# LIAL HORNSBY SCHNEIDER

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LIAL HORNSBY SCHNEIDER
COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER

1.6 Other Types of Equations and Applications Rational Expressions

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.

Solve each equation. a. Solution
SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Example 1 Solve each equation. a. 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.

Solve each equation. a. Solution Example 1
SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Example 1 Solve each equation. a. Solution Multiply by the LCD, 3(x – 1), where x ≠ 1.

Solve each equation. a. Solution Example 1
SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Example 1 Solve each equation. a. Solution Simplify on both sides. Multiply.

Solve each equation. a. Solution Example 1
SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Example 1 Solve each equation. a. Solution Multiply. Subtract 3x2; combine terms. Solve the linear equation. The restriction x ≠ 1 does not affect this result.

Solve each equation. a. Solution Example 1
SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Example 1 Solve each equation. a. Solution The restriction x ≠ 1 does not affect this result.

Solve each equation. b. Solution Example 1
SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Example 1 Solve each equation. b. Solution Multiply by the LCD, x – 2, where x ≠ 2.

Solve each equation. b. Solution Example 1
SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Example 1 Solve each equation. b. Solution Distributive property Subtract 2x; combine like terms. Multiply by –1.

Solve each equation. b. Solution
SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Example 1 Solve each equation. b. 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 .

Solve each equation. a. Solution Example 2
SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Example 2 Solve each equation. a. Solution Factor the last denominator. Multiply by x(x – 2), x ≠ 0, 2.

Solve each equation. a. Solution Example 2
SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Example 2 Solve each equation. a. Solution Multiply by x(x – 2), x ≠ 0, 2. Distributive property

Solve each equation. a. Solution Example 2
SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Example 2 Solve each equation. a. Solution Distributive property Standard form Factor.

Set each factor equal to 0.
SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Example 2 Solve each equation. a. Solution Factor. Zero-factor property Set each factor equal to 0. or Proposed solutions or

Because of the restriction x ≠ 0, the only valid solution is – 1.
SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Example 2 Solve each equation. a. Solution Because of the restriction x ≠ 0, the only valid solution is – 1. The solution set is {– 1}.

Solve each equation. b. Solution Example 2
SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Example 2 Solve each equation. b. Solution Factor.

Solve each equation. b. Solution Example 2
SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Example 2 Solve each equation. b. Solution

Solve each equation. b. Solution Example 2
SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Example 2 Solve each equation. b. Solution Distributive property Standard form Divide by – 4.

Solve each equation. b. Solution or or Example 2
SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Example 2 Solve each equation. b. Solution Divide by – 4. Factor. Zero-factor property or or Proposed solutions

Solve each equation. b. Solution or
SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Example 2 Solve each equation. b. Solution or Proposed solutions Since the restrictions on x are x ≠ 1, – 1, neither proposed solution is valid, so the solution set is .

Work Rate Problems rate  time = portion of the job completed.
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

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

SOLVING A WORK RATE PROBLEM
Example 3 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?

SOLVING A WORK RATE PROBLEM
Example 3 Solution Step 1 Read the problem. We must find the time it would take each computer working alone to do the job.

SOLVING A WORK RATE PROBLEM
Example 3 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, and

Part of the Job Accomplished
SOLVING A WORK RATE PROBLEM Example 3 Solution Step 2 Assign a variable. Rate Time Part of the Job Accomplished Faster Computer 2 Slower Computer A = r t

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 Part of the job done by the slower computer One whole job + =

Solution Step 4 Solve. SOLVING A WORK RATE PROBLEM Example 3
Multiply both sides by x. Distributive property Multiply. Add.

SOLVING A WORK RATE PROBLEM
Example 3 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.

Rate Problem Note The sum of the rates of the individual computers is equal to their rate working together. Multiply both sides 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 number.

Caution Be very careful when using the power property
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.

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.

Solve Solution SOLVING AN EQUATION CONTAINING A RADICAL Example 4
Isolate the radical. Square both sides.

Solve Solution: or or SOLVING AN EQUATION CONTAINING A RADICAL
Example 4 Solve Solution: Solve the quadratic equation. Factor. or Zero-factor property or Proposed solutions

Only 3 is a solution, giving the solution set {3}.
SOLVING AN EQUATION CONTAINING A RADICAL Example 4 Solve Solution: or Proposed solutions Only 3 is a solution, giving the solution set {3}.

