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Hawkes Learning Systems:
College Algebra
Section 2.3: Quadratic Equations in One
Variable
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Objectives
o Solving quadratic equations by factoring.
o Solving “perfect square” quadratic equations.
o Solving quadratic equations by completing the
square.
o The quadratic formula.
o Interlude: gravity problems.
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Quadratic Equations
A quadratic equation in one variable, say the variable
x is an equation that can be transformed into the form
ax  bx  c  0
2
Where a , b , and c are real numbers and a  0 . Such
equations are also called second-degree equations, as
x appears to the second power. The name quadratic
comes from the Latin word quadrus, meaning
“square”.
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Solving Quadratic Equations by Factoring
Zero –Factor Property
Let A and B represent algebraic expressions. If
the product of A and B is 0 , then at least one of
A and B itself is 0 . That is,
AB  0  A  0 or B  0
For example,
x   x  1   0  x  0 or
 x  1  0
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Solving Quadratic Equations by Factoring
o The key to using factoring to solve a quadratic
equation is to rewrite the equation so that 0 appears
by itself on one side of the equation.
o If the trinomial ax  bx  c can be factored, it can be
written as a product of two linear factors A and B .
2
o The Zero-Factor Property then implies that the only
way for ax 2  bx  c to be 0 is if one (or both) of A
and B is 0 .
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Example 1: Solving Quadratic Equations by
Factoring
Solve the quadratic equation by factoring.
x 
2
Step 1: Multiply both
sides by 5 .
Step 2: Subtract 2 from
both sides so 0 is
on one side.
Step 3: Factor and solve the
two linear equations.
3x

5
2
5
5x  3x  2
2
5x  3x  2  0
2
(5 x  2)( x  1)  0
5x  2  0
x
2
5
or
x 1  0
or
x 1
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Example 2: Solving Quadratic Equations by
Factoring
Solve the quadratic equation by factoring.
x  16  8 x
2
x  8 x  16  0
2
 x  4
x40
2
0
or
x40
x4
In this example, the two linear factors are the same. In
such cases, the single solution is called a double solution
or a double root.
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Example 3: Solving Quadratic Equations by
Factoring
Solve the quadratic equation by factoring.
6 x  18 x  0
2
6 x ( x  3)  0
6x  0
x0
x30
x  3
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Solving “Perfect Square” Quadratic Equations
In some cases where the factoring method is
unsuitable, the solution to a second-degree polynomial
can be obtained by using our knowledge of square
roots.
If A is an algebraic expression and if c is a constant:
A  c implies A   c
2
If a given quadratic equation can be written in the
form A 2  c we can use the above observation to
obtain two linear equations that can be easily solved.
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Example 4: “Perfect Square” Quadratic
Equations
Solve the quadratic equation by taking square roots.
(5 x  2)  3
2
Step 1: Take the square root
of each side.
Step 2: Subtract 2 from
both sides.
Step 3: Divide both sides by
5.
5x  2   3
5 x  2 
3
2 3
x
5
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Example 5: “Perfect Square” Quadratic
Equations
Solve the quadratic equation by taking square roots.
x  7
2
 25  0
x  7
2
  25
x  7    25
x  7   5i
x  7  5i
In this example, taking square roots leads to two
complex number solutions.
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Solving Quadratic Equations by Completing the
Square
Method of 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 2
is 1 : x 
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 6: Completing the Square
Solve the quadratic equation by completing the square.
Step 1: Move the constant
term to the other side
of the equation.
Step 2: Divide by 2 , the
coefficient of x.
Step 3: Add 4 to both sides.
Step 4: Factor the trinomial,
and solve.
2 x  8 x  10  0
2
2 x  8 x  10
2
x  4x  5
2
x  4x  4  5  4
2
( x  2)  9
2
x  2  3
x  2  3
x   5, 1
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Example 7: Completing the Square
Solve the quadratic equation by completing the square.
2 x  24 x  216  0
2
2 x  24 x  216
2
x  12 x  108
2
x  12 x  36  108  36
2
 x  6
2
 144
x  6   12
x   6  12
x  6,  18
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The Quadratic Formula
The method of completing the square can be used to derive
the quadratic formula, a formula that gives the solution to
any equation of the form ax 2  bx  c  0 .
ax  bx  c  0
2
Step 1: Move the constant to
the other side of the
equation.
ax  bx   c
2
Step 2: Divide by a .
x 
2
x
a
b
Step 3: Divide a by 2
square the result, and
add it to both sides of
the equation.
b
x 
2
b
a
x
b
c
a
2
4a
2

c
a

b
2
4a
2
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The Quadratic Formula
Deriving the quadratic formula (cont.).
2
Step 4: Factor the
2
b 
4ac
b


trinomial, and
x
 
2
2
2a 
4a
4a

solve for x .
2
b 
b  4ac

x
 
2
2
a
4
a


x
b
2
b  4 ac
2

2a
x
2a
b 
b  4 ac
2
2a
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The Quadratic Formula
The solutions of the equation
x
b 
ax  bx  c  0
2
are:
b  4 ac
2
2a
Note:
• The equation has real solutions if b 2  4 a c  0 .
• The equation has a double solution if b 2  4 a c  0.
• The equation has complex solutions (which are
conjugates of one another) if b 2  4 ac  0 .
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Example 8: The Quadratic Formula
Solve using the quadratic formula.
x
b 
b  4ac
2
2a
4 x  7 x  15
2
c
a
b
2
4 x  7 x  15  0
x
x
  7  
7
289
8
x
7  17
x  3,
8
5
4
 7   4  4   15 
24
2
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Example 9: The Quadratic Formula
Solve using the quadratic formula.
t  22 t  96  0
2
t
t
 22 
22  4 1   96 
2
2 1 
 22 
100
2
t
 22  10
2
t   6,  16
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Interlude: Gravity Problems
o When an object near the surface of the Earth is
moving under the influence of gravity alone, its
height above the surface of the earth is described by
a quadratic polynomial in the variable t where t
stands for time and is usually measured in seconds.
o In some applications involving this formula, one of
the two solutions must be discarded as meaningless
in the given problem.
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Interlude: Gravity Problems
If we let h represent the height at time t ,
h
1
2
g t  v 0 t  h0 ,
2
where g , v 0 , and h0 are all constants:
• g  gravitat ional fo r ce (32.2 f t/s , or 9.81 m /s )
2
• v 0  initial velocity
• h0  initial height
• t  tim e, m easured in seconds
2
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Example 1: Gravity Problems
Luke stands on a tier of seats in a baseball stadium, and
throws a ball out onto the field with a vertical upward velocity
of 60 ft/s . The ball is 50 ft above the ground at the moment he
releases it. When does the ball land?
0   16 t  60 t  50
2
0  8 t  30 t  25
2
t
Since we know the
answer cannot be
negative, this
solution is discarded.
t
30 
  30 
2
 4  8    25 
2 8 
30  10 17
16
t   0.70, 4.45