Download 1.3 Solving Equations in One Variable Using a Graphing Utility

24
CHAPTER 1
1.3
Graphs
Solving Equations in One Variable Using a Graphing Utility
PREPARING FOR THIS SECTION
Before getting started, review the following:
• Linear, Rational, and Quadratic Equations (Appendix, Section A.5, pp. 984–997)
Now work the ‘Are You Prepared?’ problems on page 26.
OBJECTIVE
1
Solve Equations in One Variable Using a Graphing Utility
1 Solve Equations in One Variable Using a Graphing Utility
✓
In this text, we present two methods for solving equations: algebraic and graphical.We
shall see as we proceed through this book that some equations can be solved using algebraic techniques that result in exact solutions. For other equations, however, there
are no algebraic techniques that lead to an exact solution. For such equations, a graphing utility can often be used to investigate possible solutions. When a graphing utility
is used to solve an equation, usually approximate solutions are obtained.
One goal of this text is to determine when equations can be solved algebraically. If
an algebraic method for solving an equation exists, we shall use it to obtain an exact solution. A graphing utility can then be used to support the algebraic result. However, if
an equation must be solved for which no algebraic techniques are available, a graphing
utility will be used to obtain approximate solutions. Unless otherwise stated, we shall
follow the practice of giving approximate solutions rounded to two decimal places.
The ZERO (or ROOT) feature of a graphing utility can be used to find the solutions of an equation when one side of the equation is 0. In using this feature to
solve equations, we make use of the fact that the x-intercepts (or zeros) of the graph
of an equation are found by letting y = 0 and solving the equation for x. Solving an
equation for x when one side of the equation is 0 is equivalent to finding where the
graph of the corresponding equation crosses or touches the x-axis.
EXAMPLE 1
Using ZERO (or ROOT) to Approximate Solutions of an Equation
Find the solution(s) of the equation x3 - x + 1 = 0. Round answers to two decimal places.
Solution
The solutions of the equation x3 - x + 1 = 0 are the same as the x-intercepts of
the graph of Y1 = x3 - x + 1. We begin by graphing the equation.
Figure 32 shows the graph. From the graph there appears to be one x-intercept
(solution to the equation) between -2 and -1.
Using the ZERO (or ROOT) feature of our graphing utility, we determine that
the x-intercept, and thus the solution to the equation, is x = -1.32, rounded to two
decimal places. See Figure 33.
Figure 32
Figure 33
10
–3
10
–3
3
–5
NOW WORK PROBLEM
3
–5
5.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
SECTION 1.3
Solving Equations in One Variable Using a Graphing Utility
25
A second method for solving equations using a graphing utility involves the
INTERSECT feature of the graphing utility. This feature is used most effectively
when one side of the equation is not 0.
EXAMPLE 2
Using INTERSECT to Approximate Solutions of an Equation
Find the solution(s) to the equation 4x4 - 3 = 2x + 1. Round answers to two decimal places.
Solution
Figure 34
10
–4
4
–10
We begin by graphing each side of the equation as follows: graph Y1 = 4x4 - 3 and
Y2 = 2x + 1. See Figure 34. At the point of intersection of the graphs, the value of
the y-coordinate is the same. Thus, the x-coordinate of the point of intersection represents the solution to the equation. Do you see why?
The INTERSECT feature on a graphing utility determines the point of intersection of the graphs. Using this feature, we find that the graphs intersect at
1-0.87, -0.732 and 11.12, 3.232, rounded to two decimal places. See Figures 35(a)
and (b). The solutions of the equation are x = -0.87 and x = 1.12, rounded to two
decimal places.
Figure 35
10
–4
10
4 –4
–10
(a)
4
–10
(b)
NOW WORK PROBLEM
7.
The steps to follow for solving equations graphically are given next.
Steps for Solving Equations Graphically Using ZERO (or ROOT)
STEP 1: Write the equation in the form 5expression in x6 = 0.
STEP 2: Graph Y1 = 5expression in x6.
STEP 3: Use ZERO (or ROOT) to determine each x-intercept of the graph.
Steps for Solving Equations Graphically Using INTERSECT
STEP 1: Graph Y1 = 5expression in x on the left side of the equation6;
Y2 = 5expression in x on the right side of the equation6.
STEP 2: Use INTERSECT to determine the x-coordinate of each point of
intersection.
In the next example, we solve an equation both algebraically and graphically.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
26
CHAPTER 1
Graphs
EXAMPLE 3
Solving an Equation Algebraically and Graphically
Solve the equation:
31x - 22 = 51x - 12
Algebraic Solution
31x - 22
3x - 6
3x - 6 - 5x
-2x - 6
-2x - 6 + 6
-2x
-2 x
-2
Graphing Solution
51x - 12
5x - 5
5x - 5 - 5x
-5
-5 + 6
1
1
=
-2
1
x = 2
=
=
=
=
=
=
Graph Y1 = 31x - 22 and Y2 = 51x - 12.
See Figure 36. Using INTERSECT, we find
the point of intersection to be 1-0.5, -7.52.
The solution of the equation is x = -0.5.
Remove the parentheses.
Subtract 5x from each side.
Simplify.
Add 6 to each side.
Simplify.
Figure 36
Divide each side by -2.
5
–5
Simplify.
1
in the expression in x on the left side of the
2
1
equation and simplify. Let x = - in the expression in x on the right
2
side of the equation and simplify. If the two expressions are equal, the solution checks.
5
✔ CHECK: Let x = -
–15
1
5
15
- 2b = 3 a - b = 2
2
2
1
3
15
51x - 12 = 5a - - 1b = 5 a - b = 2
2
2
31x - 22 = 3a -
Since the two expressions are equal, the solution x = -
1
checks.
2
NOW WORK PROBLEM
19.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall