Review for Test 2
The Imaginary Unit i and its Properties
The Imaginary Unit i
The imaginary unit i is defined as i  1 . In other
words, i has the property that its square is 1 :
.
i 2  1
Square Roots of Negative Numbers
If a is a positive real number, a  i a . Note that
by this definition, and 2by a logical
extension of
2
2
i
a

i
a  a .
exponentiation,
 
 
Powers of i
The Powers of i follow a repeating pattern:
i1 = i
i2 = -1
i3 = -i
i4 = 1
The Algebra of Complex Numbers
When faced with a complex number, the goal is to
write it in the form a  bi .
Simplifying Complex Expressions
Step 1: Add, subtract, or multiply the complex
numbers, as required, by treating every complex
number a  bi as a polynomial expression. Remember,
though, that i is not actually a variable. Treating a  bi
as a binomial in i is just a handy device.
Step 2: Complete the simplification by using the fact
that i 2  1.
Example: The Algebra of Complex Numbers
Simplify the following complex number
The product of two complex numbers leads
expressions.
 2  i  1  2i 
 4  3i
to four products via the distributive
property.
The Algebra of Complex Numbers
Division of Complex Numbers:
In order to rewrite a quotient in the standard form
a  bi , we make use of the following observation:
 a  bi  a  bi   a 2  abi  abi  b2i 2  a 2  b2
Given any complex number a  bi , the complex
number a  bi is called its complex conjugate.
We simplify the quotient of two complex numbers by
multiplying the numerator and denominator of the
fraction by the complex conjugate of the
denominator.
Types of Equations
There are three types of equations:
1. A conditional equation has a countable
number of solutions. For example, x + 7 = 12
has one solution, 5. The solution set is {5}.
2. An identity is true for all real numbers and has
an infinite number
of solutions. For example,
2
x( x  1)  x  x is true for all real number
values of x . The solution set is R.
3. A contradiction is never true and has
no solution. For example, x  6  x is false
for any value of x . The solution set is Ø.
Solving Linear Equations
To solve a linear equation (in x):
1. Simplify each side of the equation separately by removing any
grouping symbols and combining like terms.
2. Add or subtract the same expression(s) on both sides of the
equation in order to get the variable term(s) on one side and the
constant term(s) on the other side of the equation and simplify.
3. Multiply or divide by the same nonzero quantity on both sides of
the equation in order to get the numerical coefficient of the
variable term to be one.
4. Check your answer by substitution in the original equation.
Solving Absolute Value Equations
The absolute value of any quantity is either the
original quantity or its negative (opposite).
This means that, in general, every occurrence of an
absolute value term in an equation leads to two
equations with the absolute value signs removed,
if c > 0. Note: if c < 0, it has no solution.
ax  b  c means
ax  b  c or ax + b = -c
Example: Solving Absolute Value Equations
3x  2  5
Solve:
Step 1: Rewrite the
absolute value
equation without
absolute values.
Step 2: Solve the
two equations
3x  2  5
or
3x – 2 = -5
3x  7
or
3x = -3
7
3
or
x  1
x
Example: Solving Formulas
for One Variable
S  2 r 2  2 rh. Solve for h .
S  2 r  2 rh
2
S  2 r 2
h
2 r
S  2 r 2
h
2 r
Distance and Interest Problems
Good examples of linear equations arise
from certain distance and simple interest
problems.
The basic distance formula is d  rt where
d is distance traveled at rate r for time t .
The simple interest formula is I  Prt
where I is the interest earned on
principal P invested at rate r for time t .
Example: Distance Problems
Two trucks leave a warehouse at the same time. One travels due
west at an average speed of 61 miles per hour, and the other
travels due east at an average speed of 53 miles per hour. After
how many hours will the two trucks be 456 miles apart?
d1  d 2  456
r1  61
r2  53
d1  d 2  rt
1 1  r2t2
Given.
456  61t  53t
Plug values in.
456  114t
Combine like terms, and solve.
t  4 hours
Solving Linear Inequalities
Cancellation Properties for Inequalities
Throughout this table, A, B and C, represent algebraic expressions. These properties
are true for all inequalities.
Property
A B  AC  B C
Description
Adding the same quantity to both sides of
an inequality results in an equivalent
inequality.
If C  0, A  B  A  C  B  C
If both sides of an inequality are
multiplied by a positive quantity, the
sense of the inequality is unchanged.
If C  0, A  B  A  C  B  C
If both sides of an inequality are
multiplied by a negative quantity, the
sense of the inequality is reversed.
Example: Linear Inequalities
Solve the following linear inequality.
10  4( x  2)  (2  x)
Step 1: Distribute.
Step 2: Combine like terms.
Step 3: Divide by 5 . Note
the reversal of the
inequality sign.
10  4 x  8  2  x
4 x  18  2  x
5 x  20
x4
Solution is  4, 
Solving Compound Linear Inequalities
A compound inequality is a statement containing two
inequality symbols, and can be interpreted as two
distinct inequalities joined by the word “and”.
For example, in a course where the grade depends
solely on the grades of 5 exams, the following
compound inequality could be used to determine the
final exam grade needed to score a B in the course.
92  65  71  80  x
80 
 90
5
Example: Solving Compound Inequalities
Solve the compound inequality from the previous slide.
92  65  71  80  x
80 
 90
5
400  308  x  450
Step 1: Multiply all
sides by 5.
Step 2: Subtract 308
from all sides.
92  x  142
Solution is 92,142 
Note: If this compound inequality relates to test scores, as
indicated on the previous slide, the solution set is 92,100,
assuming 100 is the highest score possible.
Solving Absolute Value Inequalities
An absolute value inequality is an inequality in which
some variable expression appears inside absolute value
symbols.
x can be interpreted as the distance between x and
zero on the real number line. This means that absolute
value inequalities can be written without absolute
values as follows, assuming a is a positive real number:
x  a  a  x  a
and
x  a  x  a or x  a
Example: Solving Absolute Value Inequalities
Solve the following absolute value inequality.
5y  3  2  9
Step 1: Subtract 2.
Step 2: Rewrite the
inequality without
absolute values.
Step 3: Solve as compound
inequality.
5y  3  7
7  5 y  3  7
4  5 y  10
4
 y2
5
 4 
Solution is  ,2 
5 
Example: Solving Absolute Value Inequalities
Solve the following absolute value inequality and graph the solution.
6  2x  8
6  2 x  8
2 x  14
x7
or
6  2x  8
or
2 x  2
x  1
or
Solution is  , 1   7,  

