Chapter 16
Quadratic
Equations
Chapter Sections
16.1 – Solving Quadratic Equations by the Square Root
Property
16.2 – Solving Quadratic Equations by Completing the
Square
16.3 – Solving Quadratic Equations by the Quadratic
Formula
16.4 – Graphing Quadratic Equations in Two Variables
16.5 – Interval Notation, Finding Domains and Ranges from
Graphs and Graphing Piecewise-Defined Functions
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§ 16.1
Solving Quadratic
Equations by the Square
Root Property
Square Root Property
We previously have used factoring to solve
quadratic equations.
This chapter will introduce additional
methods for solving quadratic equations.
Square Root Property
If b is a real number and a2 = b, then
a b
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Square Root Property
Example
Solve x2 = 49
x   49  7
Solve 2x2 = 4
x2 = 2
x 2
Solve (y – 3)2 = 4
y  3   4  2
y=32
y = 1 or 5
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Square Root Property
Example
Solve x2 + 4 = 0
x2 = 4
There is no real solution because the square root
of 4 is not a real number.
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Square Root Property
Example
Solve (x + 2)2 = 25
x  2   25  5
x = 2 ± 5
x = 2 + 5 or x = 2 – 5
x = 3 or x = 7
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Square Root Property
Example
Solve (3x – 17)2 = 28
3x – 17 =  28   2 7
3x  17  2 7
17  2 7
x
3
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§ 16.2
Solving Quadratic
Equations by Completing
the Square
Completing the Square
In all four of the previous examples, the constant in the
square on the right side, is half the coefficient of the x
term on the left.
Also, the constant on the left is the square of the
constant on the right.
So, to find the constant term of a perfect square
trinomial, we need to take the square of half the
coefficient of the x term in the trinomial (as long as the
coefficient of the x2 term is 1, as in our previous
examples).
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Completing the Square
Example
What constant term should be added to the following
expressions to create a perfect square trinomial?
x2 – 10x
add 52 = 25
x2 + 16x
add 82 = 64
x2 – 7x
2
49
7
add   
4
2
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Completing the Square
Example
We now look at a method for solving quadratics
that involves a technique called completing the
square.
It involves creating a trinomial that is a perfect
square, setting the factored trinomial equal to a
constant, then using the square root property
from the previous section.
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Completing the Square
Solving a Quadratic Equation by Completing
a Square
1) If the coefficient of x2 is NOT 1, divide both
sides of the equation by the coefficient.
2) Isolate all variable terms on one side of the
equation.
3) Complete the square (half the coefficient of the
x term squared, added to both sides of the
equation).
4) Factor the resulting trinomial.
5) Use the square root property.
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Solving Equations
Example
Solve by completing the square.
y2 + 6y = 8
y2 + 6y + 9 = 8 + 9
(y + 3)2 = 1
y+3=± 1=±1
y = 3 ± 1
y = 4 or 2
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Solving Equations
Example
Solve by completing the square.
y2 + y – 7 = 0
y2 + y = 7
y2 + y + ¼ = 7 + ¼
(y +
½)2
=
29
4
1
29
29
y 

2
4
2
1
29  1  29
y 

2
2
2
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Solving Equations
Example
Solve by completing the square.
2x2 + 14x – 1 = 0
2x2 + 14x = 1
x2 + 7x = ½
x2 + 7x +
49
4
=½+
49
4
=
51
4
7 2
51
(x + ) =
2
4
x
7
51
51


2
4
2
7
51  7  51
x 

2
2
2
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§ 16.3
Solving Quadratic
Equations by the
Quadratic Formula
The Quadratic Formula
Another technique for solving quadratic
equations is to use the quadratic formula.
The formula is derived from completing the
square of a general quadratic equation.
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The Quadratic Formula
A quadratic equation written in standard
form, ax2 + bx + c = 0, has the solutions.
 b  b  4ac
x
2a
2
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The Quadratic Formula
Example
Solve 11n2 – 9n = 1 by the quadratic formula.
11n2 – 9n – 1 = 0, so
a = 11, b = -9, c = -1
9  (9)  4(11)( 1) 9  81  44 9  125
n



22
22
2(11)
2
95 5
22
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The Quadratic Formula
Example
Solve
1
8
x2
+x–
5
2
= 0 by the quadratic formula.
x2 + 8x – 20 = 0 (multiply both sides by 8)
a = 1, b = 8, c = 20
 8  (8) 2  4(1)( 20)  8  64  80  8  144
x



2(1)
2
2
 8  12 20
4

or ,  10 or 2
2
2
2
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The Quadratic Formula
Example
Solve x(x + 6) = 30 by the quadratic formula.
x2 + 6x + 30 = 0
a = 1, b = 6, c = 30
 6  (6)  4(1)(30)  6  36  120  6   84
x


