Transcript 10TH EDITION COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER 1.6 - 1 1.6 Other Types of Equations and Applications Rational Expressions Work Rate Problems Equations with Radicals Equations Quadratic in Form 1.6 - 2 Rational Equations A.

10TH EDITION
COLLEGE ALGEBRA
LIAL
HORNSBY
SCHNEIDER
1.6 - 1
1.6
Other Types of Equations and
Applications
Rational Expressions
Work Rate Problems
Equations with Radicals
Equations Quadratic in Form
1.6 - 2
Rational Equations
A rational equation is an equation that has a
rational expression for one or more terms.
Because a rational expression is not
defined when its denominator is 0, values
of the variable for which any
denominator equals 0 cannot be
solutions of the equation.
To solve a rational equation, begin by
multiplying both sides by the least common
denominator of the terms of the equation.
1.6 - 3
Example 1
SOLVING RATIONAL EQUATIONS THAT
LEAD TO LINEAR EQUATIONS
Solve each equation.
3 x  1 2x
a.

x
3
x 1
Solution
The least common denominator is 3(x – 1),
which is equal to 0 if x = 1. Therefore, 1
cannot possibly be a solution of this
equation.
1.6 - 4
Example 1
SOLVING RATIONAL EQUATIONS THAT
LEAD TO LINEAR EQUATIONS
Solve each equation.
3 x  1 2x
a.

x
3
x 1
Solution
3 x  1 2x

x
3
x 1
 3 x  1
 2x 
3( x  1) 
  3( x  1) 
  3( x  1)x
 3 
 x  1
Multiply by the LCD,
3(x – 1), where x ≠ 1.
1.6 - 5
Example 1
SOLVING RATIONAL EQUATIONS THAT
LEAD TO LINEAR EQUATIONS
Solve each equation.
3 x  1 2x
a.

x
3
x 1
Solution
 3 x  1
 2x 
3( x  1) 
  3( x  1) 
  3( x  1)x
 3 
 x  1
on
( x  1)(3 x  1)  3(2x )  3 x( x  1) Simplify
both sides.
3x  4x  1 6x  3x  3x
2
2
Multiply.
1.6 - 6
Example 1
SOLVING RATIONAL EQUATIONS THAT
LEAD TO LINEAR EQUATIONS
Solve each equation.
3 x  1 2x
a.

x
3
x 1
Solution
2
2
3x  4x  1 6x  3x  3x
1 10x  3x
1  7x
1
x
7
Multiply.
Subtract 3x2;
combine terms.
Solve the linear
equation.
The restriction x ≠ 1 does
not affect this result.
1.6 - 7
Example 1
SOLVING RATIONAL EQUATIONS THAT
LEAD TO LINEAR EQUATIONS
Solve each equation.
3 x  1 2x
a.

x
3
x 1
Solution
1
x
7
The restriction x ≠ 1 does
not affect this result.

1
The solution set is
.
7
1.6 - 8
Example 1
SOLVING RATIONAL EQUATIONS THAT
LEAD TO LINEAR EQUATIONS
Solve each equation.
b. x  2  2
x 2 x 2
Solution
 x 
 2 
( x  2) 
  ( x  2) 
  ( x  2)2
 x  2
 x  2
Multiply by the LCD, x – 2, where x ≠ 2.
x  2  2( x  2)
1.6 - 9
Example 1
SOLVING RATIONAL EQUATIONS THAT
LEAD TO LINEAR EQUATIONS
Solve each equation.
b. x  2  2
x 2 x 2
Solution
x  2  2( x  2)
x  2  2x  4
 x  2
x 2
Distributive property
Subtract 2x; combine
like terms.
Multiply by –1.
1.6 - 10
Example 1
SOLVING RATIONAL EQUATIONS THAT
LEAD TO LINEAR EQUATIONS
Solve each equation.
b. x  2  2
x 2 x 2
Solution
The only proposed solution is 2. However,
the variable is restricted to real numbers
except 2; if x = 2, then multiplying by x – 2 in
the first step is multiplying both sides by 0,
which is not valid. Thus, the solution set is .
1.6 - 11
Example 2
SOLVING RATIONAL EQUATIONS THAT
LEAD TO QUADRATIC EQUATIONS
Solve each equation.
a. 3 x  2  1  2
2
x  2 x x  2x
Solution
3x  2 1
2
 
x  2 x x( x  2)
Factor the last
denominator.
 2 
 3x  2 
 1
x( x  2) 
  x( x  2)    x( x  2) 

 x 2 
x
 x( x  2) 
Multiply by x(x –2), x ≠ 0, 2.
1.6 - 12
Example 2
SOLVING RATIONAL EQUATIONS THAT
LEAD TO QUADRATIC EQUATIONS
Solve each equation.
a. 3 x  2  1  2
2
x  2 x x  2x
Solution
 2 
 3x  2 
 1
x( x  2) 
  x( x  2)    x( x  2) 

x

2
x
x
(
x

2)


