Pre Calculus

NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 6 ALGEBRA I Lesson 6 Solving Basic OneVariable Quadratic Equations S32 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM4TE130092015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Lesson 6 Solving Basic OneVariable Quadratic Equations Classwork Example 1 A physics teacher put a ball at the top of a ramp and let it roll down toward the floor The class determined that the height of the ball could be represented by the equation ℎ 16𝑡2 4 where the height ℎ is measured in feet from the ground and time 𝑡 is measured in seconds a What do you notice about the structure of the quadratic expression in this problem b In the equation explain what the 4 represents c Explain how you would use the equation to determine the time it takes the ball to reach the floor d Now consider the two solutions for 𝑡 Which one is reasonable Does the final answer make sense based on this context Explain NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 6 ALGEBRA I Lesson 6 Solving Basic OneVariable Quadratic Equations S33 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM4TE130092015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Example 2 Lord Byron is designing a set of square garden plots so some peasant families in his kingdom can grow vegetables The minimum size for a plot recommended for vegetable gardening is at least 2 m on each side Lord Byron has enough space around the castle to make bigger plots He decides that each side should be the minimum 2 m plus an additional 𝑥 m a What expression can represent the area of one individual garden based on the undecided additional length 𝑥 b There are 12 families in the kingdom who are interested in growing vegetables in the gardens What equation can represent the total area 𝐴 of the 12 gardens c If the total area available for the gardens is 300 sq m what are the dimensions of each garden d Find both values for 𝑥 that make the equation in part c true the solution set What value of 𝑥 does Lord Byron need to add to the 2 m NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 6 ALGEBRA I Lesson 6 Solving Basic OneVariable Quadratic Equations S34 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM4TE130092015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Exercises Solve each equation Some of them may have radicals in their solutions 1 3𝑥2 9 0 2 𝑥 32 1 3 4𝑥 32 1 4 2𝑥 32 12 NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 6 ALGEBRA I Lesson 6 Solving Basic OneVariable Quadratic Equations S35 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM4TE130092015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License 5 Analyze the solutions for Exercises 24 Notice how the questions all had 𝑥 32 as a factor but each solution was different radical mixed number whole number Explain how the structure of each expression affected each problemsolution pair 6 Peter is a painter and he wonders if he would have time to catch a paint bucket dropped from his ladder before it hits the ground He drops a bucket from the top of his 9foot ladder The height ℎ of the bucket during its fall can be represented by the equation ℎ 16𝑡2 9 where the height is measured in feet from the ground and the time since the bucket was dropped 𝑡 is measured in seconds After how many seconds does the bucket hit the ground Do you think he could catch the bucket before it hits the ground NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 6 ALGEBRA I Lesson 6 Solving Basic OneVariable