D. The Quadratic Formula and the Discriminant
Tải bản đầy đủ - 0trang
cob19545_ch03_293-313.qxd
8/10/10
6:07 PM
Page 300
College Algebra G&M—
300
3–20
CHAPTER 3 Quadratic Functions and Operations on Functions
On the left, our final solutions are x ϭ Ϫ1 or x ϭ Ϫ32. The general solution is
called the quadratic formula, which can be used to solve any equation belonging to
the quadratic family.
Quadratic Formula
If ax2 ϩ bx ϩ c ϭ 0, with a, b, and c ʦ ޒand a 0, then
Ϫb ؊ 2b2 Ϫ 4ac
Ϫb ؉ 2b2 Ϫ 4ac
xϭ
or
xϭ
;
2a
2a
Ϫb ؎ 2b2 Ϫ 4ac
.
also written x ϭ
2a
ᮣ
CAUTION
EXAMPLE 5
ᮣ
It’s very important to note the values of a, b, and c come from an equation written in standard form. For 3x2 Ϫ 5x ϭ Ϫ7, a ϭ 3 and b ϭ Ϫ5, but c Ϫ7! In standard form we have
3x2 Ϫ 5x ϩ 7 ϭ 0, and note the value for use in the formula is actually c ϭ 7.
Solving Quadratic Equations Using the Quadratic Formula
Solve 4x2 ϩ 1 ϭ 8x using the quadratic formula. State the solution(s) in both exact
and approximate form. Check one of the exact solutions in the original equation.
Solution
ᮣ
Begin by writing the equation in standard form and identifying the values of a, b,
and c.
4x2 ϩ 1 ϭ 8x
4x Ϫ 8x ϩ 1 ϭ 0
a ϭ 4, b ϭ Ϫ8, c ϭ 1
2
xϭ
Ϫ1Ϫ82 Ϯ 21Ϫ82 2 Ϫ 4142112
2142
8 Ϯ 148
8 Ϯ 164 Ϫ 16
ϭ
8
8
8 Ϯ 4 13
8
4 13
xϭ
ϭ Ϯ
8
8
8
13
13
xϭ1ϩ
or x ϭ 1 Ϫ
2
2
or x Ϸ 0.13
x Ϸ 1.87
xϭ
Check
ᮣ
4x2 ϩ 1 ϭ 8x
13 2
13
4 a1 ϩ
b ϩ 1 ϭ 8 a1 ϩ
b
2
2
4 c 1 ϩ 2a
13
3
b ϩ d ϩ 1 ϭ 8 ϩ 4 13
2
4
4 ϩ 4 13 ϩ 3 ϩ 1 ϭ 8 ϩ 4 13
8 ϩ 4 13 ϭ 8 ϩ 4 13 ✓
original equation
standard form
substitute 4 for a, ؊8 for b, and 1 for c
simplify
simplify radical (see following CAUTION)
exact solutions
approximate solutions
original equation
substitute 1 ϩ
13
2
for x
square binomial; distribute
distribute
result checks
A graphing calculator check is also shown.
Now try Exercises 61 through 90
ᮣ
cob19545_ch03_293-313.qxd
8/10/10
6:07 PM
Page 301
College Algebra G&M—
3–21
301
Section 3.2 Solving Quadratic Equations and Inequalities
1
CAUTION
ᮣ
For
8 Ϯ 413
8 Ϯ 413
ϭ 1 Ϯ 413.
, be careful not to incorrectly “cancel the eights” as in
8
8
1
No! Use a calculator to verify that the results are not equivalent. Both terms in the numerator are divided by 8 and we must either rewrite the expression as separate terms (as
above) or factor the numerator to see if the expression simplifies further:
1
8 Ϯ 413
2 Ϯ 13
4 12 Ϯ 132
13
ϭ
ϭ
, which is equivalent to 1 Ϯ
.
