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 to eliminate fractions, and then solve the resulting equation.
A value of the variable that appears to be a solution after each side of a rational equation is multiplied by a variable expression (the LCD) is called a proposed solution. 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.
Be sure to check all proposed solutions in the original equation.
■ Rational Equations
■ Work Rate Problems
■ Equations with Radicals
■ Equations with Rational Exponents
■ Equations Quadratic in Form
EXAMPLE 1 Solving Rational Equations That Lead to Linear Equations Solve each equation.
(a) 3x- 1
3 - 2x
x- 1 = x (b) x
x- 2 = 2 x- 2 + 2 SOLUTION
(a) The least common denominator is 31x - 12, which is equal to 0 if x= 1.
Therefore, 1 cannot possibly be a solution of this equation.
3x- 1
3 - 2x
x- 1 = x 31x -12a3x- 1
3 b - 31x- 12a 2x
x- 1b = 31x -12x Multiply by the LCD.
31x-12, where x≠1.
1x- 1213x- 12- 312x2= 3x1x -12 Divide out common factors.
3x2 - 4x+ 1- 6x= 3x2 - 3x Multiply.
1 -10x= -3x Subtract 3x2. Combine like terms.
1= 7x Solve the linear equation.
x= 1
7 Proposed solution
The proposed solution 17 meets the requirement that x≠1 and does not cause any denominator to equal 0. Substitute to check for correct algebra.
CHECK 3x -1
3 - 2x
x - 1 =x Original equation 3A17B - 1
3 - 2A17B
1
7 -1≟1
7 Let x=
1 7 .
- 4
21 - a-1 3b≟1
7 Simplify the complex fractions.
1 7 = 1
7 ✓ True The solution set is E17F.
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1.6 Other Types of Equations and Applications
(b) x
x- 2 = 2 x- 2 + 2 1x- 22a x
x- 2b = 1x- 22a 2
x- 2b + 1x - 222 Multiply by the LCD, x-2, where x≠2.
x = 2+ 21x- 22 Divide out common factors.
x = 2+ 2x- 4 Distributive property
-x = -2 Solve the linear equation.
x = 2 Proposed solution
The proposed solution is 2. However, the variable is restricted to real num- bers except 2. If x= 2, then not only does it cause a zero denominator, but also multiplying by x- 2 in the first step is multiplying both sides by 0, which is not valid. Thus, the solution set is ∅.
■✔ Now Try Exercises 17 and 19.
x1x- 22a3x+ 2
x- 2 b + x1x- 22a1
xb = x1x- 22a -2 x1x- 22b
Multiply by x1x-22, x≠0, 2.
x13x+ 22 +1x - 22= -2 Divide out common factors.
3x2 +2x +x - 2= -2 Distributive property 3x2 + 3x= 0 Standard form 3x1x + 12= 0 Factor.
3x =0 or x + 1= 0 Zero-factor property x =0 or x = -1 Proposed solutions Set each factor
equal to 0.
Because of the restriction x≠0, the only valid proposed solution is -1.
Check -1 in the original equation. The solution set is 5-16.
(b) -4x
x -1 + 4
x +1 = -8 x2 - 1 -4x
x -1 + 4
x +1 = -8
1x + 121x - 12 Factor.
The restrictions on x are x≠ {1. Multiply by the LCD, 1x+ 121x -12. EXAMPLE 2 Solving Rational Equations That Lead to Quadratic
Equations Solve each equation.
(a) 3x + 2 x - 2 + 1
x = -2
x2 - 2x (b) -4x
x- 1 + 4
x+ 1 = -8 x2 -1 SOLUTION
(a) 3x+ 2
x- 2 + 1
x = -2 x2- 2x 3x+ 2
x- 2 + 1
x = -2 x1x - 22
Factor the last denominator.
-4x1x+ 12 + 41x - 12= -8 Divide out common factors.
-4x2 - 4x+ 4x- 4= -8 Distributive property -4x2 + 4= 0 Standard form
x2 - 1= 0 Divide by -4.
1x +121x - 12= 0 Factor.
x+ 1= 0 or x - 1= 0 Zero-factor property x= -1 or x = 1 Proposed solutions Neither proposed solution is valid, so the solution set is ∅.
