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College Algebra
Fifth Edition
James Stewart  Lothar Redlin

Saleem Watson
1
Equations and
Inequalities
1.3
Basic Equations
Linear Equations vs. Quadratic Equations
Linear equations are first-degree equations,
such as:
2x + 1 = 5 or 4 – 3x = 2
Quadratic equations are second-degree
equations, such as:
x2 + 2x – 3 = 0 or 2x2 + 3 = 5x
Quadratic Equation—Definition
A quadratic equation is an equation
of the form
ax2 + bx + c = 0
where a, b, and c are
real numbers with a ≠ 0.
Solving Quadratic Equations
by Factoring
Solving Quadratic Equations
Some quadratic equations can be solved
by factoring and using the following
basic property of real numbers.
Zero-Product Property
AB = 0 if and only if
A = 0 or B = 0
• This means that, if we can factor the LHS
of a quadratic (or other) equation, then we can
solve it by setting each factor equal to 0 in turn.
• This method works only when the RHS is 0.
E.g. 1—Solving a Quadratic Equation by Factoring
Solve the equation
x2 + 5x = 24
• We must first rewrite the equation
so that the RHS is 0.
E.g. 1—Solving a Quadratic Equation by Factoring
x2 + 5x = 24
x2 + 5x – 24 = 0
(Subtract 24)
(x – 3)(x + 8) = 0
(Factor)
x – 3 = 0 or x + 8 = 0
(Zero-Product Property)
x=3
x = –8
(Solve)
Solving a Quadratic Equation by Factoring
Do you see why one side of the equation
must be 0 in Example 1?
• Factoring the equation as x(x + 5) = 24
does not help us find the solutions.
• 24 can be factored in infinitely many ways,
such as 6 . 4, ½ . 48, (–2/5) . (–60), and so on.
Solving Quadratic Equations
by Completing the Square
Solving Simple Quadratics
As we saw in Section 1.1, Example 5(b), if a
quadratic equation is of the form (x ± a)2 = c,
we can solve it by taking the square root of
each side.
• In an equation of this form, the LHS is a perfect
square: the square of a linear expression in x.
Completing the Square
So, if a quadratic equation does not factor
readily, we can solve it using the technique
of completing the square.
Completing the Square
2
x2
b

+ bx a perfect square, add   ,
2
To make
the square of half the coefficient of x.
This gives the perfect
square
2
b 
b

x  bx      x  
2
2 
2
2
Completing the Square
To compete the square, we add a constant to
an expression to make it a perfect square.
• For example, to make
x2 – 6x
a perfect square, we must add (6/2)2 = 9. Then
x2 – 6x + 9 = (x – 3)2
is a perfect square.
Completing the Square
The table gives some more examples of
completing the square.
E.g. 2—Solving by Completing the Square
Solve each equation.
(a) x2 – 8x + 13 = 0
(b) 3x2 – 12x + 6 = 0
E.g. 2—Completing the Square
Example (a)
x2 – 8x + 13 = 0
x2 – 8x = –13
(Subtract 13)
x2 – 8x + 16 = –13 + 16
(Complete the square)
(x – 4)2 = 3
(Perfect square)
x–4=± 3
x=4± 3
(Take square root)
(Add 4)
E.g. 2—Completing the Square
Example (b)
First, we subtract 6 from each side.
Then, we factor the coefficient of x2 (the 3)
from the left side.
• This puts the equation in the correct form
for completing the square.
E.g. 2—Completing the Square
3x2 – 12x + 6 = 0
3x2 – 12x = –6
(Subtract 6)
3(x2 – 4x) = –6
(Factor 3 from LHS)
E.g. 2—Completing the Square
Now, we complete the square by adding
(–2)2 = 4 inside the parentheses.
• Since everything inside the parentheses is
multiplied by 3, this means that we are actually
adding 3 . 4 = 12 to the left side of the equation.
• Thus, we must add 12 to the right side as well.
E.g. 2—Completing the Square
3(x2 – 4x + 4) = –6 + 3 . 4
(Complete the square)
3(x – 2)2 = 6
(Perfect square)
(x – 2)2 = 2
(Divide by 3)
x 2 = 2
(Take square root)
x = 2 2
(Add 2)
The Quadratic Formula
Deriving a Formula for the Roots
We can use the technique of completing
the square to derive a formula for the roots
of the general quadratic equation
ax2 + bx + c = 0
The Quadratic Formula
The roots of the quadratic equation
ax2 + bx + c = 0
where a ≠ 0, are:
b  b  4ac
x
2a
2
The Quadratic Formula—Proof
First, we divide each side of the equation
by a and move the constant to the right
side, giving:
b
c
x  x
a
a
2
The Quadratic Formula—Proof
We now complete the square by adding
(b/2a)2 to each side of the equation.
2
2
b
b 
c  b 

