Download Intermediate Algebra

Chapter 1
Real Numbers and Algebra
1.1 Describing Data with Set of
numbers

Natural Numbers are counting numbers and
can be expressed as
N = { 1, 2, 3, 4, 5, 6, …. }
Set braces { }, are used to enclose the elements
of a set.
 A whole numbers is a set of numbers, is given
by
W = { 0, 1, 2, 3, 4, 5, ……}
…Continued

The set of integers include both natural and
the whole numbers and is given by
I = { …, -3, -2, -1, 0, 1, 2, 3, ….}

A rational number is any number can be
written as the ratio of two integers p/q,
where q = 0. Rational numbers can be written
as fractions and include all integers.
Some examples of rational numbers are
8/1, 2/3, -3/5, -7/2, 22/7, 1.2, and 0.
…continued
 Rational numbers may be expressed in decimal
form that either repeats or terminates.
The fraction 1/3 may be expressed as 0.3, a
repeating decimal, and the fraction ¼ may be
expressed as 0.25, a terminating decimal. The
overbar indicates that 0.3 = 0.3333333….
 Some real numbers cannot be expressed by
fractions. They are called irrational numbers.
2, 15, and  are examples of irrational
numbers.
 Identity Properties
For any real number a,
• a + 0 = 0 + a = a,
0 is called the additive identity and
• a . 1 = 1 . a = a,
The number 1 is called the multiplicative identity.
 Commutative Properties
For any real numbers a and b,
a+b=b+a
a.b = b.a
(Commutative Properties of
addition)
(Commutative Properties of
multiplication)
…Continued
Associative Properties
For any real numbers a, b, c,
(a + b) + c = a + (b + c) (Associative Properties of
addition)
(a.b) . c = a . (b . c) (Associative Properties for
multiplication)
Distributive Properties
For any real numbers a, b, c,
a(b + c) = ab + ac
a(b- c) = ab - ac
and
1.2 Operation on Real
Numbers
The Real Number Line
-3
-2
-1
0
1
2
3
Origin
-2
-3
-2
-1
2
0
Origin
1
-2 = 2
2
3
2 =2
Absolute value
cannot be
negative
…Continued
If a real number a is located to the left of a real
number b on the number line, we say that a is
less than b and write a<b.
Similarly, if a real number a is located to the right
of a real number b, we say that a is greater than
b and write a>b.
Absolute value of a real number a, written a , is
equal to its distance from the origin on the
number line. Distance may be either positive
number or zero, but it cannot be a negative
number.
Arithmetic Operations
Addition of Real Numbers
To add two numbers that are either both positive or both
negative, add their absolute values. Their sum has the
same sign as the two numbers.
Subtraction of real numbers
For any real numbers a and b, a-b = a + (-b).
Multiplication of Real Numbers
The product of two numbers with like signs is positive. The
product of two numbers with unlike signs is negative.
Division of Real Numbers
For real numbers a and b, with b = 0, a/b = a . 1/b
That is, to divide a by b, multiply a by the reciprocal of b.
1.3 Bases and Positive Exponents
Squared
4 Cubed
4
4
4
4
4.4=
4
42
Exponent
Base
4 . 4. 4 = 43
Powers of Ten
Power of 10
103
102
101
100
10-1
10-2
10-3
Value
1000
100
10
1
1/10 = 0.1
1/ 100 = 0.01
1/ 1000 = 0.001
1.3
Integer Exponents
Let a be a nonzero real number and n be a positive
integer. Then
an = a. a. a. a……a (n factors of a )
a0 = 1, and a –n = 1/an
a -n b m
b -m = a n
a
b
-n
=
b
a
n
… cont
The Product Rule

For any non zero number a and integers m and n,
am . an = a m+n
 The Quotient Rule
For any nonzero number a and integers m and n,
am
= a m–n
an
Raising Products To Powers
For any real numbers a and b and integer n,
(ab) n
=anbn
Raising Powers to Powers
For any real number a and integers m and n,
(am)n = a mn
Raising Quotients to Powers
For nonzero numbers a and b and any integer n.
a
b
n
=
an
bn
…Continued
A positive number a is in scientific notation
when a is written as b x 10n, where 1 < b <
10 and n is an integer.
Scientific Notation
Example : 52,600 = 5.26 x 104 and 0.0068 =
6.8 x 10 -3
1.4
Variables, Equations , and Formulas
A variable is a symbol, such as x, y, t, used
to represent any unknown number or
quantity.
An algebraic expression consists of
numbers, variables, arithmetic symbols,
paranthesis, brackets, square roots.
Example 6, x + 2, 4(t – 1)+ 1, X + 1
…cont
 An equation is a statement that says two
mathematical expressions are equal.
Examples of equation
3 + 6 = 9, x + 1 = 4, d = 30t, and x + y = 20
 A formula is an equation that can be used to
calculate one quantity by using a known value of
another quantity.
The formula y = x/3 computes the no. of yards in x
feet. If x= 15, then y=15/3= 5.
Square roots
The number b is a square root of a number a if b2 = a.
Example - One square root of 9 is 3 because 32 = 9. The other
square root of 9 is –3 because (-3)2 = 9. We use the symbol to
9 denote the positive or principal square root of 9. That
is, 9 = +3. The following are examples of how to evaluate
the square root symbol. A calculator is sometimes needed to
approximate square roots,
4= +2
The symbol ‘ + ‘ is read ‘plus or minus’. Note that 2
represents the numbers 2 or –2.
Cube roots
The number b is a cube root of a number a if b3 = a
The cube root of 8 is 2 because 23 = 8, which may
be written as 3 8 = 2. Similarly 3 –27 = -3 because
(- 3)3 = - 27. Each real number has exactly one
cube root.
1.5 Introduction to graphing
Relations is a set of Ordered pairs.
If we denote the ordered pairs in a relation
(x,y),
then the set of all x-values is called the
domain of the relation and the set of all y
values is called the range.
Example 1.
Find the domain and range for the relation given
by
S = {( -1, 5), (0,1), (2, 4), (4,2), (5,1)} Solution
The domain D is determined by the first element
in each ordered pair, or
D ={-1, 0, 2, 4,5}
The range R is determined by the second element
in each ordered pair, or R = {1,2,4,5}
The Cartesian Coordinate
System
Quadrant II y
Quadrant I
y
(1, 2)
2
1
Origin
2
1
x
-2
-1 0
-1
-2
Quadrant III
The xy – plane
1 2
-2
-1
1
2
-1
Quadrant IV
-2
Plotting a point
x
Scatterplots and Line Graphs
If distinct points are plotted in the xy- plane, the
resulting graph is called a scatterplot.