■✔ Now Try Exercises 25 and 27.
1x + 121x - 12a -4x
x - 1b + 1x+ 121x- 12a 4
x+ 1b = 1x +121x - 12a -8 1x + 121x- 12b
Work Rate Problems If a job can be completed in 3 hr, then the rate of work is 13 of the job per hr. After 1 hr the job would be 13 complete, and after 2 hr the job would be 23 complete. In 3 hr the job would be 33 complete, meaning that 1 complete job had been accomplished.
PROBLEM-SOLVING HINT If a job can be completed in t units of time, then the rate of work, r, is 1t of the job per unit time.
r = 1 t
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 worked
or A=rt
EXAMPLE 3 Solving a Work Rate Problem
One printer can do a job twice as fast as another. Working together, both print- ers 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.
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, 1
x = the rate of the faster printer (job per hour)
and 1
2x = the rate of the slower printer (job per hour).
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1.6 Other Types of Equations and Applications
The time for the printers to do the job together is 2 hr. Multiplying each rate by the time will give the fractional part of the job completed by each.
Rate Time
Part of the Job Completed Faster Printer 1x 2 2A1xB =2x
Slower Printer 2x1 2 2A2x1B= 1x
Step 3 Write an equation. The sum of the two parts of the job completed is 1 because one whole job is done.
Part of the job Part of the job
done by the + done by the = One whole faster printer slower printer job (111111)111111* (111111)111111* (1111)1111*
2
x + 1
x = 1
Step 4 Solve. xa2
x + 1
xb = x112 Multiply each side by x, where x≠0.
xa2
xb + xa1
xb = x112 Distributive property
2 + 1= x Multiply.
3= x Add.
Step 5 State the answer. The faster printer would take 3 hr to do the job alone.
The slower printer would take 2132= 6 hr. Give both answers here.
Step 6 Check. The answer is reasonable because the time working together (2 hr, as stated in the problem) is less than the time it would take the faster printer working alone (3 hr, as found in Step 4).
■✔ Now Try Exercise 39.
A=rt
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.
Because the printers can complete the job together in 2 hr, their combined rate is 12 of the job per hr.
1 x + 1
2x = 1 2 2xa1
x + 1
2xb = 2xa1
2b Multiply each side by 2x.
2+ 1 = x Distributive property 3 =x Same solution found earlier
Equations with Radicals To solve an equation such as x- 215 - 2x= 0,
in which the variable appears in a radicand, we use the following power property to eliminate the radical.
When the power property is used to solve equations, the new equation may have more solutions than the original equation. For example, the equation
x = -2 has solution set 5-26.
If we square each side of the equation x= -2, we obtain the new equation x2 = 4, which has solution set 5-2, 26.
Because the solution sets are not equal, the equations are not equivalent. When we use the power property to solve an equation, it is essential to check all pro- posed solutions in the original equation.
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.
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.
EXAMPLE 4 Solving an Equation Containing a Radical (Square Root) Solve x - 215- 2x= 0.
SOLUTION
x- 215- 2x= 0
Step 1 x= 215- 2x Isolate the radical.
Step 2 x2 = A215- 2x B2 Square each side.
x2 = 15- 2x A1a B2=a, for aÚ0.
Step 3 x2 + 2x- 15= 0 Write in standard form.
1x+ 521x -32 = 0 Factor.
x +5 = 0 or x- 3 =0 Zero-factor property x = -5 or x =3 Proposed solutions
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.
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1.6 Other Types of Equations and Applications
Step 4
CHECK x- 215- 2x= 0 Original equation
-5 - 215- 21-52≟0 Let x= -5. 3 - 215- 2132≟0 Let x=3.
-5- 225≟0 3 - 29≟0
-5 -5≟0 3- 3≟0
-10 = 0 False 0 = 0 ✓ True
As the check shows, only 3 is a solution, so the solution set is 536.
■✔ Now Try Exercise 45.
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 5 Solving an Equation Containing Two Radicals Solve 22x+ 3- 2x+ 1 = 1.
SOLUTION
22x+ 3- 2x + 1= 1
Step 1 22x +3 = 1+ 2x +1 Isolate 12x+3.
