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

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

2 1.6 Other Types of Equations and Applications Rational Equations Work Rate Problems Equations with Radicals Equations with Rational Exponents Equations Quadratic in Form

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 3 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.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 4 Example 1 SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Solve each equation. Solution (a) 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.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 5 Example 1 SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Solve each equation. Solution (a) Multiply by the LCD, 3(x – 1), where x ≠ 1.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 6 Example 1 SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Solve each equation. Solution (a) Divide out common factors. Multiply. Subtract 3x 2 ; combine like terms. Solve the linear equation.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 7 Example 1 SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Solve each equation. Solution (a) Proposed solution The proposed solution meets the requirement that x ≠ 1 and does not cause any denominator to equal 0. Substitute to check for correct algebra.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 8 Example 1 SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Solve each equation. Solution (b) Multiply by the LCD, x – 2, where x ≠ 2. Divide out common factors.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 9 Example 1 SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Solve each equation. Solution (b) Distributive property Solve the linear equation. Proposed solution.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 10 Example 1 SOLVING RATIONAL EQUATIONS THAT LEAD TO LINEAR EQUATIONS Solve each equation. Solution (b) 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.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 11 Example 2 SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Solve each equation. Solution (a) Factor the last denominator. Multiply by x(x – 2), x ≠ 0, 2.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 12 Example 2 SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Solve each equation. Solution (a) Distributive property Divide out common factors. Standard form Factor.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 13 Example 2 SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Solve each equation. Solution (a) or Zero-factor property Set each factor equal to 0. or Proposed solutions Because of the restriction x ≠ 0, the only valid solution is – 1. The solution set is {– 1}.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 14 Example 2 SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Solve each equation. Solution (b) Factor. Multiply by (x + 1)(x – 1), x ≠ ±1.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 15 Example 2 SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Solve each equation. Solution (b) Distributive property Standard form Divide by – 4. Divide out common factors.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 16 Example 2 SOLVING RATIONAL EQUATIONS THAT LEAD TO QUADRATIC EQUATIONS Solve each equation. Solution (b) Factor. or Zero-factor property or Proposed solutions Neither proposed solution is valid, so the solution set is.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 17 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 worked or

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 18 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?

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 19 Example 3 SOLVING A WORK RATE PROBLEM Solution Step 1 Read the problem. We must find the time it would take each printer, working alone, to do the job.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 20 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 printer, working alone, to do the job. The time for the slower printer to do the job alone is then 2x hours. Therefore, and

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 21 Example 3 SOLVING A WORK RATE PROBLEM Solution Step 2 Assign a variable. RateTimePart of the Job Accomplished Faster Printer 2 Slower Printer 2 A = r t

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 22 Example 3 SOLVING A WORK RATE PROBLEM 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

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 23 Example 3 SOLVING A WORK RATE PROBLEM Solution Multiply each side by x, where x ≠ 0. Step 4 Solve. Distributive property Multiply. Add.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 24 Example 3 SOLVING A WORK RATE PROBLEM 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).

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 25 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.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 26 Rate Problem Note Multiply each side by 2x. Same solution found earlier Distributive property

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 27 Power Property If P and Q are algebraic expressions, then every solution of the equation P = Q is also a solution of the equation P n = Q n, for any positive integer n.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 28 Caution Be very careful when using the power property. It does not say that the equations P = Q and P n = Q n are equivalent; it says only that each solution of the original equation P = Q is also a solution of the new equation P n = Q n.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 29 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.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 30 Example 4 SOLVING AN EQUATION CONTAINING A RADICAL (SQUARE ROOT) Solve Solution Isolate the radical. Square each side. Solve the quadratic equation.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 31 Example 4 SOLVING AN EQUATION CONTAINING A RADICAL (SQUARE ROOT) Solve Solution Factor. or Zero-factor property or Proposed solutions Only 3 is a solution, so the solution set is {3}.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 32 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. Square each side.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 33 Example 5 SOLVING AN EQUATION CONTAINING TWO RADICALS Solve Solution Don’t forget this term when squaring. Be careful! Isolate the remaining radical. Square again.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 34 Example 5 SOLVING AN EQUATION CONTAINING TWO RADICALS Solve Solution Solve the quadratic equation. or Proposed solutions Apply the exponents. Distributive property Factor Zero-factor property

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 35 Example 5 SOLVING AN EQUATION CONTAINING TWO RADICALS Solve Solution or Proposed solutions 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.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 36 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.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 37 Example 6 SOLVING AN EQUATION CONTAINING A RADICAL (CUBE ROOT) Solve Solution Isolate a radical. Cube each side. Solve the quadratic equation. Apply the exponents.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 38 Example 6 SOLVING AN EQUATION CONTAINING A RADICAL (CUBE ROOT) Solve Solution or Proposed solutions Factor. Zero-factor property Both are valid solutions, and the solution set is {1/4,1}.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 39 Example 7 SOLVING EQUATIONS WITH RATIONAL EXPONENTS Solve each equation. Solution Raise each side to the power 5/3, the reciprocal of the exponent of x. (a) The solution set is {243}.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 40 Example 7 SOLVING EQUATIONS WITH RATIONAL EXPONENTS Solve each equation. 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) Both proposed solutions check in the original equation, so the solution set is {–60,68}.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 41 Equation Quadratic in Form An equation is said to be quadratic in form if it can be written as where a ≠ 0 and u is some algebraic expression.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 42 Example 8 SOLVING EQUATIONS QUADRATIC IN FORM Solve Solution Substitute. Factor. or Zero-factor property (a) Since

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 43 Example 8 SOLVING EQUATIONS QUADRATIC IN FORM or Cube each side. or Proposed solutions Solve (a) Don’t forget this step. Both proposed solutions check in the original equation, so the solution set is {– 2, 7}.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 44 Example 8 SOLVING EQUATIONS QUADRATIC IN FORM Solve Solution Subtract 2 from each side. Factor. or Zero-factor property (b) Remember to substitute for u.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 45 Example 8 SOLVING EQUATIONS QUADRATIC IN FORM Solve Solution or (b) or Substitute again. or x -1 is the reciprocal of x.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 46 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.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 47 Example 9 SOLVING AN EQUATION QUADRATIC IN FORM Solve Solution Let u = x 2. Then u 2 = x 4. Solve the quadratic equation. or Zero-property factor

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 48 Example 9 SOLVING AN EQUATION QUADRATIC IN FORM Solve Solution or Square root property or Replace u with x 2.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 49 Example 9 SOLVING AN EQUATION QUADRATIC IN FORM Solve Solution or Simplify radicals. or

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 50 Equation Quadratic in Form Note Some equations that are quadratic in form are simple enough to avoid using the substitution variable technique. To solve we could factor directly as (3x 2 – 2)(4x 2 – 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.