Transcript College Algebra

College Algebra
Exam 2 Material
Quadratic Applications
• Application problems may give rise to all
types of equations, linear, quadratic and
others
• Here we take a look at two that lead to
quadratic equations
Example
Two boys have two way radios with a range of 5 miles, how long can
they communicate if they leave from the same point at the same
time with one traveling north at 10 mph and the other traveling east
at 7 mph?
D
=
R
T
N
N boy
R  10 mph
5 mi
10 x
E
R  7 mph
7x
E boy
10 x
7x
10
7
x
x
For right tria ngle with legs a and b and hypotenuse c : a 2  b 2  c 2
25
2
5 149
7 x 2  10 x 2  52
x 
x
hr
149
149
49 x 2  100 x 2  25
5
x
149 x 2  25
x  0.4096 hr
149
Example
• A rectangular piece of metal is 2 inches longer than it is
wide. Four inch squares are cut from each corner to
make a box with a volume of 32 cubic inches. What
were the original dimensions of the metal?
4
4
x
4
x2
4
Unknowns
L Rec x  2
W Rec x
L Box x  2  8
W Box x  8
Height Box  4
V  LWH
32  x  6x  8 4
32  x 2  14 x  484
32  4 x 2  56 x  192
2
0  4 x  56 x  160
0  x 2  14 x  40
0  x 10x  4
x 10  0 OR x  4  0
x4
Impossible
x  10
W  10 in.
.
L  12 in.
Homework Problems
• Section: 1.5
• Page: 130
• Problems: 5 – 9, 21 – 22
• MyMathLab Assignment 22 for practice
Other Types of Equations
• Thus far techniques have been discussed
for solving all linear and quadratic
equations and some higher degree
equations
• Now address techniques for identifying
and solving many other types of
equations
Solving Higher Degree
Polynomial Equations
• So far methods have been discussed for solving
first and second degree polynomial equations
• Higher degree polynomial equations may
sometimes be solved using the “zero factor
method” or, the “zero factor method” in
combination with the “quadratic formula” or the
“square root property”
• Consider two examples:
x3  8
x  x  9x  9  0
3
2
Example One
x3  8
Make one side zero:
x 8  0
3
Factor non-zero side:
x  2x 2  2 x  4  0
Apply zero factor property and solve:
x2  0
x2
x  2x  4  0
2
 2  2  414  2   12
x

21
2
 2  2i 3
x
 1  i 3
2
Note : 3 solutions, 1 real and 2 non - real complex
OR
2
Example Two
x3 + x2 – 9x – 9 = 0
One side is already zero, so factor non-zero side
x3 + x2 – 9x – 9 = 0
x2(x + 1) – 9(x + 1) = 0
(x + 1)(x2 – 9) = 0
Apply zero factor property and solve:
x + 1 = 0 OR x2 – 9 = 0
x = -1
x2 = 9
x=±3
Homework Problems
• Section: 1.4
• Page: 124
• Problems: 59 – 62
• MyMathLab Assignment 23 for practice
Rational Equations
• Technical Definition: An equation that
contains a rational expression
• Practical Definition: An equation that has
a variable in a denominator
• Example:
1
5
2


2
x  2x  3 x 1 x  3
Solving Rational Equations
1. Find “restricted values” for the equation by
setting every denominator that contains a
variable equal to zero and solving
2. Find the LCD of all the fractions and multiply
both sides of equation by the LCD to eliminate
fractions
3. Solve the resulting equation to find apparent
solutions
4. Solutions are all apparent solutions that are
not restricted
Example
RV
1
5
2
x  2x  3  0


2
x 1x  3  0
x  2x  3 x 1 x  3
x 1  0 OR x  3  0
x  3
1
5
2
x 1


x 1  0 Already Solved
x  1x  3 x  1 x  3
x  3  0 Already Solved

LCD
1
5
2  LCD
x 1x  3
 x  1x  3  x  1  x  3  1


2
1  5x  3  2x 1
1  5x  15  2x  2
1  3x  17
16  3x
16
x
3
Not RV!
Homework Problems
• Section:
• Page:
• Problems:
1.6
144
Odd: 1 – 25
• MyMathLab Assignment 24 for practice
“Quadratic in Form” Equation
• An equation is “quadratic in form” if the
same algebraic expression is found
twice where one time the exponent on
the expression is twice as big as it is the
other time
• Examples:
m6 – 7m3 – 8 = 0
8(x – 4)4 – 10(x – 4)2 + 3 = 0
Solving Equations that are
Quadratic in Form
1. Make a substitution by letting “u” equal
the repeated expression with exponent
that is half of the other
2. Solve the resulting quadratic equation
for “u”
3. Make a reverse substitution for “u”
4. Solve the resulting equation
Example of Solving an Equation
that is Quadratic in Form
m  7m  8  0
Let u  m 3
u 2  7u  8  0
6
3
u  8u  1  0
u 8  0
u 8
3
m 8
OR
m 8  0
3
u 1  0
u  1
3
m  1
m3  1  0
m  2m