x  x         (Complete the square)
a
a  2a 
 2a 
2
 x  b   4ac  b


2
2a 
4a

2
b
b 2  4ac
x

2a
2a
b  b  4ac
x
2a
2
(Perfect square)
(Take square root)
2
(Subtract
b
2a
)
The Quadratic Formula
The quadratic formula could be used to
solve the equations in Examples 1 and 2.
• You should carry out the details
of these calculations.
E.g. 3—Using the Quadratic Formula
Find all solutions of each equation.
(a) 3x2 – 5x – 1 = 0
(b) 4x2 + 12x + 9 = 0
(c) x2 + 2x + 2 = 0
E.g. 3—Using Quadratic Formula
In 3x2 – 5x – 1 = 0,
a = 3 b = –5
Example (a)
c = –1
By the quadratic formula,
  5    5   4  3  1 5  37
x

2 3
6
2
E.g. 3—Using Quadratic Formula
Example (a)
If approximations are desired, we can use
a calculator to obtain:
5  37
x
 1.8471
6
5  37
x
 0.1805
6
E.g. 3—Using Quadratic Formula
Example (b)
Using the quadratic formula
with a = 4, b = 12, and c = 9 gives:
12  12   4  4  9 12  0
3
x


24
8
2
2
• This equation has only one solution, x = –3/2.
E.g. 3—Using Quadratic Formula
Example (c)
Using the quadratic formula
with a = 1, b = 2, and c = 2 gives:
2  2  4  2 2  4 2  2 1
x


2
2
2
 1  1
2
• Since the square of a real number is nonnegative,
1 is undefined in the real number system.
• The equation has no real solution.
Complex Number System
In the next section, we study the complex
number system, in which the square roots
of negative numbers do exist.
• The equation in Example 3(c) does have
solutions in the complex number system.
The Discriminant
Discriminant
The quantity b2 – 4ac that appears under
the square root sign in the Quadratic Formula
is called the discriminant of the equation
ax2 + bx + c = 0.
• It is given the symbol D.
D<0
If D < 0, then b  4ac is undefined.
2
• The quadratic equation has no real
solution—as in Example 3(c).
D = 0 and D > 0
If D = 0, then the equation has only one real
solution—as in Example 3(b).
Finally, if D > 0, then the equation has two
distinct real solutions—as in Example 3(a).
Discriminant
The following box summarizes these
observations.
E.g. 4—Using the Discriminant
Use the discriminant to determine how many
real solutions each equation has.
(a) x2 + 4x – 1 = 0
(b) 4x2 – 12x + 9 = 0
(c) 1/3x2 – 2x + 4 = 0
E.g. 4—Using the Discriminant
Example (a)
x2 + 4x – 1 = 0
The discriminant is:
D = 42 – 4(1)(–1) = 20 > 0
• So, the equation has two distinct real solutions.
E.g. 4—Using the Discriminant
Example (b)
4x2 – 12x + 9 = 0
The discriminant is:
D = (–12)2 – 4 . 4 . 9 = 0
• So, the equation has exactly one real solution.
E.g. 4—Using the Discriminant
Example (c)
1/3x2 – 2x + 4 = 0
The discriminant is:
D = (–2)2 – 4(1/3)4 = –4/3 < 0
• So, the equation has no real solution.
Modeling with
Quadratic Equations
Quadratic Equations in Real Life
Let’s consider a real-life situation that
can be modeled by a quadratic
equation.
• The principles discussed in Section 1.2
for setting up equations as models are
useful here as well.
E.g. 5—Dimensions of a Building Lot
A rectangular building lot is 8 ft
longer than it is wide and has an area
of 2900 ft2.
• Find the dimensions of the lot.
E.g. 5—Dimensions of a Building Lot
We are asked to find the width and
length of the lot.
• So, let:
w = width of lot
E.g. 5—Dimensions of a Building Lot
Then, we translate
the information
in the problem
into the language
of algebra.
In Words
In Algebra
Width of lot
w
Length of lot
w+8
E.g. 5—Dimensions of a Building Lot
Now, we set up the model.
Width of Lot . Length of Lot
= Area of Lot
E.g. 5—Dimensions of a Building Lot
w(w + 8) = 2900
w2 + 8w = 2900
(Expand)
w2 + 8w – 2900 = 0
(w – 50)(w + 58) = 0
w = 50
or
w = –58
(Factor)
(Zero-Product Property)
E.g. 5—Dimensions of a Building Lot
Since the width of the lot must be a positive
number, we conclude that:
w = 50 ft
The length of the lot is:
w + 8 = 50 + 8 = 58 ft
E.g. 6—A Distance-Speed-Time Problem
A jet flew from New York (NY) to Los Angeles
(LA), a distance of 4,200 km.
The speed for the return trip was 100 km/h
faster than the outbound speed.
• If the total trip took 13 hours, what was the jet’s
speed from NY to LA?
E.g. 6—A Distance-Speed-Time Problem
We are asked for the speed of the jet
from NY to LA.
• So, let:
s = speed from NY to LA
• Then,
s + 100 = speed from LA to NY
E.g. 6—A Distance-Speed-Time Problem
Now, we organize the information in a table.
First, we fill in the “Distance” column—as
we know that the cities are 4,200 km apart.
E.g. 6—A Distance-Speed-Time Problem
Then, we fill in the “Speed” column—as
we have expressed both speeds (rates)
in terms of the variable s.
E.g. 6—A Distance-Speed-Time Problem
Finally, we calculate the entries for
the “Time” column, using:
distance
time =
rate
E.g. 6—A Distance-Speed-Time Problem
The total trip took 13 hours.
So, we have the model:
Time from NY to LA
+ Time from LA to NY
= Total time
4200
4200