Y
7
(4, 6)
6
(3, 5)
5
(5, 4)
(2, 4)
4
(6, 3)
3
(1, 2)
2
1
0
1 2 3 4
5
6 7
X
Using Technology
Ex 9 page 44
Make a table for y = x 2 / 9, starting at x = 10 and incrementing by 10 and compare
The table for example 4 ( pg 41)
Go to Y= and enter x 2 /9
Go to 2nd then table set and enter
Go to 2nd then table
Viewing Rectangle ( Page 57 )
Ymax
}Ysc1
Xmin
Xmax
Xsc1
Ymin
[ -2, 3, 0.5] by [-100, 200, 50]
Making a scatterplot with a graphing calculator
Plot the points (-2, -2), (-1, 3), (1, 2) and (2, -3) in [ -4, 4, 1] by [-4, 4, 1]
(Example 10, page 58)
Go to Stat Edit then enter points
Go to 2nd then stat plot
Scatter plot
[ -4, 4, 1] by [-4, 4, 1]
Example 11
Cordless Phone Sales
Year
1987
1990
1993
1996
2000
Phones
(millions)
6.2
9.9
18.7
22.8
33.3
Go to Stat edit and enter data
Enter datas in window
Enter line graph
Hit graph
[1985, 2002, 5] by [0, 40, 10]
Chapter 2
Linear Functions and Models
Ch 2.1 Functions and Their
Representations
A function is a set of ordered pairs (x, y), where each xvalue corresponds to exactly one y-value.
Input x
Function f
(x, y)
Input
Output
Output y
…continued
y is a function of x because the output y is determined
by and depends on the input x. As a result, y is called
the dependent variable and x is the independent variable
To emphasize that y is a function of x, we use the
notation y = f(x) and is called a function notation.
A function f forms a relation between inputs x and
outputs y that can be represented verbally (Words) ,
numerically (Table of values) , symbolically (Formula),
and graphically (Graph).
Representation of Function
y
Table of Values
Graph
20
16
x
(yard
s
1
2
3
4
5
6
7
y= 3x
12
y(fee 3
t)
6
9
12
15
18
21
8
4
x
0
Numerically
4
8
12
16
20
Graphically
24
Diagrammatic Representation
Function
x
1
2
3
Not a function
y
x
3
y
(1, 3), (2, 6), (3, 9)
6
1
9
4
5
1
2
3
2
4
5
(1, 4), (2, 5), (2, 6)
(1,4), (2, 4), (3, 5)
Example 1, pg - 77
6
Domain and Range Graphically
The domain of f is the set of all x- values, and
the range of f is the set of all y-values
y
3
2
Range R includes all y – values
satisfying 0 < y < 3
Range
x
1
-3 -2 -1 0 1 2 3
Domain
Pg 79 Ex-5
Domain D includes all x values
Satisfying –3 < x < 3
Vertical Line Test
If each vertical line intersects the graph at most once, then it is a graph of a function
5
4
3
2
Not a function
(-1, 1)
-4
-3
-2
(-1, -1)
1
-1
0
-1
-2
-3
-4
1
2
3
4
…Continued
Not a function
4
3
(-1, 1)
2
1
-3
-2
-1
0
1
2
3
-1
(1, -1)
-2
-3
Example 9 page - 81
Using Technology
Graph of y = 2x - 1
Hit Y and enter 2x - 1
x
y
-1
-3
0
-1
1
1
2
3
Hit 2nd and hit table and enter data
[ - 10, 10, 1] by [ - 10, 10, 1]
2.2 Linear Function
A function f represented by f(x) = ax + b,
where a and b are constants, is a linear
function.
100
100
90
90
80
80
70
70
60
60
0
1
2
3
4
Scatter Plot
5
6
f(x) = 2x + 80
0
1
2
3
4
5
6
A Linear Function
Modeling data with Linear Functions
Example 7
1500
1250
1000
750
Cost
(dollars)
500
250
0
4
8
12
16
20
Credits
Symbolic Representation f(x) = 80x + 50
Numerical representation
4
8
12
16
$ 370 $ 690 $1010 $1330
x
Using a graphing calculator
Example 5
Give a numerical and graphical representation f(x) = ½ x
-2
Numerical representation – Y1 = .5x – 2 starting x = -3
Graphical representation –
[ -10, 10, 1] by [-10, 10, 1]
2.3 The Slope of a line
Y
8
Cost
7
(dollars)
6
5
Rise = 3
4
3
Slope = Rise
Run = 2
2
Run
=3
2
1
1
2
3
4
5
6
x
Gasoline (gallons )
Cost of Gasoline
Every 2 gallons purchased the cost increases by $3
2.3
Slope
rise
m=
y2
y1
run = x
(x2, y2)
(x1, y1)
x2 –x1
y2 - y1
2
-x
1
y2 –y1 Rise
Run
The Slope m of the line passing through the points
(x1 y1 ) and (x2, y2) is
m= y2 –y1/x2 –x1
Where x1 = x2. That is, slope equals rise over run.
Ex- 1, 2 pg 104
4
4
3
3
2
-1
2
m=2>0
1
-4
-2
1
2
3
4
-1
-2
-3
2
m=-½<0
2
-4
1
-2
0
1 2
-1
Positive slope
-2
Negative slope
m=0
m is undefined
Zero slope
Undefined slope
Example 2 - Sketch a line passing through the point (0, 4) and
having slope - 2/3
4
( 0, 4)
y - values
decrease
2 units each times x- values
increase by 3
(0 + 3, 4 – 2)
= (3, 2)
-4 -3
3
Rise = -2
2
( 3, 2)
1
-2 1
0
-1
1
-2
-3
-4
2
3
4
Slope-Intercept Form
The line with slope m and y = intercept b is
given by
y= mx + b
The slope- Intercept form of a line
Example – 3, and 4, 5, 6, 7 pg - 106
Example - 3
3
Y=½x
2
Y=½x+2
1
-3
-2
-1
-1
1
2
Y=½x -2
-2
-3
Analyzing Growth in Walmart
Example 9
Year 1997 1999 2002 2007
Employees
3.0
0.7
1.1
1.4 2.2
2.5
Employees
(millions)
2.0
1.5
m3
m2
1.0
m1
0.5
0
1999
2003
2007
Years
m1 = 1.1 – 0.7 = 0.2
m2 = 1.4 - 1.1 = 0.1 and
1999 – 1997
2002 – 1999
m3 = 2.2 - 1.4 = 0.16
Average increase rate
2007 - 2002
2.4 Point- slope form
The line with slope m passing through
the point (x1 , y1 ) is given by
y = m ( x - x 1 ) + y1
(x, y)
Or equivalently,
(x1, y1)
y – y1
x–x
y – y1 = m (x –x1)
The point- slope form of a line
1
m = y – y1 / x – x1
Ex 1, 2,3 pg 117
Horizontal and Vertical Lines
Equation of vertical line
y
y
x=h
x
x
b
y= b
Equation of Horizontal Line
h
…Continued
Parallel Lines
Two lines with the same slope are parallel.
