College Algebra
Sixth Edition
James Stewart  Lothar Redlin

Saleem Watson
1
Equations and
Inequalities
1.4
Solving Equations
Graphically
Solving Equations Graphically
In this section, we use graphs to solve
equations.
• To do this, we must first draw a graph using
a graphing device.
• So, we begin by giving a few guidelines to
help us use graphing devices effectively.
Solving Equations Algebraically
In Chapter 1, we learned how to solve
equations.
To solve an equation like 0 = 3x – 5,
we used the algebraic method.
• This means we used the rules of algebra
to isolate x on one side of the equation.
Solving Equations Algebraically
We view x as an unknown and we use
the rules of algebra to isolate it on one side of
the equation.
• For this equation we add 5 and then divide by
3 to obtain the solution x = 5/3.
Solving Equations Graphically
We can also solve this equation by
the graphical method.
• We view x as a variable and sketch the graph
of the equation y = 3x – 5.
• Different values for x give different values for y.
• Our goal is to find the value of x for which y = 0.
Solving Equations Graphically
From the graph,
we see that y = 0
when x ≈ 1.7.
• Thus, the solution is
x ≈ 1.7.
• Note that, from
the graph, we obtain
an approximate solution.
Solving Equations Graphically
We summarize these methods here.
Algebraic Method—Advantages
The advantages of the algebraic method
are that:
• It gives exact answers.
• The process of unraveling the equation to arrive
at the answer helps us understand the algebraic
structure of the equation.
Algebraic Method—Disadvantage
On the other hand, for many
equations, it is difficult or impossible
to isolate x.
Graphical Method—Advantages
The graphical method gives a numerical
approximation to the answer.
• This is an advantage when a numerical answer
is desired.
• For example, an engineer might find an answer
expressed as x ≈ 2.6 more immediately useful
than x = 7.
Graphical Method—Advantages
Also, graphing an equation helps us
visualize how the solution is related
to other values of the variable.
E.g. 1—Solving an Equation Algebraically and Graphically
Solve the quadratic equations algebraically
and graphically.
x2 – 7 = 0
E.g. 1—Solving Algebraically
x 7  0
2
x 7
2
x 7
• There is no real solution.
(Add 7)
E.g. 1—Solving Graphically
We graph the equations
y = x2 – 7
By determining the x-intercepts of the graphs,
we find the following solutions.
E.g. 1—Solving Graphically
We see that the graph crosses the x ≈ 2.6
and x ≈ –2.6
E.g. 2—Another Graphical Method
Solve the equation algebraically
and graphically:
5 – 3x = 8x – 20
E.g. 2—Algebraic Solution
5 – 3x = 8x – 20
– 3x = 8x – 25
(Subtract 5)
–11x = –25
(Subtract 8x)
25
3
x=
 2 11
11
(Divide by  11 and simplify)
E.g. 2—Graphical Solution
We could:
•
Move all terms to one side of the equal sign.
•
Set the result equal to y.
•
Graph the resulting equation.
E.g. 2—Graphical Solution
However, to avoid all that algebra, we
graph two equations instead:
y1 = 5 – 3x
and
y2 = 8x – 20
• The solution of the original equation will be
the value of x that makes y1 equal to y2.
• That is, the solution is the x-coordinate
of the intersection point of the two graphs.
E.g. 2—Graphical Solution
Using the TRACE feature or the intersect
command on a graphing calculator, we see
that the solution is
x ≈ 2.27.
Solving Equations Graphically
In the next example, we use the graphical
method to solve an equation that is extremely
difficult to solve algebraically.
E.g. 3—Solving an Equation in an Interval
Solve the equation x 3  6x 2  9x  x
in the interval [1, 6].
• We need to find all solutions x that satisfy 1 ≤ x ≤ 6.
• So, we will graph the equation in a viewing
rectangle for which the x-values are restricted
to this interval.
x3  6x 2  9x  x
x3  6x 2  9x  x  0
(Subtract
x)
E.g. 3—Solving an Equation in an Interval
The figure shows the graph of the equation
in the viewing rectangle [1, 6] by [–5, 5].
• There are two x-intercepts in this rectangle.
E.g. 3—Solving an Equation in an Interval
Zooming in, we see that the solutions
are:
x ≈ 2.18 and x ≈ 3.72
Solving an Equation in an Interval
The equation in Example 3 actually
has four solutions.
• You are asked to find the other two
in Exercise 32.
E.g. 4—Intensity of Light
Two light sources are 10 m apart.
• One is three times as intense as the other.
• The light intensity L (in lux) at a point x meters
from the weaker source
is given by:
10
30
L 2 
x
10  x 2


E.g. 4—Intensity of Light
Find the points at which the light
intensity is 4 lux.
10
30
• We need to solve the equation 4  2 
2
x
10

x


E.g. 4—Intensity of Light
The graphs of
10
30
y1  4 and y 2  2 
2
x
10

x


are shown.
• Zooming in (or
using the intersect
command), we find
two solutions:
x ≈ 1.67431
and x ≈ 7.1927193
E.g. 4—Intensity of Light
So, the light intensity is 4 lux at
points that are 1.67 m and 7.19 m
from the weaker source.