1

7
Solving Quadratic Equations by Factoring
• The key to using factoring to solve a quadratic
equation is to rewrite the equation so that 0 appears
by itself on one side of the equation.
• If the trinomial ax  bx  c can be factored, it can be
written as a product of two linear factors A and B.
2
• The Zero-Factor Property then implies that the only
way for ax 2  bx  c to be 0 is if one (or both) of A
and B is 0.
Example: Solving Quadratic Equations by Factoring
Solve the quadratic equation by factoring.
3x 2
x  
5 5
2
Step 1: Multiply both
sides by 5 .
Step 2: Subtract 2 from
both sides so 0 is
on one side.
Step 3: Factor and solve the
two linear equations.
5 x 2  3x  2
5 x 2  3x  2  0
(5 x  2)( x  1)  0
5x  2  0
2
x
5
or
x 1  0
or
x 1
Solving “Perfect Square” Quadratic Equations
In some cases where the factoring method is
unsuitable, the solution to a second-degree polynomial
can be obtained by using our knowledge of square
roots.
If A is an algebraic expression and if c is a constant:
A  c implies A   c
2
If a given quadratic equation can be written in the
2
form A  c we can use the above observation to
obtain two linear equations that can be easily solved.
Example: “Perfect Square” Quadratic Equations
Solve the quadratic equation by taking square roots.
 x  7
2
 25  0
 x  7
2
 25
x  7   25
x  7  5i
x  7  5i
In this example, taking square roots leads to two
complex number solutions.
The Quadratic Formula
The solutions of the equation ax 2  bx  c  0 are:
b  b 2  4ac
x
2a
Note:
2
• The equation has 2 real solutions if b  4ac  0.
• The equation has 1 real solution if b2  4ac  0.
• The equation has 2 complex solutions (which are
conjugates of one another) if b2  4ac  0 .
Example: The Quadratic Formula
Solve using the quadratic formula.
4 x 2  7 x  15
c
a2 b
4x  7 x  15  0
b  b 2  4ac
x
2a
x
  7  
7  289
x
8
7  17
x
8
5
x  3,
4
 7   4  4  15
2  4
2
Solving Quadratic-Like Equations
An equation is quadratic-like, or quadratic in
form, if it can be written in the form
aA2  bA  c  0
Where a, b , and c are constants, a  0, and A is
an algebraic expression. Such equations can be
solved by first solving for A and then solving for
the variable in the expression A . This is the
Substitution Method.
Example: Solving Quadratic-Like Equations
Solve the quadratic-like equation.
Step 1: Let
A  x 2  3x
and factor.
Step 2: Replace
A with
x 2  3x
and solve
for x .
x
2