8
8
2
2
2
The Discriminant of the Quadratic Formula
The conclusions we reached graphically in Figure 3.14 regarding the nature of number
of quadratic roots can now be seen algebraically through a closer look at the quadratic
formula. For any real-valued expression X, recall that 1X represents a real number
only for X Ն 0. Since the quadratic formula contains the radical 2b2 Ϫ 4ac, the
expression b2 Ϫ 4ac, called the discriminant, will determine the nature (real or complex) and the number of solutions to a given quadratic equation.
The Discriminant of the Quadratic Formula
For f 1x2 ϭ ax2 ϩ bx ϩ c, where a, b, c ʦ ޒand a
1. If b Ϫ 4ac 7 0,
there are two
real roots
2
1. f 1x2 ϭ x2 Ϫ 4x
a ϭ 1, b ϭ Ϫ4, c ϭ 0
b2 Ϫ 4ac ϭ 1Ϫ42 2 Ϫ 4112102
ϭ 16 Ϫ 0
ϭ 16
16 7 0, f has two real roots
0,
3. If b2 Ϫ 4ac 6 0,
there are two
nonreal roots
These ideas are further illustrated here, by comparing the value of the discriminant
with the graph of the related function shown below each calculation.
2. g1x2 ϭ x2 Ϫ 4x ϩ 4
a ϭ 1, b ϭ Ϫ4, c ϭ 4
b2 Ϫ 4ac ϭ 1Ϫ42 2 Ϫ 4112142
ϭ 16 Ϫ 16
ϭ0
0 ϭ 0, g has one real root
10
Ϫ5
2. If b Ϫ 4ac ϭ 0,
there is one real
(repeated) root
2
3. h1x2 ϭ x2 Ϫ 4x ϩ 5
a ϭ 1, b ϭ Ϫ4, c ϭ 5
b2 Ϫ 4ac ϭ 1Ϫ42 2 Ϫ 4112152
ϭ 16 Ϫ 20
ϭ Ϫ4
Ϫ4 6 0, h has two nonreal roots
10
10
Ϫ5
10
10
Ϫ5
Ϫ5
Ϫ5
10
Ϫ5
Finally, we note from (3) and the structure of the quadratic formula, that the
complex solutions must occur in conjugate pairs.
Complex Solutions
The complex solutions of a quadratic equation with real coefficients
must occur in conjugate pairs.
If a ϩ bi is a solution, then a Ϫ bi is also a solution.
cob19545_ch03_293-313.qxd
8/10/10
6:07 PM
Page 302
College Algebra G&M—
302
3–22
CHAPTER 3 Quadratic Functions and Operations on Functions
Further analysis of the discriminant reveals even more concerning the nature of
quadratic solutions. Namely, if a, b, and c are rational and the discriminant is
1. zero, the original equation is a perfect square trinomial.
2. a perfect square, there will be two rational roots which means
the original equation can be solved by factoring.
3. not a perfect square, there will be two irrational roots.
See Exercises 91 through 102.
EXAMPLE 6
ᮣ
Solving Quadratic Equations Using the Quadratic Formula
Solve: 2x2 Ϫ 6x ϩ 5 ϭ 0.
Solution
ᮣ
With a ϭ 2, b ϭ Ϫ6, and c ϭ 5, the discriminant becomes 1Ϫ62 2 Ϫ 4122152 ϭ Ϫ4,
showing there will be two complex roots. The quadratic formula then yields
xϭ
xϭ
Ϫb Ϯ 2b2 Ϫ 4ac
2a
Ϫ1Ϫ62 Ϯ 1Ϫ4
2122
6 Ϯ 2i
xϭ
4
3
1
xϭ Ϯ i
2
2
quadratic formula
b 2 Ϫ 4ac ϭ Ϫ4, substitute 2 for a and ؊6 for b
simplify, write in i form
solutions are complex conjugates
A calculator check is shown in Figures 3.25 and 3.26.
