Section 2.3
2.3
Quadratic
Equations,
Functions,
and Models
•
Quadratic Equations, Functions, and Models
191
Find zeros of quadratic functions and solve quadratic equations by using
the principle of zero products, by using the principle of square roots, by
completing the square, and by using the quadratic formula.
Solve equations that are reducible to quadratic.
Solve applied problems using quadratic equations.
Quadratic Equations and Quadratic Functions
In this section, we will explore the relationship between the solutions of
quadratic equations and the zeros of quadratic functions. We define quadratic equations and functions as follows.
Quadratic Equations
A quadratic equation is an equation equivalent to
ax 2 bx c 0, a 0,
where a, b, and c are real numbers.
Quadratic Functions
A quadratic function f is a function that can be written in the form
fx ax 2 bx c ,
a 0,
where a, b, and c are real numbers.
zeros of a function
review section 2.1.
A quadratic equation written in the form ax 2 bx c 0 is said to be
in standard form.
The zeros of a quadratic function fx ax 2 bx c are the solutions of the associated quadratic equation ax 2 bx c 0. (These solutions are sometimes called roots of the equation.) Quadratic functions
can have real-number or imaginary-number zeros and quadratic equations can have real-number or imaginary-number solutions. If the zeros
or solutions are real numbers, they are also the first coordinates of the
x-intercepts of the graph of the quadratic function.
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Chapter 2
• Functions, Equations, and Inequalities
The following principles allow us to solve many quadratic equations.
Equation-Solving Principles
The Principle of Zero Products: If ab 0 is true, then a 0 or
b 0, and if a 0 or b 0, then ab 0.
The Principle of Square Roots: If x 2 k , then x k or
x k .
factoring trinomials
EXAMPLE 1 Solve: 2x 2 x 3.
review section R.4.
Algebraic Solution
Visualizing the Solution
We have
2x 2 x 3
2x 2 x 3 0
Subtracting 3 on
both sides
x 1 2x 3 0
Factoring
x10
or 2x 3 0
x 1 or
x 1 or
CHECK :
2x 3
3
x 2.
Using the
principle of
zero products
For x f(x) 2x 2 x 3
y
5
4
3
For x 1:
2x 2 x 3
212 1 ? 3
211
21
3 3
The solutions of the equation 2x 2 x 3, or the
equivalent equation 2x 2 x 3 0, are the
zeros of the function fx 2x 2 x 3. They
are also the first coordinates of the x-intercepts
of the graph of fx 2x 2 x 3.
2
(1, 0)
1
3
(−,
0)
2
5 4 3 2 1
1
1
2
3
4
5 x
2
TRUE
3
4
5
3
2:
2x 2 x 3
3
The solutions are 1 and 2 .
2 2 2 2 ? 3
3
3
9
3
9
2
2
242
3
6
2
3
3
The solutions are 1 and
TRUE
3
2.
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Section 2.3
• Quadratic Equations, Functions, and Models
193
EXAMPLE 2 Solve: 2x 2 10 0.
Algebraic Solution
Visualizing the Solution
We have
2x 2 10 0
2x 2 10
The solutions of the equation 2x 2 10 0 are
the zeros of the function fx 2x 2 10. Note
that they are also the first coordinates of the
x-intercepts of the graph of fx 2x 2 10.
Adding 10 on
both sides
x2 5
Dividing by 2 on
both sides
x 5 or x 5.
Using the principle
of square roots
y
6
4
2x 2 10 0
CHECK :
2 5 2 10 ? 0
2 5 10
10 10
0 0
(5, 0)
2
5 4 3 2 1
2
We can check
both solutions
at once.
(5, 0)
1
2
3
4
5 x
4
6
TRUE
8
The solutions are 5 and 5, or 5.
10
f(x) 2x 2 10
The solutions are 5 and 5.
f (x) x 2 3x 4
y
(1, 0)
5
4
3
2
1
5432
(4, 0)
1 2 3
5 x
We have seen that some quadratic equations can be solved by factoring
and using the principle of zero products. For example, consider the equation
x 2 3x 4 0:
x 2 3x 4 0
x 1 x 4 0
x10
or x 4 0
x 1 or
4
5
6
7
x 4.
