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Hawkes Learning Systems:
Intermediate Algebra
Section 7.1a: Quadratic Equations: The Square
Root Method
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Objectives
o Solve quadratic equations by factoring
o Solve quadratic equations by using the definition of a
square root.
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Solving Quadratic Equations by Factoring
When the solutions of a polynomial equation can be
found by factoring, the method depends on the
zero-factor property.
Zero-Factor Property
If the product of two factors is 0, then one or
both of the factors must be 0. For factors a and b,
If a  b  0, then a  0 or b  0 or both.
For example,
 x  1  x  2   0
x 1 0
x20
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Solving Quadratic Equations by Factoring
An equation that can be written in the form
ax  bx  c  0
2
Where a, b, and c are real numbers and a
is called a quadratic equation.
0
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To Solve an Equation by Factoring
1. Add or subtract terms so that one side of the
equation is 0.
2. Factor the polynomial expression.
3. Set each factor equal to 0 and solve each of
the resulting equations.
Note: If two of the factors are the same, then the
d solution is said to be a double root or a root
d of multiplicity two.
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Example 1: Solving Quadratic Equations by
Factoring
Solve the following quadratic equation by factoring.
x  3 x  18
2
x  3 x  18  0
2
 x  3  x  6   0
x30
x  3
or
x60
x6
Subtract 18 from both sides.
One side must be zero.
Factor the left-hand side.
Set each factor equal to 0.
Solve each linear equation.
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Example 2: Solving Quadratic Equations by
Factoring
Solve the following quadratic equation by factoring.
2 x  16 x   32
2
2 x  16 x  32  0
2
2  x  8 x  16   0
2
2 x  4  0
2
x40
x4
The solution is a double root.
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Solving Quadratic Equations by Factoring
o Quadratic equations may have nonreal complex
solutions.
o In particular, the sum of two squares can be factored
as the product of complex conjugates.
o For example,
x  81   x  9 i   x  9 i  .
2
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Example 3: Quadratic Equations Involving the
Sum of Two Squares
Solve the following equation by factoring.
x  49  0
2
 x  7i   x  7i   0
x  7i  0
x  7i  0
x  7i
Check:
 7i 
2
 49  0
x  7i
 7i 
2
 49  0
49 i  49  0
49 i  49  0
49  49  0
49  49  0
00
00
2
2
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Square Root Property
vIf
x  c , then x   c .
2
If,  x  a 
2
c
then or
xa c
, or
xa
c.
Note: If c is negative, then the solutions will be
nonreal.
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Using the Definition of Square Root and the
Square Root Property
Consider the equation
x 8
2
Allowing that the variable, x, might be positive or
negative, we use the definition of square root, x 2
 x.
Taking the square root of both sides of the equation
gives:
x 
8
So we have two solutions,
x
8 and x   8 , or x   8 .
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Using the Definition of Square Root and the
Square Root Property
Similarly, for the equation
 x  4
2
3
the definition of square root gives
x  4   3.
This leads to the two equations and the two solutions,
as follows:
x4
x4 3
3
x  4 
3
x  4 
x  4 
3
3
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Example 4: Using the Square Root Property
Solve the following quadratic equations by using the
Square Root Property.
a.  x  5 
2
3
x5  3
x 5
b.
3
y   16
2
y    16
y   4i
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Hawkes Learning Systems:
Intermediate Algebra
Section 7.1b: Quadratic Equations:
Completing the Square
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Objectives
o Solve quadratic equations by completing the square
o Find polynomials with given roots.
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Completing the Square
o Recall that a perfect square trinomial is the result of
squaring a binomial.
o Our objective here is to find the third term of a
perfect square trinomial when the first two terms are
given. This is called completing the square.
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Completing the Square
Perfect Square
Trinomials
Equal Factors
Square of a
Binomial
x  12 x  36
 x  6 x  6
 x  6
2
x  8 x  16
 x  4 x  4
 x  4
2
x  2 ax  a
2
 x  a x  a
x  a
2
x  2 ax  a
2
 x  a x  a
x  a
2
2
2
2
2
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Completing the Square
Step 1: Write the equation ax 2  bx  c  0 in the form
2
ax  bx   c .
Step 2: Divide by a if a  1 , so that the coefficient of
x is 1 : x 
2
2
b
a
x
c
a
.
Step 3: Divide the coefficient of x by 2 , square the
result, and add this to both sides.
Step 4: The trinomial on the left side will now be a
perfect square. That is, it can be written as the
square of an algebraic expression.
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Example 1: Completing the Square
Add the constant that will complete the square for the
expression. Then write the new expression as the
square of a binomial.
x  18 x
2
Solution: Since the leading coefficient is 1, we can begin
to complete the square. We take half of the
coefficient of the x term and square the result.
1
2
18   9
9
2
 81
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Example 1: Completing the Square
Adding 81 to the expression x 2  18 x yields the perfect
square trinomial
x  18 x  81
2
This can be factored as  x  9  .
2
Thus completing the square gives us:
x  18 x  8 1   x  9  .
2
2
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Example 2: Completing the Square
Solve the following quadratic equation by completing
the square.
x  6 x  10
2
1
2


 
2
 3
Take half of the
coefficient of the x
term.
9
Square the result.
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Example 2: Completing the Square (Cont.)
Adding 9 to both sides of the original equation will
result in a perfect square trinomial.
2
2
Add 9 to both sides
x x6 x 6 9x  10  9
(completing the square).
2
Factor the left-hand side.
 x  3   19
Use the Square Root Property to solve.
x3
19
x  3
19
or
x  3   19
or
x  3
19
We write these two solutions as: x  3  19
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Example 3: Completing the Square
Solve the following quadratic equation by completing
the square.
2 x  8 x  24  0
2
x  4 x  12  0
2
x  4 x  12
2
1
2
  2
 
