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114 • Chapter 1 • Equations, Inequalities, and Mathematical Models
SECTION 1.5 Quadratic Equations
Objectives
Serpico, 1973, starring Al Pacino, is a
movie about police corruption.
In 2000, a police scandal shocked Los Angeles. A police officer who had been
convicted of stealing cocaine held as evidence described how members of his
unit behaved in ways that resembled the gangs they were targeting, assaulting
and framing innocent people.
Is police corruption on the rise? The graph in Figure 1.14 shows the number
of convictions of police officers throughout the United States for seven years.
Convictions of Police Officers
600
Number of Convictions
1. Solve quadratic equations
by factoring.
2. Solve quadratic equations
by the square root
method.
3. Solve quadratic equations
by completing the square.
4. Solve quadratic equations
using the quadratic
formula.
5. Use the discriminant to
determine the number
and type of solutions.
6. Determine the most
efficient method to use
when solving a quadratic
equation.
7. Solve problems modeled
by quadratic equations.
560
500
380
400
300
200
246
143
149
135
83
100
1994 1995 1996 1997 1998 1999 2000
Year
Figure 1.14
Source: F.B.I.
The data can be modeled by the formula
N = 23.4x2 - 259.1x + 815.8
where N is the number of police officers convicted of felonies x years after 1990.
If present trends continue, in which year will 1000 police officers be convicted of
felonies? To answer the question, it is necessary to substitute 1000 for N in the
formula and solve for x, the number of years after 1990:
1000 = 23.4x2 - 259.1x + 815.8.
Do you see how this equation differs from a linear equation? The exponent on
x is 2. Solving such an equation involves finding the set of numbers that make
the equation a true statement. In this section, we study a number of methods for
solving equations in the form ax2 + bx + c = 0. We also look at applications of
these equations.
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Section 1.5 • Quadratic Equations • 115
The General Form of a Quadratic Equation
We begin by defining a quadratic equation.
Definition of a Quadratic Equation
A quadratic equation in x is an equation that can be written in the general
form
ax2 + bx + c = 0
where a, b, and c are real numbers, with a Z 0. A quadratic equation in x is
also called a second-degree polynomial equation in x.
An example of a quadratic equation in general form is x2 - 7x + 10 = 0.
The coefficient of x2 is 1(a = 1), the coefficient of x is -7(b = -7), and the
constant term is 10(c = 10).
1
Solve quadratic equations
by factoring.
Solving Quadratic Equations by Factoring
We can factor the left side of the quadratic equation x2 - 7x + 10 = 0. We obtain
(x - 5)(x - 2) = 0. If a quadratic equation has zero on one side and a factored
expression on the other side, it can be solved using the zero-product principle.
The Zero-Product Principle
If the product of two algebraic expressions is zero, then at least one of the
factors is equal to zero.
If AB = 0,
then A = 0 or B = 0.
For example, consider the equation (x - 5)(x - 2) = 0. According to the zeroproduct principle, this product can be zero only if at least one of the factors is zero.
We set each individual factor equal to zero and solve each resulting equation for x.
(x - 5)(x - 2) = 0
x - 5 = 0
x - 2 = 0
or
x = 5
x = 2
We can check each of these proposed solutions in the original quadratic
equation, x2 - 7x + 10 = 0.
Check 5:
52 - 7 5 + 10 0
25 - 35 + 10 0
0 = 0
✓
Check 2:
22 - 7 2 + 10 0
4 - 14 + 10 0
0 = 0
✓
The resulting true statements, indicated by the checks, show that the solutions
are 5 and 2. The solution set is {5, 2}. Note that with a quadratic equation, we can
have two solutions, compared to the conditional linear equation that had one.
Solving a Quadratic Equation by Factoring
1. If necessary, rewrite the equation in the form ax2 + bx + c = 0,
moving all terms to one side, thereby obtaining zero on the other side.