Factoring
Using the principle of
zero products
The equation x 3x 4 0 has two real-number solutions, 1 and 4.
These are the zeros of the associated quadratic function fx x 2 3x 4
and the first coordinates of the x-intercepts of the graph of this function.
(See Fig. 1.)
2
Two real-number zeros
Two x-intercepts
FIGURE 1
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194
Chapter 2
• Functions, Equations, and Inequalities
g(x) x 2 6x 9
y
x 2 6x 9 0
x 3 x 3 0
x 3 0 or x 3 0
9
8
7
6
5
4
3
2
1
21
1
Next, consider the equation x 2 6x 9 0. Again, we factor and use
the principle of zero products:
x 3 or
1 2 3 4 5 6 7 8 9 x
(3, 0)
One real-number zero
One x-intercept
FIGURE 2
x 3.
Factoring
Using the principle of
zero products
The equation x 2 6x 9 0 has one real-number solution, 3. It is the
zero of the quadratic function gx x 2 6x 9 and the first coordinate
of the x-intercept of the graph of this function. (See Fig. 2.)
The principle of square roots can be used to solve quadratic equations
like x 2 13 0:
x 2 13 0
x 2 13
x 13
x 13i.
Using the principle of square roots
The equation has two imaginary-number solutions, 13i and 13i .
These are the zeros of the associated quadratic function hx x 2 13.
Since the zeros are not real numbers, the graph of the function has no
x-intercepts. (See Fig. 3.)
h(x) x 2 13
y
20
18
16
12
10
8
6
4
2
54321
1 2 3 4 5 x
No real-number zeros
No x-intercepts
FIGURE 3
Completing the Square
Neither the principle of zero products nor the principle of square roots would
yield the exact zeros of a function like fx x 2 6x 10 or the exact
solutions of the associated equation x 2 6x 10 0. If we wish to find
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Section 2.3
Technology
Connection
Approximations for the zeros
of the quadratic function
fx x 2 6x 10 in
Example 3 can be found using
the Zero method.
y x2 6x 10
10
10
10
EXAMPLE 3 Find the zeros of fx x 2 6x 10 by completing the
square.
Solution We find the values of x for which fx 0. That is, we solve the
associated equation x 2 6x 10 0. Our goal is to find an equivalent
equation of the form x 2 bx c d in which x 2 bx c is a perfect
square. Since
b
2
2
x
x 2 6x 10 0
x 2 6x
10
x 2 6x 9 10 9
y x2 6x 10
10
10
x 6x 9 19.
2
Zero
X 1.358899 Y 0
20
(3 √19, 0)
5432
Yscl 5
x 32 19
x 3 19
x 3 19.
2
4
6
10
12
14
16
18
20
(3 √19, 0)
1 2 3 4 5 6
2
,
Adding 10
Adding 9 to complete the square:
b 2
6 2
32 9
2
2
Because x 2 6x 9 is a perfect square, we are able to write it as x 32,
the square of a binomial. We can then use the principle of square roots to
finish the solution:
y
6
4
2
b
2
the number c is found by taking half the coefficient of the x-term and
squaring it. Then for the equation x 2 6x 10 0, we have
Yscl 5
10
195
exact zeros or solutions, we can use a procedure called completing the
square and then use the principle of square roots. (Recall that we completed the square in order to write the equation of a circle in standard
form in Section 1.1. Here we complete the square in order to solve quadratic equations.)
x 2 bx Zero
X 7.3588989
Y0
20
• Quadratic Equations, Functions, and Models
8 9 10 x
Factoring
Using the principle of square roots
Adding 3
Therefore, the solutions of the equation are 3 19 and 3 19, or
simply 3 19. The zeros of fx x 2 6x 10 are also 3 19
and 3 19, or 3 19.
Decimal approximations for 3 19 can be found using a calculator:
3 19 7.359 and 3 19 1.359.
f (x) x 2 6x 10
The zeros are approximately 7.359 and 1.359.
Before we can complete the square, the coefficient of the x 2-term
must be 1. When it is not, we divide both sides of the equation by the
x 2-coefficient.
