Download complex number

SAT Problem of the Day
5.6 Quadratic Equations and Complex Numbers
Objectives:
•Classify and find all roots of a quadratic equation
Solutions of a Quadratic Equation
The expression b2 – 4ac is called the discriminant.
Let ax2 + bx + c = 0, where a = 0.
If b2 – 4ac > 0, then the quadratic equation has 2
distinct real solutions.
If b2 – 4ac = 0, then the quadratic equation has 1 real
solution.
If b2 – 4ac < 0, then the quadratic equation has 0 real
solutions.
Example 1
Find the discriminant for each equation. Then
determine the number of real solutions.
a) 3x2 – 6x + 4 = 0
b2 – 4ac = (-6)2 – 4(3)(4) = 36 – 48 = -12
no real solutions
b) 3x2 – 6x + 3 = 0
b2 – 4ac = (-6)2 – 4(3)(3) = 36 – 36 = 0
one real solution
c) 3x2 – 6x + 2 = 0
b2 – 4ac = (-6)2 – 4(3)(2) = 36 – 24 = 12
two real solutions
Practice
Identify the number of real solutions:
1) -3x2 – 6x + 15 = 0
Imaginary Numbers
The imaginary unit is defined as i  1 and i2 = -1.
If r > 0, then the imaginary number
follows:
r is defined as
r  1  r  i r
10  1  10  i 10
Example 2
Solve 6x2 – 3x + 1 = 0.
b  b2  4ac
x
2a
3  32  4(6)(1)
x
2(6)
x
3  9  24
12
3  15
x
12
3  i 15
x
12
Practice
Solve -4x2 + 5x – 3 = 0.
Homework
Lesson 5.6 exercises 19-35 Odd
SAT Problem of the Day
5.6 Quadratic Equations and Complex Numbers
Objectives:
•Graph and perform operations on complex numbers
Imaginary Numbers
A complex number is any number that can be written as
a + bi, where a and b are real numbers and i  1;
a is called the real part and b is called the imaginary part.
3 + 4i
real
part
imaginary
part
Example 1
Find x and y such that -3x + 4iy = 21 – 16i.
Real parts
Imaginary parts
-3x = 21
x = -7
x = -7 and y = -4
4y = -16
y = -4
Example 2
Find each sum or difference.
a) (-10 – 6i) + (8 – i)
= (-10 + 8) + (-6i – i)
= -2 – 7i
b) (-9 + 2i) – (3 – 4i)
= (-9 – 3) + (2i + 4i)
= -12 + 6i
Example 3
Multiply.
(2 – i)(-3 – 4i)
= -6 - 8i + 3i + 4i2
= -6 - 5i + 4(-1)
= -10 – 5i
Conjugate of a Complex Number
The conjugate of a complex number a + bi is a – bi.
The conjugate of a + bi is denoted a + bi.
Example 4
3  2i
Simplify
. Write your answer in standard form.
4  i
3  2i 4  i


4  i 4  i
multiply by 1, using the
conjugate of the denominator
(3 – 2i) (-4 – i)
=
(-4 + i) (-4 - i)
-12 - 3i + 8i + 2i2
=
16 + 4i - 4i - i2
-12 + 5i + 2(-1)
-14 + 5i
=
=
16 - (-1)
17
Practice
3  4i
Simplify
. Write your answer in standard form.
2i
Homework
Lesson 5.6 Exercises 49-57 odd, 65, 67, 71,
75