Figure 3.25, 3.26
WORTHY OF NOTE
While it’s possible to solve by
b
completing the square if is a
a
fraction or an odd number (see
Example 4), the process is usually
b
most efficient when is an even
a
number. This is one observation
you could use when selecting a
solution method.
D. You’ve just seen how
we can solve quadratic
equations using the quadratic
formula and the discriminant
Now try Exercises 103 through 108
ᮣ
Summary of Solution Methods for ax2 ؉ bx ؉ c ؍0
1. If b ϭ 0, ax2 ϩ c ϭ 0: isolate x2 and use the square root property of equality.
2. If c ϭ 0, ax2 ϩ bx ϭ 0: factor out the GCF and use the zero product property.
3. If no coefficient is zero, you can attempt to solve by
a. factoring b. completing the square c. using the quadratic formula
d. using the intersection-of-graphs method e. using the zeroes method
cob19545_ch03_293-313.qxd
11/25/10
3:28 PM
Page 303
College Algebra G&M—
3–23
303
Section 3.2 Solving Quadratic Equations and Inequalities
E. Quadratic Inequalities
The study of quadratic inequalities is simply an extension of our earlier work in analyzing functions (Section 2.1). While we’ve developed the ability to graph a variety of
new functions, the solution set for an inequality will still be determined by analyzing
the behavior of the function at its zeroes. The key idea is to recognize the following
statements are synonymous:
1. f 1x2 7 0.
2. Outputs are positive.
3. The graph is above the x-axis.
Similar statements can be made using the other inequality symbols.
Solving a quadratic inequality only requires that we (a) locate any real zeroes of
the function and (b) determine whether the graph opens upward or downward. If there
are no x-intercepts, the graph is entirely above the x-axis (output values are positive),
or entirely below the x-axis (output values are negative), making the solution either all
real numbers or the empty set.
EXAMPLE 7
ᮣ
Analytical Solution
ᮣ
Solving a Quadratic Inequality
For f 1x2 ϭ x2 ϩ x Ϫ 6, solve f 1x2 7 0.
The graph of f will open upward since a 7 0. Factoring gives f 1x2 ϭ 1x ϩ 321x Ϫ 22 ,
with zeroes at Ϫ3 and 2. Using the x-axis alone (since graphing the function is not
our focus), we plot (Ϫ3, 0) and (2, 0) and visualize a parabola opening upward
through these points (Figure 3.27).
Figure 3.27
When Ϫ3 Ͻ x Ͻ 2,
the graph is below
the x-axis,
f(x) Ͻ 0.
Ϫ5
Ϫ4
Ϫ3
Ϫ2
Ϫ1
0
aϾ0
1
2
x
When x Ͼ 2,
the graph is above
the x-axis,
f(x) Ͼ 0.
When x Ͻ Ϫ3,
the graph is above
the x-axis,
f(x) Ͼ 0.
The diagram clearly shows the graph is above the x-axis (outputs are positive)
when x 6 Ϫ3 or when x 7 2. The solution is x ʦ 1Ϫq, Ϫ32 ´ 12, q 2 .
Graphical Solution
ᮣ
The complete graph of f shown in Figure 3.28
confirms the analytical solution. For the intervals
of the domain shown in bold 1Ϫq, Ϫ32 ´ 12, q 2 ,
the graph is above the x-axis 1 f 1x2 7 02 . For the
portion shown in red, the graph is below the
x-axis 1 f 1x2 6 02 .
Figure 3.28
5
Ϫ5
y f(x) ϭ x2 ϩ x Ϫ 6
5
x
Ϫ5
Now try Exercises 109 through 120
ᮣ
When solving general inequalities, zeroes of multiplicity continue to play a role.
In Example 7, the zeroes of f were both of multiplicity 1, and the graph crossed the
x-axis at these points. In other cases, the zeroes may have even multiplicity.
cob19545_ch03_293-313.qxd
8/10/10
6:07 PM
Page 304
College Algebra G&M—
304
3–24
CHAPTER 3 Quadratic Functions and Operations on Functions
EXAMPLE 8
ᮣ
Solving a Quadratic Inequality
Solve the inequality Ϫx2 ϩ 6x Յ 9.