2
2
2(1)
2
So there is no real solution.
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The Discriminant
The expression under the radical sign in the
formula (b2 – 4ac) is called the discriminant.
The discriminant will take on a value that is
positive, 0, or negative.
The value of the discriminant indicates two
distinct real solutions, one real solution, or no
real solutions, respectively.
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The Discriminant
Example
Use the discriminant to determine the number and
type of solutions for the following equation.
5 – 4x + 12x2 = 0
a = 12, b = –4, and c = 5
b2 – 4ac = (–4)2 – 4(12)(5)
= 16 – 240
= –224
There are no real solutions.
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Solving Quadratic Equations
Steps in Solving Quadratic Equations
1) If the equation is in the form (ax+b)2 = c, use
the square root property to solve.
2) If not solved in step 1, write the equation in
standard form.
3) Try to solve by factoring.
4) If you haven’t solved it yet, use the quadratic
formula.
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Solving Equations
Example
Solve 12x = 4x2 + 4.
0 = 4x2 – 12x + 4
0 = 4(x2 – 3x + 1)
Let a = 1, b = -3, c = 1
3  (3)  4(1)(1) 3  9  4 3  5
x


2
2
2(1)
2
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Solving Equations
Example
Solve the following quadratic equation.
5 2
1
m m 0
8
2
5m 2  8m  4  0
(5m  2)( m  2)  0
5m  2  0 or m  2  0
2
m  or m  2
5
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§ 16.4
Graphing Quadratic
Equations in Two
Variables
Graphs of Quadratic Equations
We spent a lot of time graphing linear equations
in chapter 3.
The graph of a quadratic equation is a parabola.
The highest point or lowest point on the parabola
is the vertex.
Axis of symmetry is the line that runs through
the vertex and through the middle of the
parabola.
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Graphs of Quadratic Equations
Example
y
Graph y = 2x2 – 4.
x
y
2
4
1
–2
0
–4
–1
–2
–2
4
(–2, 4)
(2, 4)
x
(–1, – 2)
(1, –2)
(0, –4)
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Intercepts of the Parabola
Although we can simply plot points, it is helpful
to know some information about the parabola
we will be graphing prior to finding individual
points.
To find x-intercepts of the parabola, let y = 0 and
solve for x.
To find y-intercepts of the parabola, let x = 0 and
solve for y.
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Characteristics of the Parabola
If the quadratic equation is written in standard
form, y = ax2 + bx + c,
1) the parabola opens up when a > 0 and
opens down when a < 0.
b
2) the x-coordinate of the vertex is  .
2a
To find the corresponding y-coordinate, you
substitute the x-coordinate into the equation
and evaluate for y.
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Graphs of Quadratic Equations
Example
Graph y = –2x2 + 4x + 5.
Since a = –2 and b = 4, the
graph opens down and the
x-coordinate of the vertex
4
is 
1
y
(0, 5)
(1, 7)
(2, 5)
2(2)
x
y
3
–1
2
5
1
7
0
5
–1
–1
(–1, –1)
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x
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§ 16.5
Interval Notation, Finding
Domain and Ranges from
Graphs, and Graphing
Piecewise-Defined Functions
Domain and Range
Recall that a set of ordered pairs is also called
a relation.
The domain is the set of x-coordinates of the
ordered pairs.
The range is the set of y-coordinates of the
ordered pairs.
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Domain and Range
Example
Find the domain and range of the relation {(4,9), (–4,9),
(2,3), (10, –5)}
• Domain is the set of all x-values, {4, –4, 2, 10}
• Range is the set of all y-values, {9, 3, –5}
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Domain and Range
y
Example
Domain
Find the domain and
range of the function
graphed to the right.
Use interval notation.
Domain is [–3, 4]
x
Range
Range is [–4, 2]
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Domain and Range
y
Example
Find the domain and
range of the function
graphed to the right.
Use interval notation.
Range
x
Domain is (– , )
Range is [– 2, )
Domain
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Domain and Range
Example
Find the domain and range of the following relation.
Input (Animal)
Output (Life Span)
• Polar Bear
20
• Cow
• Chimpanzee
15
• Giraffe
• Gorilla
10
• Kangaroo
• Red Fox
7
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Domain and Range
Example continued
Domain is {Polar Bear, Cow, Chimpanzee,
Giraffe, Gorilla, Kangaroo, Red Fox}
Range is {20, 15, 10, 7}
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Graphing Piecewise-Defined Functions
Example
3x  2 if x  0
.
Graph f (x)  
 x  3 if x  0
Graph each “piece” separately.
Values  0.
x
f (x) = 3x – 1
x
f (x) = x + 3
0
– 1(closed circle)
1
4
2
5
3
6
–1 – 4
Values > 0.
–2 – 7
Continued.
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Graphing Piecewise-Defined Functions
Example continued
y
x
f (x) = 3x – 1
0
– 1(closed circle)
–1 – 4
–2 – 7
(3, 6)
Open circle
(0, 3)
(0, –1)
x
f (x) = x + 3
1
4
2
5
3
6
x
(–1, 4)
(–2, 7)
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