2
4
Divide each term by 2
so that the leading
coefficient will be 1.
Isolate the constant
term.
Take half of the
coefficient of the x
term, and square it.
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Example 3: Completing the Square (Cont.)
Add 4 to both sides of x 2  4 x  1 2 .
x  4 x  4  12  4
2
 x  2
2
 16
x2
16
x24
or
x  2  4
x2
or
x  6
Factor the left-hand side
of the equation.
Use the Square Root
Property.
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Writing Equations with Known Roots
To find the quadratic equation that has the given
roots x  4  i and x  4  i , first get 0 on one side of
each equation.
x4i 0
x4i0
Set the product of the two factors equal to 0 and
simplify.
 x  4  i x  4  i  0
Continued on the next slide…
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Writing Equations with Known Roots
Regroup the terms to present the product of complex
conjugates. This makes the multiplication easier.
 x  4  i x  4  i  0
  x  4   i    x  4   i   0
 x  4  i  0
2
2
x  8 x  16  1  0
2
x  8 x  17  0
2
i   1.
2
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Example 4: Equations with Known Roots
Find the quadratic equation that has the given roots.
x7
x7
x7
5
5 0
Get 0 on one side
of each equation.
5 x  7 

5  0

Set the product of
the two factors
equal to 0.
5   x  7  

5  0

5 0
x  7 

 x  7  

5
and
x7
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Example 4: Equations with Known Roots
(Cont.)
Simplify and solve.
 x  7  

5   x  7  

x  7
2

5  0

 5
2
0
x  14 x  49  5  0
2
x  14 x  44  0
2
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Hawkes Learning Systems
Intermediate Algebra
Section 7.2: Quadratic Equations: The Quadratic
Formula
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Objectives
o Write quadratic equations in standard form.
o Identify the coefficients of quadratic equations in
standard form.
o Solve quadratic equations using the quadratic
formula.
o Determine the nature of the solutions (one real, two
real, or two non-real) for quadratic equations using
the discriminant.
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Standard Form of the Quadratic Equation
The standard form of the quadratic equation is
ax  bx  c  0
2
where a , b , and c are real numbers and a  0 .
We are interested in developing a formula that can
solve quadratic equations of any form. This formula
will always work, though other techniques may be
more convenient to use.
We want to solve the standard form equation of the
quadratic formula for x in terms of a, b and c.
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Development of the Quadratic Formula
ax  bx  c  0
2
Standard form.
ax  bx   c
2
x 
2
b
x
a
x 
2
b
a
x
b
Add –c to both sides.
c
Divide both sides by a so that the
leading coefficient is 1.
a
2
4a
2
2

c
a

b
2
4a
Complete the square.
2
2
b 
c
b

x
  
2
2
a
a
4
a


Simplify.
Continued on the next page…
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Development of the Quadratic Formula
2
2
b 
4ac
b


x
 
2
2
2a 
4a
4a

2
b 
b  4ac

x
 
2
2a 
4a

x
b
2
b  4 ac
Combine the fractions on the right
side of the equation.
2

2a
x
Find LCM of the denominators on
the right side.
2a
b 
b  4 ac
Square Root method.
2
2a
The quadratic formula.
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Development of the Quadratic Formula
The quadratic formula,
x
b 
b  4 ac
2
,
2a
can ALWAYS solve quadratic equations of any form.
Because it is so useful, you should memorize the
quadratic formula.
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Discriminant
The expression b 2  4 ac is called the discriminant. The
discriminant determines the number of solutions to the
given quadratic equation.
If the discriminant is:
Discriminant
Number of
Solutions
Solution(s) Real
or Non-real
Positive
Zero
Negative
b  4 ac  0
b  4 ac  0
b  4 ac  0
2
1
2
Real
Real
Non-real
2
2
2
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Solve Quadratic Equations Using the
Quadratic Formula
Ex: Solve the quadratic equation using the quadratic
formula.
1x  2 x  1  0
a  1, b = 2 , c = 1
2
2
x
x
2
Compare it to the standard quadratic
2
ax  bx  c  0 equation to find a, b and c.
Solve the discriminant by plugging a, b
and c into b 2  4 a c . Since the discriminant
 4  1   1   4  4  0 is 0, there is one real solution.
2 
2 1 
2
2
 1
Use the quadratic formula
0
x
b 
b  4 ac
2
to find the solution, x.
2a
We know that the discriminant is 0, so we
can just plug that in.
Solution: ̶ 1
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Example 1: Solve Quadratic Equation Using
the Quadratic Formula
Solve the quadratic equation using the quadratic formula.
x  x 1  0
a   1, b = 1 , c =  1
2
Compare it to the standard quadratic
2
ax  bx  c  0 equation to find a, b and
c.
Discriminant:
1   4   1    1 
2
Solve the discriminant by plugging a, b
and c into b 2  4 a c .
1 4
 3
Number of solutions: 2
Real or non-real: non  real
What does it mean for a discriminant to
be negative?
Continued on the next page…
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Example 1: Solve Quadratic Equation Using
the Quadratic Formula (Cont.)
 1  
x
x
Use the quadratic formula
3
x
2   1
1 
x
3
2