SOLVING AN EQUATION CONTAINING TWO RADICALS
Example 5 Solve Solution When an equation contains two radicals, begin by isolating one of the radicals on one side of the equation. Square both sides.

Don’t forget this term when squaring.
SOLVING AN EQUATION CONTAINING TWO RADICALS Example 5 Solve Solution Square both sides. Be careful! Don’t forget this term when squaring. Isolate the remaining radical.

Solve Solution SOLVING AN EQUATION CONTAINING TWO RADICALS Example 5
Isolate the remaining radical. Square again.

Solve Solution or or SOLVING AN EQUATION CONTAINING TWO RADICALS
Example 5 Solve Solution Solve the quadratic equation. or Proposed solutions or

SOLVING AN EQUATION CONTAINING TWO RADICALS
Example 5 Solve Solution Proposed solutions or Both 3 and – 1 are solutions of the original equation, so {3, – 1} is the solution set.

Caution Remember to isolate a radical in Step 1
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.

Solve Solution Example 6
SOLVING AN EQUATION CONTAINING A RADICAL (CUBE ROOT) Example 6 Solve Solution Isolate a radical. Cube both sides. Solve the quadratic equation.

Solve Solution or or Example 6
SOLVING AN EQUATION CONTAINING A RADICAL (CUBE ROOT) Example 6 Solve Solution Solve the quadratic equation. or or Proposed solutions

Both are valid solutions, and the solution set is {¼ ,1}.
SOLVING AN EQUATION CONTAINING A RADICAL (CUBE ROOT) Example 6 Solve Solution or Proposed solutions Both are valid solutions, and the solution set is {¼ ,1}.

An equation is said to be quadratic in form if it can be written as where a ≠ 0 and u is some algebraic expression.

a. Solve Solution Since or or SOLVING EQUATIONS QUADRATIC IN FORM
Example 7 a. Solve Solution Since Substitute. Factor. Zero-factor property or or

a. Solve or or or or SOLVING EQUATIONS QUADRATIC IN FORM Example 7
Don’t forget this step. Zero-factor property or or Cube each side. or Proposed solutions or

a. Solve Solution or The solution set is {– 2, 7}.
SOLVING EQUATIONS QUADRATIC IN FORM Example 7 a. Solve Solution or Proposed solutions The solution set is {– 2, 7}.

Remember to substitute for u.
SOLVING EQUATIONS QUADRATIC IN FORM Example 7 b. Solve Solution Subtract. Factor. or Zero-factor property Remember to substitute for u. or

b. Solve Solution or or or SOLVING EQUATIONS QUADRATIC IN FORM
Example 7 b. Solve Solution or or Substitute again. x-1 is the reciprocal of x. or

b. Solve Solution or SOLVING EQUATIONS QUADRATIC IN FORM Example 7
x-1 is the reciprocal of x.

Solve Solution or SOLVING AN EQUATION QUADRATIC IN FORM Example 8
Let u = x2; thus u2 = x4 Solve the quadratic equation. or Zero-property factor

Solve Solution or or or SOLVING AN EQUATION QUADRATIC IN FORM
Example 8 Solve Solution Zero-property factor or or or Replace u with x2.

Solve Solution or or or SOLVING AN EQUATION QUADRATIC IN FORM
Example 8 Solve Solution or Replace u with x2. or Square root property Simplify radicals. or

Solve Solution or SOLVING AN EQUATION QUADRATIC IN FORM Example 8

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

Solve Solution Example 9
SOLVING AN EQUATION THAT LEADS TO ONE THAT IS QUADRATIC IN FORM Example 9 Solve Solution Raise both sides to the fourth power. Power rule for exponents. Let u = x2; thus u2 = x4.

Solve Solution or or Example 9
SOLVING AN EQUATION THAT LEADS TO ONE THAT IS QUADRATIC IN FORM Example 9 Solve Solution Let u = x2; thus u2 = x4. Factor. Zero-property factor or or

Solve Solution or or or Example 9
SOLVING AN EQUATION THAT LEADS TO ONE THAT IS QUADRATIC IN FORM Example 9 Solve Solution or or u = x2 Square root property or Checking the four proposed solutions in the original equation shows that only are solutions, since the left side of the equation cannot represent a negative number.

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.