m1 = m2
Perpendicular Lines
Two lines with nonzero slopes m1 and m2 are
perpendicular
if m1 m2 = -1
Examples – 7, 8
Page - 123
m2 = -1
m1 = 1
m2 = - 1/2
m1 = 2
m2 = - 1/m1
m1
Chapter 3
Linear Equations and
Inequalities
3.1 Linear Equation in One
Variable
A Linear Equation in one variable is an equation
that can be written in the form
ax + b = 0
Where a = 0
A linear function can be written as f(x) = ax + b
Examples of Linear equation.
2x –1 = 0, -5x = 10 + x, and 3x + 8 = 5
Properties of Equality
Addition Property of Equality
If a, b, c are real numbers, then
a = b is equivalent to a+ c = b + c.
Multiplication Property of Equality
If a, b, c are real numbers with c = 0, then
a = b is equivalent to ac = bc.
Example 1,2,3,4 pg 146
Example 7 Solving a Linear equation graphically
using technology
[ -6, 6, 1] by [-4, 4, 1]
Example 8
[ 1984, 1991, 1] by [0, 350, 50] in 1987
In 1987 CD and LP record sales were both 107 million
Standard form of a line
An equation for a line is in standard form when it
is written as
ax + b = c, where a, b, c are constants
With a, b and c are constants with a and b not
both 0
 To find x-intercept of a line, let y = 0 in the
equation and solve for x
 To find y-intercept of a line, let x = 0 in the
equation and solve for x
3.2 Linear Inequality in One Variable
 A linear inequality in one variable is an
inequality that can be written in the form
ax + b > 0, where a = 0. ( The symbol > may be
replaced with >, <, or > )
 There are similarities among linear functions,
equations, and inequalities. A linear function is
given by f(x) = ax + b, a linear equation by ax +
b = 0, and a linear inequality by ax + b > 0.
Examples of linear inequalities are
2x + 1< 0, 1-x > 6, and 5x + 1 < 3 – 2x
A solution to an inequality is a value of the variable that
makes the statement true. The set of all solutions is
called the solution set. Two inequalities are equivalent if
they have the same solution set.
Inequalities frequently have infinitely many solutions. For
example, the solution set to the inequality x- 5> 0
includes all real numbers greater than 5, which can be
written as x > 5. Using set builder notation, we can
write the solution set as { x x > 5 }.
Meaning
This expression is read as “ the set of all real numbers x
such that x is greater than 5. “
3.2 Solving a problem
Using a variable to an unknown quantity
Number problem
n is the smallest even integers
Three consecutive three even integers
n, n + 2, n +4
n is the smallest odd integers
Three consecutive odd integers
n, n + 2, n + 4
Mixing acid ( Ex 8 pg 163)
20 %
2 liters
+
Step 1
x: liters of 60% sulphuric acid
x + 2: Liters of 50% sulphuric acid
Step 2
Concentration
0.20 (20%)
0.60 (60%)
0.50(50%)
Equation
0.20(2)
+
(pure acid in 20% sol.)
Step 3
Solution Amount
2
x
x+2
60%
x liters
=
50%
x + 2 liters
Pure Acid
0.20(2)
0.60x
0.50(x + 2)
0.60x =
(pure acid in 60% sol.)
0.50(x + 2)
(pure acid in 50% sol.)
Solve for x
0.20(2) + 0.60x = 0.50(x + 2)
2 (2) + 6x = 5(x + 2)
Multiply by 10
4 + 6x = 5x + 10 (Distributive Property) Subtract 5x and 4 from each side x = 6
Six liters of the 60% acid solution should be added to the 2 liters of 20% acid solution.
Step 4
If 6 liters of 60% acid solution are added to 2 liters of 20% solution, then there
will be 8 liters of acid solution containing 0.60(6) + 0.20(2) = 4 liters of pure acid.
Check The mixture represents a 4/8 = 0.50 or 50% mixture
Ex 46 (Pg 166) Anti freeze mixture
A radiator holds 4 gallons of fluid
• x represents the amount of antifreeze that is drained and
replaced
• The remaining amount is 4 – x
• 20 % of
+ 70 % of solution of = 50% of solution
4 – x gallons
x gallons
4 gallons
•
•
•
•
•
0.20(4-x) + 0.70x = 0.50(4)
0.8 – 0.2x + 0.7x = 2
0.5x = 1.2
x = 2.4
The amount of antifreeze to be drained and replaces is 2.4
gallons
Geometric Formulas
Perimeter of triangle P = a+b+c unit
Area of triangle
A = ½ bh sq.unit
Area of Rectangle =LW sq.unit
Perimeter P = 2(L + W) unit
Area of Parallelogram
A = bh sq.unit
a
c
h
b
W
L
h
b
Area & Volume of cylinder
A = 2 rh sq.unit
h
V =  r2 h cu.unit
Area & Volume of a cube
r
A = 6a2 sq.unit
V = a3 cu.unit
a
a
a
3.3 Properties of Inequalities
Let a, b, c be real numbers.
 a < b and a+c < b+c are equivalent
( The same number may be added to or subtracted from
both sides of an inequality.)
 If c > 0, then a < b and ac < bc are equivalent.
(Both sides of an inequality may be multiplied or divided
by the same positive number)
 If c< 0, then a < b and ac > bc are equivalent.
Each side of an inequality may be multiplied or divided by
the same negative number provided the inequality
symbol is reversed.
