Download 7.7 Quadratic Equations and Applications

7.7
7.7
Quadratic Equations and Applications
381
Quadratic Equations and Applications
Quadratic Equations Recall that a linear equation is one that can be written
in the form ax b c, where a, b, and c are real numbers, a 0.
Quadratic Equation
An equation that can be written in the form
ax2 bx c 0
where a, b, and c are real numbers, with a 0, is a quadratic equation.
(Why is the restriction a 0 necessary?) A quadratic equation written in the form
ax 2 bx c 0 is in standard form.
The simplest method of solving a quadratic equation, but one that is not always
easily applied, is by factoring. This method depends on the following property.
Zero-Factor Property
If ab 0, then a 0 or b 0 or both.
EXAMPLE
1
Solve 6x 2 7x 3.
6x2 7x 3
6x2 7x 3 0
3x 12x 3 0
Standard form
Factor.
By the zero-factor property, the product 3x 12x 3 can equal 0 only if
3x 1 0
3x 1
1
x
3
The solve feature gives the two
solutions of the equation in
Example 1. Notice that the guess 0
yields the solution 13, while the
guess 5 yields the solution
32. Compare to Example 1.
Check by first substituting
set is
or
or
or
2x 3 0
2x 3
3
x .
2
1
3
and then in the original equation. The solution
3
2
1
3
,
.
3
2
A quadratic equation of the form x 2 k, k 0, can be solved by factoring.
x2 k
x2 k 0
x k x k 0
x k 0
x k 0
or
x k
x k
or
This proves the square root property for solving equations.
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382
CHAPTER 7
The Basic Concepts of Algebra
Square Root Property
If k 0, then the solutions of x 2 k are x k.
If k 0, the equation x 2 k has two real solutions. If k 0, there is only one
solution, 0. If k 0, there are no real solutions. (However, in this case, there are
imaginary solutions. Imaginary numbers are discussed briefly in the Extension on
complex numbers at the end of Chapter 6.)
Completing the square, used in
deriving the quadratic formula, has
important applications in algebra.
To transform the expression
x 2 kx into the square of a
binomial, we add to it the square
of half the coefficient of x; that is,
12k 2 k 24. We then get
x 2 kx k2
k
x
4
2
2
.
For example, to make x 2 6x the
square of a binomial, we add 9,
since 9 126 2. This results
in the trinomial x 2 6x 9,
which is equal to x 32.
The Greeks had a method of
completing the square
geometrically. For example, to
complete the square for x 2 6x ,
begin with a square of side x. Add
three rectangles of width 1 and
length x to the right side and the
bottom. Each rectangle has area 1x
or x, so the total area of the figure
is now x 2 6x . To fill in the
corner (that is, “complete the
square”), we add 9 1-by-1 squares
as shown.
x+3
x
x
x+3
The new, larger square has sides
of length x 3 and area
x 32 x 2 6x 9.
EXAMPLE 2
for real solutions.
Use the square root property to solve each quadratic equation
(a) x 2 25
Since 25 5, the solution set is 5, 5, which may be abbreviated 5.
(b) r 2 18
r 18
Square root property
r 9 2
r 9 2
Product rule for square roots
r 32
9 3
The solution set is 32 .
(c) z2 3
Since 3 0, there are no real roots, and the solution set is 0.
(d) x 42 12
Use a generalization of the square root property, working as follows.
x 42 12
x 4 12
x 4 12
x 4 4 3
x 4 23
The solution set is 4 23 .
The Quadratic Formula By using a procedure called completing the square
(see the margin note) we can derive one of the most important formulas in algebra,
the quadratic formula. We begin with the standard quadratic equation
ax 2 bx c 0, a 0.
b
c
x2 x 0
Divide by a.
a
a
b
c
c
x2 x Add .
a
a
a
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7.7
Quadratic Equations and Applications
383
The polynomial on the left will be the square of a binomial if we add b24a2
to both sides of the equation.
x2 b
b2
b2
c
x 2 2
a
4a
4a
a
x
b
2a
2
b2 4ac
4a2
b
2a
x
b
b2 4ac
2a
4a2
x
3
3
n
2
n
2
n
2
2
m
3
n
2
2
x
Subtract
b
.
2a
Combine terms.
b b2 4ac
.
2a
Notice that the fraction bar in the quadratic formula extends under the b term in
the numerator.
EXAMPLE 3
Solve x 2 4x 2 0.
Here a 1, b 4, and c 2. Substitute these values into the quadratic formula to obtain
x
3
.
b b2 4ac
2a
4 42 412
21
4 16 8
x
2
4 22
x
2
2 2 2 x
2
x 2 2 .
x
.
Try solving for one solution of the
equation x 3 9x 26 using this
formula. (The solution given by the
formula is 2.)
b b2 4ac
2a
Quotient rule for square roots
The solutions of ax 2 bx c 0, a 0, are
3
m
3
b
b2 4ac
2a
2a
Square root property
Quadratic Formula
x
b2 4ac
4a2
x
x
For centuries mathematicians
wrestled with finding a formula
that could solve cubic (thirddegree) equations. A story from
sixteenth-century Italy concerns
two main characters, Girolamo
Cardano and Niccolo
Tartaglia. In those days,
mathematicians often participated
in contests. Tartaglia had
developed a method of solving a
cubic equation of the form
x 3 mx n and had used it in
one of these contests. Cardano
begged to know Tartaglia’s method
and after he was told was sworn to
secrecy. Nonetheless, Cardano
published Tartaglia’s method in his
1545 work Ars Magna (although
he did give Tartaglia credit).
The formula for finding one
real solution of the above equation
is
Factor on the left; combine terms on the right.
a 1, b 4, c 2
16 8 8 22
Factor out a 2 in the numerator.
Lowest terms
The solution set is 2 2, 2 2 , abbreviated 2 2 .
An Addison-Wesley product. Copyright © 2004 Pearson Education, Inc.
384
CHAPTER 7
The Basic Concepts of Algebra
EXAMPLE 4
Solve 2x 2 x 4.
To find the values of a, b, and c, first rewrite the equation in standard form as
2x 2 x 4 0. Then a 2, b 1, and c 4. By the quadratic formula,
A Radical Departure from the
Other Methods of Evaluating
the Golden Ratio Recall from
a previous chapter that the golden
ratio is found in numerous places
in mathematics, art, and nature. In
a margin note there, we showed
that
1
1
1
1
1
1
1…
The solution set is
is equal to the golden ratio,
1 5 2. Now consider this
“nested” radical:
1 33
.
4
Applications
When solving applied problems that lead to quadratic equations,
we might get a solution that does not satisfy the physical constraints of the problem.
For example, if x represents a width and the two solutions of the quadratic equation are
9 and 1, the value 9 must be rejected, since a width must be a positive number.
1 1 1 … .
Let x represent this radical.
Because it appears “within itself,”
we can write
x 1 x
x 2 1 x Square
EXAMPLE 5
Two cars left an intersection at the same time, one heading due
north, and the other due west. Some time later, they were exactly 100 miles apart.
The car headed north had gone 20 miles farther than the car headed west. How far
had each car traveled?
both
sides.
x2 x 1 0 .
1 12 424
22
1 1 32
x
4
1 33
x
.
4
x
Write in
standard
Step 1: Read the problem carefully.
Step 2: Assign a variable. Let x be the distance traveled by the car headed west.
Then x 20 is the distance traveled by the car headed north. See Figure 13. The cars are 100 miles apart, so the hypotenuse of the right triangle
equals 100.
Step 3: Write an equation. Use the Pythagorean theorem.
form.
Using the quadratic formula, with
a 1, b 1, and c 1, it
can be shown that the positive
solution of this equation, and thus
the value of the nested radical is …
(you guessed it!) the golden ratio.
c2 a2 b2
100 2 x 2 x 202
North
Step 4: Solve.
100
10,000 x 2 x 2 40x 400
0 2x 2 40x 9600
0 2x 2 20x 4800
0 x 2 20x 4800
x + 20
Square the binomial.
Get 0 on one side.
Factor out the common factor.
Divide both sides by 2.
Use the quadratic formula to find x.
West
x
90°
Intersection
FIGURE 13
20 400 414800
2
20 19,600
2
x 60
or
x 80
x
a 1, b 20, c 4800
Use a calculator.
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7.7 Quadratic Equations and Applications
385
Step 5: State the answer. Since distance cannot be negative, discard the negative
solution. The required distances are 60 miles and 60 20 80 miles.
Step 6: Check. Since 602 802 1002, the answer is correct.
EXAMPLE 6
If a rock on Earth is thrown upward from a 144-foot building
with an initial velocity of 112 feet per second, its position (in feet above the ground)
is given by s 16t 2 112t 144, where t is time in seconds after it was thrown.
When does it hit the ground?
When the rock hits the ground, its distance above the ground is 0. Find t when
s is 0 by solving the equation
0 16t2 112t 144 .
0 t 2 7t 9
t
7 49 36
2
t
7 85
2
t 8.1
or
t 1.1
Let s 0.
Divide both sides by 16.
Quadratic formula
Use a calculator.
Since time cannot be negative, discard the negative solution. The rock will hit the
ground about 8.1 seconds after it is thrown.
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b 2 4ac
b 2 4ac 17
ax 2 bx c 0