 


Multiply by x(x –2), x ≠ 0, 2.
x(3 x  2)  ( x  2)  2
3 x 2  2x  x  2  2
Distributive
property
1.6 - 13
SOLVING RATIONAL EQUATIONS THAT
LEAD TO QUADRATIC EQUATIONS
Example 2
Solve each equation.
a. 3 x  2  1  2
2
x  2 x x  2x
Solution
3 x  2x  x  2  2
2
Distributive
property
3x 2  3x  0
Standard form
3 x( x  1)  0
Factor.
1.6 - 14
Example 2
SOLVING RATIONAL EQUATIONS THAT
LEAD TO QUADRATIC EQUATIONS
Solve each equation.
a. 3 x  2  1  2
2
x  2 x x  2x
Solution
Set each
factor equal
to 0.
3 x( x  1)  0
3x  0 or
x 1 0
x  0 or x  1
Factor.
Zero-factor
property
Proposed
solutions
1.6 - 15
Example 2
SOLVING RATIONAL EQUATIONS THAT
LEAD TO QUADRATIC EQUATIONS
Solve each equation.
a. 3 x  2  1  2
2
x  2 x x  2x
Solution
Because of the restriction x ≠ 0, the only
valid solution is –1.
The solution set is {–1}.
1.6 - 16
Example 2
SOLVING RATIONAL EQUATIONS THAT
LEAD TO QUADRATIC EQUATIONS
Solve each equation.
b.  4 x  4  8
2
x 1 x 1 x 1
Solution  4 x
4
8


x  1 x  1 ( x  1)( x  1)
Factor.
4
  4x 
( x  1)( x  1) 
  ( x  1)( x  1)
x 1
 x  1
8


 ( x  1)( x  1) 

 ( x  1)( x  1) 
1.6 - 17
Example 2
SOLVING RATIONAL EQUATIONS THAT
LEAD TO QUADRATIC EQUATIONS
Solve each equation.
b.  4 x  4  8
2
x 1 x 1 x 1
Solution ( x  1)( x  1)  4 x   ( x  1)( x  1) 4