b 
b  4 ac
2
2a
1
x. We know that the discriminant is ̶ 3,
so we can plug that in.
3
2
1 i 3
2
Solutions:
to find the solution,
x
1 i 3
2
and x 
1 i 3
2
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Example 2: Solve Quadratic Equation Using
the Quadratic Formula
Solve the quadratic equation using the quadratic formula.
x  3x  2  0
a  1, b  3 , c  2
2
Compare it to the standard quadratic
2
ax  bx  c  0 equation to find a, b and
c.
Discriminant:
 3   4 1   2 
2
Solve the discriminant by plugging a, b
and c into b 2  4 a c .
 98
1
Number of solutions: 2
Real or non-real: real
What does it mean for a discriminant to
be positive?
Continued on the next page…
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Example 2: Solve Quadratic Equation Using
the Quadratic Formula (Cont.)
x
x
 3 
2 1 
3  1
Use the quadratic formula
1
x
b 
b  4 ac
2
to find the solution,
2a
x. We know that the discriminant is 1,
so we can plug that in.
2
Solutions: x   1 and x   2
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Hawkes Learning Systems:
Intermediate Algebra
Section 7.3: Applications- Quadratic Equations
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Objective
Solve applied problems by using quadratic equations.
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Strategy for Solving Word Problems
1. Read the problem carefully.
2. Decide what is asked for and assign a variable to
the unknown quantity.
3. Draw a diagram or set up a chart whenever
possible.
4. Form an equation (or inequality) that relates the
information provided.
5. Solve the equation or inequality.
6. Check your solution with the wording of the
problem to be sure that it makes sense.
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The Pythagorean Theorem
o Problems involving right triangles often require the
use of quadratic equations.
o In a right triangle, one of the angles is a right angle
(measures 90 ), and the side opposite this (the
longest side) is called the hypotenuse.
o The other two sides are called legs.
leg
hypotenuse
leg
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The Pythagorean Theorem
In a right triangle, the square of the hypotenuse
(c ) is equal to the sum of the squares of the legs
( a and b ).
c a b
2
2
2
c
b
a
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Example 1: The Pythagorean Theorem
The width of a rectangle is 5 yards less than its length. If
one diagonal measures 25 yards, what are the dimensions
of the rectangle?
Solution: Draw a diagram for problems involving
geometric figures whenever possible.
Let
x  length of the rectangle
x  5  w idth of the rectangle
Then, by the Pythagorean Theorem,
x   x  5    25 
2
2
2
x5
x
25
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Example 1: The Pythagorean Theorem (Cont.)
Now, solve the quadratic equation.
x   x  5    25 
2
2
2
x  x  10 x  25  625
2
2
2 x  10 x  25  625  625  625
2
2 x  10 x  600  0
2
2  x  5 x  300   0
2
2  x  20   x  15   0
Length:
Width:
x  20
x  5  15
x  15
A negative
number does not
fit the conditions
of the problem.
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Projectiles
The formula h   16 t 2  v 0 t  h0 is used in physics and
relates to the height of a projectile such as a thrown
ball, a bullet or a rocket.
h  height of object, in feet.
t  time object is in the air, in seconds.
v 0  beginning velocity, in feet per second.
h0  beginning height.
h0  0 if object is initially at ground level.
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Example 2: Projectiles
A bullet is fired straight up from 6 feet above ground
level with a muzzle velocity of 420 ft per sec. When will
the bullet hit the ground?
Solution: v 0  420 ft per sec.
h0  6
The bullet hits the ground when h  0.
h   16 t  v 0 t  h0
2
0   16 t  420 t  6
2
0   8 t  210 t  3
2
Divide both sides
by 2.
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Example 2: Projectiles
0   8 t  210 t  3
2
Use the quadratic formula to solve the equation.
t
b 
b  4 ac
2
2a
t
t
 210 
 210   4   8   3 
2  8 
 210 
44196
t  26.26
2
 16
t   0.01
Therefore, the bullet
hits the ground in
26.26 seconds.
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Example 3: Cost per Person
The Ski Club is planning to charter a bus to a ski resort. The cost
will be $200 and each member will share the cost equally. At the
last minute, 30 more member decide to go on the trip. The cost
to each of the members will be $6 less. How many members are
going to the ski resort now?
Let x  the original number of club members going on the trip.
x  30  the actual number of club members that took the trip.
Final cost
Difference in cost
Initial cost per


per member
per member
member
200
x

200
x  30

6
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Example 3: Cost per Person (Cont.)
Multiplying each term by the LCM of the denominators results in
the following:
200
x
 x  x  30  
200
x  30
 x  x  30   6  x  x  30 
200  x  30   200 x  6 x  x  30 
200 x  6000  200 x  6 x  x  30 
6000  6 x  180 x
2
0  6 x  180 x  6000
2
0  x  30 x  1000
2
Factor
out 6.
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Example 3: Cost per Person (Cont.)
x  30 x  1000  0
2
Use the quadratic formula to solve.
x
b 
b  4 ac
2
2a
x
x
 30 
 30 
2
 4 1    1000 
2 1 
 30  70
2
x  20
x   50
The number of members that actually went to the ski resort is
20  30  5 0.
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Example 4: Geometry
The Smiths have a rectangular swimming pool that is 5 ft longer
than it is wide. The pool is completely surrounded by a concrete
deck that is 3 ft wide. The total area of the pool and the deck is
11750 ft 2 . Find the dimensions of the pool.
Let
3 ft
w  6  width of the pool
w5
and the deck
w  11 
length of the pool
and the deck
 w  6   w  11   1750
3 ft
3 ft w
3 ft
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Example 4: Geometry (Cont.)
 w  6   w  11   1750 
w  17 w  1684  0
2
Use the quadratic equation to solve.
w
b 
b  4 ac
2
2a
w
Therefore, the length of
the pool is 33.41 feet, and
the width of the pool is
38.41 feet.
w
17  4 1    1684 
 17 
2
2 1 
 17 
7025
2
w  33.41
w  5  38.41
or
w   50.41
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Intermediate Algebra
Section 7.4: Equations in Quadratic Form
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Objectives
o Make substitutions that allow equations to be
written in quadratic form.
o Solve equations that can be written in quadratic
form.
o Solve equations that contain rational expressions.
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Solving Equations in Quadratic Form
The general quadratic equation is ax 2  bx  c  0 where
a  0.
The equations
2
1
and x 3  4 x 3  21  0
are not quadratic equations, but they are in quadratic
form because the degree of the middle term is onehalf the degree of the first term. Specifically,
x  7 x  12  0
4
1
2
4
2