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196
Chapter 2
• Functions, Equations, and Inequalities
EXAMPLE 4 Solve: 2x 2 1 3x .
Solution
We have
2x 2 1 3x
2x 3x 1 0
2
Study Tip
The examples in the text are
carefully chosen to prepare you
for success with the exercise sets.
Study the step-by-step solutions
of the examples, noting that
substitutions and explanations
appear in red. The time you
spend studying the examples
will save you valuable time
when you do your homework.
Subtracting 3x. We are unable to
factor the result.
2x 2 3x
1
3
1
x2 x
2
2
3
9
1
9
x2 x 2
16
2
16
3 2 17
x
4
16
3
17
x 4
4
3
17
x 4
4
3 17
.
x
4
Adding 1
Dividing by 2 to make the
x 2-coefficient 1
Completing the square: 2 2 4
2
and 34 169 ; adding 169
1
3
3
Factoring and simplifying
Using the principle of square roots
and the quotient rule for radicals
Adding 34
The solutions are
3 17
3 17
and
,
4
4
or
3 17
.
4
To solve a quadratic equation by completing the square:
1. Isolate the terms with variables on one side of the equation and
arrange them in descending order.
2. Divide by the coefficient of the squared term if that coefficient is
not 1.
3. Complete the square by taking half the coefficient of the firstdegree term and adding its square on both sides of the equation.
4. Express one side of the equation as the square of a binomial.
5. Use the principle of square roots.
6. Solve for the variable.
Using the Quadratic Formula
Because completing the square works for any quadratic equation, it can
be used to solve the general quadratic equation ax 2 bx c 0 for x.
The result will be a formula that can be used to solve any quadratic equation quickly.
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Section 2.3
• Quadratic Equations, Functions, and Models
197
Consider any quadratic equation in standard form:
ax 2 bx c 0,
a 0.
For now, we assume that a 0 and solve by completing the square. As the
steps are carried out, compare them with those of Example 4.
ax 2 bx c 0
ax 2 bx c
b
c
x2 x a
a
Technology
Connection
We can solve the equation
3x 2 2x 7 in Example 5
using the Intersect method.
We graph y1 3x 2 2x and
y2 7 and use the INTERSECT
feature to find the coordinates
of the points of intersection.
The first coordinates of these
points are the solutions of
the equation y1 y2, or
3x 2 2x 7.
y1 3x2 2x, y2 7
y1
10
Half of
x2 b
b
b
is and
a 2a
2a
Dividing by a
b2
b2
.
Thus
we
add
:
4a2
4a2
b 2 4ac
4a2
b
b 2 4ac
x
2a
2a
x
b
2a
2
Intersection
X 1.896805
1
10
Adding b
2a
The Quadratic Formula
The solutions of ax 2 bx c 0, a 0, are given by
y1
x
Y7
Using the principle of square roots
and the quotient rule for radicals.
Since a 0, 4a 2 2a.
It can also be shown that this result holds if a 0.
y2
Intersection
X 1.2301386
b2
to complete the square
4a 2
Factoring and finding a common
denominator:
4a c
4ac
c
2
a
4a a
4a
b
b 2 4ac
2a
2a
2
b b 4ac
.
x
2a
5
Y7
Adding
x
y1 3x2 2x, y2 7
5
Adding c
b
c
b2
b2
x 2 2
a
4a
a
4a
2
b
4ac
b2
x
2 2
2a
4a
4a
y2
5
2
Standard form
b b 2 4ac
.
2a
5
1
The solutions are
approximately 1.897 and
1.230. We could also write the
equation in standard form,
3x 2 2x 7 0, and use
the Zero method.
EXAMPLE 5 Solve 3x 2 2x 7. Find exact solutions and approximate
solutions rounded to the nearest thousandth.
Solution After finding standard form, we are unable to factor, so we
identify a, b, and c in order to use the quadratic formula:
3x 2 2x 7 0;
a 3, b 2, c 7.