Quadratic Equations S36 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM4TE130092015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Problem Set 1 Factor completely 15𝑥2 40𝑥 15 Solve each equation 2 4𝑥2 9 3 3𝑦2 8 13 4 𝑑 42 5 5 4𝑔 12 6 13 6 12 25 𝑘2 20 7 Mischief is a toy poodle that competes with her trainer in the agility course Within the course Mischief must leap through a hoop Mischiefs jump can be modeled by the equation ℎ 16𝑡2 12𝑡 where ℎ is the height of the leap in feet and 𝑡 is the time since the leap in seconds At what values of 𝑡 does Mischief start and end the jump Lesson Summary By looking at the structure of a quadratic equation missing linear terms perfect squares factored expressions you can find clues for the best method to solve it Some strategies include setting the equation equal to zero factoring out the GCF or common factors and using the zero product property Be aware of the domain and range for a function presented in context and consider whether answers make sense in that context M1 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 14 ALGEBRA I Lesson 14 Solving Inequalities S73 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Lesson 14 Solving Inequalities Classwork Exercise 1 1 Consider the inequality 𝑥2 4𝑥 5 a Sift through some possible values to assign to 𝑥 that make this inequality a true statement Find at least two positive values that work and at least two negative values that work b Should your four values also be solutions to the inequality 𝑥𝑥 4 5 Explain why or why not Are they c Should your four values also be solutions to the inequality 4𝑥 𝑥2 5 Explain why or why not Are they d Should your four values also be solutions to the inequality 4𝑥 𝑥2 6 1 Explain why or why not Are they e Should your four values also be solutions to the inequality 12𝑥 3𝑥2 15 Explain why or why not Are they M1 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 14 ALGEBRA I Lesson 14 Solving Inequalities S74 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Example 1 What is the solution set to the inequality 5𝑞 10 20 Express the solution set in words in set notation and graphically on the number line Exercises 23 2 Find the solution set to each inequality Express the solution in set notation and graphically on the number line a 𝑥 4 7 b 𝑚 3 8 9 c 8𝑦 4 7𝑦 2 d 6𝑥 5 30 e 4𝑥 3 2𝑥 2 M1 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 14 ALGEBRA I Lesson 14 Solving Inequalities S75 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License 3 Recall the discussion on all the strange ideas for what could be done to both sides of an equation Lets explore some of the same issues here but with inequalities Recall in this lesson we have established that adding or subtracting and multiplying through by positive quantities does not change the solution set of an inequality Weve made no comment about other operations a Squaring Do 𝐵 6 and 𝐵2 36 have the same solution set If not give an example of a number that is in one solution set but not the other b Multiplying through by a negative number Do 5 𝐶 2 and 5 𝐶 2 have the same solution set If not give an example of a number that is in one solution set but not the other c Bonzos ignoring exponents Do 𝑦2 52 and 𝑦 5 have the same solution set M1 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 14 ALGEBRA I Lesson 14 Solving Inequalities S76 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Example 2 Jojo was asked to solve 6𝑥 12 3𝑥 6 for 𝑥 She answered as follows 6𝑥 12 3𝑥 6 6𝑥 2 3𝑥 2 Apply the