 13
• This gives:
s
s  100
E.g. 6—A Distance-Speed-Time Problem
Multiplying by the common denominator,
s(s + 100), we get:
4200  s  100   4200s  13s  s  100 
8400s  420,000  13s 2  1300s
0  13s  7100s  420,000
2
• Although this equation does factor,
with numbers this large, it is probably quicker
to use the quadratic formula and a calculator.
E.g. 6—A Distance-Speed-Time Problem
s
7100 
 7100 
2
 4 13  420,000 
2 13 
7100  8500

26
s  600 or
1400
s
 53.8
26
• As s represents speed, we reject the negative
answer and conclude that the jet’s speed from
NY to LA was 600 km/h.
E.g. 7—The Path of a Projectile
An object thrown or fired straight upward
at an initial speed of v0 ft/s will reach a height
of h feet after t seconds, where h and t are
related by the formula
h = –16t2 + v0t
E.g. 7—The Path of a Projectile
Suppose that a bullet is shot straight
upward with an initial speed of 800 ft/s.
a) When does the bullet fall back
to ground level?
b) When does it reach a height
of 6,400 ft?
c) When does it reach a height
of 2 mi?
d) How high is the highest point
the bullet reaches?
E.g. 7—The Path of a Projectile
The initial speed is v0 = 800 ft/s.
Thus, the formula is:
h = –16t2 + 800t
E.g. 7—The Path of a Projectile
Example (a)
Ground level corresponds
to h = 0.
So, we must solve:
0 = –16t2 + 800t
(Set h = 0)
0 = –16t(t – 50)
(Factor)
• Thus, t = 0 or t = 50.
• This means the bullet starts (t = 0) at ground level
and returns to ground level after 50 s.
E.g. 7—The Path of a Projectile
Setting h = 6400 gives:
6400 = –16t2 + 800t
(Set h = 6400)
16t2 – 800t + 6400 = 0
(All terms to LHS)
t2 – 50t + 400 = 0
(Divide by 16)
Example (b)
E.g. 7—The Path of a Projectile
(t – 10)(t – 40) = 0
(Factor)
t = 10 or t = 40
(Solve)
• The bullet reaches 6400 ft
after 10 s (on its ascent)
and again after 40 s
(on its descent to earth).
Example (b)
E.g. 7—The Path of a Projectile
Example (c)
Two miles is:
2 x 5,280 = 10,560 ft
10,560 = –16t2 + 800t
(Set h = 10,560)
16t2 – 800t + 10,560 = 0
(All terms to LHS)
t2 – 50t + 660 = 0 (Divide by 16)
E.g. 7—The Path of a Projectile
The discriminant of this
equation is:
D = (–50)2 – 4(660) = –140
• It is negative.
• Thus, the equation has
no real solution.
• The bullet never reaches
a height of 2 mi.
Example (c)
E.g. 7—The Path of a Projectile
Each height the bullet
reaches is attained twice—
once on its ascent and once
on its descent.
• The only exception is
the highest point of its path,
which is reached only once.
Example (d)
E.g. 7—The Path of a Projectile
Example (d)
This means that, for
the highest value of h,
the following equation has
only one solution for t:
h = –16t2 + 800t
16t2 – 800t + h = 0
• This, in turn, means that the discriminant
of the equation is 0.
E.g. 7—The Path of a Projectile
So,
D = (–800)2 – 4(16)h = 0
640,000 – 64h = 0
h = 10,000
• The maximum height
reached is 10,000 ft.
Example (d)