Step 2 A22x+ 3 B2 = A 1+ 2x + 1 B2 Square each side.
2x+ 3= 1+ 22x + 1+ 1x+ 12 Be careful:
1a+b22=a2+2ab+b2
Step 1 x+ 1 = 22x+ 1 Isolate the remaining radical.
Step 2 1x+ 122 = 1 22x + 122 Square again.
x2+ 2x+ 1 = 41x + 12 Apply the exponents.
x2+ 2x+ 1 = 4x+ 4 Distributive property
Step 3 x2- 2x- 3 = 0 Write in standard form.
1x- 321x+ 12 = 0 Factor.
x- 3 =0 or x +1 = 0 Zero-factor property x =3 or x = -1 Proposed solutions Step 4
Don’t forget this term when squaring.
Isolate one of the radicals on one side of the equation.
CHECK 22x+ 3- 2x + 1= 1 Original equation
22132+ 3- 23 + 1≟1 Let x=3. 221-12+ 3- 2-1+ 1≟1 Let x= -1.
29- 24≟1 21 - 20≟1
3- 2≟1 1 - 0≟1
1= 1 ✓ True 1= 1 ✓ True
Both 3 and -1 are solutions of the original equation, so 5-1, 36 is the solution set.
■✔ Now Try Exercise 57.
1ab22=a2b2
EXAMPLE 6 Solving an Equation Containing a Radical (Cube Root) Solve 23 4x2- 4x+ 1 - 23x = 0.
SOLUTION
234x2- 4x+ 1 - 23 x= 0
Step 1 234x2 - 4x+ 1= 23 x Isolate a radical.
Step 2 A234x2 -4x +1 B3 = A23x B3 Cube each side.
4x2 - 4x+ 1= x Apply the exponents.
Step 3 4x2 - 5x+ 1= 0 Write in standard form.
14x -121x - 12= 0 Factor.
4x- 1= 0 or x- 1 =0 Zero-factor property x= 1
4 or x =1 Proposed solutions Step 4
33 4A14B2 - 4A14B + 1- 33 14≟0 Let x=14 .
33 14 - 33 14≟0
0 =0 ✓ True
2341122 - 4112 + 1- 231≟0 Let x=1..
231 - 23 1≟0
0 =0 ✓ True Both are valid solutions, and the solution set is E14 , 1F.
■✔ Now Try Exercise 69.
Equations with Rational Exponents An equation with a rational expo- nent contains a variable, or variable expression, raised to an exponent that is a rational number. For example, the radical equation
A25 x B3 = 27 can be written with a rational exponent as x3/5= 27 and solved by raising each side to the reciprocal of the exponent, with care taken regarding signs as seen in Example 7(b).
EXAMPLE 7 Solving Equations with Rational Exponents Solve each equation.
(a) x3/5 =27 (b) 1x - 422/3= 16 SOLUTION
(a) x3/5= 27
1x3/525/3=275/3 Raise each side to the power 53 , the reciprocal of the exponent of x.
x =243 275/3=A2327 B5=35=243 CHECK 23 4x2 -4x + 1- 23 x= 0 Original equation
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1.6 Other Types of Equations and Applications
Equations Quadratic in Form Many equations that are not quadratic equations can be solved using similar methods. The equation
1x + 122/3-1x + 121/3- 2= 0 is not a quadratic equation in x. However, with the substitutions
u = 1x+ 121/3 and u2 = 31x+ 121/342 = 1x +122/3, the equation becomes
u2 -u - 2= 0,
which is a quadratic equation in u. This quadratic equation can be solved to find u, and then u= 1x+ 121/3 can be used to find the values of x, the solutions to the original equation.
Equation Quadratic in Form
An equation is quadratic in form if it can be written as au2 + bu+c = 0, where a≠0 and u is some algebraic expression.
(b) 1x - 422/3=16 Raise each side to the power 32 . Insert { because this involves an even root, as indicated by the 2 in the denominator.
CHECK Let x = 243 in the original equation.
x3/5= 2433/5= A25243 B3 = 33= 27 ✓ True
The solution set is 52436.
1-60- 422/3≟16 Let x= -60.