Fig. 3.14 Graphical Solutions (Pg 172)
y1
350
D
i
s
t
Distance
a
(miles) n
c
e
300
250
200
150
100
50
When x = 2 , y1 = y2, ie car 1 and car 2 both are 150 miles
From Chicago
y2
y2
y1
0
1
x=2
2
3
4
(
Time ( hours )
m
i
l
e
Distances of two cars
y1 < y2 when x< 2 car 1 is closer to Chicago than car 2
y1 is below the graph of y2
y1 > y2 when x > 2 Car 1 is farther from Chicago than Car 2
Y1 above the graph of y2
…Continued
Ex 5 (Pg 173)
Solving an inequality graphically
Solve 5 – 3x < x – 3
y1 = 5 – 3x and y2 = x – 3 Intersect at the point (2, -1)
5
4
3
2
1
-4
-3
-2
-1 0
-1
-2
-3
X=2
y1
y2
1
2
3
(2, -1)
4
y1 = y2 when x = 2
y1 < y2 when x > 2 , y1 is below the graph of y2
Combining the above result
y1 < y2 when x > 2
Thus 5 – 3x < x – 3 is satisfied when x > 2.
The solution set is {x / x> 2 }
Using Technology
Hit window
Hit Y
Enter inequality
Enter
[ -5, 5, 1] by [-5, 5, 1]
Hit Graph
Ex 80 ( Pg 178) Sales of CD and LP records
Hit Y , enter equations
Enter table set
Enter window
Hit Table
Hit graph
1987 or after CD sales were greater than or equal to LP records
3.4 Compound inequalities
A compound inequality consists of two
inequalities joined by the words and or or.
The following are two examples of compound
inequalities.
2x > -3 and 2x < 5
Example 1
2(1) > -3 and 2(1) < 5
1 is a solution
True
True
5 + 2 > 3 or 5 – 1 < -5
5 is a solution
Example 2
x + 2 > 3 or x – 1 < -5
True
False
…. Cont.
If a compound inequality contains the word and,
solution must satisfy both inequalities.
For example, x = 1 is a solution of the first compound
inequality because
2 (1) > -3 and 2 (1) < 5
True
True
are both true statements.
If a compound inequality contains the word or,
solution must satisfy atleast one of the two
inequalities. Thus x = 5 is a solution to the second
compound inequality.
5 + 2 > 3 or 5 – 1 < -5
True
False
Symbolic Solutions and Number Lines
]
x<6
-8
-6
-4
-2
0
2
4
6
8
-6
(
-4
-2
0
2
4
6
8
x> - 4
-8
(
x < 6 and
x> -4
-8
-6
-4
- 4< x < 6
]
-2
0
2
4
6
8
Three-part inequality
(
-1 0 1 2 3 4
5
]
6
7
8 9
10
5 < x < 10
Sometimes compound inequality containing the
word and can be combined into a three part
inequality. For example, rather than writing
x > 5 and x < 10
We could write the three-part inequality
5 < x < 10
Example 5
Page 184
Solve x + 2 < -1 or x + 2 > 1
x < -3 or x > -1 ( subtract 2 )
The solution set for the compound inequality results from taking the
union of the first two number lines. We can write] the solution,
using builder notation, as
{ x x < - 3} U { x x > - 1 } or
{x x < - 3 or x > - 1}
x<-3
-4
)
-3
-2
-1
0
1
2
3
4
-3
-2
(
-1
0
1
2
3
4
-4
)
-3
-2
(
-1
0
1
x>-1
-4
x < - 3 or x > -1
2
3
4
Interval Notation
Table 3.5 (Page 186)
Inequality
Interval Notation
-1<x<3
( - 1, 3)
-3<x<2
( - 3, 2]
-2<x<2
Number line Graph
-4 -3
(
-4 -3
[ - 2, 2 ]
-4 -3
x < - 1 or x > 2
(
)
-2 -1 0 1 2 3 4
-2
[
-2
( -  , - 1) U (2,  )
x>-1
( - 1,  )
x<2
( - , 2 ]
]
-1 0 1 2 3 4
]
-1 0 1 2 3 4
)
(
-4 -3 -2 -1 0 1 2 3 4
(
-4 -3 -2 -1 0 1 2 3 4
-4 -3
-2
]
-1 0 1 2 3 4
Example 6 (Page 185)
Solving a compound inequality numerically and graphically Using Technology
 Tution at private colleges and universities from 1980 to 1997
 Can be modelled by f(x) = 575(x – 1980) + 3600
 Estimate when average tution was between $8200 and $10,500.
Hit Y and enter equation
Hit Window and enter
Hit 2nd and Table
[ 1980, 1997, 1] by [3000, 12000, 3000 ]
Hit 2nd and calc and go to Intersect and enter 4 times to get intersection
Ex 85 (Page 189)
Solve graphically and numerically. Write your answer in interval notation
x + 1 < -1 or x + 1 > 1
Y1 = -1, Y2 = x + 1, Y3 = 1
[ -5, 5, 1] by [ -5, 5, 1]
x + 1 < - 1 or x + 1 > 1
Solution in interval notation is ( -  -2) U (0,  )
School Enrollment
Example 92 (Pg 190)
•
70
•
60
•
55
•
50
1970 1980
Year
1990
2000
Chapter 4
Systems of Linear Equations
4.1 Systems of Linear Equations in
two variables
The system of linear equation with two variables. Each
equation contains two variables, x and y.
Example
x+y=4
x–y=8
An ordered pair (x, y) is a solution to linear equation
if the values for x and y satisfy both equations.
The standard form is
ax + by = c
dx + ey = k
Where a, b, c, d, e, k are constants.
Types of Linear Equations in two
variables
A system of linear equations in two variables can
be represented graphically by two lines in the
xy-plane
 The lines intersect at a single point, which
represents a unique solution. Consistent system
The equations are called independent equation
 If the two lines are parallel it is an inconsistent
system and no solution
 If two lines are identical and every point on the
line represents solution and give infinitely
many solutions. The equations are called
dependent equations
..cont.
Example
The equations x + y = 1 and 2x + 2y = 2 are equivalent.
If we divide the second equation by 2 we obtain the first
equation. As a result, their graphs are identical and
every point on the line represents a solution.Thus there
are infinitely many solutions, and the system of
equations is a dependent system.
y
y
y
x
x
Inconsistent
x
dependent
Unique
The Substitution Method
Consider the following system of equations.
2x + y = 5
3x – 2y = 4
It is convenient to solve the first equation for y to obtain y = 5 – 2x.
