```Chapter 3
Functions and
Equations
1
3.2
Equations and
Problem Solving
♦
equations
♦ Use factoring, the square root property,
completing the square, and the quadratic
♦ Understand the discriminant
♦ Solve problems involving quadratic equations
2
A quadratic equation in one variable is an
equation that can be written in the form
ax2 + bx + c
where a, b, and c are constants with a ≠ 0.
3
Quadratic equations can have no real
solutions, one real solution, or two real
solutions.
The are four basic symbolic strategies in
which quadratic equations can be solved.
• Factoring
• Square root property
• Completing the square
4
Factoring
Factoring is a common technique used to
solve equations. It is based on the zeroproduct property, which states that if
ab = 0, then a = 0 or b = 0 or both. It is
important to remember that this property
works only for 0. For example, if ab = 1,
then this equation does not imply that
either a = 1 or b = 1. For example, a = 1/2
and b = 2 also satisfies ab = 1 and neither
a nor b is 1.
5
equations with factoring
Solve the quadratic equation 12t2 = t + 1. Check
Solution
2
12t  t  1
12t  t  1 0
2
3t  14t  1 0
3t  1 0 or 4t  1 0
1
t
or
3
1
t
4
6
equations with factoring
Solution continued
Check:
2
 1
1
12     1
3
 3
2
 1
1
12       1
4
 4
4 4

3 3
3 3

4 4
7
The Square Root Property
Let k be a nonnegative number. Then the
solutions to the equation
x2 = k
are given by
x   k.
8
Example: Using the square root
property
If a metal ball is dropped 100 feet from a
water tower, its height h in feet above the
ground after t seconds is given by
h(t) = 100 – 16t2. Determine how long it takes
the ball to hit the ground.
Solution
The ball strikes the ground when the equation
100 – 16t2 = 0 is satisfied.
9
Example: Using the square root
property
Solution continued 100  16t 2  0
100  16t
2
100
t 
16
2
100
10
t

16
4
The ball strikes the ground after 10/4, or 2.5,
seconds.
10
Completing the Square
Another technique that can be used to
solve a quadratic equation is completing
the square. If a quadratic equation is
written in the form x2 + kx =d, where k and
d are constants, then the equation can be
solved using
2
2
k

k
x  kx      x   .
2
 2

2
11
Example: Completing the Square
Solve 2x2 – 8x = 7.
Solution
Divide each side by 2:
7
2
x  4x 
2
7
2
x  4x  4   4
2
2
15
x2 
2


15
x2 
2
15
x  2
2
12
Symbolic, Numerical and Graphical
Solutions
symbolically, numerically, and graphically.
The following example illustrates each
technique for the equation x(x – 2) = 3.
13
Symbolic Solution
14
Numerical Solution
15
Graphical Solution
16
The solutions to the quadratic equation
ax2 + bx + c = 0, where a ≠ 0, are given by
b  b  4ac
x
.
2a
2
17
formula
Solve the equation 3x2 – 6x + 2 = 0.
Solution
Let a = 3, b = 6, and c = 2.
b  b 2  4ac
x
2a
x
  6  
 6   4  3  2 
2 3
2
6  12
1
1
x
 1
4  3  1
3
6
6
3
18
The Discriminant
If the quadratic equation ax2 + bx + c = 0 is
solved graphically, the parabola
y = ax2 + bx + c can intersect the x-axis zero,
one, or two times. Each x-intercept is a real
b 2  4ac  0
b 2  4ac  0
b 2  4ac  0
19
To determine the number of real solutions to
ax2 + bx + c = 0 with a ≠ 0, evaluate the
discriminant b2 – 4ac.
1. If b2 – 4ac > 0, there are two real
solutions.
2. If b2 – 4ac = 0, there is one real solution.
3. If b2 – 4ac < 0, there are no real
solutions.
20
Example: Using the discriminant
Use the discriminant to find the number of
solutions to 4x2 – 12x + 9 = 0. Then solve the
equation by using the quadratic formula.
Solution
Let a = 4, b = –12, and c = 9
b2 – 4ac = (–12)2 – 4(4)(9) = 0
Discriminant is 0, there is one solution.
21
Example: Using the discriminant
Solution continued
b  b  4ac
x
2a
2
x
  12   0
8
3
x
2
The only solution is 3/2.
22
Example: Using the discriminant
Solution continued
The graph suggests there is only one
intercept 3/2.
23
Problem Solving
Many types of applications involve
problems, we use the steps for “Solving
Application Problems” from Section 2.2.
24
Example: Solving a construction
problem
A box is being constructed by cutting
2-inch squares from the corners of a rectangular
piece of cardboard that is 6 inches longer than it is
wide. If the box is to have a volume of 224 cubic
inches, find the dimensions of the piece of
cardboard.
25
Example: Solving a construction
problem
Solution
Step 1: Let x be the width and x + 6 be the
length.
Step 2: Draw a picture.
26
Example: Solving a construction
problem
Since the height times the width times the length
must equal the volume, or 224 cubic inches, the
following can be written
2(x – 4)(x + 2) = 224
Step 3: Write the quadratic equation in
the form ax2 + bx + c = 0 and factor.
x  2x  8  112
2
x 2  2x  120  0
x  12x  10 0
x  12 or x  10
27
Example: Solving a construction
problem
The dimensions can not be negative, so the
width is 12 inches and the length is 6
inches more, or 18 inches.
Step 4:
After the 2-inch-square corners are cut out,
the dimensions of the bottom of the box are
12 – 4 = 8 inches by 18 – 4 = 14 inches.
The volume of the box is then 2•8•14 = 224
cubic inches, which checks.
28
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