x 1
 x  1
8
 ( x  1)( x  1)
( x  1)( x  1)
4 x( x  1)  4( x  1)  8
1.6 - 18
Example 2
SOLVING RATIONAL EQUATIONS THAT
LEAD TO QUADRATIC EQUATIONS
Solve each equation.
b.  4 x  4  8
2
x 1 x 1 x 1
Solution
 4 x( x  1)  4( x  1)  8
 4 x 2  4 x  4 x  4  8
 4x  4  0
2
x 1 0
2
Distributive
property
Standard form
Divide by –4.
1.6 - 19
SOLVING RATIONAL EQUATIONS THAT
LEAD TO QUADRATIC EQUATIONS
Example 2
Solve each equation.
b.  4 x  4  8
2
x 1 x 1 x 1
Solution
x 1 0
2
( x  1)( x  1)  0
x  1  0 or
x  1
or
x 1 0
x 1
Divide by –4.
Factor.
Zero-factor
property
Proposed
solutions
1.6 - 20
Example 2
SOLVING RATIONAL EQUATIONS THAT
LEAD TO QUADRATIC EQUATIONS
Solve each equation.
b.  4 x  4  8
2
x 1 x 1 x 1
Solution
x  1
or
x 1
Proposed
solutions
Since the restrictions on x are x ≠ 1, –1,
neither proposed solution is valid, so the
solution set is .
1.6 - 21
Work Rate Problems
Problem Solving If a job can be done in t units
of time, then the rate of work is 1/t of the job per
time unit. Therefore,
rate  time = portion of the job completed.
If the letters r, t, and A represent the rate at which
work is done, the time, and the amount of work
accomplished, respectively, then A  r t .
1.6 - 22
Work Rate Problems
Problem Solving Amounts of work are often
measured in terms of the number of jobs
accomplished. For instance, if one job is
accomplished in t time units, then A = 1 and
1
r
t
1.6 - 23
Example 3
SOLVING A WORK RATE
PROBLEM
One computer can do a
job twice as fast as
another. Working
together, both
computers can do the
job in 2 hr. How long
would it take each
computer, working
alone, to do the job?
1.6 - 24
Example 3
SOLVING A WORK RATE
PROBLEM
Solution
Step 1 Read the problem. We must find the
time it would take each computer working
alone to do the job.
1.6 - 25
Example 3
SOLVING A WORK RATE
PROBLEM
Solution
Step 2 Assign a variable. Let x represent
the number of hours it would take the faster
computer, working alone, to do the job. The
time for the slower computer to do the job
alone is then 2x. Therefore,
1
 rate of the faster computer (job per hour)
x
and
1
 rate of the slower computer (job per hour).
2x
1.6 - 26
SOLVING A WORK RATE
PROBLEM
Example 3
Solution
Step 2 Assign a variable.
Rate
Faster
Computer
Slower
Computer
1
x
1
2x
Time
2
2
Part of the Job
Accomplished
 1 2
2  
x x
 1 1
2  
 2x  x
A = rt
1.6 - 27
SOLVING A WORK RATE
PROBLEM
Example 3
Solution
Step 3 Write an equation. The sum of the
two parts of the job together is 1, since one
whole job is done.
Part of the
job done by
the faster
+
computer
2
x

Part of the
job done by
the slower
computer
1
x
=
One
whole job

1
1.6 - 28
Example 3
SOLVING A WORK RATE
PROBLEM
Solution
Step 4 Solve.
 2 1
x     x(1)
x x
2
 1
x    x    x(1)
x
x
2 1 x
3x
Multiply both sides by x.
Distributive property
Multiply.
Add.
1.6 - 29
Example 3
SOLVING A WORK RATE
PROBLEM
Solution
Step 5 State the answer. The faster
computer would take 3 hr to do the job
alone, while the slower computer would take
2(3) = 6 hr. Be sure to give both answers
here.
Step 6 Check. The answer is reasonable,
since the time working together (2 hr) is less
than the time it would take the faster
computer working alone.
1.6 - 30
Rate Problem
Note The sum of the rates of the
individual computers is equal to their rate
working together. 1 1 1


x 2x 2
Multiply both
1 1 
 1
2x  
  2x  
sides by 2x.
 x 2x 
 2
2 1 x
Same solution
x  3 found earlier
1.6 - 31
Power Property
If P and Q are algebraic expressions,
then every solution of the equation
P = Q is also a solution of the equation
Pn = Qn , for any positive number.
1.6 - 32
Caution Be very careful when using
the power property. It does not say that
the equations P = Q and Pn = Qn are
equivalent; it only says that each solution of
the original equation P = Q is also a
solution of the new equation Pn = Qn.
1.6 - 33
Solving an Equation Involving
Radicals
Step 1 Isolate the radical on one side of the equation.
Step 2 Raise each side of the equation to a power that
is the same as the index of the radical so that
the radical is eliminated.
If the equation still contains a radical, repeat
Steps 1 and 2.
Step 3 Solve the resulting equation.
Step 4 Check each proposed solution in the original
equation.
1.6 - 34
SOLVING AN EQUATION
CONTAINING A RADICAL
Example 4
Solve x  15  2 x  0
Solution
x  15  2 x
x 
2

15  2x
Isolate the radical.

2
Square both sides.
x 2  15  2x
1.6 - 35
SOLVING AN EQUATION
CONTAINING A RADICAL
Example 4
Solve x  15  2 x  0
Solution:
x  15  2x
2
x  2x  15  0
2
( x  5)( x  3)  0
x 5 0
x  5
or x  3  0
or
x 3
Solve the quadratic
equation.
Factor.
Zero-factor property
Proposed solutions
1.6 - 36
SOLVING AN EQUATION
CONTAINING A RADICAL
Example 4
Solve x  15  2 x  0 
Solution:
x  5
or
x 3
Proposed solutions
Only 3 is a solution, giving the
solution set {3}.
1.6 - 37
SOLVING AN EQUATION
CONTAINING TWO RADICALS
Example 5
Solve
2x  3  x  1  1
Solution When an equation contains two
radicals, begin by isolating one of the radicals
on one side of the equation.
2x  3  x  1  1
2x  3  1  x  1