Now substitute ( 5 – 2x) for y into the second equation.
3x – 2(y) = 4
Second equation
3x – 2(5-2x) = 4, Substitute
to obtain a linear equation in one variable.
3x – 2(5 – 2x) = 4
3x – 10 + 4x = 4
Distributive property
7x – 10 = 4
Combine like terms
7x = 14
Add 10 to both sides
x=2
Divide both sides by 7
To determine y we substitute x = 2 into y = 5 – 2x to obtain
y = 5 – 2(2) = 1
The solution is (2, 1).
Elimination Method
The elimination method is the second way to solve linear systems
symbolically. This method is based on the property that ‘ equals
added to equals are equal.’ That is, if
a = b and c = d
Then a + c = b + d
Note that adding the two equations eliminates the variable y
Example
2x – y = 4
x+y=1
3x = 5
or x = 5/3 and solve for x
Substituting x = 5/3 into the second equation gives
5/3 + y = 1 or y = - 2/3
The solution is (5/3, - 2/3)
Solve the system of equations using Technology
4.1 Pg 226
Ex 43
No solution
Ex 44
Infinitely many solutions
Burning calories
Ex 74 pg 239
During strenuous exercise an athelete can burn on
Rowing machine
Stair climber
10 calories per minute
11.5 calories per minute
In 60 minute an athelete burns 633 calories by using both exercise
machines
Let x minute in rowing machine, y minute in stair climber
The equations are
x + y =60
10x + 11.5 y = 633
Find x and y
-10x - 10 y = -600 (Multiply by -10)
10x + 11.5 y = 633
Add 1.5y = 33, y = 33/1.5= 22 minute in stair climber
and x = 38 minute in rowing machine
Mixing acids Ex 76 ( Pg 239 )
x represents the amount of 10% solution of Sulphuric acid
y represents the amount of 25 % solution of Sulphuric acid
According to statement x + y = 20
10% of x + 25% of y = 0.18(20)
0.10x + .25 y = 3.6
.1x + .25y = 3.6 Multiply by 10
x + 2.5y = 36
x + y = 20
Subtract
1.5 y = 16
y = 16/1.5 = 10.6
x = 20 – y = 20 – 10.6 = 9.4
Mix 9.4 ml of 10 % acid with 10.6 ml of 25% acid
River current ( Ex 84, pg 240 )
x
y
Speed of tugboat
Speed of current
Distance = Speed x Time
165 = (x – y) 33 Upstream
165 = (x + y) 15 Downstream
By elimination method
x–y=5
x + y = 11
2x = 16, x = 8
y=3
The tugboat travels at a rate of 8 mph and river flows at a rate of 3
mph
Solving Linear inequalities in Two
variables
x<1
-2
-1
x>1
1
2
3
-2
-1
1
2
-2
-1
0
1 2 3
4
-1
-1
y < 2x - 1
Choose a test point
Let x = 0, y = 0
x - 2y < 4
0 - 2(0) < 4
0 < 4 which is true statement
Shade containing (0, 0)
-2
x – 2y < 4
x –intercept = 4 (4, 0)
y –intercept = -2 (0, -2)
Solving System of Linear Inequalities ( Pg 244)
x+y<4
y>x
Testing point
x+y<4
x = 1, y = 2
1 + 2 < 4 ( True )
4
3
Shaded region
Testing point
2 > 1 ( True)
(1, 2)
2
(1, 2)
y>x
1
-4
-3
-2 -1
0 1 2
3
4
Shaded region
Shaded region
( 1, 2)
To solve inequalities
Modeling target heart rates (Ex 5 Pg 246 )
For Aerobic Fitness
A person’s Maximum heart rate ( MHR) = 220 - A
200
Heart
Rate
175
T = - 0.8A + 196 ( Upper Line )
Beats 150
Per
minute 125
T = - 0.7 A + 154 (Lower Line )
100
75
50
25
0 20 30
40 50 60 70 80
Age in years
(30, 150) is a solution
-0.8 (40) + 196 <= 165 When A = 40 yrs
- 0.7 (40) + 196> = 125
150 < - 0.8 (30) + 196 = 172
150 > - 0.7 (30) + 196 = 133
True
True
Solving a system of linear inequalities with technology
Ex 6 ( Pg 247 )
Shade a solution set for the system of inequalities, using
graphing calculator
- 2x + y > 1 or y> 2x + 1
2x + y < 5 or y < 5 – 2x
[ - 15, 15, 5 ] by [ - 10, 10, 5 ]
Hit 2nd and Draw then go to Shade
[ - 15, 15, 5 ] by [ - 10, 10, 5 ]
Linear Functions and Polynomial
Functions
Every linear function can be written as
f(x) = ax +b and is an example of a polynomial
function. However, polynomial functions of
degree 2 or higher are nonlinear functions. To
model nonlinear data we use polynomial
functions of degree 2 or higher.
Chapter 5
Polynomial Expressions and
Functions
5.1 Polynomial Functions
The following are examples of polynomial functions.
f(x) = 3
f(x) = 5x – 3
f(x) = x2 – 2x –1
f(x) = 3x3 + 2x2 –6
Degree = 0
Degree = 1
Degree = 2
Degree = 3
Constant
Linear
Quadratic
Cubic
Polynomial Expressions
A term is a number, a variable, or a product of numbers
and variables raised to powers.