distributive property 6 3 Multiply through by 1 𝑥2 a Since the final line is a false statement she deduced that there is no solution to this inequality that the solution set is empty What is the solution set to 6𝑥 12 3𝑥 6 b Explain why Jojo came to an erroneous conclusion Example 3 Solve 𝑞 7 for 𝑞 M1 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 14 ALGEBRA I Lesson 14 Solving Inequalities S77 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Exercises 47 4 Find the solution set to each inequality Express the solution in set notation and graphically on the number line a 2𝑓 16 b 𝑥 12 1 4 c 6 𝑎 15 d 32𝑥 4 0 5 Use the properties of inequality to show that each of the following is true for any real numbers 𝑝 and 𝑞 a If 𝑝 𝑞 then 𝑝 𝑞 b If 𝑝 𝑞 then 5𝑝 5𝑞 Recall the properties of inequality Addition property of inequality If 𝐴 𝐵 then 𝐴 𝑐 𝐵 𝑐 for any real number 𝑐 Multiplication property of inequality If 𝐴 𝐵 then 𝑘𝐴 𝑘𝐵 for any positive real number 𝑘 M1 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 14 ALGEBRA I Lesson 14 Solving Inequalities S78 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License c If 𝑝 𝑞 then 003𝑝 003𝑞 d Based on the results from parts a through c how might we expand the multiplication property of inequality 6 Solve 4 2𝑡 14 18𝑡 6 100𝑡 for 𝑡 in two different ways first without ever multiplying through by a negative number and then by first multiplying through by 1 2 7 Solve 𝑥 4 8 1 2 for 𝑥 in two different ways first without ever multiplying through by a negative number and then by first multiplying through by 4 M1 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 14 ALGEBRA I Lesson 14 Solving Inequalities S79 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Problem Set 1 Find the solution set to each inequality Express the solution in set notation and graphically on the number line a 2𝑥 10 b 15𝑥 45 c 2 3 𝑥 1 2 2 d 5𝑥 1 10 e 13𝑥 91 𝑥 2 Find the mistake in the following set of steps in a students attempt to solve 5𝑥 2 𝑥 2 5 for 𝑥 What is the correct solution set 5𝑥 2 𝑥 2 5 5 𝑥 2 5 𝑥 2 5 factoring out 5 on the left side 5 1 dividing by 𝑥 2 5 So the solution set is the empty set 3 Solve 𝑥 16 1 5𝑥 2 for 𝑥 without multiplying by a negative number Then solve by multiplying through by 16 4 Lisa brought half of her savings to the bakery and bought 12 croissants for 1420 The amount of money she brings home with her is more than 200 Use an inequality to find how much money she had in her savings before going to the bakery Write the inequality that represents the situation and solve it M1 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 5 ALGEBRA I Lesson 5 Two Graphing Stories S20 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Problem Set 1 Reread the story about Maya and Earl from Example 1 Suppose that Maya walks at a constant rate of 3 ft every second and Earl walks at a constant rate of 4 ft every second starting from 50 ft away Create equations for each persons distance from Mayas door and determine exactly when they meet in the hallway How far are they from Mayas door at this time 2 Consider the story May June and July were running at the track May started first and ran at a steady pace of 1 mi every 11 min June started 5 min later than May and ran at a steady pace of 1 mi every 9 min July started 2 min after June and ran at a steady pace running the first lap 1 4 mi in 15 min She maintained this steady pace for 3 more laps and then slowed down to 1 lap every 3 min a Sketch May June and Julys distanceversustime graphs on