2x  3
  1
2
Isolate

x 1
2x  3
2
Square both sides.
1.6 - 38
SOLVING AN EQUATION
CONTAINING TWO RADICALS
Example 5
Solve
2x  3  x  1  1
Solution

2x  3
  1
2

x 1
2
2x  3  1  2 x  1  ( x  1)
Square
both
sides.
Be careful!
Don’t forget this
term when squaring.
x 1 2 x 1
Isolate the
remaining
radical.
1.6 - 39
SOLVING AN EQUATION
CONTAINING TWO RADICALS
Example 5
2x  3  x  1  1
Solve
Solution
Isolate the remaining
radical.
x 1 2 x 1
 x  1   2 x  1
2
2
Square again.
x  2 x  1  4( x  1)
2
x  2x  1  4 x  4
2
1.6 - 40
Example 5
SOLVING AN EQUATION
CONTAINING TWO RADICALS
2x  3  x  1  1
Solve
Solution
x  2x  1  4 x  4
2
Solve the
quadratic
equation.
x  2x  3  0
2
( x  3)( x  1)  0
x  3  0 or
x 3
or
x 1 0
x  1
Proposed
solutions
1.6 - 41
Example 5
Solve
SOLVING AN EQUATION
CONTAINING TWO RADICALS
2x  3  x  1  1
Solution
x 3
or
x  1
Proposed
solutions
Both 3 and –1 are solutions of the
original equation, so {3, –1} is the
solution set.
1.6 - 42
Caution Remember to isolate a radical
in Step 1. It would be incorrect to square
each term individually as the first step in
Example 5.
1.6 - 43
SOLVING AN EQUATION CONTAINING
A RADICAL (CUBE ROOT)
Example 6
3
Solve
4x  4x  1  x  0
2
3
Solution
3

3
4x 2  4x  1  3 x
Isolate a radical.
  x
3
4x  4x  1 
2
3
3
Cube both sides.
4x  4x  1  x
2
4x 2  5x  1  0
Solve the quadratic
equation.
1.6 - 44
SOLVING AN EQUATION CONTAINING
A RADICAL (CUBE ROOT)
Example 6
Solve
3
4x  4x  1  x  0
2
3
Solution
4x 2  5x  1  0
Solve the
quadratic
equation.
(4 x  1)( x  1)  0
4x  1  0 or
1
x
or
4
x 1 0
x 1
Proposed
solutions
1.6 - 45
SOLVING AN EQUATION CONTAINING
A RADICAL (CUBE ROOT)
Example 6
Solve
3
Solution
4x  4x  1  x  0
2
1
x
or
4
3
x 1
Proposed
solutions
Both are valid solutions, and
the solution set is {¼ ,1}.
1.6 - 46
Equation Quadratic in Form
An equation is said to be quadratic in
form if it can be written as
au  bu  c  0,
2
where a ≠ 0 and u is some algebraic
expression.
1.6 - 47
SOLVING EQUATIONS
QUADRATIC IN FORM
Example 7
a. Solve
( x  1)  ( x  1)  2  0
2
1
3
3
Solution Since
2
( x  1)  ( x  1)  , let u  ( x  1)
2
3
1
3
u u 2  0
2
1
3
Substitute.
(u  2)(u  1)  0
Factor.
u 20
or
u 1 0
u2
or
u  1
Zero-factor
property
1.6 - 48
SOLVING EQUATIONS
QUADRATIC IN FORM
Example 7
( x  1)  ( x  1)  2  0
2
a. Solve
1
3
3
Don’t forget
this step.
u2
or
( x  1)  2
1
3
or
u  1
( x  1)  1
1
3
Zero-factor
property
Replace u with
( x  1) 3 .
1
( x  1)   2 or ( x  1)   ( 1)3
Cube each
side.
x  7 or x  2
Proposed
solutions
1
3
3
3
1
3
3
1.6 - 49
SOLVING EQUATIONS
QUADRATIC IN FORM
Example 7
a. Solve
( x  1)  ( x  1)  2  0
2
3
Solution
x  7 or x  2
1
3
Proposed
solutions
The solution set is {–2, 7}.
1.6 - 50
Example 7
SOLVING EQUATIONS
QUADRATIC IN FORM
b. Solve 6 x 2  x 1  2
Solution
2
1
6x  x  2  0
Subtract.
Let u  x 1;
6u  u  2  0
2
thus u 2  x 2.
Factor.
(3u  2)(2u  1)  0
3u  2  0 or
Remember
to
substitute
for u.
2
u
3
or
2u  1  0
Zero-factor
property
1
u
2
1.6 - 51
Example 7
SOLVING EQUATIONS
QUADRATIC IN FORM
b. Solve 6 x 2  x 1  2
Solution
2
u
3
2
x 
3
3
x
2
1
or
1
u
2
or
1
or
x
x 2
1