 2m  4  0
m  2  0 or m2  2m  4  0
m  2 or m  1  i 3
2
m  1m2  m  1  0
m  1  0 or m2  m  1  0
1
3
m  1 or m  
i
2 2
Example of Solving an Equation
that is Quadratic in Form
8 x  4   10 x  4   3  0
4
2
LET u   x  4
2
8u 2  10u  3  0
4u  32u 1  0
4u  3  0 OR 2u 1  0
2u  1
4u  3
1
3
u
u
2
4
3
x  4 
4
3
x4 
2
2
3
x  4
2
1
2
x  4 
2
1
x4  
2
2
x  4
2
Homework Problems
• Section:
• Page:
• Problems:
1.6
145
All: 61 – 64, 73 – 74
• MyMathLab Assignment 25 for practice
“Negative Integer Exponent”
Equation
• An equation is a “negative integer
exponent equation” if it has a variable
expression with a negative integer
exponent
• Examples:
x  1 1  2
6x
2
x
1
3
Quadratic in Form
This one can also be classified as what other type ?
Solving “Negative Integer
Exponent” Equations
• If the equation is “quadratic in form”,
begin solution by that method
• Otherwise, use the definition of
negative exponent to convert the
equation to a rational equation and solve
by that method
Example of Solving Equation
With Negative Integer Exponents
x  1
1
2
1
2
x 1
 1
 LCD
 x  1  2 1
1  2x  2
3  2x
3
x
Not RV
2
RV
x 1  0
x 1
LCD
x 1
Example of Solving Equation
With Negative Integer Exponents
x
2
x
1
2
Let u  x
1
u2  u  2
u u 2  0
u  2u 1  0
2
u  2  0 OR u 1  0
u 1
u  2
1
x  2
x1  1
1
RV
 2
x0
x
LCD
1
 LCD
x


2
 x
 1
1  2 x
1
x
Not RV
2
1
1
x
x 1
Not RV
Homework Problems
• Section:
• Page:
• Problems:
1.6
145
75, 76
• MyMathLab Assignment 26 for practice
• MyMathLab Homework Quiz 5/6 will be due for a
grade on the date of our next class meeting
Radical Equations
• An equation is called a radical equation if it
contains a variable in a radicand
• Examples:
x  x 3  5
x  x 5 1
3
x  4  3 2x  0
Solving Radical Equations
1. Isolate ONE radical on one side of the equal
sign
2. Raise both sides of equation to power
necessary to eliminate the isolated radical
3. Solve the resulting equation to find “apparent
solutions”
4. Apparent solutions will be actual solutions if
both sides of equation were raised to an odd
power, BUT if both sides of equation were
raised to an even power, apparent solutions
MUST be checked to see if they are actual
solutions
Why Check When Both Sides are
Raised to an Even Power?
• Raising both sides of an equation to a power does not always result
in equivalent equations
• If both sides of equation are raised to an odd power, then resulting
equations are equivalent
• If both sides of equation are raised to an even power, then resulting
equations are not equivalent (“extraneous solutions” may be
introduced)
• Raising both sides to an even power, may make a false statement
true:
2
2
4
4
 2  2 , however : - 2  2 , - 2  2 , etc.
• Raising both sides to an odd power never makes a false statement
true:
3
3
5
5
 2  2 , and : - 2  2 , - 2  2 , etc.
.
Example of Solving
Radical Equation
Check x  4
x  x 3  5
x 5  x 3
 x  5
2


x 3

2
x  10 x  25  x  3
x 2  11x  28  0
x  4x  7  0
x  4  0 OR x  7  0
x  4 OR x  7
2
4 43  5?
4 1 5?
35
x  4 is NOT a solution
Check x  7
7 73  5?
7 4 5?
55
x7
IS a solution
Example of Solving
Radical Equation

x  x 5 1
x  5  1 x
x5
  1  x 
2
Check x  4
4  4  5 1?
2
x  5  1 2 x  x
4  2 x
2  x
2
2
 2  x
 
4 x
4  9 1?
2  3 1?
5 1
x  4 is NOT a solution
Equation has No Solution!

