1.6 - 1 10 TH EDITION Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 1 Equations and...

1.6 - 1

10TH EDITION

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1

1Equations and Inequalities

Copyright © 2013, 2009, 2005 Pearson Education, Inc.


1.6Other Types of Equations and Applications

• Rational Equations• Work Rate Problems• Equations with Radicals• Equations with Rational Exponents• Equations Quadratic in Form


Rational Equations

A rational equation is an equation that has a rational expression for one or more terms. To solve a rational equation, multiply each side by the least common denominator (LCD) of the terms of the equation and then solve the resulting equation. Because a rational expression is not defined when its denominator is 0, proposed solutions for which any denominator equals 0 are excluded from the solution set.


Example 1 SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS

Solve each equation.

Solution

(a) 3 1 23 1

x xx

x

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.




Solution

(a) 3 1 23 1

x xx

x

3 1 2

3 1x x

xx

3( 1) 3( 1) 3( 13

)1 2

3 1x x

xxx

x x

Multiply by the LCD, 3(x – 1), where x ≠ 1.




Solution

(a) 3 1 23 1

x xx

x

Divide out common factors.

( 1)(3 1) 3(2 ) 3 ( 1)x x x x x 2 23 4 1 6 3 3x x x x x Multiply.

1 10 3x x Subtract 3x2; combine like terms.

1 7x Solve the linear equation.




Solution

(a) 3 1 23 1

x xx

x

17

x Proposed solution

1The solution set is .

7

The proposed solution meets the requirement that x ≠ 1 and does not cause any denominator to equal 0. Substitute to check for correct algebra.




Solution

(b) 22

2 2x

x x

( 2) ( 22

) ( 2)22 2

xx x

x x x

Multiply by the LCD, x – 2, where x ≠ 2.

2 2( 2)x x Divide out common factors.




Solution

(b) 22

2 2x

x x

2 2 4x x Distributive property

2x Solve the linear equation.

2x Proposed solution.




Solution

(b) 22

2 2x

x x

The proposed solution is 2. However, the variable is restricted to real numbers except 2. If x = 2, then not only does it cause a zero denominator, but multiplying by x – 2 in the first step is multiplying both sides by 0, which is not valid. Thus, the solution set is . 0


Example 2 SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS


Solution

(a)2

3 2 1 22 2

xx x x x

( 2)3 2 1 2

2xx x x x

Factor the last denominator.

3 2 1 22 ( 2)

( 2) ( 2) ( 2)xx x

x x x x x xx x

Multiply by x(x – 2), x ≠ 0, 2.




Solution

(a)2

3 2 1 22 2

xx x x x

(3 2) ( 2) 2x x x 23 2 2 2x x x Distributive property


23 3 0x x Standard form

3 ( 1) 0x x Factor.




Solution

(a)2

3 2 1 22 2

xx x x x

3 0x or 1 0x Zero-factor property

Set each factor equal

to 0.

0x or 1x Proposed solutions

Because of the restriction x ≠ 0, the only valid solution is – 1. The solution set is {– 1}.




Solution

(b)2

4 4 81 1 1x

x x x

( 1)( 14 8

1 )4

1x

x x x x

Factor.

( 1)( 14 4

1 1) ( 1)( 1)

xx

x xx

x x

( 1)(8

( 1) 11)

( )xx x

x

Multiply by (x + 1)(x – 1), x ≠ ±1.




Solution

(b)

4 ( 1) 4( 1) 8x x x 24 4 4 4 8x x x Distributive

property24 4 0x Standard form

2 1 0x Divide by – 4.

2

4 4 81 1 1x

x x x





Solution

(b)

( 1)( 1) 0x x Factor.

1 0x 1 0x or Zero-factor property

1x 1x or Proposed solutions

2

4 4 81 1 1x

x x x

Neither proposed solution is valid, so the solution set is . 0


Work Rate Problems

Problem Solving If a job can be done in t units of time, then the rate of work, r, is 1/t of the job per unit time.