a coordinate plane b Create linear equations that represent each girls mileage in terms of time in minutes You will need two equations for July since her pace changes after 4 laps 1 mi c Who was the first person to run 3 mi d Did June and July pass May on the track If they did when and at what mileage e Did July pass June on the track If she did when and at what mileage 3 Suppose two cars are travelling north along a road Car 1 travels at a constant speed of 50 mph for two hours then speeds up and drives at a constant speed of 100 mph for the next hour The car breaks down and the driver has to stop and work on it for two hours When he gets it running again he continues driving recklessly at a constant speed of 100 mph Car 2 starts at the same time that Car 1 starts but Car 2 starts 100 mi farther north than Car 1 and travels at a constant speed of 25 mph throughout the trip a Sketch the distanceversustime graphs for Car 1 and Car 2 on a coordinate plane Start with time 0 and measure time in hours b Approximately when do the cars pass each other c Tell the entire story of the graph from the point of view of Car 2 What does the driver of Car 2 see along the way and when d Create linear equations representing each cars distance in terms of time in hours Note that you will need four equations for Car 1 and only one for Car 2 Use these equations to find the exact coordinates of when the cars meet Lesson Summary The intersection point of the graphs of two equations is an ordered pair that is a solution to both equations In the context of a distance or elevation story this point represents the fact that both distances or elevations are equal at the given time Graphing stories with quantities that change at a constant rate can be represented using piecewise linear equations M1 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 5 ALGEBRA I Lesson 5 Two Graphing Stories S21 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License 4 Suppose that in Problem 3 above Car 1 travels at the constant speed of 25 mph the entire time Sketch the distanceversustime graphs for the two cars on a graph below Do the cars ever pass each other What is the linear equation for Car 1 in this case 5 Generate six distinct random whole numbers between 2 and 9 inclusive and fill in the blanks below with the numbers in the order in which they were generated 𝐴 0 𝐵 𝐶 10 𝐷 0 𝐸 10 Link to a random number generator httpwwwmathgoodiescomcalculatorsrandomnocustomhtml a On a coordinate plane plot points 𝐴 𝐵 and 𝐶 Draw line segments from point 𝐴 to point 𝐵 and from point 𝐵 to point 𝐶 b On the same coordinate plane plot points 𝐷 and 𝐸 and draw a line segment from point 𝐷 to point 𝐸 c Write a graphing story that describes what is happening in this graph Include a title 𝑥 and 𝑦axis labels and scales on your graph that correspond to your story M1 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 5 ALGEBRA I Lesson 5 Two Graphing Stories S22 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License 6 The following graph shows the revenue or income a company makes from designer coffee mugs and the total cost including overhead maintenance of machines etc that the company spends to make the coffee mugs a How are revenue and total cost related to the number of units of coffee mugs produced b What is the meaning of the point 0 4000 on the total cost line c What are the coordinates of the intersection point What is the meaning of this point in this situation d Create linear equations for revenue and total cost