Example of Solving
Radical Equation
3
x  4  3 2x  0
3

3
x  4  3 2x
x4
   2x 
3
3
3
x  4  2x
4 x
(No need to check)
Homework Problems
• Section:
• Page:
• Problems:
1.6
144
Odd: 27 – 51, 55 – 57
• MyMathLab Assignment 27 for practice
Rational Exponent Equations
• An equation in which a variable
expression is raised to a “fractional
power”
Example:
x 1  9x
2
3
1
3
0
Solving
Rational Exponent Equations
•
•
1.
2.
3.
4.
If the equation is quadratic in form, solve that way
Otherwise, solve essentially like radical equations
Isolate ONE rational exponent expression
Raise both sides of equation to power necessary to
change the fractional exponent into an integer
exponent
Solve the resulting equation to find “apparent
solutions”
Apparent solutions will be actual solutions if both sides
of equation were raised to an odd power, but if both
sides of equation were raised to an even power,
apparent solutions MUST be checked to see if they
are actual solutions
Example
x  2  9x
2
3
 x  2
1
3
2
3
x  4 OR x  1
 9 x 
1
3
3
x  2    9 x  

 

2
3
x  4  0 OR x 1  0
0
x  22  9 x
x 2  4 x  4  9 x
x 2  5x  4  0
x  4x 1  0
1
3
3
No reason to have to check!
Homework Problems
• Section:
• Page:
• Problems:
1.6
145
All: 53 – 54, 59 – 60, 65 – 72
• MyMathLab Assignment 28 for practice
Definition of Absolute Value
• “Absolute value” means “distance away from zero” on
a number line
• Distance is always positive or zero
• Absolute value is indicated by placing vertical parallel
bars on either side of a number or expression
Examples:
The distance away from zero of -3 is shown as:
3  3
The distance away from zero of 3 is shown as:
3
3
The distance away from zero of u is shown as:
u
Can' t be simplified , because value of " u" is unknown. However, its value is zero or positive.
Absolute Value Equation
• An equation that has a variable contained
within absolute value symbols
• Examples:
| 2x – 3 | + 6 = 11
| x – 8 | – | 7x + 4 | = 0
| 3x | + 4 = 0
Solving Absolute Value
Equations
• Isolate one absolute value that contains an algebraic
expression, | u |
– If the other side is a negative number there is no solution
(distance can’t be negative)
If 2 x  5  4, then :
No solution
– If the other side is zero, then write:
u = 0 and Solve If 2 x  5  0, then : 2 x  5  0
– If the other side is “positive n”, then write:
u = n OR u = - n and Solve
If 2 x  5  3, then : 2x  5  3 or 2x  5  3
– If the other side is another absolute value expression, | v |, then
write:
u = v OR u = - v and Solve
If 2 x  5  x  3 , then : 2x  5  x  3 or 2x  5  x  3
Example of Solving
Absolute Value Equation
2 x  3  6  11
2x  3  5
2x  3  5 OR 2x  3  5
2 x  2
2x  8
x  1
x4
Example of Solving
Absolute Value Equation
x  9  7x  4  0
x  9  7x  4
x  9  7x  4
13  6x
 13
x
6
OR
x  9  7 x  4
x  9  7 x  4
8x  5
5
x
8
Example of Solving
Absolute Value Equation
3x  6  2
3x  4
?
This says distance is negative - NOT POSSIBLE!
Equation has NO SOLUTION!

Homework Problems
• Section:
• Page:
• Problems:
1.8
164
Odd: 9 – 23, 41 – 43,
67 – 69
• MyMathLab Assignment 29 for practice
• MyMathLab Homework Quiz 7 will be due
for a grade on the date of our next class
meeting
Inequalities
• An equation is a comparison that says two algebraic
expressions are equal
• An inequality is a comparison between two or three
algebraic expressions using symbols for:
greater than:
greater than or equal to:
less than:
less than or equal to:
• Examples:


x 15  3x  3
1
 3   x  4   1
2


Two part inequality
Three part inequality
.
Inequalities
• There are lots of different types of
inequalities, and each is solved in a
special way
• Inequalities are called equivalent if they
have exactly the same solutions
• Equivalent inequalities are obtained by
using “properties of inequalities”
Properties of Inequalities
•
Adding or subtracting the same number to all parts of an
inequality gives an equivalent inequality with the same sense
(direction) of the inequality symbol
Add 3
x  3  2 is equivalent to : x  5
•
Multiplying or dividing all parts of an inequality by the same
POSITIVE number gives an equivalent inequality with the
same sense (direction) of the inequality symbol
Divide by 3
3x  6 is equivalent to : x  2
•
Multiplying or dividing all parts of an inequality by the same
NEGATIVE number and changing the sense (direction) of the
inequality symbol gives an equivalent inequality
Divide by - 2
 2 x  8 is equivalent to : x  4
Solutions to Inequalities
• Whereas solutions to equations are usually sets
of individual numbers, solutions to inequalities
are typically intervals of numbers
• Example:
Solution to x = 3 is {3}
Solution to x < 3 is every real number that is less
than three
• Solutions to inequalities may be expressed in:
– Standard Notation
– Graphical Notation
– Interval Notation
Two Part Linear Inequalities
• A two part linear inequality is one that
looks the same as a linear equation
except that an equal sign is replaced by
inequality symbol (greater than, greater
than or equal to, less than, or less than or
equal to)
• Example:
x 15  3x  3
Expressing Solutions to Two
Part Inequalities
• “Standard notation” - variable appears alone on left side of
inequality symbol, and a number appears alone on right side:
x2
• “Graphical notation” - solutions are shaded on a number line
using arrows to indicate all numbers to left or right of where shading
ends, and using a parenthesis to indicate that a number is not
included, and a square bracket to indicate that a number is
included
]2
• “Interval notation” - solutions are indicated by listing in order the
smallest and largest numbers that are in the solution interval,
separated by comma, enclosed within parenthesis and/or square
bracket. If there is no limit in the negative direction, “negative
infinity symbol” is used, and if there is no limit in the positive
direction, a “positive infinity symbol” is used. When infinity
symbols are used, they are always used with a parenthesis.
(, 2]
Solving
Two Part Linear Inequalities
• Solve exactly like linear equations
EXCEPT:
– Always isolate variable on left side of
inequality
– Correctly apply principles of inequalities
(In particular, always remember to reverse
sense of inequality when multiplying or
dividing by a negative)
Example of Solving Two Part
Linear Inequalities
x 15  3x  3
x 15  3x  9
 2x  6
x  3
When dividing by a negative, reverse sense of inequality !
Standard Notation Solution
3
]
(,  3]
Graphical Notation Solution
Interval Notation Solution
Homework Problems
• Section:
• Page:
• Problems:
1.7
156
Odd: 13 – 23
• MyMathLab Assignment 30 for practice
Three Part Linear Inequalities
• Consist of three algebraic expressions compared with
two inequality symbols
• Both inequality symbols MUST have the same sense
(point the same direction) AND must make a true
statement when the middle expression is ignored
• Good Example:
1
 3   x  4   1
2
• Not Legitimate:
1
 3   x  4   1 Inequality Symbols Don' t Have Same Sense
2
.
1
 3   x  4   1 - 3 is NOT  -1
2
Expressing Solutions to Three
Part Inequalities
• “Standard notation” - variable appears alone
in the middle part of the three expressions
being compared with two inequality symbols:
2 x 3
• “Graphical notation” – same as with two part
inequalities:
2
3
(
]
• “Interval notation” – same as with two part
inequalities:
(2, 3]
Solving
Three Part Linear Inequalities
• Solved exactly like two part linear
inequalities except that:
– solution is achieved when variable is
isolated in the middle
– all three parts must be kept balanced by doing
the same operation on all parts
Example of Solving
Three Part Linear Inequalities
1
 x  4   1
2
1
 3  x  2  1
2
3
 6  x  4  2
2 x  2
Standard Notation Solution
2
2
[
)
Graphical Notation Solution
[2, 2) Interval Notation Solution
Homework Problems
• Section:
• Page:
• Problems:
1.7
156
Odd: 23 – 33
• MyMathLab Assignment 31 for practice
Quadratic Inequalities
• Looks like a quadratic equation
EXCEPT that equal sign is replaced by
an inequality symbol
• Example:
x x2
2
Solving Quadratic Inequalities
1. Put quadratic inequality in standard form (make right side zero and
put trinomial in descending powers)
2. Change quadratic inequality to a quadratic equation and solve to
find “critical points”
3. Graph “critical points” on a number line and draw a vertical line
through each one to divide number line into intervals
4. Pick a “test point” in each interval (a “nice” number that is close
to zero)
5. Evaluate the “trinomial” described in step 1 with each “test
point” to determine whether the result is positive or negative
and write the appropriate + or - above each test point
6. Now graph the solution to the inequality by shading all the
intervals of the number line for which the + or – satisfies the
inequality written in step 1
Example of
Solving Quadratic Inequality
x x2
2
x x20
2
x x20
x  2x 1  0
x  2  0 OR x  1  0
x  2 OR x  1
2
Critical Numbers
Evaluate Test Numbers : x 2  x  2
 22   2  2
02  0  2
422
Test Numbers :

