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MM218 - Unit 7
Seminar Topics
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Imaginary Numbers
Complex Numbers
Values of in
Solving Quadratic
Equations by Factoring
Imaginary Numbers
• The imaginary number i is defined as follows:
i = √(-1) and i2 = -1
• Definition: For all positive real numbers a,
√(-a) = √(-1) √(a) = i √(a)
Example 1: √(-16) = i√(16) = 4i
Example 2: √(-54) = i√(54) = i√(9*6) = 3i√(6)
Complex Numbers
• Definition: A number that can be written in the
form a+bi where a and b are real numbers, is a
complex number. We say that a is the real part
and bi is the imaginary part.
Example 1: 6 + 2i is complex
Example 2: -5 - 7i is complex
Example 3: 4i is complex (because we can
write it as: 0 + 4i)
Adding/Subtracting
Complex Numbers
• Add (or subtract) the real parts separately from
the imaginary parts
Example 1: Add 4+2i and 6 + 3i
4 + 2i + 6 + 3i
= (4 + 6) + (2i + 3i)
= 10 + 5i
Example 2: Subtract (4 + 2i) – (6 + 3i)
4 + 2i - 6 - 3i
= (4 - 6) + 2i - 3i
= -2 - i
Values of in
i=i
i2 = -1
i3 = i2 * I = -i
i4 = i2 * i2 =1
i5 = i
i6 = -1
i7 = -i
i8 = 1
Example: Evaluate i30
= (i4)7 (i2)
= (1)7(-1)
= -1
i9 = i
i10 = -1
i11 = -i
i12 = 1
EVALUATE powers of i
i0 = 1
Note the pattern, or cycle,
i1 = i
1, i, -1 –i, …
i2 = -1
i3 = i2 * I = -i
RULE: DIVIDE exponent by 4, and the
i4 = i2 * i2 = -1 * -1 = 1
remainder will give the power of i,
that is, in = ir, where r is the
remainder of division by 4.
Example: Evaluate i24
= i(24/4)
= i(6 R 0)
= i0
=1
Multiplying
Complex Numbers
• Apply the distributive property
Example: 2i (5 – 2i)
2i (5 – 2i)
= (2i * 5) + (2i * -2i)
= 10i + (-4) i2
= 10i + (-4)*(-1) {Replace i2 with -1}
= 4 + 10i
{WRITE as Complex Number}
Multiplying
Complex Numbers
• Multiply TWO complex numbers, using FOIL
Example: (4+2i) * (6 + 3i)
F
O
I
L
= (4*6) + (4)*(3i) + (2i)*(6) + (2i)*(3i)
= 24 + 12i + 12i + 6 i2
= 24 + 24i + 6(-1)
{ADD like terms,
replace i2 with -1}
= 24 + 24i – 6
= 18 + 24i
Dividing Complex Numbers
Multiply numerator & denominator by CONJUGATE
(May not keep an i in the denominator)
Example: (4+i) / (6 + 3i)
(4 + i) * (6 - 3i) =
(6 + 3i) * (6 - 3i)
F O I
L
24 - 12i + 6i - 3i2
36 - 18i + 18i - 9i2
=
3(9 - 2i)
3*15
=
9 - 2i
15
=
24 – 6i - 3(-1)
36 - 9(-1)
=
27 - 6i
45
Quadratic Equation
• A quadratic equation has this format
(standard form):
ax2 + bx + c = 0
• Where a, b, and c are real numbers,
and the value of a cannot be 0 (there
MUST BE a square term)
Solving Quadratics by Factoring
Given the quadratic in standard form:
x2 + 3x + 2 = 0
To solve this equation, we will use the Zero Factor
Property (which states that if a product is zero,
than one of the two factors must be zero)
In symbol form, property states that
if a*b = 0 then a = 0 or b = 0
Solving Quadratics by Factoring
The steps for using the Zero Factor
Property are outlined in your text also
1. Factor, if possible, the quadratic
expression that equals zero.
2. Set each factor equal to 0.
3. Solve the resulting equations to find
each root.
4. Check each root by substitution.
Solving Quadratics by Factoring
x2 + 3x + 2 = 0
(x + 1)(x + 2) = 0
Now, either
x + 1 = 0 or x + 2 = 0
GIVEN
FACTOR
ZERO factor
property
Hence, x = -1, x = -2
(CHECK both answers back in ORIGINAL equation)
Solving Quadratics by Factoring
4x2 = 9
4x2 – 9 = 0
GIVEN
Write in STANDARD FORM
(2x + 3)(2x – 3) = 0
FACTOR
2x + 3 = 0 or 2x -3 =0
Apply ZERO factor
property
Hence, x = -3/2, x = 3/2
(CHECK answers)
Solving Quadratics by Factoring
x2 + 5x + 6 = 0
GIVEN
Solving Quadratics by Factoring
x2 - 16 = 0
(x + 4)(x - 4) = 0
x + 4 = 0 or x - 4 =0
GIVEN
FACTOR
Apply ZERO
factor property
Hence, x = -4, x = 4
Solving Quadratics by Factoring
Example: Solve x2 + 5x - 14=0
Factor the left
Use Zero Property
Solve each
(x + 7)(x - 2) = 0
x + 7 = 0 or x - 2 = 0
x = -7 and x = 2
Check by substituting into original equation:
x = -7: x2 + 5x – 14 = 0
x = 2: x2 + 5x – 14 = 0
(-7)2 + 5(-7) – 14 = 0
(2)2 + 5(2) – 14 = 0
49 - 35 -14 =0
4 + 10 – 14 = 0
49 - 49 = 0
14 - 14 = 0
0 = 0 true
0 = 0 true
Solving Quadratics by Factoring
Example: Solve 6x2 - 13x – 5 = 0
Factor the left
(Outer product, -30
= -10*3, 10*3, -5*6, 5*-6, -15*2, 15*-2)
6x2 - 13x – 5 = 0
6x2 - 15x + 2x – 5 = 0
3x(2x - 5) + 1(2x - 5) = 0
(3x + 1) (2x – 5) = 0
Apply ZERO FACTOR principle, 3x + 1= 0 OR 2x - 5 = 0
AND solve each
3x = -1
x = -1/3
2x = 5
x = 5/2
Practice Exercises