1A: Simplifying Square Roots of Negative
Numbers
Express the number in terms of i.
Factor out –1.
Product Property.
Simplify.
Multiply.
Express in terms of i.
Example 1B: Simplifying Square Roots of Negative
Numbers
Express the number in terms of i.
Factor out –1.
Product Property.
Simplify.
4 6i  4i 6
Express in terms of i.
Check It Out! Example 1a
Express the number in terms of i.
A complex number is a
number that can be written
in the form a + bi, where a
and b are real numbers and
i=
. The set of real
numbers is a subset of the
set of complex numbers C.
Every complex number has a real part a and an
imaginary part b.
Example 3A: Adding and Subtracting Complex
Numbers
Add or subtract. Write the result in the form
a + bi.
(4 + 2i) + (–6 – 7i)
(4 – 6) + (2i – 7i)
–2 – 5i
Add real parts and imaginary
parts.
Example 3B: Adding and Subtracting Complex
Numbers
Add or subtract. Write the result in the form
a + bi.
(5 –2i) – (–2 –3i)
(5 – 2i) + 2 + 3i
(5 + 2) + (–2i + 3i)
7+i
Distribute.
Add real parts and imaginary
parts.
Check It Out! Example 3b
Add or subtract. Write the result in the form
a + bi.
2i – (3 + 5i)
(2i) – 3 – 5i
(–3) + (2i – 5i)
–3 – 3i
Distribute.
Add real parts and imaginary
parts.
Example 5A: Multiplying Complex Numbers
Multiply. Write the result in the form a + bi.
–2i(2 – 4i)
–4i + 8i2
Distribute.
–4i + 8(–1)
Use i2 = –1.
–8 – 4i
Write in a + bi form.
Example 5B: Multiplying Complex Numbers
Multiply. Write the result in the form a + bi.
(3 + 6i)(4 – i)
12 + 24i – 3i – 6i2
Multiply.
12 + 21i – 6(–1)
Use i2 = –1.
18 + 21i
Write in a + bi form.
Example 5C: Multiplying Complex Numbers
Multiply. Write the result in the form a + bi.
(2 + 9i)(2 – 9i)
4 – 18i + 18i – 81i2
Multiply.
4 – 81(–1)
Use i2 = –1.
85
Write in a + bi form.
Assignment #45
p. 253
#9-12, 18 – 26
Ch. 4.6 : I Can define and use imaginary
and complex numbers and solve quadratic
equations with complex roots
Entry Task
The base of a triangle is 3 cm longer than
its altitude. The area of the triangle is 35
cm2. Find the altitude. Hint: draw the
triangle and use formula.
Success Criteria:
 I can use complex numbers
 I can apply the conjugates
Today’s Agenda
 Notes
 Assignment
The solutions
and
are related.
These solutions are a complex conjugate pair.
Their real parts are equal and their imaginary
parts are opposites. The complex conjugate of
any complex number a + bi is the complex
number a – bi.
If a quadratic equation with real coefficients has
nonreal roots, those roots are complex conjugates.
Helpful Hint
When given one complex root, you can always
find the other by finding its conjugate.
Example 5: Finding Complex Zeros of Quadratic
Functions
Find each complex conjugate.
B. 6i
A. 8 + 5i
8 + 5i
8 – 5i
Write as a + bi.
Find a – bi.
0 + 6i
0 – 6i
–6i
Write as a + bi.
Find a – bi.
Simplify.
D.
C. 9 – i
9 + (–i)
Write as a + bi.
Write as a + bi.
9 – (–i)
Find a – bi.
Find a – bi.
9+i
Simplify.
Use complex conjugates to simplify
Simplify.
Multiply by the conjugate.
Distribute.
Use i2 = –1.
Simplify.
Use complex conjugates to simplify
Simplify.
Multiply by the conjugate.
Distribute.
Use i2 = –1.
Simplify.
Example 2A: Solving a Quadratic Equation with
Imaginary Solutions
Solve the equation.
Take square roots.
Express in terms of i.
Check
x2 = –144
(12i)2 –144
144i 2 –144
144(–1) –144 
x2 =
(–12i)2
144i 2
144(–1)
–144
–144
–144
–144 
Example 2B: Solving a Quadratic Equation with
Imaginary Solutions
Solve the equation.
5x2 + 90 = 0
Add –90 to both sides.
Divide both sides by 5.
Take square roots.
Express in terms of i.
Check
5x2 + 90 = 0
0
5(18)i 2 +90 0
90(–1) +90 0 
Check It Out! Example 2a
Solve the equation.
x2 = –36
x2 + 48 = 0
x2 = –48
9x2 + 25 = 0
9x2 = –25
Example
Find the zeros of the function by completing the
square.
f(x) = x2 + 10x + 26
x2 + 10x + 26 = 0
x2 + 10x +
= –26 +
x2 + 10x + 25 = –26 + 25
(x + 5)2 = –1
g(x) = x2 + 4x + 12
x2 + 4x + 12 = 0
x2 + 4x +
= –12 +
x2 + 4x + 4 = –12 + 4
(x + 2)2 = –8
Assignment #46
Pg 273 #2,5,6-16
Lesson Quiz: Part II
Perform the indicated operation. Write the
result in the form a + bi.
1. (2 + 4i) + (–6 – 4i) –4
2. (5 – i) – (8 – 2i) –3 + i
3. (2 + 5i)(3 – 2i)
16 + 11i
4.
5. Simplify i31. –i
3+i
Lesson Quiz
1. Express
in terms of i.
Solve each equation.
2. 3x2 + 96 = 0
3. x2 + 8x +20 = 0
4. Find the complex conjugate of
Ch. 4.6 : I Can define and use imaginary
and complex numbers and solve quadratic
equations with complex roots
Entry Task
Find the zeros of the function.
g(x) = x2 – 8x + 18
f(x) = x2 + 4x + 13
Success Criteria:
 I can use complex numbers
 I can apply the conjugates
Today’s Agenda
 Review
 Assignment
Check It Out! Example 4a
Find the zeros of the function.
g(x) = x2 – 8x + 18
f(x) = x2 + 4x + 13
x2 + 4x + 13 = 0
x2 – 8x + 18 = 0
x2 + 4x +
x2 – 8x +
= –13 +
x2 + 4x + 4 = –13 + 4
(x + 2)2 = –9
x = –2 ± 3i
= –18 +
x2 – 8x + 16 = –18 + 16