in terms of units produced and sold Verify the coordinates of the intersection point e Profit for selling 1000 units is equal to revenue generated by selling 1000 units minus the total cost of making 1000 units What is the companys profit if 1000 units are produced and sold 0 2000 4000 6000 8000 10000 12000 14000 0 100 200 300 400 500 600 700 800 900 1000 Dollars Units Produced and Sold Total Cost Revenue NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 26 ALGEBRA II Lesson 26 Solving Rational Equations S130 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IIM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Lesson 26 Solving Rational Equations Classwork Exercises 12 Solve the following equations for 𝑥 and give evidence that your solutions are correct 1 𝑥 2 1 3 5 6 2 2𝑥 9 5 9 8 9 Example Solve the following equation 𝑥3 12 5 6 NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 26 ALGEBRA II Lesson 26 Solving Rational Equations S131 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IIM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Exercises 37 3 Solve the following equation 3 𝑥 8 𝑥2 4 Solve the following equation for 𝑎 1 𝑎2 1 𝑎2 4 𝑎24 5 Solve the following equation Remember to check for extraneous solutions 4 3𝑥 5 4 3 𝑥 NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 26 ALGEBRA II Lesson 26 Solving Rational Equations S132 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IIM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License 6 Solve the following equation Remember to check for extraneous solutions 7 𝑏 3 5 𝑏 3 10𝑏 2 𝑏2 9 7 Solve the following equation Remember to check for extraneous solutions 1 𝑥 6 𝑥 𝑥 2 4 𝑥2 8𝑥 12 NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 26 ALGEBRA II Lesson 26 Solving Rational Equations S133 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IIM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Problem Set 1 Solve the following equations and check for extraneous solutions a 𝑥8 𝑥4 2 b 4𝑥8 𝑥2 4 c 𝑥4 𝑥3 1 d 4𝑥8 𝑥2 3 e 1 2𝑎 2 2𝑎3 0 f 3 2𝑥1 5 4𝑥3 g 4 𝑥5 2 5𝑥 2 𝑥 h 𝑦2 3𝑦2 𝑦 𝑦1 2 3 i 3 𝑥1 2 1𝑥 1 j 4 𝑥1 3 𝑥 3 0 k 𝑥1 𝑥3 𝑥5 𝑥2 17 6 l 𝑥7 4 𝑥1 2 5𝑥 3𝑥14 m 𝑏2𝑏6 𝑏2 2𝑏12 𝑏 𝑏39 2𝑏 n 1 𝑝𝑝4 1 𝑝6 𝑝 o 1 ℎ3 ℎ4 ℎ2 6 ℎ2 p 𝑚5 𝑚2𝑚 1 𝑚2𝑚 𝑚6 𝑚1 2 Create and solve a rational equation that has 0 as an extraneous solution 3 Create and solve a rational equation that has 2 as an extraneous solution Lesson Summary In this lesson we applied what we have learned in the past two lessons about addition subtraction multiplication and division of rational expressions to solve rational equations An extraneous solution is a solution to a transformed equation that is not a solution to the original equation For rational functions extraneous solutions come from the excluded values of the variable Rational equations can be solved one of two ways 1 Write each side of the equation as an equivalent rational expression with the same denominator and equate the numerators Solve the resulting polynomial equation and check for extraneous solutions 2 Multiply both sides of the equation by an expression that is the common denominator of all terms in the equation Solve the resulting polynomial equation and check for extraneous solutions NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 26 ALGEBRA II Lesson 26 Solving Rational Equations S134 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg This file derived from ALG IIM1TE130072015 This work is licensed under a Creative Commons AttributionNonCommercialShareAlike 30 Unported License