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Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 5 Clear fractions by multiplying each term by the least common denominator, 6x(x – 5). (Note that the domain must be restricted to x ≠ 0 and x ≠ 5.) Solve. 3 x = 2 x – 5 + 5 6 EXAMPLE 1 Solving an Equation with Fractions That Leads to a Quadratic Equation 10.3 Equations Quadratic in Form 3 x = 2 x – 5 + 5 6 6x(x – 5) 18(x – 5) + 12x = 5x(x – 5) 18x – 90 + 12x = 5x 2 – 25x Distributive property 30x – 90 = 5x 2 – 25x Combine terms.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 6 Combine and rearrange terms so that the quadratic equation is in standard form. Then factor to solve the resulting equation. Solve. 3 x = 2 x – 5 + 5 6 EXAMPLE 1 Solving an Equation with Fractions That Leads to a Quadratic Equation 10.3 Equations Quadratic in Form 30x – 90 = 5x 2 – 25x 5(x 2 – 11x + 18) = 0 Factor. 5x 2 – 55x + 90 = 0 Standard form 5(x – 2)(x – 9) = 0 Factor. x – 2 = 0 or x – 9 = 0 Zero-factor property. x = 2 or x = 9 Solve each equation. Check by substituting these solutions in the original equation. The solution set is { 2, 9 }.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 7 A riverboat for tourists averages 10 mph in still water. It takes the boat 1 hr, 40 minutes to go 8 miles upstream and return. Find the speed of the current. EXAMPLE 2 Solving a Motion Problem 10.3 Equations Quadratic in Form Step 1 Read the problem carefully. Step 2 Assign a variable. Let x = the speed of the current. The current slows down the boat when it is going upstream, so the rate (or speed) upstream is the speed of the boat in still water less the speed of the current, or 10 – x. Riverboat traveling upstream – the current slows it down.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 8 A riverboat for tourists averages 10 mph in still water. It takes the boat 1 hr, 40 minutes to go 8 miles upstream and return. Find the speed of the current. EXAMPLE 2 Solving a Motion Problem 10.3 Equations Quadratic in Form Step 2 Assign a variable. Let x = the speed of the current. The current slows down the boat when it is going upstream, so the rate (or speed) upstream is the speed of the boat in still water less the speed of the current, or 10 – x. Similarly, the current speeds up the boat as it travels downstream, so its speed downstream is 10 + x. Thus, 10 – x = the rate upstream; 10 + x = the rate downstream.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 9 A riverboat for tourists averages 10 mph in still water. It takes the boat 1 hr, 40 minutes to go 8 miles upstream and return. Find the speed of the current. EXAMPLE 2 Solving a Motion Problem 10.3 Equations Quadratic in Form Step 2 Assign a variable. Let x = the speed of the current. The current slows down the boat when it is going upstream, so the rate (or speed) upstream is the speed of the boat in still water less the speed of the current, or 10 – x. 8 10 – x 8 10 + x drt Upstream8 10 – x Downstream8 10 + x

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 10 = 8 10 – x 8 10 + x A riverboat for tourists averages 10 mph in still water. It takes the boat 1 hr, 40 minutes to go 8 miles upstream and return. Find the speed of the current. EXAMPLE 2 Solving a Motion Problem 10.3 Equations Quadratic in Form Step 3 Write an equation. The total time, 1 hr and 40 min, can be written as 8 10 – x 8 10 + x drt Upstream8 10 – x Downstream8 10 + x 40 60 1+ = 2 3 1+ = 5 3 hr. + Time upstreamTime downstream Total Time

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 11 A riverboat for tourists averages 10 mph in still water. It takes the boat 1 hr, 40 minutes to go 8 miles upstream and return. Find the speed of the current. EXAMPLE 2 Solving a Motion Problem 10.3 Equations Quadratic in Form Step 4 Solve the equation. Multiply each side by 3(10 – x )(10 + x ), the LCD, and solve the resulting quadratic equation. 8 10 – x 8 10 + x += 5 3 3(10 + x )8 + 3(10 – x )8 = 5(10 – x )(10 + x ) 24(10 + x ) + 24(10 – x ) = 5(100 – x 2 ) 240 + 24 x + 240 – 24 x = 500 – 5 x 2 Distributive property