2. Factor.
(continues on the next page)
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116 • Chapter 1 • Equations, Inequalities, and Mathematical Models
Solving a Quadratic Equation by Factoring (continued)
3. Apply the zero-product principle, setting each factor equal to zero.
4. Solve the equations in step 3.
5. Check the solutions in the original equation.
EXAMPLE 1 Solving Quadratic Equations by Factoring
Solve by factoring:
a. 4x2 - 2x = 0
b. 2x2 + 7x = 4.
Solution
a. We begin with 4x2 - 2x = 0.
Step 1 Move all terms to one side and obtain zero on the other side. All terms
are already on the left and zero is on the other side, so we can skip this step.
Step 2 Factor. We factor out 2x from the two terms on the left side.
4x2 - 2x = 0
2x(2x - 1) = 0
This is the given equation.
Factor.
Steps 3 and 4 Set each factor equal to zero and solve the resulting equations.
2x = 0
or
2x - 1 = 0
x = 0
2x = 1
x =
1
2
Step 5 Check the solutions in the original equation.
Check 0:
4x2 - 2x = 0
4 02 - 2 0 0
0 - 00
0 = 0
The solution set is E0,
✓
Check 12 :
4x2 - 2x
2
4A 12 B - 2A 12 B
4A 14 B - 2A 12 B
1 - 1
0
= 0
0
0
0
= 0
✓
1
2 F.
b. Next, we solve 2x2 + 7x = 4.
Step 1 Move all terms to one side and obtain zero on the other side.
Subtract 4 from both sides and write the equation in general form.
2x2 + 7x = 4
2x2 + 7x - 4 = 4 - 4
2x2 + 7x - 4 = 0
This is the given equation.
Subtract 4 from both sides.
Simplify.
Step 2 Factor.
2x2 + 7x - 4 = 0
(2x - 1)(x + 4) = 0
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Section 1.5 • Quadratic Equations • 117
Steps 3 and 4 Set each factor equal to zero and solve each resulting equation.
2x - 1 = 0
2x = 1
x = 12
x + 4 = 0
x = -4
or
Step 5 Check the solutions in the original equation.
Check 12 :
Check -4:
2
2x2 + 7x = 4
2(-4) + 7(-4) 4
32 + (-28) 4
2x + 7x = 4
+ 7A 12 B 4
1
7 4
2 + 2
2
2A 12 B
4 = 4
The solution set is E-4,
2
✓
4 = 4
1
2 F.
✓
Solve by factoring:
Check
Point
1
a. 3x2 - 9x = 0
b. 2x2 + x = 1.
Technology
x-intercept is −4.
x-intercept is 21 .
[–5, 2, 1] by [–12, 2, 1]
2
Solve quadratic equations
by the square root method.
You can use a graphing utility to check the real solutions of a quadratic
equation. The solutions of ax 2 + bx + c = 0 correspond to the
x-intercepts of the graph of y = ax 2 + bx + c. For example, to check
the solutions of 2x2 + 7x = 4, or 2x2 + 7x - 4 = 0, graph
y = 2x2 + 7x - 4. The cuplike U-shaped graph is shown on the left.
Note that it is important to have all nonzero terms on one side of the
quadratic equation before entering it into the graphing utility. The
x-intercepts are -4 and 12 , and the graph of y = 2x2 + 7x - 4 passes
through (-4, 0) and (12, 0). This verifies that E-4, 12 F is the solution set
of 2x2 + 7x - 4 = 0.
Solving Quadratic Equations by the Square Root Method
Quadratic equations of the form u2 = d, where d 7 0 and u is an algebraic
expression, can be solved by the square root method. First, isolate the squared
expression u2 on one side of the equation and the number d on the other side.
Then take the square root of both sides. Remember, there are two numbers
whose square is d. One number is positive and one is negative.