2
Degree of the Degree of the
first term
second term
and
12
  
23
1
3
Degree of the Degree of the
first term
second term
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Solving Equations in Quadratic Form
Solving Equations in Quadratic Form by Substitution
1. Look at the middle term.
2. Substitute a first-degree variable, such as u, for the
variable expression in the middle term.
3. Substitute the square of this variable, u², for the
variable expression in the first term.
4. Solve the resulting quadratic equation for u.
5. Substitute the results back for u in the beginning
substitution and solve for the original variable.
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Example 1: Solving Equations in Quadratic Form
Solve the equation in quadratic form by substitution.
x  7 x  12  0
6
3
u  x
3
u  7 u  12  0
2
u  3u  4   0
Step 1: Look at the middle term.
Step 2: Substitute a first-degree variable, such as u,
for the variable expression in the middle term.
Step 3: Substitute the square of this variable, u²,
for the variable expression in the first term.
Step 4: Solve the resulting quadratic equation for u.
u  3 or u  4
x  3 or x  4
3
x
3
3
3 or x 
3
4
Step 5: Substitute the results back for u in the
beginning substitution and solve for the original
variable.
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Example 2: Solving Equations in Quadratic Form
Solve the equation in quadratic form by substitution.
5
x  9 x 2  18  0
5
Step 1: Look at the middle term.
5
ux
2
u  9 u  18  0
2
u  6u  3  0
Step 2: Substitute a first-degree variable, such as u,
for the variable expression in the middle term.
Step 3: Substitute the square of this variable, u²,
for the variable expression in the first term.
Step 4: Solve the resulting quadratic equation for u.
u  6 or u  3
5
5
x 2  6 or x 2  3
x  36 or x  9
5
x
5
5
36 or x 
5
9
Step 5: Substitute the results back for u in the
beginning substitution and solve for the original
variable.
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Example 3: Solving Equations in Quadratic Form
Solve the equation in quadratic form by substitution.
2
1
x 3  3x 3  2  0
u  x
Step 1: Look at the middle term.
Step 2: Substitute a first-degree variable, such as u,
for the variable expression in the middle term.
Step 3: Substitute the square of this variable, u²,
for the variable expression in the first term.
1 3
u  3u  2  0
2
 u  1  u  2   0
Step 4: Solve the resulting quadratic equation for u.
u   1 or u   2
1
1
x   1 or x   2
3
3
x    1  or x    2 
3
x   1 or x   8
3
Step 5: Substitute the results back for u in the
beginning substitution and solve for the original
variable.
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Example 4: Solving Equations in Quadratic Form
Solve the equation in quadratic form by substitution.
x
2
 10 x
1
 16  0 Step 1: Look at the middle term.
Step 2: Substitute a first-degree variable, such as u,
for the variable expression in the middle term.
2
u  10 u  16  0 Step 3: Substitute the square of this variable, u²,
for the variable expression in the first term.
u  x
1
 u  2   u  8   0 Step 4: Solve the resulting quadratic equation for u.
u  2 or u  8
x
1
 2 or x
x
1
2
1
or x 
8
1
8
Step 5: Substitute the results back for u in the
beginning substitution and solve for the original
variable.
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Example 5: Solving Equations in Quadratic Form
Solve the equation in quadratic form by substitution.
 x  6
2
 4 x  6  4  0
u  x6
u  4u  4  0
2
u  2u  2  0
u 2
 x  6  2
x  4
Step 1: Look at the middle term.
Step 2: Substitute a first-degree
variable, such as u, for the variable
expression in the middle term. Step 3:
Substitute the square of this variable,
u², for the variable expression in the
first term.
Step 4: Solve the resulting quadratic
equation for u.
Step 5: Substitute the results back for u
in the beginning substitution and solve
for the original variable.
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Solving Equations with Rational Expressions
Recall that to solve equations with rational expressions,
first multiply every term on both sides of the equation
by the least common multiple (LCM) of the
denominators. This will “clear” the equation of
fractions. Remember to check the restrictions on the
variables. That is, no denominator can have a value of 0.
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Example 6: Solving Equations with Rational
Expressions
Solve the following equation containing rational
3
2
x
expressions.


4x  2
x3
x3
This equation is not in the quadratic form. Multiply both sides of the equation by
1
the LCM, (4x – 2)(x + 3), to “clear” the fractions. The restrictions on x are x  ,  3.
2
2 
 3
 x 

 4 x  2  x  3
  4 x  2 x  3

 4x  2 x  3 
 x3
Simplify.
3 x  3  2 4 x  2   x 4 x  2 
3x  9  8x  4  4 x  2 x
2
0  4 x  13 x  5
2
Continued on next slide...
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Example 6: Solving Equations with Rational
Expressions (Cont.)
0  4 x  13 x  5
2
x
x
13 
  13   4  4    5 
24
13 
169  80
Use the quadratic formula to solve.
2
8
x
13 
249
8
There are two solutions to this
equation.
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Example 7: Solving Equations with Rational
Expressions
Solve the following equation containing rational expressions.
1
2x  3

2
x2

x
x2
This equation is not in the quadratic form. Multiply both sides of the equation by
3
, 2.
the LCM, (2x + 3)(x – 2), to “clear” the fractions. The restrictions on x are x 
2
2 
 1
 x 

 2 x  3 x  2 
  2 x  3 x  2 

 2x  3 x  2 
x2
Simplify.
1 x  2   2  2 x  3   x  2 x  3 
x  2  4 x  6  2 x  3x
2
0  2x  2x  4
2
Continued on next slide...
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Example 7: Solving Equations with Rational
Expressions (Cont.)
0  2x  2x  4
2
x
 2 
2
2
Use the quadratic formula to solve.
 4  2   4 
22
x
2
4  32

2
4
x
26
4
36
4
 1
x
or
2
3
2
, 2 , so the only solution is x   1
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Example 8: Solving Higher-Degree Equations
Solve the higher degree equation.
x  9x  0
5
x x  9  0
4
Factor out x.
x  x  3  x  3   0
2
Factor the difference of two
squares.
2
x  0 or x   3
or x  3
x  0 or x   i 3
or x   3
2
x  0, i 3 ,  i 3 ,
2
3, 
3
Solve for x.
There are five solutions for x.
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Example 9: Solving Higher-Degree Equations
Solve the higher degree equation.
x 8  0
3
 x  2 x  2x  4  0
2
 x  2 x  1  i
3
 x  1  i 3   0
Factor out (x – 2).
Factor the quadratic equation.
x  2 or x   1  i 3 or x   1  i 3 Solve for x.
x  2,  1  i 3 ,  1  i 3
There are three solutions for x.
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Intermediate Algebra
Section 7.5: Graphing Parabolas
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Objectives
o Graph a parabola (a quadratic function) and
determine its vertex, domain , range, line of
symmetry and zeros.
o Solve applied problems by using quadratic functions
and the concepts of maximum and minimum.
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Introduction to Quadratic Functions
Concepts related to various types of functions and their
graphs include domain, range and zeros.
In this section we will include a detailed analysis of
quadratic functions, functions that are represented by
quadratic expressions.
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Introduction to Quadratic Functions
What is the graph of the function y  x 2  4 x  3 ? The nature
of the graph can be investigated by plotting several points.
x
x  4x  3
2
1
  1
0
0
1
2
1
2
3
2
2
 4   1  3
8
 40  3
3
2
1
1