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Chapter 2
• Functions, Equations, and Inequalities
We then use the quadratic formula:
b b 2 4ac
2a
2 22 43 7
Substituting
23
2 4 84 2 88
6
6
2 4 22 2 222 2 1 22 6
6
23
2 1 22 1 22
.
2
3
3
x
The exact solutions are
1 22
1 22
and
.
3
3
Using a calculator, we approximate the solutions to be 1.897 and 1.230.
EXAMPLE 6
Solve: x 2 5x 8 0.
Algebraic Solution
To find the solutions, we use the quadratic
formula. Here
a 1, b 5,
b b 2 4ac
x
2a
5 52 41 8
21
5 7
2
5 7i
.
2
The solutions are c 8;
Visualizing the Solution
The graph of the function fx x 2 5x 8
has no x-intercepts.
f(x) x 2 5x 8
y
Substituting
10
8
Simplifying
6
4
2
8 6 4 2
2
2
4
6
8 x
5
5
7
7
i and i.
2
2
2
2
Thus the function has no real-number zeros
and there are no real-number solutions of the
associated equation x 2 5x 8 0.
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Section 2.3
• Quadratic Equations, Functions, and Models
199
The Discriminant
From the quadratic formula, we know that the solutions x1 and x2 of a quadratic equation are given by
x1 b b 2 4ac
and
2a
x2 b b 2 4ac
.
2a
The expression b 2 4ac shows the nature of the solutions. This expression
is called the discriminant. If it is 0, then it makes no difference whether we
b
choose the plus or the minus sign in the formula. That is, x1 x2,
2a
so there is just one solution. In this case, we sometimes say that there is one
repeated real solution. If the discriminant is positive, there will be two real
solutions. If it is negative, we will be taking the square root of a negative
number; hence there will be two imaginary-number solutions, and they will
be complex conjugates.
Discriminant
For ax 2 bx c 0:
b 2 4ac 0
b 2 4ac 0
b 2 4ac 0
One real-number solution;
Two different real-number solutions;
Two different imaginary-number solutions,
complex conjugates.
In Example 5, the discriminant, 88, is positive, indicating that there are
two different real-number solutions. If the discriminant is negative, as it is
in Example 6, we know that there are two different imaginary-number
solutions.
Equations Reducible to Quadratic
Some equations can be treated as quadratic, provided that we make a
suitable substitution. For example, consider the following:
x 4 5x 2 4 0
x 2 2 5x 2 4 0
u 2 5u 4 0.
x4 x 22
Substituting u for x 2
The equation u 2 5u 4 0 can be solved for u by factoring or using
the quadratic formula. Then we can reverse the substitution, replacing u
with x 2, and solve for x. Equations like the one above are said to be
reducible to quadratic, or quadratic in form.
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Chapter 2
• Functions, Equations, and Inequalities
EXAMPLE 7 Solve: x 4 5x 2 4 0.
Algebraic Solution
We let u x 2 and substitute:
Visualizing the Solution
Substituting u for x 2
u 2 5u 4 0
u 1 u 4 0
Factoring
u 1 0 or u 4 0
Using the
The solutions of the given equation are the zeros
of fx x 4 5x 2 4. Note that the zeros
occur at the x-values 2, 1, 1, and 2.
u1
or
u 4.
principle of
zero products
f(x) x 4 5x 2 4
y
Don’t stop here! We must solve for the original
variable. We substitute x 2 for u and solve for x:
5
x2 1
or x 2 4
x 1 or x 2.
3
4
2
Using the principle of
square roots
1
5 4 3 2 1
1
The solutions are 1, 1, 2, and 2.
1
2
3
4
5 x
2
3
4
5
Technology
Connection
We can use the Zero method to solve the equation in Example 7,
x 4 5x 2 4 0. We graph the function y x 4 5x 2 4 and
use the ZERO feature to find the zeros.
y x 4 5x 2 4
5
5
5
Zero
X 2
Y0
5
The leftmost zero is 2. Using the ZERO feature three more times,
we find that the other zeros are 1, 1, and 2.
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Section 2.3
• Quadratic Equations, Functions, and Models
201
Applications
Some applied problems can be translated to quadratic equations.