We can use factoring to verify that u2 = d has two solutions.
u2 = d
u - d = 0
2
u + 2d = 0
Au + 2dBAu - 2dB = 0
u = - 2d
or u - 2d = 0
u = 2d
This is the given equation.
Move all terms to one side and obtain
zero on the other side.
Factor.
Set each factor equal to zero.
Solve the resulting equations.
Because the solutions differ only in sign, we can write them in abbreviated
notation as u = ; 1d . We read this as “u equals positive or negative the square
root of d” or “u equals plus or minus the square root of d.”
Now that we have verified these solutions, we can solve u2 = d directly by
taking square roots. This process is called the square root method.
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118 • Chapter 1 • Equations, Inequalities, and Mathematical Models
The Square Root Method
If u is an algebraic expression and d is a positive real number, then u2 = d
has exactly two solutions:
If u2 = d, then u = 2d or u = - 2d .
Equivalently,
If u2 = d, then u = ; 2d .
EXAMPLE 2 Solving Quadratic Equations
by the Square Root Method
Solve by the square root method:
a. 4x2 = 20
b. (x - 2)2 = 6.
Solution
a. In order to apply the square root method, we need a squared expression
by itself on one side of the equation.
4x2 = 20
We want x 2
by itself.
We can get x2 by itself if we divide both sides by 4.
4x2
20
=
4
4
2
x = 5
Now, we can apply the square root method.
x = ; 25
By checking both values in the original equation, we can confirm that the
solution set is E- 15 , 15 F.
b. (x - 2)2 = 6
The squared expression
is by itself.
With the squared expression by itself, we can apply the square root method.
x - 2 = ; 26
We solve for x by adding 2 to both sides.
x = 2 ; 26
By checking both values in the original equation, we can confirm that the
solution set is E2 + 16 , 2 - 16F or E2 ; 16 F.
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Section 1.5 • Quadratic Equations • 119
Check
Point
2
3
Solve quadratic equations
by completing the square.
Solve by the square root method:
a. 3x2 = 21
b. (x + 5)2 = 11.
Completing the Square
How do we solve an equation in the form ax2 + bx + c = 0 if the trinomial
ax2 + bx + c cannot be factored? We cannot use the zero-product principle in
such a case. However, we can convert the equation into an equivalent equation
that can be solved using the square root method. This is accomplished by
completing the square.
Completing the Square
b 2
If x2 + bx is a binomial, then by adding a b , which is the square of half
2
the coefficient of x, a perfect square trinomial will result. That is,
b 2
b 2
x2 + bx + a b = a x + b .
2
2
EXAMPLE 3 Completing the Square
What term should be added to the binomial x2 + 8x so that it becomes a perfect
square trinomial? Then write and factor the trinomial.
Solution The term that should be added is the square of half the coefficient of
x. The coefficient of x is 8. Thus, we will add A 82 B = 42. A perfect square trinomial
is the result.
2
x2 + 8x + 4 2 = x2 + 8x + 16 = (x + 4)2
(half)2
Check
Point
3
What term should be added to the binomial x2 - 14x so that it becomes
a perfect square trinomial? Then write and factor the trinomial.
We can solve any quadratic equation by completing the square. If the
coefficient of the x2-term is one, we add the square of half the coefficient of x to
both sides of the equation. When you add a constant term to one side of the
equation to complete the square, be certain to add the same constant to the
other side of the equation. These ideas are illustrated in Example 4.
EXAMPLE 4 Solving a Quadratic Equation
by Completing the Square
Solve by completing the square:
x2 - 6x + 2 = 0.
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120 • Chapter 1 • Equations, Inequalities, and Mathematical Models
Solution We begin the procedure of solving x2 - 6x + 2 = 0 by isolating the
binomial, x2 - 6x, so that we can complete the square. Thus, we subtract 2 from
both sides of the equation.
x2 - 6x + 2 = 0
This is the given equation.
x2 - 6x + 2 - 2 = 0 - 2 Subtract 2 from both sides.
x2 - 6x = -2
Simplify.