4
 
 3
2
 
2
1   4 1   3
2
2  42  3
2
3  4 3  3
2
2
y
4
0
1
0
2
7
7

4
 

2
2
4
4  44  3
3
5
5
8
7

3

5
2
2
 4 5  3
5
4
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Introduction to Quadratic Functions
The complete graph of y  x 2  4 x  3 is shown below.
This curve is called a parabola.
The point  2,  1  is the “turning
point” of the parabola and is
called the vertex. The line x  2
is the line of symmetry or axis
of symmetry for the parabola.
That is, the curve is a “mirror
image” of itself with respect to
the line x  2 .
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Introduction to Quadratic Functions
Quadratic Functions
A quadratic function is any function that can be written
in the form
2
f  x   ax  bx  c
where a, b, and c are real numbers and a  0. The
graph of every quadratic function is a parabola.
Parabolas that open up or down are vertical parabolas
and parabolas that open left or right are horizontal
parabolas. Horizontal parabolas do not represent
functions.
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Introduction to Quadratic Functions
We will discuss quadratic functions in each of the
following five forms where a, b, c, h and k are constants:
y  ax
2
y  ax  k
2
y  ax  h
2
y  ax  h  k
2
y  ax  bx  c
2
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Quadratic Functions
Functions of the Form y  ax 2
The graph lies
Vertex
Domain
Range
If a  0
If a  0
Below the x-axis
Above the x-axis
 0, 0  This is the only point where
the graph touches the x-axis.
(All real numbers)
y0
y0
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Quadratic Functions
Domain: {x|x is any real number}
Range: {y|y ≥ 0}
In interval notation:
Domain:   ,   , Range:  0,  
Domain: {x|x is any real number}
Range: {y|y ≤ 0}
In interval notation:
Domain:   ,   , Range:   , 0 
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Quadratic Functions
The graphs on the previous slide illustrate the following
characteristics of quadratic equations of the form y  ax 2:
a. If a  0 , the parabola “opens upward.”
b. If a  0 , the parabola “opens downward.”
c. The bigger a is, the narrower the opening.
d. The smaller a is, the wider the opening.
e. The line x  0 (the y-axis) is the line of symmetry.
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Quadratic Functions
Functions of the Form y  ax 2  k
The y-value
Shift y  ax 2
Vertex
Line of
Symmetry
If k  0
If k  0
Decreases by k
Increases by k
Down k units
Up k units
 0, k 
x0
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Quadratic Functions
Domain: {x|x is any real number}
Range: {y|y ≥ k}
In interval notation:
Domain:   ,   , Range:  k ,  
Domain: {x|x is any real number}
Range: {y|y ≤ k}
In interval notation:
Domain:   ,   , Range:   , k 
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Quadratic Functions
Functions of the Form y  a  x  h 
The graph lies
If a  0
Below or on the
x-axis
If a  0
Above or on the
x-axis
 h, 0 
Vertex
The graph
opens
2
Down
Up
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Quadratic Functions
Domain: {x|x is any real number}
Range: {y|y ≥ 0}
In interval notation:
Domain:   ,   , Range:  0,  
Domain: {x|x is any real number}
Range: {y|y ≤ 0}
In interval notation:
Domain:   ,   , Range:   , 0 
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Quadratic Functions
For the graph of y  a  x - h  ,
2
The x-value
Shift y  ax 2
Line of
Symmetry
If h  0
Decreases by h
Left h units
xh
If h  0
Increases by h
Right h units
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Quadratic Functions
In summary,
Shift of y  ax
Graph
y  ax
2
y  ax  k
2
y  ax  h
2
2
Does not apply
Up or Down
Left or Right
Vertex
Line of
Symmetry
 0, 0 
x0
 0, k 
x0
 h, 0 
xh
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Quadratic Functions
Ex: Graph the quadratic function y  2 x 2  3 . Find the
line of symmetry and the vertex, and state the domain
and range of each function.
Solution:
Any function of the form y  ax  k
Line of symmetry is x  0 .
has x = 0 for its line of symmetry.
2
The parabola opens upward.
a = 2, so a is positive.
Vertex:  0,  3 
k   3 , so the vertex is  0,  3  .
Continued on the next slide…
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Quadratic Functions
Domain:  x | x is any real num ber 
or    ,  
Range:  y | y   3 or   3,  
True of all functions of the
form y  ax 2  k .
Because k   3 .
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Example 1: Quadratic Functions
2
5

Graph the quadratic function y    x   . Find the line
2

of symmetry and the vertex, and state the domain and
range of each function.
Solution:
5
Line of symmetry is x  .
A function of the form y  a  x  h 
has x = h for its line of symmetry.
2
2
The parabola opens dow nw ard . a = ̶ 1, so a is negative.
5
5 
5 
h
, so the vertex is  , 0  .
Vertex:  , 0 
2
2 
2 
Continued on the next slide…
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Quadratic Functions
Domain:  x | x is any real num ber 
or    ,  
Range:  y | y  0 or    , 0 
( )
True of all functions of the
2
form y  a  x  h  .
Because a   1 .
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Quadratic Functions
Functions of the form y  a  x  h   k combine both
the vertical shift of k units and the horizontal shift of h
units. The vertex is at  h , k  .
2
For example, the graph of the function y   2  x  3   5
is a shift of the graph of y   2 x 2 up 5 units and to the
right 3 units and has its vertex at  3, 5  .
2
2
The graph of
1

y x  2
2 1

of y  x 2 but is shifted left
The vertex is at   1 / 2,  2  .
2
is the same as the graph
unit and down 2 units.
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Quadratic Functions
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Quadratic Functions
To easily find the vertex, line of symmetry, range, and
to graph the parabola, we want to change the general
2
2
form y  ax  bx  c into the form y  a  x  h   k .
This can be accomplished by “completing the square”
using the following technique.
f  x   ax  bx  c
2
 2 b 
 a x  x  c
a 

