Download Ch3-Sec3.3

Chapter 3
Quadratic
Functions and
Equations
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3.3
Complex Numbers
♦
Perform arithmetic operations on complex
numbers
♦ Solve quadratic equations having complex
solutions
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Properties of the Imaginary Unit i
i  1,
i  1
2
Defining the number i allows us to say
that the solutions to the equation
x2 + 1 = 0 are i and –i.
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Complex Numbers
A complex number can be written in
standard form as a + bi where a and b are
real numbers. The real part is a and the
imaginary part is b. Every real number a
is also a complex number because it can
be written as a + 0i.
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Imaginary Numbers
A complex number a + bi with b ≠ 0 is an
imaginary number. A complex number
a + bi with a = 0 and b ≠ 0 is sometimes
called a pure imaginary number.
Examples of pure imaginary numbers
include 3i and –i.
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The Expression a
If a > 0, then
a  i a.
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Example: Simplifying expressions
Simplify each expression.
(a)
3  3
(b)
2  8
Solution
(a) 3  3  i 3  i 3  i
2
(b) 2  8  i 2  i 8  i
2
 3   13 3
2

16  1 4  1
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Example: Performing complex
arithmetic
Write each expression in standard form.
Support your results using a calculator.
a) (3 + 4i) + (5  i)
b) (7i)  (6  5i)
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c) (3 + 2i)2
d)
4i
Solution
a) (3 + 4i) + (5  i)
= 3 + 5 + 4i  i = 2 + 3i
b) (7i)  (6  5i)
= 6  7i + 5i = 6  2i
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Example: Performing complex
arithmetic
c) (3 + 2i)2 = (3 + 2i)(3 + 2i)
= 9 – 6i – 6i + 4i2
= 9  12i + 4(1)
= 5  12i
d) 17  17  17
4i
4i 4i
68  17i

16  i 2
68  17i

4i
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Quadratic Equations with Complex
Solutions
We can use the quadratic formula to solve
quadratic equations if the discriminant is
negative.
There are no real solutions, and the graph
does not intersect the x-axis.
The solutions can be expressed as imaginary
numbers.
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Example: Solving a quadratic
Solve the quadratic equation x2 + 3x + 5 = 0.
Support your answer graphically.
Solution
3  11
a = 1, b = 3, c = 5
b  b  4ac
x
2a

2

3 
3  4 15
2 1
2
2
3  i 11

2
3 i 11
 
2
2
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Example: Solving a quadratic
Solution continued
The graph does not
intersect the x-axis,
so no real solutions,
but two complex
solutions that are
imaginary.
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Example: Solving a quadratic
1 2
Solve the quadratic equation
x  17  5x.
Support your answer graphically. 2
Solution
Rewrite the equation:
1 2
x  5x  17  0.
a = 1/2, b = –5, c = 17
2
b  b 2  4ac
x
2a

5 
 5 
2
 4  0.5 17 
2  0.5 
 5  9
 5  3i
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Example: Solving a quadratic
Solution continued
The graphs do not
intersect, so no real
solutions, but two
complex solutions that
are imaginary.
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Example: Solving a quadratic
Solve the quadratic equation –2x2 = 3.
Support your answer graphically.
Solution
Apply the square root property.
2x 2  3
3
2
x 
2
3
3
x  i
x 
2
2
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Example: Solving a quadratic
Solution continued
The graphs do not
intersect, so no real
solutions, but two
complex solutions that
are imaginary.
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