We need to add a constant
to this binomial that will
make it a perfect square
trinomial.
What constant should we add? Add the square of half the coefficient of x.
x2 - 6x = -2
–6 is the coefficient of x.
2
–6
2
––
2– =(–3) =9
Study Tip
When factoring perfect square
trinomials, the constant in the
factorization is always half the
coefficient of x.
Thus, we need to add 9 to x2 - 6x. In order to obtain an equivalent equation,
we must add 9 to both sides.
x2 - 6x = -2
x2 - 6x + 9 = (x - 3)2
This is the quadratic equation with
the binomial isolated.
x2 - 6x + 9 = -2 + 9
Add 9 to both sides to complete
the square.
Half the coefficient of
x, −6, is −3.
(x - 3)2 = 7
In this step we have
converted our equation
into one that
can be solved by the
square root method.
Factor the perfect square trinomial.
x - 3 = ; 27
Apply the square root method.
x = 3 ; 27
Add 3 to both sides.
The solution set is E3 + 27 , 3 - 27 F or E3 ; 17F.
Check
Point
4
Solve by completing the square:
x2 - 2x - 2 = 0.
If the coefficient of the x2-term in a quadratic equation is not one, you
must divide each side of the equation by this coefficient before completing the
square. For example, to solve 3x2 - 2x - 4 = 0 by completing the square, first
divide every term by 3:
3x2
2x
4
0
- =
3
3
3
3
2
4
2
x - x - = 0.
3
3
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Section 1.5 • Quadratic Equations • 121
Now that the coefficient of x2 is one, we can solve by completing the square
using the method of Example 4.
Solve quadratic equations
using the quadratic formula.
4
Solving Quadratic Equations Using the Quadratic Formula
We can use the method of completing the square to derive a formula that can be
used to solve all quadratic equations. The derivation given here also shows a
particular quadratic equation, 3x2 - 2x - 4 = 0, to specifically illustrate each
of the steps.
Deriving the Quadratic Formula
General Form
of a Quadratic Equation
2
ax + bx + c = 0,
Comment
A Specific Example
This is the given equation.
a 7 0
x2 +
b
c
x +
= 0
a
a
x2 +
b
c
x = a
a
x2 +
b
b 2
c
b 2
x + a b = - + a b
a
2a
a
2a
Divide both sides by the coefficient of x2.
Isolate the binomial by adding -
c
on both sides.
a
3x2 - 2x - 4 = 0
x2 -
2
4
x - = 0
3
3
x2 -
2
4
x =
3
3
1 2 4
1 2
Complete the square. Add the square of half the 2 2
x - x + a- b = + a- b
coefficient of x to both sides.
3
3
3
3
(half)2
(half)2
b
b2
c
b2
x +
= - +
2
a
a
4a
4a2
b 2
c 4a
b2
ax +
b = - +
2a
a 4a
4a2
x2 +
b 2
-4ac + b2
b =
2a
4a2
b 2
b2 - 4ac
ax +
b =
2a
4a2
ax +
Factor on the left side and obtain a common
denominator on the right side.
Add fractions on the right side.
x2 -
2
1
4
1
x + = +
3
9
3
9
ax -
1 2
4 3
1
b = +
3
3 3
9
ax -
1 2
12 + 1
b =
3
9
ax -
1 2
13
b =
3
9
x +
b
b2 - 4ac
= ;
2a
C 4a2
Apply the square root method.
x -
13
1
= ;
3
B9
x +
2b2 - 4ac
b
= ;
2a
2a
Take the square root of the quotient,
simplifying the denominator.
x -
213
1
= ;
3
3
x =
-b
2b2 - 4ac
;
2a
2a
x =
1
213
;
3
3
x =
-b ; 2b2 - 4ac
2a
x =
1 ; 213
3
Solve for x by subtracting
b
from both sides.
2a
Combine fractions on the right.