2


 3x  2 x 2  3x  8  0
A2  2 A  8  0
 A  4 A  2  0
A4
or
A  2
x 2  3x  4
or
x 2  3x  2
x 2  3x  4  0
or
x 2  3x  2  0
 x  4 x  1  0
or  x  2  x  1  0
x  4 or x  1
or x  2 or x  1
Example: General Polynomial Equations
Solve the equation by factoring. x 4  4
x4  4  0
Step 1: Isolate
0 on one
side and
factor.
Step 2: Set both
equations
equal to 0
and solve.
2
2
x

2
x

  2   0
x2  2  0
x 2  2
x  i 2
or
or
or
x2  2  0
x2  2
x 2
Operations with Rational Expressions
• To add or subtract two rational expressions, a
common denominator must first be found.
• To multiply two rational expressions, the two
numerators are multiplied and the two denominators
are multiplied.
• To divide one rational expression by another, the first
is multiplied by the reciprocal of the second.
• No matter which operation is being considered, it is
generally best to factor all the numerators and
denominators before combining rational
expressions.
Example: Add/Subtract Rational Expressions
Subtract the rational expression.
5x  2
2x

x2  5x  6 x2  4

5x  2
2x

 x  2  x  3  x  2  x  2 
 x  2
 x  3
5x  2
2x




 x  2   x  2  x  3  x  3  x  2  x  2 
5x2  8x  4
2x2  6x


 x  2  x  2  x  3  x  2  x  2  x  3
3x 2  14 x  4

,
 x  2  x  2  x  3
x  2, 2, 3
Step 1: Factor both
denominators.
Step 2: Multiply to
obtain the least
common
denominator
(LCD) and
simplify.
Example: Multiply Rational Expressions
Multiply the rational expression.
x2  2x  3
x2
 2
x2
x  3x  4
 x  3 x  1  x  2 


 x  2   x  1 x  4 
 x  3 x  1 x  2 

 x  2  x  1 x  4 
x  3 x  2 


, x  4, 2,1
 x  2  x  4 
Solving Rational Equations
• A rational equation is an equation that contains at
least one rational expression, while any non-rational
expressions are polynomials.
• To solve these, we multiply each term in the
equation by the LCD of all the rational expressions.
This converts rational expressions into polynomials,
which we already know how to solve.
• However, values for which rational expressions in a
rational equation are not defined must be excluded
from the solution set.
Work-Rate Problems
There are two keys to solving a work-rate problem:
1. The rate of work is the reciprocal of the time
needed to complete the task. If a given job can be done
by a worker in x units of time, the worker works at a
rate of 1 jobs per unit of time.
x
2. Rates of work are “additive”. This means that two
workers working together on the same task have a
combined rate of work that is the sum of their
individual rates.
Rate 1
Rate 2
Rate Together
1
1
1


Time 1
Time 2
Time Together
Example: Work-Rate Problem
One hose can fill a swimming pool in 10 hours. The owner buys a
second hose that can fill the pool in half the time of the first one.
If both hoses are used together, how long does it take to fill the
pool?
1
The work rate of the first hose is
10
1
The work rate of the second hose is
5
1 1 1
 
Step 1: Set up the problem.
10 5 x
Step 2: Multiply both sides by the
x  2 x  10
LCD 10x , and solve.
3 x  10
1
x  3 hours
3