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 12 A riverboat for tourists averages 10 mph in still water. It takes the boat 1 hr, 40 minutes to go 8 miles upstream and return. Find the speed of the current. EXAMPLE 2 Solving a Motion Problem 10.3 Equations Quadratic in Form Step 4 Solve the equation. Multiply each side by 3(10 – x )(10 + x ), the LCD, and solve resulting quadratic equation. 480 = 500 – 5 x 2 Combine terms. 5 x 2 = 20 x 2 = 4 Divide by 5. x = 2 or x = –2 Square root property 240 + 24 x + 240 – 24 x = 500 – 5 x 2

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 13 A riverboat for tourists averages 10 mph in still water. It takes the boat 1 hr, 40 minutes to go 8 miles upstream and return. Find the speed of the current. EXAMPLE 2 Solving a Motion Problem 10.3 Equations Quadratic in Form Step 5 State the answer. The speed of the current cannot be –2, so the answer is 2 mph. Step 6 Check that this value satisfies the original problem. drt Upstream8 10 – x Downstream8 10 + x drt Upstream8 10 – 2 Downstream8 10 + 2 drt Upstream8 8 Downstream8 12 2 3 1

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 14 10.3 Equations Quadratic in Form Caution on “Solutions” CAUTION As shown in Example 2, when a quadratic equation is used to solve an applied problem, sometimes only one answer satisfies the application. Always check each answer in the words of the original problem.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 15 It takes two carpet layers 5 hr to carpet a room. If each worked alone, one of them could do the job in 2 hr less time than the other. How long would it take each carpet layer to complete the job alone? EXAMPLE 3 Solving a Work Problem 10.3 Equations Quadratic in Form Step 1 Read the problem again. There will be two answers. Step 2 Assign a variable. Let x represent the number of hours for the slower carpet layer to complete the job alone. Then the faster carpet layer could do the entire job in ( x – 2) hours. The slower person’s rate is, and the faster person’s rate is. 1 x 1 x – 2

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 16 Time Working Fractional Part Rate Together of the Job Done Slower Worker Faster Worker Step 2 (continued) Now complete the table below. The slower person’s rate is, and the faster person’s rate is. Together, they can do the job in 5 hr. It takes two carpet layers 5 hr to carpet a room. If each worked alone, one of them could do the job in 2 hr less time than the other. How long would it take each carpet layer to complete the job alone? EXAMPLE 3 Solving a Work Problem 10.3 Equations Quadratic in Form 1 x 1 x – 2 1 x 1 1 x 1 (5) 5 5

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 17 Step 3 Write an equation. The sum of the fractional parts done by the workers should equal 1 (the whole job). It takes two carpet layers 5 hr to carpet a room. If each worked alone, one of them could do the job in 2 hr less time than the other. How long would it take each carpet layer to complete the job alone? EXAMPLE 3 Solving a Work Problem 10.3 Equations Quadratic in Form 5 x 5 x – 2 Part done by slower worker Part done by faster worker + + = = 1 whole job. 1

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 18 Step 4 Solve the equation from Step 3. It takes two carpet layers 5 hr to carpet a room. If each worked alone, one of them could do the job in 2 hr less time than the other. How long would it take each carpet layer to complete the job alone? EXAMPLE 3 Solving a Work Problem 10.3 Equations Quadratic in Form 5 x 5 x – 2 +=1Multiply by the LCD. 5 x 5 x – 2 += 1 x ( x – 2) Distributive property +=5( x – 2) x ( x – 2)5x5x Distributive property +=5 x – 10 x 2 – 2 x 5x5x Standard form = x 2 – 12 x + 100