Analytical Solution
ᮣ
WORTHY OF NOTE
Begin by writing the inequality in standard form: Ϫx2 ϩ 6x Ϫ 9 Յ 0. Note this is
equivalent to g1x2 Յ 0 for g1x2 ϭ Ϫx2 ϩ 6x Ϫ 9. Since a 6 0, the graph of g will
open downward. The factored form is g1x2 ϭ Ϫ1x Ϫ 32 2, showing 3 is a zero and a
repeated root. Using the x-axis, we plot the point (3, 0) and visualize a parabola
opening downward through this point.
Figure 3.29 shows the graph is below the x-axis (outputs are negative) for all
values of x except x ϭ 3. But since this is a less than or equal to inequality, the
solution is x ʦ ޒ.
Since x ϭ 3 was a zero of
multiplicity 2, the graph “bounced
off” the x-axis at this point, with no
change of sign for g. The graph is
entirely below the x-axis, except at
the vertex (3, 0).
Graphical Solution
Figure 3.29
Ϫ1
0
1
2
3
4
5
6
7
x
aϽ0
ᮣ
The complete graph of g shown in Figure 3.30 confirms the analytical solution
(using the zeroes method). For the intervals of the domain shown in red:
1Ϫq, 32 ´ 13, q 2 , the graph of g is below the x-axis 3g1x2 6 0 4 . The point (3, 0)
is on the x-axis 3g132 ϭ 04 . As with the analytical solution, the solution to this
“less than or equal to” inequality is all real numbers. A calculator check of the
original inequality is shown in Figure 3.31.
Figure 3.30
Figure 3.31
y
10
2
Ϫ2
6
x
Ϫ2
8
g(x)
Ϫ3
Ϫ8
Now try Exercises 121 through 132
ᮣ
As an alternative to the Zeroes Method, the Interval Test Method can be used to
solve quadratic inequalities. Using the fact that all polynomials are continuous, test
values are selected from certain intervals of the domain and substituted into the original function.
Interval Test Method for Solving Inequalities
1. Find all real roots of the related equation (if they exist) and
plot them on the x-axis.
2. Select any convenient test value from each interval created
by the zeroes, and substitute these into the function.
3. The sign of the function at these test values will be the sign
of the function for all values of x in this interval.
cob19545_ch03_293-313.qxd
8/10/10
6:07 PM
Page 305
College Algebra G&M—
3–25
305
Section 3.2 Solving Quadratic Equations and Inequalities
EXAMPLE 9
ᮣ
Solving a Quadratic Inequality
Solve the inequality Ϫ2x2 Ϫ 3x ϩ 20 Յ 0 using interval tests.
Solution
ᮣ
To begin, we find the zeroes of f 1x2 ϭ Ϫ2x2 Ϫ 3x ϩ 20 by factoring.
Ϫ2x2 Ϫ 3x ϩ 20 ϭ 0
2x2 ϩ 3x Ϫ 20 ϭ 0
12x Ϫ 52 1x ϩ 42 ϭ 0
2x Ϫ 5 ϭ 0 or x ϩ 4 ϭ 0
5
x ϭ or x ϭ Ϫ4
2
related equation
multiply by ؊1
factored form
zero factor property
solutions
Plotting these intercepts creates three intervals on the x-axis (Figure 3.32).
Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1
Figure 3.32
1
0
1
2
3
4
5
6
2
7
8
7
8
3
Selecting a test value from each interval (in red) gives Figure 3.33:
Ϫ8 Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1
Figure 3.33
1
f 1Ϫ52 ϭ Ϫ15
f 1x2 6 0 in 1
When evaluating a function using the
interval test method, it’s usually
easier to use the factored form
instead of the polynomial form, since
all you really need is whether the
result will be positive or negative. For
instance, you could likely tell
f 1Ϫ52 ϭ Ϫ 321Ϫ52 Ϫ 5 4 1Ϫ5 ϩ 42 is
going to be negative, more
quickly than f 1Ϫ52 ϭ
Ϫ21Ϫ52 2 Ϫ 31Ϫ52 ϩ 20.