Extension 4 Two lengths 𝑎 and 𝑏 where 𝑎 𝑏 are in golden ratio if the ratio of 𝑎 𝑏 is to 𝑎 is the same as 𝑎 is to 𝑏 Symbolically this is expressed as 𝑎 𝑏 𝑎𝑏 𝑎 We denote this common ratio by the Greek letter phi pronounced fee with symbol 𝜑 so that if 𝑎 and 𝑏 are in common ratio then 𝜑 𝑎 𝑏 𝑎𝑏 𝑎 By setting 𝑏 1 we find that 𝜑 𝑎 and 𝜑 is the positive number that satisfies the equation 𝜑 𝜑1 𝜑 Solve this equation to find the numerical value for 𝜑 5 Remember that if we use 𝑥 to represent an integer then the next integer can be represented by 𝑥 1 a Does there exist a pair of consecutive integers whose reciprocals sum to 5 6 Explain how you know b Does there exist a pair of consecutive integers whose reciprocals sum to 3 4 Explain how you know c Does there exist a pair of consecutive even integers whose reciprocals sum to 3 4 Explain how you know d Does there exist a pair of consecutive even integers whose reciprocals sum to 5 6 Explain how you know Lesson 26 Exercises 1 x2 13 56 6x2 13 56 6 3x 2 5 3x 3 x 1 replaces x 1 in equation 12 13 3 26 56 2 2x9 59 89 2x 5 8 2x 3 x 32 replaces x 32 in equation 29 32 59 39 59 89 example x 312 56 x 3 1256 25 10 x 7 3 3x 8x2 3x 2 8x 3x 6 8x 6 5x x 65 4 Note that that a squared 4 a 2a 2 1a2 1a2 4a squared 4 a squared 41a2 1a 2 4 a 2 a 2 4 2a 4 a 2 5 93x 54 3x 12x93x 54 12x3x 16 35x 36 15x 20 x 2015 43 6 Note that b squared 9 b 3b 3 b squared 97b3 5b3 10b 2 7b 3 5b 3 10b 2 12b 6 10b 2 2b 4 b 2 7 Note that x squared 8x 12 x 6x 2 1x 6 xx 2 6 x 6x 21x 6 xx 2 6 x 2 xx 6 6 x 2 x squared 6x 6 x squared 5x 6 0 x 5 25 4162 5 992 x1 5 72 6 or x2 5 72 1 these solutions are Problem set a x 8x 4 2 x 8 2x 4 x 8 2x 8 x 2x x 0 b x 8x 2 3 3x 8 3x 6 x 8 6 2 não é uma solução da equação anula denominador c x 9x 3 1 x 9 x 3 9 3 without solutions e 12a 22a 3 0 2a2a 312a 22a 3 0 2a 3 22a 0 2a 3 a 32 f 32x 1 59x 3 39x 3 52x 1 12x 9 10x 5 2x 4 x 2 g x 5x 5 x 9x 25 x xx 5x 52x 9xx 5 2x 5x 2x 5x 5 9x squared 20x 2x squared 10x 2x squared 25 2x squared 30x 2x squared 50 30x 50 x 53 h 33y 2y 1y 23y 2 yy 1 33y 2y 1 23 3y 1y 2 3y3y 2 23y 2y 1 3y squared y 2 33y squared 2y 23y squared 5y 2 3y squared 3y 6 9y squared 6y 6y squared 10y 4 6y squared 7y 10 0 y 7 49 96102 7 2892 y1 7 172 5 or y2 7 172 12 6 x11x3x1 2x x x11x 31x 2x1 1x² 33x2x2 1 x² x² 5x 0 x x5 x0 or x5 7 x x1 6x1 3x 3 0 9x 3 x1 3x² x 0 9x3x1 3x² 3x 0 3x² 10 x 1 0 3x² 10x 1 0 x 10 100 4 3 16 10 886 x₁ 10 886 or x₂ 10 886 k 6 x3 x2 x1x3 x5x2 17 x3 x2 6 x1 x2 6 x3 x5 17 x3 x2 6 x²3x2 6 x² 2x 15 17 x² 5x 6 18x 12 12x 90 17x² 5x 6 17x² 25x 96 0 x 25 625 4 17 9634 25 715334 x₁ 25 715334 or x₂ 25 715334 l 8 3x19 x74 x12 5x3x19 2 x7 3x19 4 x1 3x19 8 5x 2 3x² 7x 98 4 3x² 11x 19 90 8x 6x² 14x 196 12x² 44x 56 90 8x 6x² 66x 180 0 x₁ 66 66² 4 6 18012 66 612 6012 5 or x₂ 66 612 7212 6 m 2b² b²b6b² 2b 12b 20² b 3920 2 b² b 6 2b 2b 12 b² 39 b 2b² 2b 12 4b² 24b b² 39 b 3b² 13b 12 0 b₁ 13 13² 4 3 12 6 13 56 3 or b₂ 13 13² 12²6 13 56 43 n p p9 1p p9 1 p6p 1 pp9 p4p6 1 p² 9 p p² 10 p 29 6p 23 p 236 o k 3k 2 1k 3 k 4 6k 2 k 3k 2 k 2 k 3k 10 k 2 k² 13 k 30 k² 12 k 32 0 x₁ 12 12² 4 322 4 or x₂ 12 42 8 P m²m m5m²m m²m 1m²m m6m1 m5 1mm6 m5 1m²6m m²5m60 m₁ 5254162 82 4 or m₂ 532 1 2 1x 2 1x 1 x21x 2x1 2x0 x0 but the equation 1x 2 1x dont have solutions 3 4x2 2 1x2 x2 but the equation 1x 2 1x dont have solutions Extension 4 φ φ1φ φ² φ 1 φ² φ 1 0 consider x²x10 x₁ 114112 152 or x₂ 152 Then φ0 φ 152 5a 1x 1x1 56 62x1 5xx1 12x 6 5x2 5x 5x2 7x 6 0 x1 2 or x 35 Yes take x 2 b 1x 1x1 39 92x1 3x2 3x 8x 9 3x2 3x 3x2 5x 9 0 x1 5 25 932 5 132 or x2 5 