Write the function.
Factor a from the first two
terms.
2
2
 2 b
b
b 
 a x  x 

 c Complete the square.
2
2 
a
4a
4a 

Continued on the next slide…
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Quadratic Functions
2
2
 2 b
b  b
 a x  x 

c
2 
a
4a  4a

2
b 
4ac  b

 a x 
 
2a 
4a

2
Move the negative factor out
of the parentheses, first
multiplying by a.
Write the square of the
binomial and simplify.
In terms of the coefficients a, b and c,
x
b
2a
is the line of symmetry
Note: the vertex must lie at
 b 4 ac  b 2 
and   ,
is the vertex.  b  b  

, f 
4a
 .

 2a

 2a  
 2a
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Quadratic Functions
Note!
Rather than memorize the formulas for the coordinates
of the vertex, you should just remember that the
x-coordinate of the vertex is x  
b
2a
. Substituting this
value for x in the function will give the y-value for the
vertex.
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Zeros of a Quadratic Equation
The points where a parabola crosses the x-axis, if any,
are the x-intercepts. This is where y  0 . These points
are also called the zeros of the function. We find these
points by substituting 0 for y and solving the resulting
quadratic equation.
2
y  ax  b x  c quadratic function
2
0  ax  bx  c quadratic equation
If the solutions are non-real complex numbers, then
the graph does not cross the x-axis.
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Zeros of a Quadratic Function
Ex: Find the zeros of the function, the line of symmetry,
the vertex, the domain, the range and graph the
parabola.
2
x  6x 1  0
Solution:
2
Quadratic Formula
x
 6 
6
4
2
x
6
32
2
x
The zeros of the function.
64 2
2
x  3 2 2
Continued on the next slide…
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Zeros of a Quadratic Equation
Change the form of the function for easier
graphing.
Add 0  9  9 inside the parenthesis.
y  x  6x 1
2
  x  6x  9  9  1
2
  x  6x  9  9  1
2
  x  3  8
3
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Zeros of a Quadratic Function
Summary
Zeros: 3  2 2
Line of Symmetry: x  3
Vertex:  3,  8 
Domain:    ,   ;
 x | x is any real num ber 
Range:   8,   ;  y | y   8
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Example 2: Zeros of a Quadratic Function
Find the zeros of the function, the line of symmetry, the
vertex, the domain, the range and graph the parabola.
x  4x  2  0
2
Solution:
Quadratic Formula.
x
x
x
4
 4 
2
 4   1  2 
2   1
4
24
2
42 6
2
The zeros of the function. x   2  6
Continued on the next slide…
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Example 2: Zeros of a Quadratic Function (Cont.)
Change the form of the function.
y  x  4x  2
2
Factor  1 from the first two terms.
 x  4x  2
Add 0  4  4 inside the parenthesis.
  x  4x  4  4  2
Put   1    4  outside of the parenthesis.
  x  4x  4  4  2
2
2
2
  x  2  6
2
Continued on the next slide…
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Example 2: Zeros of a Quadratic Function (Cont.)
Summary
Zeros:  2  6
Line of Symmetry: x   2
Vertex:   2, 6 
Domain:    ,   ;
x | x
is any real num ber 
Range:   , 6  ;  y | y  6 
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Example 3: Zeros of a Quadratic Function
Find the zeros of the function, the line of symmetry, the
vertex, the domain, the range and graph the parabola.
2x  6x  5  0
2
Solution:
x
x
 6 
6
2
 4  2 5
22
6
4
4
Quadratic Formula.
There are no real zeros
because the discriminant
is negative. The graph will
not cross the x-axis.
Continued on the next slide…
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Example 2: Zeros of a Quadratic Function (Cont.)
x
b

2a
6
22

3
2
To find the vertex, use another
approach. First find the x-value.
2
3
3
3
f    2   6   5
2
2
2
9
 95
2
1

2
3 1
 , 
2 2
Plug the x-value into the original
equation.
Simplify.
So we have the following vertex.
Continued on the next slide…
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Example 2: Zeros of a Quadratic Function (Cont.)
Summary
Zeros: N o real zeros
Line of Symmetry: x 
3 1
Vertex:  , 
2 2
Domain:    ,   ;
x | x
3
2
is any real num ber 
1
1
 
Range: ,   ;  y | y  
2
2

 
Insert graph p. 597 example
2 c.
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Applications with Maximum/Minimum Values
Note:
The vertex of a vertical parabola is either the lowest
point or the highest point on the graph of the parabola.
Therefore, the vertex can be used to determine the
maximum or minimum value of a quadratic function.
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Applications with Minimum/Maximum Values
Minimum and Maximum Values
For a parabola with its equation in the form
y  ax  h  k,
2
1. If a  0 , then the parabola opens upward and  h , k 
is the lowest point and y  k is called the minimum
value of the function.
2. If a  0 , then the parabola opens downward and
 h , k  is the highest point and y  k is called the
maximum value of the function.
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Application and Minimum/Maximum Values
If the function is in the general quadratic form
2
y  ax  bx  c , then the maximum or minimum value
b
can be found by letting x  
and solving for y.
2a
The concepts of minimum and maximum values of a
function help not only in graphing but also in solving
many types of applications.
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Example 3: Maximization/Minimization Problems
A movie theater sells regular adult tickets for $7.50
each. On average, they sell 5,000 tickets per day. The
company estimates that each time they raise ticket
prices by 50¢, they will sell 1,000 fewer tickets. What
price should they charge to maximize their revenue
(income) per day? What will be the maximum revenue?
Let x = number of 50¢ increases in price.
Then the price per ticket = 7.50  .50 x
and the number of tickets sold = 5, 000  1, 000 x .
Revenue = (price per unit)(number of units sold)
Continued on the next slide…
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Example 3: Maximization/Minimization
Problems (Cont.)
So,
R   7.50  .50 x   5, 000  1, 000 x 
 37, 500  7500 x  2500 x  500 x
2
  500 x  5000 x  37, 500
2
The revenue represented by a quadratic function and
the maximum revenue occurs at the point where
x
b
2a