The formula shown at the bottom of the left column is called the quadratic
formula. A similar proof shows that the same formula can be used to solve
quadratic equations if a, the coefficient of the x2-term, is negative.
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122 • Chapter 1 • Equations, Inequalities, and Mathematical Models
To Die at Twenty
The Quadratic Formula
The solutions of a quadratic equation in standard form ax2 + bx + c = 0,
with a Z 0, are given by the quadratic formula
x =
PHOTO NOT
AVAILABLE
-b ; 2b2 - 4ac
.
2a
x equals negative b, plus or minus
the square root of b2 – 4ac, all
divided by 2a.
To use the quadratic formula, write the quadratic equation in general form
if necessary. Then determine the numerical values for a (the coefficient of the
squared term), b (the coefficient of the x-term), and c (the constant term).
Substitute the values of a, b, and c in the quadratic formula and evaluate the
expression. The ; sign indicates that there are two solutions of the equation.
EXAMPLE 5 Solving a Quadratic Equation
Using the Quadratic Formula
Can the equations
7x5 + 12x3 - 9x + 4 = 0
and
8x6 - 7x5 + 4x3 - 19 = 0
be solved using a formula similar
to the quadratic formula? The first
equation has five solutions and the
second has six solutions, but they
cannot be found using a formula.
How do we know? In 1832, a 20year-old Frenchman, Evariste
Galois, wrote down a proof
showing that there is no general
formula to solve equations when
the exponent on the variable is 5
or greater. Galois was jailed as a
political activist several times
while still a teenager.The day after
his brilliant proof he fought a duel
over a woman. The duel was a
political setup. As he lay dying,
Galois told his brother, Alfred, of
the manuscript that contained his
proof: “Mathematical manuscripts
are in my room. On the table.Take
care of my work. Make it known.
Important. Don’t cry, Alfred. I
need all my courage—to die at
twenty.” (Our source is Leopold
Infeld’s biography of Galois,
Whom the Gods Love. Some
historians, however, dispute the
story of Galois’s ironic death the
very day after his algebraic proof.
Mathematical truths seem more
reliable than historical ones!)
2x2 - 6x + 1 = 0.
Solve using the quadratic formula:
Solution The given equation is in general form. Begin by identifying the values
for a, b, and c.
2x2 - 6x + 1 = 0
a=2
b=–6
-b ; 2b2 - 4ac
2a
-(-6) ; 2(-6)2 - 4(2)(1)
=
22
x =
c=1
Use the quadratic formula.
Substitute the values for a, b, and c: a = 2,
b = -6, and c = 1.
=
6 ; 236 - 8
4
-(-6) = 6 and (-6)2 = (-6)(-6) = 36.
=
6 ; 228
2
Complete the subtraction under the radical.
=
=
=
6 ; 227
4
2A3 ; 27B
4
3 ; 27
2
The solution set is e
Check
Point
5
228 = 24 7 = 24 27 = 227
Factor out 2 from the numerator.
Divide the numerator and denominator by 2.
3 + 27 3 - 27
3 ; 27
,
f or e
f.
2
2
2
Solve using the quadratic formula:
2x2 + 2x - 1 = 0.
We have seen that a graphing utility can be used to check the solutions of
the quadratic equation ax2 + bx + c = 0. The x-intercepts of the graph of
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Section 1.5 • Quadratic Equations • 123
y = 3x2 − 2x + 4
Figure 1.15 This graph has no
x-intercepts.
y = ax2 + bx + c are the solutions. However, take a look at the graph of
y = 3x2 - 2x + 4, shown in Figure 1.15. Notice that the graph has no x-intercepts.
Can you guess what this means about the solutions of the quadratic equation
3x2 - 2x + 4 = 0? If you’re not sure, we’ll answer this question in the next example.
EXAMPLE 6 Solving a Quadratic Equation
Using the Quadratic Formula
3x2 - 2x + 4 = 0.