Examples of terms include
- 15, y, x4, 3x3 y, x-1/2 y-2 , and 6 x-1 y3
If the variables in a term have only nonnegative integer
exponents, the term is called a monomial. Examples of
monomials include
- 4, 5y,
x2, 5x 2 z4, - x y4 and 6xy4
Monomials
x
x
x
x
x
x
Volume x3
y
y
y
Total Area xy + xy + xy = 3xy
Modeling AIDS cases in the United States
Ex 11 ( Pg 308 )
600
500
400
300
200
100
0
4
6
8
10
Year ( 1984
f(x) = 4.1x2
- 25x
12 14
1994 )
+ 46
f(7) = 4.1(7) 2- 25.7 + 46 = 71.9
f(17) = 4.1 (17) 2 – 25.7 + 46 = 805.9
Modeling
heart rate of an athelete ( ex 12 )
250
Heart Rate (bpm)
200
150
100
50
0
1
2
3
4
5
6
7
8
Time (minutes )
P(t) = 1.875 t2 – 30t + 200
Where 0 < t < 8
P(0) = 1.875(0) 2 – 30(0) + 200 = 200 (Initial Heart Rate)
P(8) = 1.875(8) 2– 30(80 )+ 200= 80 beaqts per minute (After 8 minutes)
The heart rate does not drop at a constant rate; rather, it drops rapidly at
first and then gradually begins to level off
A PC for all
(Ex 102 Pg 312)
200
150
100
50
0
1998
2000
2002
2004
Year
f(x) = 0.7868 x2 + 12x + 79.5
x = 0 corresponds to 1997
x = 1 corresponds to 1998 and so on
x = 6 corresponds to 2003
f(6) = 0.7868 x 6 2 + 12 x 6 + 79.5 = 179.8248 = 180 million(approx)
5.2 Review of Basic Properties
Using distributive properties
Multiply
4
4 (5 + x) = 4.5 + 4. x = 20 + 4x
20
5
Area : 20 + 4x
4x
x
Using Technology
[-6, 6, 1] by [ -4, 4, 1 ]
[-6, 6, 1] by [ -4, 4, 1 ]
….Cont
Multiplying Binomials
Multiply (x+ 1)(x + 3)
a)
Geometrically
b)
Symbolically
a) Geometrically
x+1
1
x
x+3
Area: (x + 1)(x + 3)
b) Symbolically
apply distributive property
x
x2
3
3x
x
3
Area: x2 + 4x + 3
(x + 1)(x + 3) = (x + 1)(x) + (x+ 1)(3) = x.x + 1.x + x.3 + 1.3 = x 2 + x + 3x + 3
= x2 + 4x + 3
Some Special products
(a + b) (a - b) = a 2 - b 2
Sum
(a + b) 2 = a 2 + 2ab + b 2
Difference
(a – b) 2 = a 2 - 2ab + b 2
Squaring a binomial
(a + b) 2 = a2 + ab + ab + b2 = a2 + 2ab + b2
a
a2
ab
b
ba
b2
a
b
(a + b)2 = a2 + 2ab + b2
5.3
Factoring Polynomials
 Common Factors
 Factoring by Grouping
 Factoring and Equations ( Zero Product
Property)
Zero- Product Property
For all real numbers a and b, if ab = 0, then
a = 0 or b = 0 ( or both)
Ex – 5
Solving the equation 4x – x2 = 0
graphically and symbolically
Graphically
Symbolically
Numerically
4x – x 2 = 0
x(4 – x) = 0 (Factor out x )
x = 0 or 4 – x = 0 ( Zero Product property)
x = 0 or x = 4
4
3
2
1
-3
-2
-1
1
2
3
4
5
6
x
y
-1
0
1
2
3
4
5
-5
0
3
4
3
0
-5
Factoring x2 + bx + c
x2 + bx + c ,
find integers m and n that satisfy
m.n = c and
m+n=b
x2 + bx + c = (x + m)(x + n)
5.4 Factoring Trinomials ax 2 + bx + c
by grouping
• Factoring Trinomials with Foil
• 3x 2+ 7x + 2 = ( 3x
+ 1
)( x + 2 )
x
+
6x
7x
If interchange 1 and 2
(3x + 2)(x + 1) = 3x2 + 5x + 2 which is incorrect
2x
+
3x = 5x
5.5 Special types of Factoring
Difference of Two Squares
(a –b) (a + b) = a2 - b2
(a + b) 2 = a2 + 2ab + b2
(a –b) 2 = a2 - 2ab + b2
Perfect Square
Sum and Difference of Two Cubes
(a + b)(a2 – ab + b2) = a3 + b3
(a –b)(a2 + ab + b2 ) = a3 – b3
Verify
( a + b) (a2 – ab + b2) = a. a2 – a . ab + a. b2 + b. a2 -b . ab + b. b2
= a3 - a2 b + a b2 + a2 b - a b2 + b3
= a3 + b3
5.6
Polynomial Equations
Solving
Quadratic Equations
Higher Degree Equations
Using Technology
[ -4.7, 4.7, 1] by [-100, 100, 25]
Y = 16x^4 – 64x^3 + 64x^2
x = 0 , y = 0,
x=2 y=0
Chapter 6
Rational Expressions and
Functions
6.1 Rational Expressions and Functions
Rational Function
•
Let p(x) and q(x) be polynomials. Then a
rational function is given by
•
f(x) = p(x)/ q(x)
•
The domain of f includes all x-values such that
q(x) = 0
Examples 4
, x ,
3x2 –6x + 1
x
x–5
3x - 7
Identify the domain of rational function
(Ex 2 pg 376)
b) g(x) = 2x
x2 - 3x + 2
Denominator = x2 - 3x + 2 = 0
(x – 1)(x –2) = 0 Factor
x = 1 or x = 2 Zero product property
Thus D = { x / x is any real number except 1 and 2 }
Using technology
( ex 57, pg 384 )
[ -4.7 , 4.7 , 1 ] by [ -3.1, 3.1, 1 ]
(ex 64, pg 384)
Highway curve ( ex 72, page 384 )
R(m) = 1600/(15m + 2)
500
400
300
200
100
0
0.2
0.4
0.6
0.8
slope
a) R(0.1) = 1600 / (15(0.1) + 2) = 457
About 457 : a safe curve with a slope of 0.1 will have a minimum radius of 457 ft
b) As the slope of banking increases , the radius of the curve decreases
c) 320 = 1600/(15m + 2) , 320( 15m + 2) = 1600 , 4800m +640 = 1600
4800m = 960, m =960/4800 = 0.2
Evaluating a rational function Ex 4, Pg -377
Evaluate f(-1), f(1), f(2)
Numerical value
x
y
-3 -2 -1 0 1 2
3/2 4/3 1 0 __ 4
3
3
4
f(x) = 2x
x-1
3
2
1
-4
-3
-2 -1
1
2
3
Vertical asymptote
f(-1) = 1
f(1) = undefined and
f(2) = 4
6.2 Products and Quotients of Rational
Expression
To multiply two rational expressions, multiply
numerators and multiply denominators.
 A/B. C/D = AC /BD
B and D are nonzero.
Example 2/3 . 5/7 = 10/21
To divide two rational expressions, multiply by
the reciprocal of the divisor.
 A/B – C/D = AC/BD
B, C, and D are
nonzero
Example 3/4 - 2/4 = 3.4/4.5 = 3/5
6.3 Sums and Differences of Rational
Expressions
To add (or subtract) two rational expressions with
like denominators, add (or subtract) their
numerators. The denominator does not change.