1-6422/3≟16
A23 -64 B2≟16
16 = 16 ✓ True
168- 422/3≟16 Let x=68.
642/3≟16 A2364 B2≟16
16= 16 ✓ True Both proposed solutions check, so the solution set is 5-60, 686.
■✔ Now Try Exercises 75 and 79.
C1x- 422/3D3/2= {163/2
x -4 = {64 {163/2= {A116 B3= {43={64 x= 4{64 Add 4 to each side.
x = -60 or x= 68 Proposed solutions
CHECK 1x- 422/3= 16 Original equation
EXAMPLE 8 Solving Equations Quadratic in Form Solve each equation.
(a) 1x + 122/3- 1x + 121/3 -2 = 0 (b) 6x-2 + x-1 = 2 SOLUTION
(a) 1x + 122/3-1x + 121/3- 2= 0 1x+122/3= 31x+121/342, so u2 - u- 2 =0 let u=1x+121/3.
1u- 221u+ 12 =0 Factor.
u- 2 =0 or u +1 = 0 Zero-factor property u =2 or u = -1 Solve each equation.
1x + 121/3=2 or 1x+ 121/3= -1 Replace u with (x+1)1/3. 31x +121/343 =23 or 31x + 121/343 = 1-123 Cube each side.
x+ 1 =8 or x +1 = -1 Apply the exponents.
x =7 or x = -2 Proposed solutions Don’t forget
this step.
CHECK 1x + 122/3- 1x + 121/3- 2= 0 Original equation 17 +122/3- 17 +121/3- 2≟0 Let x=7.
82/3 -81/3- 2≟0 4- 2- 2≟0
0 =0 ✓ True
1-2 + 122/3- 1-2+ 121/3- 2≟0 Let x= -2.
1-122/3- 1-121/3- 2≟0 1 + 1- 2≟0
0= 0 ✓ True Both proposed solutions check, so the solution set is 5-2, 76. (b) 6x-2 + x-1 =2
6x-2 + x-1- 2 =0 Subtract 2 from each side.
6u2 + u- 2 =0 Let u=x-1. Then u2=x-2. 13u+ 2212u- 12 = 0 Factor.
3u+ 2 = 0 or 2u- 1= 0 Zero-factor property u = -2
3 or u= 1
2 Solve each equation.
x-1 = -2
3 or x-1 = 1
2 Replace u with x-1. x = -3
2 or x= 2 x-1 is the reciprocal of x.
Both proposed solutions check, so the solution set is E-32 , 2F.
■✔ Now Try Exercises 93 and 99.
Don’t stop here.
Substitute for u.
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.
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1.6 Other Types of Equations and Applications
Check that the solution set is E{136 , {12F. ■✔ Now Try Exercise 87.
EXAMPLE 9 Solving an Equation Quadratic in Form Solve 12x4 - 11x2 + 2= 0.
SOLUTION 12x4- 11x2 + 2= 0
121x222- 11x2 + 2= 0 x4=1x222
12u2 - 11u + 2= 0 Let u=x2. Then u2=x4. 13u- 2214u- 12= 0 Solve the quadratic equation.
3u- 2= 0 or 4u- 1= 0 Zero-factor property u= 2
3 or u= 1
4 Solve each equation.
x2 = 2
3 or x2 = 1
4 Replace u with x2. x= {B
2
3 or x = {B 1
4 Square root property x= {22
23 # 23
23 or x = {1
2 Simplify radicals.
x= {26 3
NOTE To solve the equation from Example 9, 12x4 - 11x2+ 2 =0,
we could factor 12x4- 11x2 + 2 directly as 13x2 - 2214x2 - 12, set each factor equal to zero, and then solve the resulting two quadratic equations.
Which method to use is a matter of personal preference.
CONCEPT PREVIEW Fill in the blank to correctly complete each sentence.
1. A(n) is an equation that has a rational expression for one or more terms.
2. Proposed solutions for which any denominator equals are excluded from the solution set of a rational equation.
3. If a job can be completed in 4 hr, then the rate of work is of the job per hour.
4. When the power property is used to solve an equation, it is essential to check all proposed solutions in the .
5. An equation such as x3/2= 8 is an equation with a(n) , because it contains a variable raised to an exponent that is a rational number.