2
Substitute
again.
x-1 is the
reciprocal of x.
1.6 - 52
Example 7
SOLVING EQUATIONS
QUADRATIC IN FORM
b. Solve 6 x 2  x 1  2
Solution
3
x
2
or
x 2
x-1 is the
reciprocal of x.
 
3
The solution set is  , 2
2
1.6 - 53
SOLVING AN EQUATION
QUADRATIC IN FORM
Example 8
Solve 12x 4  11x 2  2  0
Solution
12  x

2 2
 11x  2  0
2
12u 2  11u   0
x4   x2 
2
Let u = x2; thus u2 = x4
(3u  2)(4u  1)  0
Solve the quadratic
equation.
3u  2  0
Zero-property factor
or
4u  1  0
1.6 - 54
Example 8
SOLVING AN EQUATION
QUADRATIC IN FORM
Solve 12x 4  11x 2  2  0
Solution
3u  2  0
2
u
3
4u  1  0
1
or u 
4
2
1
2
2
or x 
x 
3
4
or
Zero-property
factor
Replace
u with x2.
1.6 - 55
SOLVING AN EQUATION
QUADRATIC IN FORM
Example 8
Solve 12x 4  11x 2  2  0
Solution
2
x 
3
2
2
x
3
 2
x
3
1
or x 
4
Replace
u with x2.
1
or x  
4
Square root
property
2
3
3 or
1
x
2
Simplify radicals.
1.6 - 56
Example 8
SOLVING AN EQUATION
QUADRATIC IN FORM
Solve 12x 4  11x 2  2  0
Solution
 2
x
3
3
3 or
6
x
3
 6 1
The solution set is 
,  .
2
 3
1
x
2
Simplify
radicals.
1.6 - 57
Equation Quadratic in Form
Note Some equations that are quadratic
in form are simple enough to avoid using the
substitution variable technique.
12x 4  11x 2  2  0
Factor directly (3x2 – 2)(4x2 – 1) , setting
each factor equal to zero, and then solving
the resulting two quadratic equations. It is a
matter of personal preference as to which
method to use.
1.6 - 58
SOLVING AN EQUATION THAT LEADS
TO ONE THAT IS QUADRATIC IN FORM
Example 9
Solve (5 x  6)  x
1
4
2
Solution
 5 x 2  6    x 4


1
4
4
5x  6  x
2
4
Raise both sides to
the fourth power.
Power rule for
exponents.
x 4  5x 2  6  0
u  5u  6  0
2
Let u = x2; thus u2 = x4.
1.6 - 59
SOLVING AN EQUATION THAT LEADS
TO ONE THAT IS QUADRATIC IN FORM
Example 9
Solve (5 x  6)  x
2
Solution
1
4
u 2  5u  6  0
(u  3)(u  2)  0
u 3  0
u 3
or
or
u 20
u2
Let u = x2; thus
u2 = x4.
Factor.
Zero-property
factor
1.6 - 60
Example 9
SOLVING AN EQUATION THAT LEADS
TO ONE THAT IS QUADRATIC IN FORM
Solve (5 x  6)  x
2
Solution u  3
2
x 3
x 3
1
4
or
or
u2
2
x 2
or
x 2
u = x2
Square root
property
Checking the four proposed solutions in the original
equation shows that only 3 and 2 are solutions,
since the left side of the equation cannot represent a
negative number.
The solution set is


2, 3 .
1.6 - 61
Caution If a substitution variable is
used when solving an equation that is
quadratic in form, do not forget the step
that gives the solution in terms of the
original variable.
1.6 - 62