The amount of work completed, A, is found by multiplying the rate of work, r, and the amount of time worked, t. This formula is similar to the distance formula, d = rt.Amount of work completed = rate of work × amount of time workedor .A r t

1r

t


Example 3 SOLVING A WORK RATE PROBLEM

One printer can do a job twice as fast as another. Working together, both printers can do the job in 2 hr. How long would it take each printer, working alone, to do the job?



Solution

Step 1 Read the problem. We must find the time it would take each printer, working alone, to do the job.



Solution Step 2 Assign a variable. Let x represent the number of hours it would take the faster printer, working alone, to do the job. The time for the slower printer to do the job alone is then 2x hours. Therefore,

rate of the faster printer (job per h1

our)x

rate of the slower printer (job per ho1

).2

urx

and



Solution Step 2 Assign a variable.

Rate Time Part of the Job Accomplished

Faster Printer 2

Slower Printer 2

1x1

2x

12

2x x

1

22

1x x

A = r t



Solution Step 3 Write an equation. The sum of the two parts of the job accomplished is 1, since one whole job is done.

Part of the job done by the faster printer

+

Part of the job done by the slower printer

= One whole job

2x

1x

1



Solution

Multiply each side by x, where x ≠ 0.

Step 4 Solve.2 1

(1)x x

x x

2 1(1)

x xx x x

Distributive property

2 1 x Multiply.

3 x Add.



Solution

Step 5 State the answer. The faster printer would take 3 hr to do the job alone, and the slower printer 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 (3 hr).


Rate Problem

Note Example 3 can also be solved by using the fact that the sum of the rates of the individual printers is equal to their rate working together. Since the printers can complete the job together in 2 hr, their combined rate is 1/2 of the job per hour.


Rate Problem

Note

1 1 12 2x x

1 1 12 2

2 2x x

x x

2 1 x 3x

Multiply each side by 2x.

Same solution found earlier



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 integer n.


Caution Be very careful when using the power property. It does not say that the equations P = Q and Pn = Qn are equivalent; it says only that each solution of the original equation P = Q is also a solution of the new equation Pn = Qn.


Solving an Equation Involving RadicalsStep 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.


Example 4 SOLVING AN EQUATION CONTAINING A RADICAL (SQUARE ROOT)

Solve

Solution

15 2 0.x x

15 2x x Isolate the radical.

22 15 2x x Square each side.

2 15 2x x Solve the quadratic equation.

2 2 15 0x x


Example 4 SOLVING AN EQUATION CONTAINING A RADICAL (SQUARE ROOT)

Solve

Solution

15 2 0.x x

( 5)( 3) 0x x Factor.

5 0x or 3 0x Zero-factor property


Only 3 is a solution, so the solution set is {3}.


Example 5 SOLVING AN EQUATION CONTAINING TWO RADICALS

Solve

Solution When an equation contains two radicals, begin by isolating one of the radicals on one side of the equation.

2 3 1 1.x x

2 3 1 1x x

2 3 1 1x x

Square each side.

Isolate 2 3.x

2 212 3 1xx



Solve

Solution

2 3 1 1.x x

2 1)3 1 (2 1x xx Don’t forget this

term when squaring.

Be careful!

1 2 1x x Isolate the remaining radical.

221 2 1x x Square

again.



Solve

Solution

2 3 1 1.x x

2 2 1 4( 1)x x x 2 2 1 4 4x x x

2 2 3 0x x Solve the quadratic equation.

( 3)( 1) 0x x

3 0x 1 0x or


Apply the exponents.


Factor

Zero-factor property



Solve

Solution

2 3 1 1.x x


Check each proposed solution in the original equation. Both 3 and – 1 are solutions of the original equation, so {– 1,3} is the solution set.


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.


Example 6 SOLVING AN EQUATION CONTAINING A RADICAL (CUBE ROOT)

Solve

Solution

3 2 34 4 1 0x x x

3 2 34 4 1x x x Isolate a radical.