Solve using the quadratic formula:
Solution The given equation is in general form. Begin by identifying the values
for a, b, and c.
3x2 - 2x + 4 = 0
a=3
x =
Study Tip
Checking
irrational
and
complex imaginary solutions
can be time-consuming. The
solutions
given
by
the
quadratic formula are always
correct, unless you have made
a careless error. Checking for
computational errors or errors
in simplification is sufficient.
b=–2
-b ; 2b2 - 4ac
2a
Use the quadratic formula.
=
-(-2) ; 2(-2)2 - 4(3)(4)
2(3)
=
2 ; 24 - 48
6
=
2 ; 2-44
6
=
2 ; 2i211
6
c=4
Substitute the values for a, b, and c: a = 3,
b = -2, and c = 4.
-(-2) = 2 and (-2)2 = (-2)(-2) = 4.
Subtract under the radical.
Because the number under the radical
sign is negative, the solutions will not be
real numbers.
2-44 = 24(11)(-1)
= 2i211
2A1 ; i211B
=
6
1 ; i211
=
3
1
211
= ; i
3
3
Factor 2 from the numerator.
Divide numerator and denominator by 2.
Write the complex numbers in standard form.
The solutions are complex conjugates,
1
111 1
111
1
111
e + i
, - i
f or e ; i
f.
3
3 3
3
3
3
and
the
solution
set
is
If ax2 + bx + c = 0 has complex imaginary solutions, the graph of
y = ax2 + bx + c will not have x-intercepts. This is illustrated by the imaginary
solutions of 3x2 - 2x + 4 = 0 in Example 6 and the graph in Figure 1.15.
Check
Point
6
Solve using the quadratic formula:
x2 - 2x + 2 = 0.
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124 • Chapter 1 • Equations, Inequalities, and Mathematical Models
5
Use the discriminant to
determine the number and
type of solutions.
The Discriminant
The quantity b2 - 4ac, which appears under the radical sign in the quadratic
formula, is called the discriminant. In Example 5 the discriminant was 28, a
positive number that is not a perfect square. The equation had two solutions that
were irrational numbers. In Example 6, the discriminant was -44, a negative
number. The equation had solutions that were complex imaginary numbers.
These observations are generalized in Table 1.3.
Table 1.3
The Discriminant and the Kinds of Solutions
to ax 2 + bx + c = 0
Discriminant
b 2 - 4ac
Kinds of Solutions
to ax 2 + bx + c = 0
Graph of
y = ax 2 + bx + c
b2 - 4ac 7 0
Two unequal real solutions;
if a, b, and c are rational numbers and
the discriminant is a perfect square,
the solutions are rational. If the
discriminant is not a perfect square,
the solutions are irrational.
y
x
Two x-intercepts
b2 - 4ac = 0
One solution(a repeated solution)
that is a real number; If a, b, and c
are rational numbers, the repeated
solution is also a rational number.
y
x
One x-intercepts
b2 - 4ac 6 0
No real solution; two complex
imaginary solutions; The solutions
are complex conjugates.
y
x
No x-intercepts
EXAMPLE 7 Using the Discriminant
Compute the discriminant of 4x2 - 8x + 1 = 0. What does the discriminant
indicate about the number and type of solutions?
Solution Begin by identifying the values for a, b, and c.
4x2 - 8x + 1 = 0
a=4
b=–8
c=1
Substitute and compute the discriminant:
b2 - 4ac = (-8)2 - 441 = 64 - 16 = 48.
The discriminant is 48. Because the discriminant is positive, the equation
4x2 - 8x + 1 = 0 has two unequal real solutions.
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Section 1.5 • Quadratic Equations • 125
Check
Point
7
Determine the most
efficient method to use
when solving a quadratic
equation.
6
Compute the discriminant of 3x2 - 2x + 5 = 0. What does the
discriminant indicate about the number and type of solutions?