A/C + B/C = (A + B)/C
Example - 1/5 + 2/5 = (1 + 2) / 5 = 3/5
A/C – B/C = (A – B) /C
3/5 - 2/ 5 = (3 – 2)/5 = 1/5
6.4 Solving rational equations graphically and
numerically ( Ex- 3 (a) pg 409 )
1/2 + x/3 = x/5
Solution- The LCD for 2,3, and 5 is their product, 30.
30( 1/2 + x/3) = x/5 . 30 Multiply by the LCD.
30/2 +30x/3 = 30x/5
Distributive property
15 + 10x = 6x
Reduce
4x = -15
Subtract 6x and 15
x = -15/4
Solve
Graphically Y1 = 1/2 + x/3
Y2 = X/5
[ -9, 9, 1] by [ -6, 6, 1]
Determining the time required to empty a
pool ( pg 415, no.68)
• A pump can empty a pool in in 40 hours. It can empty 1/40 of the
pool in 1 hour.
• In 2 hour, can empty a pool in 2/40 th of the pool
• Generally in t hours it can empty a pool in t/40 of the pool.
• Second pump can empty the pool in 70 hours. So it can empty a
pool in t/70 of the pool in t hours.
• Together the pumps can empty
t/40 + t/70 of the pool in t hours.
The job will complete when the fraction of the pool is empty equals 1.
The equation is
t/40 + t/ 70 = 1
Multiply (40)(70)
(40)(70) t/40 + t/ 70 = 1 (40)(70)
70t + 40t = 2800
110t = 2800
t = 2800/110 = 25.45 hr Two pumps can empty a pool in 25.45 hr
Example 6( pg – 412)
x = speed of slower runner
x + 2 = the speed of the winner
d = rt , t = d/r
The time for slower runner = 3/x, ran 3 miles at x miles per hour
The winning time is 3/(x + 2 ) = , the winner ran 3 miles at x+ 2 miles per hr
Add 3 minutes = 3/60 = 1/20 hr to the winner’s time, as finishes race 3
minutes ahead of another runner which equals the slower runner’s time
3/(x+2) + 1/20 = 3/x
Multiply each side by the LCD, which is 20x(x + 2)
20x(x + 2)( 3/x + 2) + 1/20 = 20x(x + 2)3/x
60x + x(x + 2) = 60(x + 2)
Distributive property
60x + x 2 + 2x = 60x + 120
,,
x 2 + 2x – 120 = 0
(x + 12)(x – 10) = 0
Factor
x = - 12 or x = 10
Zero product property
Running speed cannot be negative. The slower runner is running at 10 miles
per hour, and the faster runner is running at 12 miles per hour.
Ex 73 Pg 416
A tugboat can travel 15 miles per hour in still water
36 miles upstream ( 15 – x)
Total time 5 hours
downstream (15 + x)
t = d/r
So the equation is 36/(15 – x) + 36/ (15 + x) = 5
The LCD is (15-x)(15 + x)
Multiply both sides we get
(15 – x)(15 + x)[36/(15 – x) + 36/ (15 + x)] = 5 (15 – x)(15 + x)
540 + 36x + 540 - 36x = 1125 – 5x 2
5x2 – 45 = 0
5x2 = 45
x = + 9, x = 3 mph
Modeling electrical resistance (Ex- 8, pg –413)
R1 = 120 ohms
R
R2 = 160 ohms
1/R = 1/R1 + 1/R2
= 1/120 +1/ 160
= 1/120 . 4/4 + 1/160 . 3/3
= 4/480 + 3/480
= 7/480
R = 480/7 = 69 ohms
LCD = 480
6.6 Proportion
a c is equivalent to ad = bc
b d
Example 6
8
5
x
6x = 40
or x = 40/6 = 20/3
6 feet
4 feet
h feet
44 feet
h/44 = 6/4
h = 6.44/4
= 66 feet
Modeling AIDS cases
[ 1980, 1997, 2] by [-10000, 800000, 100000]
Y = 1000 (x
– 1981)2
Chapter 7
Radical Expressions and
Functions
Chapter 7
Square Root
The number b is a square root of a if b2 = a
Example
100 = 102 = 10
radical sign
Under radical sign the expression is called radicand
Expression containing a radical sign is called a radical expression.
Radical expressions are
6,
5 + x + 1 , and
3x
2x - 1
Cube Root
The number b is a cube root of a if b3 = a
Example – Find the cube root of 27
3
27
=
3 33
=3
Estimating a cellular phone transmission
distance
R
The circular area A is covered by one transmission tower is A =
2
The total area covered by 10 towers are 10
R
R2
, which must
equal to 50 square miles
Now solve R
R = 1.26, Each tower must broadcast with a minimum radius of
approximately 1.26 miles
Expression
 For every real number
If n is an integer greater than 1, then a1 n = n a
Note : If a < 0 and n is an even positive integer, then
a1 n is not a real number.
 If m and n are positive integer with m/n in lowest terms,
then
a mn = n a m = ( n a ) m
Note : If a < 0 and n is an even integer, then a m n
is not a real number.
If m and n are positive integer with m/n in lowest terms, then
a - m n = 1/ a m n
a =0
Properties of Exponent
Let p and q be rational numbers. For all real numbers a and b
for which the expressions are real numbers the following
properties hold. Page -467
1
2
3
4
5
6
7
a p . a q = a p+q
a - p = 1/ a p
a/b -p = b a p
a p = a p-q
aq
a p q = a pq
ab p= a p b p
a
b
p =
ap
bp
Product rule
Negative exponents
Negative exponents for quotients
Quotient rule for exponents
Power rule for exponents
Power rule for products
Power rule for products
Power rule for quotients
Product Rule for Radical Expressions
Let a and b be real numbers, where n a and
n b are both defined. Then
n
a . n b = n a. b
To multiply radical expressions with the
same index, multiply the radicands.
Quotient Rule for Radical Expressions
Let a and b be real numbers, where a and b are
both defined and b = 0.
a/b = a/b
The Expression –a
If a>0, then –a = a
Square Root Property
Let k be a nonnegative number. Then the solutions to
the equation.
x2 = k
are x = + k. If k < 0. Then this equation has no real
solutions.