3 33 2 34 4 1x x x Cube each side.

24 4 1x x x 24 5 1 0x x Solve the quadratic

equation.

Apply the exponents.


Example 6 SOLVING AN EQUATION CONTAINING A RADICAL (CUBE ROOT)

Solve

Solution

3 2 34 4 1 0x x x

(4 1)( 1) 0x x

4 1 0x

or

1 0x 14

x 1x

or

Proposed solutions

Factor.


Both are valid solutions, and the solution set is {1/4 ,1}.


Example 7 SOLVING EQUATIONS WITH RATIONAL EXPONENTS


Solution

Raise each side to the power 5/3, the reciprocal of the exponent of x.

(a) 3 5 27x

3 5 27x

5 33 5 5 327x

243x 55 3 5327 27 3 243

The solution set is {243}.


Example 7 SOLVING EQUATIONS WITH RATIONAL EXPONENTS


Solution

Raise each side to the power 3/2. Insert ± since this involves an even root, as indicated by the 2 in the denominator.

Proposed solutions

(b) 2 34 16x

3 22 3 3 24 16x

4 64x 33 2 316 16 4 64

4 64x or 60 68x x

Both proposed solutions check in the original equation, so the solution set is {–60,68}.


Equation Quadratic in Form

An equation is said to be quadratic in form if it can be written as

2 0,au bu c

where a ≠ 0 and u is some algebraic expression.


Example 8 SOLVING EQUATIONS QUADRATIC IN FORM

Solve

Solution

2 1 13 3 3

2( 1) ( 1) , let ( 1) . x x u x

2 2 0u u Substitute.

( 2)( 1) 0u u Factor.

2 0u 1 0u or

2u or 1u


(a)2 1

3 3( 1) ( 1) 2 0.x x

Since



2u or 1u 13( 1) 2x or

13( 1) 1x 1

3

Replace with

( 1) .

u

x 13

3 3( 1) 2x or13

3 3( 1) ( 1)x Cube each side.


Solve 2 1

3 3( 1) ( 1) 2 0.x x (a)Don’t forget

this step.

Both proposed solutions check in the original equation, so the solution set is {– 2, 7}.



Solve

Solution

26 2 0u u

Subtract 2 from each side.

(3 2)(2 1) 0u u Factor.

3 2 0u 2 1 0u or23

u or12

u


(b) 2 16 2.x x

2 16 2 0x x 1 2 2Let . Then .u x u x

Remember to

substitute for u.



Solve

Solution

23

u or12

u

(b) 2 16 2.x x

1 23

x or 1 12

x Substitute again.

32

x or 2x x-1 is the reciprocal of x.

Both proposed solutions check in the original

3equation, so the solution set is , 2 .

2


Caution When using a substitution variable in solving an equation that is quadratic in form, do not forget the step that gives the solution in terms of the original variable.


Example 9 SOLVING AN EQUATION QUADRATIC IN FORM

Solve

Solution

4 212 11 2 0.x x

2 2 212 11 2 0x x 24 2x x

212 11 2 0u u Let u = x2. Then u2 = x4.

(3 2)(4 1) 0u u Solve the quadratic equation.

or3 2 0u 4 1 0u Zero-property factor



Solve

Solution

4 212 11 2 0.x x

23

x or14

x Square root property

23

u or14

u

2 23

x or 2 14

x Replace u with x2.



Solve

Solution

4 212 11 2 0.x x

2 3

3 3x

or12

x Simplify radicals.

63

x

6 1The solution set is , .

3 2

or12

x


Equation Quadratic in Form

Note Some equations that are quadratic in form are simple enough to avoid using the substitution variable technique. To solve

4 212 11 2 0,x x

we could factor directly as (3x2 – 2)(4x2 – 1), set each factor equal to zero, and then solve the resulting two quadratic equations. Which method to use is a matter of personal preference.

1.6 - 1 10 TH EDITION Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 1 Equations and...

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