Determining Which Method to Use
All quadratic equations can be solved by the quadratic formula. However, if an
equation is in the form u2 = d, such as x2 = 5 or (2x + 3)2 = 8, it is faster to
use the square root method, taking the square root of both sides. If the equation
is not in the form u2 = d, write the quadratic equation in general form
(ax2 + bx + c = 0). Try to solve the equation by the factoring method. If
ax2 + bx + c cannot be factored, then solve the quadratic equation by the
quadratic formula.
Because we used the method of completing the square to derive the quadratic
formula, we no longer need it for solving quadratic equations. However, we will use
completing the square later in the book to help graph certain kinds of equations.
Table 1.4 summarizes our observations about which technique to use
when solving a quadratic equation.
Table 1.4 Determining the Most Efficient Technique to Use When Solving a Quadratic Equation
Description and Form of
the Quadratic Equation
2
Most Efficient
Solution Method
2
ax + bx + c = 0 and ax + bx + c
can be factored easily.
Factor and use the
zero-product
principle.
ax2 + c = 0
The quadratic equation has no
x-term. (b = 0)
Solve for x2 and
apply the square
root method.
Example
3x2 + 5x - 2 = 0
(3x - 1)(x + 2) = 0
3x - 1 = 0 or x + 2 = 0
1
x = -2
x =
3
4x2 - 7 = 0
4x2 = 7
7
x2 =
4
x = ;
(ax + c)2 = d; ax + c is a
first-degree polynomial.
Use the square
root method.
ax2 + bx + c = 0 and ax2 + bx + c
cannot be factored or the factoring is
too difficult.
Use the quadratic formula:
x =
-b ; 2b2 - 4ac
.
2a
27
2
(x + 4)2 = 5
x + 4 = ; 15
x = -4 ; 15
x2 - 2x - 6 = 0
a=1
b = −2
c = −6
- A-2B ; 2A-2B 2 - 4(1)(-6)
x =
2(1)
2 ; 24 - 4(1)(-6)
=
2(1)
=
2 ; 24 27
2 ; 228
=
2
2
=
2(1 ; 27)
2 ; 2 27
=
2
2
= 1 ; 27
Copyright © 2005 Pearson Education, Inc., publishing as Pearson Prentice Hall
126 • Chapter 1 • Equations, Inequalities, and Mathematical Models
Solve problems modeled by
quadratic equations.
7
Convictions of Police Officers
Number of Convictions
600
Applications
We opened this section with a graph (Figure 1.14, repeated in the margin)
showing the number of convictions of police officers throughout the United
States from 1994 through 2000. The data can be modeled by the formula
560
N = 23.4x2 - 259.1x + 815.8
500
380
400
300
200
246
143
149
135
83
100
1994 1995 1996 1997 1998 1999 2000
Year
Figure 1.14, repeated
where N is the number of police officers convicted of felonies x years after 1990.
Notice that this formula contains an expression in the form ax2 + bx + c on the
right side. If a formula contains such an expression, we can write and solve a
quadratic equation to answer questions about the variable x. Our next example
shows how this is done.
EXAMPLE 8 Convictions of Police Officers
Source: F.B.I.
Use the formula N = 23.4x2 - 259.1x + 815.8 to answer this question: In
which year will 1000 police officers be convicted of felonies?
Solution Because we are interested in 1000 convictions, we substitute 1000 for
N in the given formula. Then we solve for x, the number of years after 1990.
N = 23.4x2 - 259.1x + 815.8
1000 = 23.4x2 - 259.1x + 815.8
2
0 = 23.4x - 259.1x - 184.2
a = 23.4
b = −259.1
c = −184.2
This is the given formula.
Substitute 1000 for N.
Subtract 1000 from both sides
and write the quadratic equation in
general form.