2
)
1

0
0
32  3  2
9 3 2
(
3

2
x2  x  2  0
Numbers that make trinomal  are solutions
 , 1  2, 
Homework Problems
• Section:
• Page:
• Problems:
1.7
157
39 – 51
• MyMathLab Assignment 32 for practice
Rational Inequality
• An inequality that involves a rational
expression (variable in a denominator)
• Example:
2
3
x 1
Solving a Rational Inequality
1.
2.
3.
4.
5.
6.
7.
8.
Make right side of inequality zero
Perform math operations on left side to end up with a single rational
expression (the rational inequality will now be in “standard form”
Factor numerator and denominator of rational expression
Find “critical points” by putting every factor that contains a variable
equal to zero and solving
Graph “critical points” on a number line and draw a vertical line through
each one to divide number line into intervals
Pick a “test point” in each interval (a “nice” number that is close to
zero)
Evaluate the left side of “standard form” described in step 1 with each
“test point” to determine whether the result is positive or negative and
write the appropriate + or - above each test point
Now graph the solution to the problem by shading all the intervals of
the number line for which the + or – satisfies inequality found in step 1
Example of
Solving a Rational Inequality
2
1
x 1
2
1  0
x 1
2
1 x  1

0
x  1 1 x  1
2  x 1
0
x 1
 x 1
0
x 1
x 1  0
 x 1  0
x  1
x 1
Critical Numbers

2
 2   1
2 1
 0   1
0 1
  2   1
 2 1
)
1

0
0
[
1
(,  1)  [1, )

2
Homework Problems
• Section:
• Page:
• Problems:
1.7
158
Odd: 69 – 85
• MyMathLab Assignment 33 for practice
Absolute Value Inequality
• Looks like an absolute value equation
EXCEPT that an equal sign is replaced
by one of the inequality symbols
• Examples:
| 3x | – 6 > 0
| 2x – 1 | + 4 < 9
| 5x - 3 | < -7
Properties of Absolute Value
• | u | < 5, means that u’s distance from zero must be less than 5.
Therefore, u must be located between what two numbers?
between -5 and 5
How could you say this with a three part inequality?
-5 < u < 5
• Generalizing: | u | < n , where “n” is positive, always translates to:
-n < u < n
• | u | > 3, means that u must be less than what, or greater than what?
less than -3, or greater than 3
How could you say this with two inequalities?
u < -3 or u > 3
• Generalizing: | u | > n , where “n” is positive, always translates to:
u < -n or u > n
Solving Absolute Value
Inequalities
1. Isolate the absolute value on the left side
to write the inequality in one of the forms:
| u | < n or | u | > n (where n is positive)
2. If | u | < n, then solve:
If | u | > n, then solve:
-n < u < n
u < -n or u > n
3. Write answer in interval notation
Example
Solve:
3x  6  0
3x  6
Equivalent Inequality:
3x  6 or 3x  6
x  2 or x  2
(,  2)  (2, )
Example
Solve:
2x 1  4  9
2x 1  5
Equivalent Inequality:
 5  2 x 1  5
 4  2x  6
2 x 3
 2, 3
Solving Other Absolute Value
Inequalities
• If isolating the absolute value on the left
does not result in a positive number on the
right side, we have to use our
understanding of the definition of absolute
value to come up with the solution as
indicated by the following examples:
Absolute Value Inequality
with No Solution
• How can you tell immediately that the following
inequality has no solution?
5x  7  2
• It says that absolute value (or distance) is
negative – contrary to the definition of absolute
value
• Absolute value inequalities of this form always
have no solution:
u  n (where  n represents a negative number)

Does this have a solution?
2x  5  0
• At first glance, this is similar to the last example,
because “ < 0 “ means negative, and:
2 x  5 can' t be less than a negative number !
• However, notice the symbol is: 
• And it is possible that: 2 x  5  0
• We have previously learned to solve this as:
2x  5  0
2x  5
5
x
2
5
Solution is : x 
2
Solve this:
4x  5  0
• This means that 4x – 5 can be anything except
zero:
4x  5  0 or 4x  5  0
• Solving these two inequalities gives the solution:
4x  5  0 or 4x  5  0
4x  5 or 4x  5
5
5
x
or x 
4
4
5 5 

  ,    ,  
4 4 

Homework Problems
• Section:
• Page:
• Problems:
1.8
164
Odd: 27 – 39, 45 – 61
• MyMathLab Assignment 34 for practice
• MyMathLab Homework Quiz 8 will be due for a
grade on the date of our next class meeting