 5000
2   500 
 5
For x   5 , price per ticket = 7.50  . 50   5   $5
And the maximum revenue =  5, 000  1, 000   5    $ 5 
 $50, 000
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Example 4: Maximization/Minimization Problems
A farmer plans to use 2500 feet of spare
fencing material to form a rectangular area
for cows to graze alongside a river, using
the river as one side of the rectangular
area. How should he split up the fencing
among the other three sides in order to
maximize the rectangular area?
MOO!
x
x
2500  2 x
If we let x represent the length of one side of the rectangular area
then the dimensions of the rectangular area are x feet by 2500-2x
feet (see image above). We will let A be the name of our function
that we wish to maximize in this problem, so we want to find the
maximum possible value of A  x   x  2500  2 x  .
Continued on the next slide…
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Example 4: Maximization/Minimization
Problems (Cont.)
Note: If we multiply out the formula for A, we get a
quadratic function. A  x   x  2500  2 x    2 x 2  2500 x
We know this function is a parabola opening down. We also
know that the vertex is the maximum point on this graph.
Remember, the vertex is the point
 b 4 ac  b 2 
 b
 b 
,
, f 
.

 or  
4a
 2a  
 2a
 2a

So, plugging in the values, we get the vertex   625, A  62 5   .
Therefore, the maximum possible area is A(625):
A (625)  781250 square fee t.
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Hawkes Learning Systems:
Intermediate Algebra
Section 7.6a: Solving Quadratic Inequalities
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Objectives
o Solve quadratic inequalities.
o Solve higher degree inequalities.
o Graph the solutions for inequalities on real number
lines.
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Quadratic Inequalities
For values of x on either side of a number a, the sign for
an expression of the form  x  a  changes.
If
x  a,
then  x  a  is positive.
If x  a , then  x  a  is negative.
xa0
any x  a
xa 0
a
any x  a
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To Solve Quadratic or Higher Degree Inequalities
1. Arrange the terms so that one side of the
inequality is 0.
2. Factor the algebraic expression, if possible, and
find the points where each factor is 0. (Use the
quadratic formula, if necessary, to find the
points where a quadratic expression is 0.)
3. Mark these points on a number line. Consider
these points as endpoints of intervals
Continued on the next slide…
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To Solve Quadratic or Higher Degree
Inequalities (Cont.)
4. Test one point from each interval to determine
the sign of the expression for that interval.
5. The solution consists of those intervals where
the test points satisfy the original inequality.
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Example 1: Quadratic Inequalities
Solve the following inequality by factoring and using a
number line. Then, graph the solution set on a number
line.
2
x  2 x  15
x  2 x  15  1 5  1 5
2
Step 1: Add -15 to both sides so
that the right-hand side is 0.
x  2 x  15  0
2
 x  5 x  3  0
Step 2: Factor the left-hand side.
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Example 1: Quadratic Inequalities (Cont.)
 x  5 x  3  0
x50
x3 0
x3
x  5
Step 3: Set each factor equal to
0 and solve to find the
endpoints.
Step 4: Mark these points on a number line and test one point
from each of the intervals.

5

3
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Example 1: Quadratic Inequalities (Cont.)
  ,  5 

7
  5, 3 

5
 3,  



0
3
5
Step 5:
Test one point
from each interval.
 x  5 x  3  0
Test x   7
  7  5    7  3     2    10   20  0  x  5   x  3   0
if x   5
Test x  0
 0  5   0  3    5    3    15  0  x  5   x  3   0
if  5  x  3
Test x  5
 5  5   5  3   10   2   20  0
 x  5  x  3  0
if x  3
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Example 1: Quadratic Inequalities (Cont.)
x  2 x  15
2
We have found that:
 x  5  x  3  0
if x   5
 x  5  x  3  0
if  5  x  3
 x  5  x  3  0
if x  3
Therefore, the solution is: x   5 or x  3
or x is in   ,  5    3,   .
)
(
5
3
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Example 2: Cubic Inequality
Solve the following inequality by using a number line
and graph each solution set on a number line.
x  8x  9x  0
3
2
x  x  8x  9  0
2
Factor the left-hand side.
x  x  1  x  9   0
x0
x0
x 1 0
x  1
x9 0
x9
Set each factor equal to 0
and solve to find the
endpoints.
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Example 2: Cubic Inequality (Cont.)
Mark the endpoints on a number line and test one point from
each of the intervals found.
Endpoints:
x  1
  ,  1    1, 0 
  
2 1
0

x0
 0, 9 
x9
 9,  

 
3
9 10
1
2
x  x  1  x  9   0
Test x   2
  2    2  1    2  9     2    1    11    22  0
Test x  3
 3   3  1   3  9    3   4    6    72  0
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Example 2: Cubic Inequality (Cont.)
x  x  1  x  9   0
  
2  1 0

 

3
1
9
10
2
Test x  
1
2
 1  1
 1




1


9




 2  2
 2

 1   1   19
    
 2  2  2
Test x  10
1 0  10  1  10  9   10  11  1   110  0
 19
0

 8
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Example 2: Cubic Inequality (Cont.)
x  8x  9x  0
3
2
We have found that:
x  x  1   x  9   0 if x   1
x  x  1   x  9   0 if 0  x  9
x  x  1   x  9   0 if x  9
x  x  1   x  9   0 if  1  x  0
Therefore, the solution is x is in    ,  1   0, 9  .
 