Because the trinomial on the right side of the equation is prime, we solve using
the quadratic formula.
x =
Technology
=
-(-259.1) ; 2(-259.1)2 - 4(23.4)(-184.2)
2(23.4)
=
259.1 ; 184,373.93
46.8
On most calculators, here is
how to approximate
259.1 + 184,373.93
:
46.8
Many Scientific Calculators
( 259.1 + 84373.93 1
) ÷ 46.8 =
-b ; 2b2 - 4ac
2a
) ÷ 46.8
84373.93
ENTER
.
Similar keystrokes can be
used to approximate the
other irrational solution in
Example 8.
Substitute the values for a, b,
and c: a = 23.4, b = -259.1,
c = -184.2.
Use a calculator to simplify
the radicand.
Thus,
259.1 + 184,373.93
259.1 - 184,373.93
or x =
46.8
46.8
x L 12
x L -1
Use a calculator and round
x =
to the nearest integer.
Many Graphing Calculators
( 259.1 + 1
Use the quadratic formula.
The model describes the number of convictions x years after 1990. Thus, we are
interested only in the positive solution, 12. This means that approximately 12 years
after 1990, in 2002, 1000 police officers will be convicted of felonies.
Check
Point
8
Use the formula in Example 8 to answer this question: In which year
after 1993 were 250 police officers convicted of felonies? How well
does the formula model the actual number of convictions for that
year shown in Figure 1.14?
Copyright © 2005 Pearson Education, Inc., publishing as Pearson Prentice Hall
Section 1.5 • Quadratic Equations • 127
Area:
25
square
units
Area:
16 square
units
54
3
In our next example, we will be using the Pythagorean Theorem to obtain
a verbal model. The ancient Greek philosopher and mathematician Pythagoras
(approximately 582–500 B.C.) founded a school whose motto was “All is
number.” Pythagoras is best remembered for his work with the right triangle, a
triangle with one angle measuring 90°. The side opposite the 90° angle is called
the hypotenuse. The other sides are called legs. Pythagoras found that if he
constructed squares on each of the legs, as well as a larger square on the
hypotenuse, the sum of the areas of the smaller squares is equal to the area of
the larger square. This is illustrated in Figure 1.16.
This relationship is usually stated in terms of the lengths of the three sides
of a right triangle and is called the Pythagorean Theorem.
B
The Pythagorean Theorem
Area:
9 square
units
Figure 1.16 The area of the large
square equals the sum of the areas of
the smaller squares.
The sum of the squares of the lengths of the legs
of a right triangle equals the square of the length
of the hypotenuse.
If the legs have lengths a and b, and the hypotenuse
has length c, then
c
Hypotenuse
A
a2 + b2 = c2.
b
Leg
a
Leg
C
EXAMPLE 9 Using the Pythagorean Theorem
In a 25-inch television set, the length of the screen’s diagonal is 25 inches. If the
screen’s height is 15 inches, what is its width?
w
25 inches
15 inches
Solution Figure 1.17 shows a right triangle that is formed by the height, width,
and diagonal.We can find w, the screen’s width, using the Pythagorean Theorem.
(Leg)2
plus
(Leg)2
equals
w2
+
152
=
(Hypotenuse)2
252
Figure 1.17 A right triangle is
formed by the television’s height,
width, and diagonal.
This is the equation resulting
from the Pythagorean Theorem.
The equation w2 + 152 = 252 can be solved most efficiently by the square root
method.
w2 + 152 = 252
This is the equation that models the verbal
conditions.
w2 + 225 = 625
Square 15 and 25.
2
w + 225 - 225 = 625 - 225
2
w = 400
Isolate w2 by subtracting 225 from both sides.
Simplify.
w = ; 2400
Apply the square root method.
w = ; 20
Simplify.
Because w represents the width of the television’s screen, this dimension must
be positive. We reject -20. Thus, the width of the television is 20 inches.
Check
Point
9
What is the width in a 15-inch television set whose height is 9 inches?
Copyright © 2005 Pearson Education, Inc., publishing as Pearson Prentice Hall