Ex 2, 3, 4, 5, 6
Technology
[ 5, 13, 1] by [0, 100, 10]
To find cube root technologically
Technologically
Y1 = x2
[ -6, 6, 1] by [-4, 4, 1]
7.2 Let a and b be real numbers where
are both defined
n
a
n
, b
Product rule for radical expression (Pg – 472)
n
a
n
b = n
ab
Quotient rule for radical expression where b = 0 (Pg 475)
n
a
=
b
n
n
a
b
Pg -476
Rationalizing Denominators having square roots
1
3
3
3
=
3
3
7.3 Operations on Radical Expressions
Addition
10 11 + 4
11=
(10 + 4)
11 = 14 11
53 6  3 6  (5  1)3 6  63 6
Subtraction
10
11 -
4
11=
(10 - 4)
11
=6
11
53 6  3 6  (5  1)3 6  43 6
Rationalize the denominator (Pg 484)
7.6 Complex Numbers
Pg 513
x2 +1=0
x 2 = -1
x =+ - 1
Square root property
Now we define a number called the imaginary unit, denoted by i
Properties of the imaginary unit i
i=
-1
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 imaginary part is b
Pg 513
a + ib
Complex Number
-3 + 2i 5 -3i
-1 + 7i - 5 – 2i 4 + 6i
Real part a
-3
5
0
-1
-5
4
2
0
-3
7
-2
6
Imaginary Part b
Complex numbers contains the set of real
numbers
Complex numbers
a +bi
a and b real
Real numbers
a +bi
Rational Numbers
-3, 2/3, 0 and –1/2
Imaginary Numbers
b=0
a +bi
Irrational numbers
3 And
- 11
b =0
Sum or Difference of Complex
Numbers
Let a + bi and c + di be two complex numbers. Then
Sum
( a + bi ) + (c +di) = (a + c) + (b + d)i
Difference
(a + bi) – (c + di) = (a - c) + (b – d)i
Chapter 8
Quadratic Functions and
Equations
QuadraticFunction
A quadratic equation is an equation that
can be written as
f(x) = ax2 + bx + c ,
where a, b, c are real numbers, with a = 0.
Axis of symmetry
(0, 2)
-2
1
2
1
(0, 0)
0
-2
-1 0 1 2
-1
(2, -1)
Vertex
x=2
Vertex Formula
• The x-coordinate of the vertex of the
graph of
y = ax2 +bx +c, a = 0, is given by
x = -b/2a
• To find the y-coordinate of the vertex,
substitute this x-value into the equation
Example 1 (pg 531)
Graph the equation f(x) = x2 -1 whether it is increasing
or decreasing and Identify the vertex and axis of symmetry
vertex
x
y = x2 -1
-2
3
-1
0
0
-1
1
0
2
3
Equal
3
2
1
0
-1
Vertex
The graph is decreasing when x < 0
And Increasing when x > 0
Example 1(c) pg 532
x
-5
-4
-3
Vertex -2
-1
0
1
y = x2 + 4x + 3
8
3
0
-1
0
3
Axis of symmetry
Equal
x = -2
8
Vertex (-2, -1)
Find the vertex of a parabola
f(x) = 2x2 - 4x + 1
Symbolically f(x) = 2x2 – 4x + 1
a=2 , b=-4
x = -b/2a = - (-4)/2.2 = 4/4 = 1
To find the y-value of the vertex,
Substitute x = 1 in the given formula
f(1) = 2. 12 - 4.1 + 1= -1
The vertex is (1, -1)
Graphically
[ -4.7, 4.7, 1] , [-3.1, 3.1, 1]
Example 5 (Pg 535)
Maximizing Revenue
The regular price of a hotel room is $ 80, Each room rented
the price decreases by $2
900
800
700
600
500
400
(20, 800)
Maximum revenue
0 5 10 15 20 25 30 35 40
If x rooms are rented then the price of each room is 80 – 2x
The revenue equals the number of rooms rented times the price of each room.
Thus f(x) = x(80 – 2x) = 80x - 2x2 = -2x2 + 80x
The x-coordinate of the vertex x = - b/2a = - 80/ 2(-2) = 20
Y coordinate f(20) = -2(20)2 + 80 (20) = 800
8.2 Vertical and Horizontal Translations
Translated upward and downward
y2 = x2 + 1
y1=
y1= x2
y1= (x-1)2
x2
y3 = x2 - 2
Translated
horizontally to the
right 1 unit
y1= x2
y2= (x + 2 )2
Translated horizontally to the left 2 units
Vertical and Horizontal Translations Of
Parabolas (pg 542)
Let h , k be positive numbers.
To graph
y = x2 + k
y = x2 – k
y = (x – k)2
y = (x +k)2
Ex 1 pg 543
shift the graph of y = x2 by k
units
upward
downward
right
left
Vertex Form of a Parabola (Pg 543)
The vertex form of a parabola with vertex (h, k) is
y = a (x – h)2 + k, where a = 0 is a constant.
If a > 0, the parabola opens upward;
if a < 0, the parabola opens downward.
Ex- 2, 3, 4
Quadratic Equation
A quadratic equation is an equation that
can be written as
ax2 +bx +c= 0, where a, b, c are real
numbers with a = 0
Quadratic Equations and Solutions
y = x2 + 25
No Solution
y = 4x2 – 20x + 25
One Solution
y = 3x2 + 11x - 20
Two Solutions
Quadratic Formula
The solutions of the quadratic equation
ax2 + bx + c = 0, where a, b, c are real
numbers with a = 0
No x intercepts
One x – intercepts
Two x - intercepts
Ex 1
Modeling Internet Users
Use of the Internet in Western Europe has increased dramatically
shows a scatter plot of online users in Western Europe with
function f given by
f(x) = 0.976 x2 - 4.643x + 0.238 x = 6 corresponds to 1996 and so on
until x = 12 represents 2002
90
80
70
60
50
40
30
20
10
0
f(10) = 0.976(10) 2 - 4.643(10) + 0.238 = 51.4
6 7 8 9 10 11 12 13
8.4 Quadratic Formula
• The solutions to ax 2 + bx + c = 0 with a = 0 are
given by
- b + b2 – 4ac
X=
Ex – 1, 2, 3, 4
2a
The Discriminant and Quadratic Equation
To determine the number of solutions to
ax2 + bx + c = 0 , evaluate the discriminant
2
b – 4ac > 0,
If
If
If
2
b – 4ac > 0, there are two real solutions
2
b – 4ac = 0, there is one real solution
2
b – 4ac
< 0, there are no real solutions , but two complex
solution
Ex 5, 6, 7, 8, 9