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 19 Step 4 Solve the equation. (continued) It takes two carpet layers 5 hr to carpet a room. If each worked alone, one of them could do the job in 2 hr less time than the other. How long would it take each carpet layer to complete the job alone? EXAMPLE 3 Solving a Work Problem 10.3 Equations Quadratic in Form 0 = x 2 – 12 x + 10 This equation cannot be solved by factoring, so use the quadratic formula. ( a = 1, b = –12, c = 10) – b b 2 – 4 ac 2a2a x = + 12 144 – 40 2 x = +

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 20 Step 4 Solve the equation. (continued) It takes two carpet layers 5 hr to carpet a room. If each worked alone, one of them could do the job in 2 hr less time than the other. How long would it take each carpet layer to complete the job alone? EXAMPLE 3 Solving a Work Problem 10.3 Equations Quadratic in Form or 12 144 – 40 2 x = + x ≈ 11.1 12 144 – 40 2 x = – x ≈.9or

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 21 Step 5 State the answer. Only the solution 11.1 makes sense in the original problem. (Why?) Thus, the slower worker can do the job in about 11.1 hr and the faster in about 11.1 – 2 = 9.1 hr. It takes two carpet layers 5 hr to carpet a room. If each worked alone, one of them could do the job in 2 hr less time than the other. How long would it take each carpet layer to complete the job alone? EXAMPLE 3 Solving a Work Problem 10.3 Equations Quadratic in Form Step 6 Check that these results satisfy the original problem.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 23 Solve each equation. EXAMPLE 4 Solving Radical Equations That Lead to Quadratic Equations 10.3 Equations Quadratic in Form (a) n =–2 n + 15 Recall from Section 9.6 that squaring both sides of a radical equation can introduce extraneous solutions that do not satisfy the original equation. All potential solutions must be checked in the original (not the squared) equation.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 24 Solve each equation. EXAMPLE 4 Solving Radical Equations That Lead to Quadratic Equations 10.3 Equations Quadratic in Form (a) n =–2 n + 15 Check: If n = –5, then n =–2 n + 15 –5 =–2(–5) + 15 ? –5 =25 If n = 3, then n =–2 n + 15 3 =–2(3) + 15 ? 3 =9 –5 = 5False3 = 3True Only the solution 3 checks, so the solution set is { 3 }.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 25 Solve each equation. EXAMPLE 4 Solving Radical Equations That Lead to Quadratic Equations 10.3 Equations Quadratic in Form (b) 3 e + e 10 = Isolate the radical on one side. 3 e 10 – e = Square both sides. 9e9e 100 – 20 e + e 2 = Standard form 0 e 2 – 29 e + 100 = Factor. 0( e – 4)( e – 25) = Zero-factor property e – 4 = 0 or e – 25 = 0 Potential solutions e = 4 or e = 25

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 26 Solve each equation. EXAMPLE 4 Solving Radical Equations That Lead to Quadratic Equations 10.3 Equations Quadratic in Form (b) 3 e + e 10 = Check both potential solutions, 4 and 25, in the original equation. Check: If e = 4, thenIf e = 25, then 3 e + e 10 = 3 4 + 410 ? = 6 + 410 ? = 1010 True = 3 e + e 10 = 3 25 + 2510 ? = 15 + 2510 ? = 4010 False = Only the solution 4 checks, so the solution set is { 4 }.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 27 Solve each equation. EXAMPLE 6 Solving Equations That are Quadratic in form 10.3 Equations Quadratic in Form (a) m 4 – 26 m 2 + 25 = 0. m 4 – 26 m 2 + 25 = 0 Because m 4 = ( m 2 ) 2, we can write this equation in quadratic form with u = m 2 and u 2 = m 4. (Instead of u, any letter other than m could be used.) ( m 2 ) 2 – 26 m 2 + 25 = 0 m 4 = ( m 2 ) 2 u 2 – 26 u + 25 = 0 Let u = m 2. ( u – 1)( u – 25) = 0 Factor. u = 1 or u = 25 Solve. u – 1 = 0 or u – 25 = 0 Zero-factor property

