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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 2  bx  c  0
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.
3x 2
x  
5 5
2
Step 1: Multiply both
sides by 5 .
Step 2: Subtract 2 from
both sides so 0 is
on one side.
Step 3: Factor and solve the
two linear equations.
5 x 2  3x  2
5 x 2  3x  2  0
(5 x  2)( x  1)  0
5x  2  0
2
x
5
or
x 1  0
or
x 1
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Example 2: Solving Quadratic Equations by
Factoring
Solve the quadratic equation by factoring.
x 2  16  8 x
x 2  8 x  16  0
 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 2  18 x  0
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
2
form A  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)2  3
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
ax 2  bx  c.
Step 2: Divide by a if a  1 , so that the coefficient of x 2
b
c
2
is 1: x  x   .
a
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.
2 x 2  8 x  10  0
Step 1: Move the constant
term to the other side
2 x 2  8 x  10
of the equation.
Step 2: Divide by 2 , the
x2  4 x  5
coefficient of x.
2
x  4x  4  5  4
Step 3: Add 4 to both sides.
Step 4: Factor the trinomial,
( x  2)2  9
and solve.
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 2  24 x  216  0
2 x 2  24 x  216
x 2  12 x  108
x 2  12 x  36  108  36
 x  6
 144
x  6  12
x  6  12
x  6, 18
2
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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.
Step 1: Move the constant to
the other side of the
equation.
Step 2: Divide by a .
b
Step 3: Divide a by 2
square the result, and
add it to both sides of
the equation.
ax 2  bx  c  0
ax 2  bx  c
b
c
x  x
a
a
2
2
2
b
b
c
b
x2  x  2    2
a
4a
a 4a
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The Quadratic Formula
Deriving the quadratic formula (cont.).
2
Step 4: Factor the
b 
4ac b 2

trinomial, and
x
  2  2
2a 
4a
4a

solve for x .
2
2
b  b  4ac

x
 
2
2
a
4
a


b
b 2  4ac
x

2a
2a
b  b 2  4ac
x
2a
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The Quadratic Formula
The solutions of the equation ax 2  bx  c  0 are:
b  b 2  4ac
x
2a
Note:
• The equation has real solutions if b2  4ac  0.
• The equation has a double solution if b2  4ac  0.
• The equation has complex solutions (which are
conjugates of one another) if b2  4ac  0 .
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Example 8: The Quadratic Formula
Solve using the quadratic formula.
4 x 2  7 x  15
c
a2 b
4x  7 x  15  0
b  b 2  4ac
x
2a
x
  7  
7  289
x
8
7  17
x
8
5
x  3,
4
 7   4  4  15
2  4
2
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Example 9: The Quadratic Formula
Solve using the quadratic formula.
t 2  22t  96  0
t
22  222  4 1 96 
2 1
22  100
t
2
22  10
t
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 ,
1 2
h   gt  v0t  h0 ,
2
where g , v0 , and h0 are all constants:
2
2
g

gravitat
ional
fo
r
ce
(32.2
f
t/s
,
or
9.81
m
/s
)
•
• v0  initial velocity
• h0  initial height
• t  time, measured in seconds
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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  16t 2  60t  50
0  8t 2  30t  25
t
Since we know the
answer cannot be
negative, this
solution is discarded.
30 
 30 
2
 4  8  25 
2 8
30  10 17
t
16
t  0.70,4.45