EXAMPLE 8 Time of a Free Fall. The Petronas Towers in Kuala Lumpur,
Malaysia, are 1482 ft tall. How long would it take an object dropped from the
top to reach the ground?
Solution
1. Familiarize. The formula s 16t 2 is used to approximate the distance s, in feet, that an object falls freely from rest in t seconds. In this
case, the distance is 1482 ft.
2. Translate. We substitute 1482 for s in the formula:
1482 16t 2.
3. Carry out. We use the principle of square roots:
1482 16t 2
1482
t2
16
1482
t
16
9.624 t .
Dividing by 16
Taking the positive square root. Time
cannot be negative in this application.
4. Check. In 9.624 sec, a dropped object would travel a distance of
169.6242, or about 1482 ft. The answer checks.
5. State. It would take about 9.624 sec for an object dropped from the top
of the Petronas Towers to reach the ground.
EXAMPLE 9 Bicycling Speed. Logan and Cassidy leave a campsite,
Logan biking due north and Cassidy biking due east. Logan bikes 7 kmh
slower than Cassidy. After 4 hr, they are 68 km apart. Find the speed
of each bicyclist.
N
68 km
E
Campsite
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Chapter 2
• Functions, Equations, and Inequalities
Solution
1. Familiarize. We let r Cassidy’s speed, in kilometers per hour. Then
r 7 Logan’s speed, in kilometers per hour. We will use the motion
formula d rt , where d is the distance, r is the rate (or speed), and t
is the time. Then, after 4 hr, Cassidy has traveled 4r km and Logan
has traveled 4r 7 km. We add these distances to the drawing, as
shown below.
68
4(r 7)
4r
the pythagorean theorem
review section R.6.
2. Translate. We use the Pythagorean theorem, a2 b 2 c 2, where a and
b are the lengths of the legs of a right triangle and c is the length of the
hypotenuse:
4r2 4r 7 2 68 2.
3. Carry out. We solve the equation:
4r2 4r 7 2 68 2
16r 2 16r 2 14r 49 4624
16r 2 16r 2 224r 784 4624
32r 2 224r 3840 0
Subtracting 4624
2
r 7r 120 0
Dividing by 32
r 8 r 15 0
Factoring
r80
or r 15 0
Principle of
r 8
or
r 15.
zero products
4. Check. Since speed cannot be negative, we need to check only 15. If
Cassidy’s speed is 15 kmh, then Logan’s speed is 15 7, or 8 kmh. In
4 hr, Cassidy travels 4 15, or 60 km, and Logan travels 4 8, or 32 km.
Then they are 60 2 322, or 68 km apart. The answer checks.
5. State. Cassidy’s speed is 15 kmh, and Logan’s speed is 8 kmh.
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Section 2.3
• Quadratic Equations, Functions, and Models
CONNECTING THE CONCEPTS
zeros, solutions, and intercepts
The zeros of a function y fx are also the solutions of the equation
fx 0, and the real-number zeros are the first coordinates of the
x-intercepts of the graph of the function.
zeros of the function;
solutions of the equation
function
Linear Function
fx 2x 4, or
y 2x 4
To find the zero of fx, we
solve fx 0:
2x 4 0
2x 4
x 2.
The solution of 2x 4 0
is 2. This is the zero of the
function fx 2x 4. That
is, f2 0.
x-intercepts of the graph
The zero of fx is the first
coordinate of the x-intercept
of the graph of y fx.
y
f(x) 2x 4
6
4
2
6 4 2
4
6 x
x-intercept
(2, 0)
2
4
Quadratic Function
gx x 2 3x 4, or
y x 2 3x 4
To find the zeros of gx, we
solve gx 0:
x 2 3x 4 0
x 1 x 4 0
x10
or x 4 0
x 1 or
x 4.
The solutions of x 2 3x 4 0 are 1 and 4. They are
the zeros of the function gx.
That is, g1 0 and
g4 0.
The real-number zeros of gx
are the first coordinates of the
x-intercepts of the graph of
y gx.
y
6
g(x) x2 3x 4
4
(1, 0)
2
4 2
(4, 0)
2
4
8 x
x-intercepts
6
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203