1
2
3
4
5
6
2
x ϭ Ϫ5
WORTHY OF NOTE
0
3
xϭ0
xϭ5
f 102 ϭ 20
f 152 ϭ Ϫ45
f 1x2 7 0 in 2
f 1x2 6 0 in 3
The interval tests show Ϫ2x2 Ϫ 3x ϩ 20 Յ 0
for x ʦ 1Ϫq, Ϫ42 ´ 1 52, q 2 , which is
supported by the graph shown in Figure 3.34.
Figure 3.34
30
Ϫ6
5
Ϫ15
Now try Exercises 133 through 144
ᮣ
The need to solve a quadratic inequality occurs in a variety of contexts. Here, the
solution is used to find the domain of a radical function.
EXAMPLE 10
ᮣ
Solving a Quadratic Inequality to Determine the Domain
Find the domain of g1x2 ϭ 29 Ϫ x2.
Solution
ᮣ
From our earlier work, the radicand must be nonnegative and we are essentially
asked to solve the inequality 9 Ϫ x2 7 0. The related equation is 9 Ϫ x2 ϭ 0 and
by inspection (or factoring), the solutions are x ϭ Ϫ3 and x ϭ 3. Plotting these
solutions creates three intervals on the x-axis (Figure 3.35).
Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1
Figure 3.35
1
0
2
1
2
3
4
5
3
6
cob19545_ch03_293-313.qxd
8/10/10
6:07 PM
Page 306
College Algebra G&M—
306
3–26
CHAPTER 3 Quadratic Functions and Operations on Functions
Selecting a test value from each interval (in red) gives Figure 3.36:
Figure 3.36
Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1
1
x ϭ Ϫ4
9 Ϫ 1Ϫ42 ϭ Ϫ7
2
9 Ϫ x2 6 0 in 1
0
1
2
2
3
4
5
6
3
xϭ0
9 Ϫ 102 ϭ 9
2
9 Ϫ x2 7 0 in 2
The interval tests show that 9 Ϫ x2 Ն 0 for
x ʦ 3Ϫ3, 34 , which is the domain of
g1x2 ϭ 29 Ϫ x2. This is the same relation
we graphed in Section 1.1/Example 3, a
semicircle with endpoints at x ϭ Ϫ3 and 3.
See Figure 3.37.
xϭ4
9 Ϫ 142 2 ϭ Ϫ7
9 Ϫ x2 6 0 in 3
Figure 3.37
4
Ϫ6.1
6.1
Ϫ4
E. You’ve just seen how
we can solve quadratic
inequalities
Now try Exercises 145 through 154
ᮣ
F. Applications of Quadratic Functions and Inequalities
A projectile is any object that is thrown, shot, or projected upward with no sustaining
source of propulsion. The height of the projectile at time t is modeled by the equation
h ϭ Ϫ16t2 ϩ vt ϩ k, where h is the height of the object in feet, t is the elapsed time in
seconds, and v is the initial velocity in feet per second. The constant k represents the
initial height of the object above ground level, as when a person releases an object 5 ft
above the ground in a throwing motion 1k ϭ 52, or when a rocket runs out of fuel at an
altitude of 240 ft 1k ϭ 2402 .
EXAMPLE 11
ᮣ
Solving an Application of Quadratic Equations — Rocketry
Figure 3.38
A model rocketry club is testing a newly developed engine. A few
seconds after liftoff, at a velocity of 160 ft/sec and a height of 240 ft,
it runs out of fuel and becomes a projectile (see Figure 3.38).
a. How high is the rocket 3 sec later?
b. For how many seconds was the height of the rocket greater
than or equal to 496 ft?