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 28 Solve each equation. EXAMPLE 6 Solving Equations That are Quadratic in form 10.3 Equations Quadratic in Form (a) m 4 – 26 m 2 + 25 = 0. To find m, we substitute m 2 for u. u = 1 or u = 25 m 2 = 1 or m 2 = 25 The equation m 4 – 26 m 2 + 25 = 0, a fourth-degree equation, has four solutions. * The solution set is { –5, –1, 1, 5 }. Check by substitution. * In general, an equation in which an n th-degree polynomial equals 0 has n solutions, although some of them may be repeated. m = 1 or m = 5 Square root property ++

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 29 Solve each equation. EXAMPLE 6 Solving Equations That are Quadratic in form 10.3 Equations Quadratic in Form (b) c 4 = 10 c 2 – 2. c 4 – 10 c 2 + 2 = 0 or ( c 2 ) 2 – 10 c 2 + 2 = 0, First write the equation as u 2 – 10 u + 2 = 0. which is quadratic in form with u = c 2.Substitute u for c 2 and u 2 for c 4 to get

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 30 Since this equation cannot be solved by factoring, use the quadratic formula. Solve each equation. EXAMPLE 6 Solving Equations That are Quadratic in form 10.3 Equations Quadratic in Form (b) c 4 = 10 c 2 – 2. u 2 – 10 u + 2 = 0 – b b 2 – 4 ac 2a2a x = + 10 100 – 8 2 u = + a = 1, b = –10, c = 2 10 2 23 2 u = + 10 92 2 u = + 92=4 ·=232

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 31 Solve each equation. EXAMPLE 6 Solving Equations That are Quadratic in form 10.3 Equations Quadratic in Form (b) c 4 = 10 c 2 – 2. Factor. 2 5 23 2 u = + 5 23 + Lowest terms c 2 = 5 23 + or c 2 = 5 23 – Substitute c 2 for u. or Square root property 5 23 c = + + 5 23 c = – +

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 32 Solve each equation. EXAMPLE 6 Solving Equations That are Quadratic in form 10.3 Equations Quadratic in Form (b) c 4 = 10 c 2 – 2. The solution set contains four numbers: 5 23 + + – – – –,,,

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 33 10.3 Equations Quadratic in Form Note on Solving Equations NOTE Some students prefer to solve equations like those in Example 6 (a) by factoring directly. For example, m 4 – 26 m 2 + 25 = 0Example 6(a) equation ( m 2 – 1)( m 2 – 25) = 0Factor. ( m + 1)( m – 1)( m + 25)( m – 25) = 0.Factor again. Using the zero-factor property gives the same solutions obtained in Example 6(a). Equations that cannot be solved by factoring (as in Example 6(c)) must be solved by substitution and the quadratic formula.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 34 Solve 3(2 x – 1) 2 + 8(2 x – 1) – 35 = 0. EXAMPLE 7 Solving Equations That are Quadratic in form 10.3 Equations Quadratic in Form 3(2 x – 1) 2 + 8(2 x – 1) – 35 = 0 Because of the repeated quantity 2 x – 1, this equation is quadratic in form with u = 2 x – 1. 3 u 2 + 8 u – 35 = 0 Let 2 x – 1 = u. (3 u – 7)( u + 5) = 0 Factor. 3 u – 7 = 0 or u + 5 = 0 Zero-factor property u = or u = –5 Zero-factor property 7 3

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 35 Solve 3(2 x – 1) 2 + 8(2 x – 1) – 35 = 0. EXAMPLE 7 Solving Equations That are Quadratic in form 10.3 Equations Quadratic in Form u = or u = –5 7 3 2 x – 1 = or 2 x – 1 = –5 Substitute 2 x – 1 for u. 7 3 2 x = or 2 x = –4 Solve for x. 10 3 x = or x = –2 Solve for x. 5 3 Check that the solution set of the original equation is –2,. 5 3

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 10.3 - 36 10.3 Equations Quadratic in Form Caution CAUTION A common error when solving problems like those in Examples 6 and 7 is to stop too soon. Once you have solved for u, remember to substitute and solve for the values of the original variable.