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Section 5.6—Quadratic Equations and Complex Numbers
I.
The Discriminant
A. Remember the quadratic formula
2
x= -b √b – 4ac
2a
B. The b2 – 4ac under the radical is called The Discriminant
C. The value of the discriminant tells you how mant real solutions
there are to a quadratic equation
2
D. If b – 4ac > 0 there are exactly 2 solutions
b2 – 4ac = 0 there is one solution
b2 – 4ac < 0 there are no real solutions
E. Remember, negative numbers have no real square roots!
II.
Imaginary Numbers
A. Leonhard Euler (1707-1783) defined √-1 as the imaginary unit i
B.
i = √-1
and
i2 = -1
C. This is useful when dealing with negative square roots in your
discriminant√-5 = (√-1)(√5) = i√5
D. Example: Solve –4x2 + 5x – 3 = 0
a=-4 b=5 c=-3
Step 1:
x= -5 √(5)2 – 4 (-4)(-3)
2(-4)
Step 2:
x= -5
Step 3:
x=-5
Step 4:
x = 5 + (√-1)(√23) OR
8
-8
x = 5 - (√-1)(√23)
8
-8
Solution:
x= 5 + i√23
8
-8
x= 5 - i√23
8
-8
√25 – 48
-8
√-23
-8
OR
These are called “complex numbers”
III.
Complex Numbers
A. Definition: a complex number is a number with a real part and
an imaginary part—just like the solution to the equation above
B. Example:
x= 5 - i√23
This is a Complex Number
8
-8
Real Part
Imaginary Part
IV.
Operations with Complex Numbers
A. Complex numbers can be added, subtracted, multiplied and
divided if you know how…
B. In general you match the real and imaginary parts of the
complex numbers like this
C. Addition/Subtraction: add (-10 – 6i) + (8 – i)
Add real numbers first: -10 + 8 = -2
Add imaginary numbers next: -6i + -i = -7i
Combine your answer: (-10 – 6i) + (8 – i) = -2 – 7i
D. Multiplication: Multiply (6 – 4i)(5 – 4i)
We use the distributive property and F.O.I.L
F.O.I.L
(6 – 4i)(5 – 4i)= 30 – 24i – 20i + 16i2
combine like terms: 30 – 44i +16i2
simplify: 30 – 44i + (16)(-1) (REMEMBER i2 = -1)
solution: 30 – 44i – 16 = 14 – 44i
E. Conjugate of a Complex Number
1. The “conjugate” (or opposite) of a complex number
a + bi is defined as a – bi
2. This is helpful when we want to divide complex numbers
F. Division:
divide 3 – 4i
2+i
Step 1: multiply top and bottom by the conjugate of the denominator
(3 – 4i) (2 – i) = 6 – 3i – 8i + 4i2
(2 + i) (2 – i)
4 – i2
Step 2: Replace i2 with –1
6 – 11i – 4
4+1
Step 3: Simplify 2 – 11i = 2 - 11i
This is your solution
5
5
5
(it is a complex number)
V.
Graphing Complex Numbers and Absolute value
We can graph complex numbers on the complex plane
2i
1i
This is the real axis
1
2
3
This is the imaginary axis
B. We plot complex numbers like this: 3 + 2i
1.
graph the real number on the real axis
2. graph the imaginary number on the imaginary axis
C. Absolute value is defined as the distance from the origin
1. Pythagorean theorem tells us that a2 + b2 = c2
2. You can see that the distance from the origin will be
c2 = 32 + 22 = 9 + 4
c = √13
3. Therefore, absolute value of a complex number is
defined as:
2
2
VI.
la + bil = √a + b
Solving algebraic problems involving complex numbers
A. To solve algebraic problems involving complex numbers simply
solve for the real part and then for the imaginary part
B. Example: Find x and y so that 7x – 2iy = 14 + 6i
Real part: 7x = 14
x=2
Imaginary part: -2iy = 6i
y = 6i = -3
-2i