Projectile
phase
c. How many seconds until the rocket returns to the ground?
Solution
ᮣ
a. Using the information given, the function h modeling
the rocket’s height in the projectile phase, is
h1t2 ϭ Ϫ16t2 ϩ 160t ϩ 240. For its height at t ϭ 3 we
have
h132 ϭ Ϫ16132 2 ϩ 160132 ϩ 240
ϭ Ϫ16192 ϩ 480 ϩ 240
ϭ 576
Three seconds later, the rocket was at an altitude of 576 ft.
240 ft
Power
phase
cob19545_ch03_293-313.qxd
8/10/10
6:08 PM
Page 307
College Algebra G&M—
3–27
Section 3.2 Solving Quadratic Equations and Inequalities
307
b. Since the height h(t) must be greater than or equal to 496 ft, we use the
function h1t2 ϭ Ϫ16t2 ϩ 160t ϩ 240 to write the inequality Ϫ16t2 ϩ 160t ϩ
240 Ն 496. In standard form, we obtain Ϫ16t2 ϩ 160t Ϫ 256 Ն 0 (subtract
496 from both sides). We begin by finding the zeroes of Ϫ16t2 ϩ 160t Ϫ 256 ϭ 0,
noting the related graph opens downward since the leading coefficient is
negative.
Ϫ16t2 ϩ 160t Ϫ 256 ϭ 0
t2 Ϫ 10t ϩ 16 ϭ 0
1t Ϫ 22 1t Ϫ 82 ϭ 0
t Ϫ 2 ϭ 0 or t Ϫ 8 ϭ 0
t ϭ 2 or t ϭ 8
related equation
divide by ؊16
factor
zero factor theorem
result
This shows the rocket is at exactly 496 ft after 2 sec (on its ascent) and after
8 sec (during its descent). We conclude the rocket’s height was greater than
496 ft for 8 Ϫ 2 ϭ 6 sec.
c. When the rocket hits the ground, its height is h ϭ 0. Substituting 0 for h1t2 and
solving gives
h1t2 ϭ Ϫ16t2 ϩ 160t ϩ 240
0 ϭ Ϫ16t2 ϩ 160t ϩ 240
0 ϭ t2 Ϫ 10t Ϫ 15
original function
substitute 0 for h(t )
divide by ؊16
The equation is nonfactorable, so we use the quadratic equation to solve, with
a ϭ 1, b ϭ Ϫ10, and c ϭ Ϫ15:
tϭ
ϭ
Ϫb Ϯ 2b2 Ϫ 4ac
2a
Ϫ1Ϫ102 Ϯ 21Ϫ102 2 Ϫ 41121Ϫ152
2112
10 Ϯ 2160
2
10
4 210
ϭ
Ϯ
2
2
ϭ
quadratic formula
substitute 1 for a, ؊10 for b, ؊15 for c
simplify
2160 ϭ 4 210
ϭ 5 Ϯ 2 210
simplify
Since we need the time t in seconds, we
use the approximate form of the answer,
obtaining t Ϸ Ϫ1.32 and t Ϸ 11.32. The
rocket will return to the ground in just over
11 sec (since t represents time, the solution
t ϭ Ϫ1.32 does not apply). A calculator
check is shown in Figure 3.39 using the
Zeroes Method.
Figure 3.39
800
0
15
Ϫ200
Now try Exercises 157 through 166
EXAMPLE 12
ᮣ
ᮣ
Solving Applications of Inequalities Using the Quadratic Formula
For the years 1995 to 2006, the amount A of annual international telephone traffic
(in billions of minutes) can be modeled by A ϭ 0.17x2 ϩ 8.43x ϩ 64.58 where
x ϭ 0 represents the year 1995 [Source: Data from the 2009 Statistical Abstract of
the United States, Table 1344, page 846]. If this trend continues, in what year will
the annual number of minutes reach or surpass 275 billion?