1
0

9
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Example 3: Quadratic Inequality
Solve the following inequality by using a number line and graph
each solution set on a number line.
x  6x  7  0
2
Solution: Since this quadratic equation will not factor with
integer coefficients, use the quadratic formula to find the
endpoints for the intervals. The test points can be integers.
x
b 
b  4 ac
2
x
6
  6   4 1   7 
2
2 1 
2a
x
6
36  28
2
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Example 3: Quadratic Inequality (Cont.)
x
6
36  28
2
x
6
8
2
x
62 2
2
x  3
2

0
3

2
3
2
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Example 3: Quadratic Inequality (Cont.)
x  6x  7  0
2


0 3

2 3 3
Test x  0
0
2
 6 0  7  0  0  7  7  0
Test x  3
3
2
 6  3   7  9  18  7   2  0
Test x  7
 7   6  7   7  49  42  7  14  0
2


2
7
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Example 3: Quadratic Inequality (Cont.)
x  6x  7  0
2
We have
If x  3 
2 then x  6 x  7  0
If x  3 
2 then x  6 x  7  0
2
2
The solution is: x is in
  , 3 


3
 
2  3
2
3
2

2, .
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Hawkes Learning Systems:
Intermediate Algebra
Section 7.6b: Solving Inequalities with
Rational Expressions
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Objectives
o Solve inequalities containing rational expressions.
o Graph the solutions for inequalities containing
rational expressions.
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Solving Inequalities with Rational Expressions
o For linear inequalities, the solution always contains
the points to one side or the other of the point
where the inequality has the value 0.
o A rational inequality may involve the product or
quotient of several first-degree expressions. For
example,
x4
x3
0
involves the two first-degree expressions x  4 and x  3 .
o For (x  a ), if x  a , then x  a is negative; if x  a , x  a
is positive.
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Solving Inequalities with Rational Expressions
Procedure for Solving Polynomial Inequalities with
Rational Expressions
1. Simplify the inequality so that one side is 0 and
the other side has a single fraction with both the
numerator and denominator in factored form.
2. Find the points that cause the factors in the
numerator or in the denominator to be 0.
3. Mark each of these points on a number line.
These are interval endpoints.
Continued on the next slide…
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Solving Inequalities with Rational Expressions
4. Test one point from each interval to determine
the sign of the polynomial expression for all
points in that interval.
5. The solution consists of those intervals where
the test points satisfy the original inequality.
6. Mark a bracket for an endpoint that is included
and a parenthesis for an endpoint that is not
included. Remember that no denominator can
be 0.
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Solving Inequalities with Rational Expressions
The steps are as follows:
x4
x3
0
a. Find the points where each linear factor has the
v value 0.
x40
x3 0
x  4
x3
b. Mark each of these points on a number line.
C Consider these points as endpoints of intervals.

4

0
3
The three intervals formed are    ,  4  ,   4, 3  ,  3,   .
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Solving Inequalities with Rational Expressions
c. Choose any number from each interval as a test value
to determine the sign of the expression for all values
in that interval. Remember, we are not interested in
the value of the expression, only whether it is
positive or negative.
  , 4 

6
  4, 3 

4
 3,  



0
3 4
We have chosen the test values  6 , 0 , and 4 .
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Solving Inequalities with Rational Expressions
Substituting the values  6 , 0 , and 4 into the original inequality
gives:
Results
x4
x3
x4
x3
x4
x3



 6   4
 6   3
0  4
0  3
4  4
4  3
Explanation
2

9

The expression is positive
for all x in (ng
, 4 .
 inf,neg4)
2
9
The expression is negative


for all x in (-4,3)
  4, 3  .
3
3
4

8
4
The expression is positive
for all x in (3,posinf)
 3,   .
8
1


4

3
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Solving Inequalities with Rational Expressions
x4
x3
0
d. The solution to the inequality consists of all the
intervals that indicate the desired sign: + (for x  0 ) or
– (for x  0 ). The solution for the above expression is:
all x in
  ,  4 
or
 3,   .
In algebraic notation: x   4 or x  3.
In interval notation: x    ,  4    3,   .
Graphically:


4
3
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Example 1: Solving and Graphing Inequalities
Solve and graph the solution for the following inequality:
x 1
x5
0
Solution: Set each linear expression equal to 0 to find
the interval endpoints.
x 1 0
x  1
x5 0
x5
Test a value from each of the intervals:
  ,  1  ,   1, 5  , an d  5,  

3
 
1 0
 
5 6
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Example 1: Solving and Graphing
Inequalities (Cont.)
Substituting the values  3 ,0 , and 6 into the original inequality
gives:
Results
x 1
x5
x 1
x5
x 1
x5



Explanation
  3   1  2 1 The expression is positive


for all x in (neginf,-1).
  ,  1  .
 3  5 8 4
0  1 1
1 The expression is negative


for all x in (-1,5).
  1, 5  .
5
 0   5 5
6  1

6  5
7
1
The expression is positive
for all x in (5,
 .
 5, inf).
7


1

5
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Example 1: Solving and Graphing
Inequalities (Cont.)
x 1
x5
0
We can conclude that the solution for the above
inequality in interval notation is:   1, 5  .
In algebraic notation:  1  x  5.
Graphically:
(
1
)
5
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Example 2: Solving and Graphing Inequalities
Solve and graph the solution for the following inequality:
x6
x2
1 0
Solution: Simplify to get one fraction.
x6
x2

x2
x2
2x  4
x2
0
0
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Example 2: Solving and Graphing
Inequalities (Cont.)
Set each linear expression equal to 0 to find the interval
endpoints.
2x  4  0
x20
x2
x  2
Test a value from each of the intervals:
  ,  2  ,   2, 2  , an d  2,  

4

2

0

2

4
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Example 2: Solving and Graphing
Inequalities (Cont.)
Substituting the values -4, 0, and 4 into the original inequality
gives:
Results
Explanation
2x  4
2  4   4
4
2x  4
2 0  4
4
The expression is positive



for all x in (neginf,
.
  ,  2 -4)
x2
 4   2 6 3
x2
2x  4
x2


0  2
24  4
4  2



2
12
2
 2
The expression is negative
for all x in (-2,2).
  2, 2  .
6
The expression is positive
for all x in (2,
inf).
 2, 
.
2


2
2
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Example 2: Solving and Graphing
Inequalities (Cont.)
x6
x2
1 0
We can conclude that the solution for the above
inequality in interval notation is:   ,  2    2,   .
In algebraic notation: x   2 or x  2.
Graphically:

2
(
2
Note: 2 from the denominator is not included in the
solution set because the quotient is undefined there.