132 No we dont have x ℤ c 12j 12j1 34 1j 1j1 32 22j1 3j2 3j 3j2 j 2 0 Yes take 2 12 19 39 d 12j 12j1 56 32j1 5j2 5j 5j2 j 3 0 j1 1 1 95310 1 6110 or j2 1 6110 No we dont have solutions Lesson 6 Example 1 a Is a quadratic equation without the term Bx b 4 represents the height of the ramp c Just find the roots of the equation thats where the ball hits the ground d 0 h 16t2 9 t2 916 14 t 12 The solution t 12 s dont make sense because the time is was started in t0 Then t 12 s Example 2 a A x22 b 12A 12x22 c and d 12A 300 12x22 300 x22 25 x2 5 x 3m or x 7m make no sense here Load Byron needs adds 3m Exercises 1 3x2 9 0 x2 93 x2 3 x 3 2 x32 1 x3 1 x 3 1 x1 4 and x2 2 3 9x32 1 x32 14 x3 12 x1 72 or x2 52 4 2x32 12 x32 6 x3 6 x1 3 6 or x2 3 6 5 In each exercise we have at some point some equality of the type x32 a the nature of the solution depends on a re a is a square of a integer then the solution will be a integer if it is a free number of squares then the solution will be a root radical 6 Hits the ground when h 0 0 h 16t2 9 t2 916 t 34 We need t 0 then t 34 075 s He may not be able to catch the bucket before he falls Problem set 1 First we find the roots by Bhaskara 15x2 90x 15 0 3x2 8x 3 0 x 8 69 936 8 106 x1 3 or x2 13 then 15x2 90x 15 15x3x 13 2 4x2 9 x2 94 x 32 3 3y2 8 13 3y2 21 y2 7 y 7 4 d 92 5 d 9 5 d 9 5 5 4g 12 6 13 4g 12 7 g 12 74 g 1 72 g 1 72 6 20 25 k2 12 20 12 25 k2 82 5 k2 5 k 2 S 2 K 7 just find the roots 0 16t2 12t t16t 12 t 0 or 16t 12 0 1216 t 34 starts the jump at t 0 and ends it at t 34 075 s lesson 14 1 a Take x1 1 and x2 2 as positive values and x3 5 x4 6 as negative values x12 4x1 1 4 5 5 x22 4x2 4 8 12 5 x32 4x3 25 20 5 5 x42 4x4 36 24 12 5 b yes there was only one factorization xx4 x2 4x 5 c yes there was just a lot of order because 4x x2 x2 4x 5 d yes add 6 we have 4x x2 6 6 1 6 4x x2 5 e yes just divide by 3 1312x 3x2 153 4x x2 5 Example 1 We have 5q 10 20 5q 10 q 2 the solution to the inequality is all real numbers greater than 2 S q R q 2 o 2 2 a x 4 7 x 7 4 x 3 S x R x 3 0 3 b m3 8 9 m3 9 8 m3 1 m 3 S m R m 3 0 3 c 8y 6 7y 2 8y 7y 2 4 y 6 S y R y 6 6 0 d 6x 5 30 divide por 6 x 5 5 x 10 S x R x 10 0 10 e 4x 3 2x 2 4x 3 x 2 2x 6 x 2 2x x 6 2 x 4 S x R x 4 0 4 3 a Take B 7 7 6 but 72 62 49 36 is not valid b Dont have the same solution set take c 1 then in the first inequality 5 1 4 2 ok but in the second 5 1 4 2 is not valid c Dont have the same solution set because in first inequality we have y2 52 y2 52 0 y 5y 5 0 y 5 0 or y 5 0 y 5 or y 5 then S1 y R 5 y 5 but in the second S2 y R y 5 Example 2 a 6x 12 3x 6 6x 3x 6 12 3x 6 x 2 b The error is in the part about multiplying by 1x2 which is a term that varies in sign and can have a null denominator 3 q 7 7 q q 7 7 q 7 q or q 7 4 a 2 f 16 2 f 16 f 8 S f ε R f 8 b x 12 14 x 12 14 x 124 x 3 S x R x 3 c 6 a 15 6 15 a 6 15 a 9 a or a 9 d 3 2x 4 0 3 2x 4 0 6x 12 0 x 126 x 2 S x R x 2 5 a adding p q p p q q p q q p or p q b adding p q p p q q p q q p or p q multiplying by 5 5 p 5 q c adding p q and multiplying by 003 003p p q 003q p q 003 q 003 p or 003 p 003 q d given k 0 if a b then k a k b 6 9 2t 19 18t 6 100t 18 16t 6 100t 100t 16t 18 6 84t 12 t 1284 321 17 12 9 2t 19 18t 12 6 100t 2 t 7 9t 3 50t 8t 9 3 50t 9 3 50t 8t 6 42t 642 t t 642 17 7 x4 8 12 8 12 x4 1º method 4 8 12 x 32 2 x 2º method 30 x or x 30 4 x4 8 4 12 x 32 2 x 32 2 x 30 Problem set 1 a 2x 10 x 5 S x R x 5 b 15x 45 x 4515 3 S x R x 3 c 23 x 12 2 32 23 x 32 12 2 x 32 52 x 152 S x R x 152 d 5 x 1 10 x 1 2 x 3 S x R x 3 e 13x 9 1 x 13x 9 9x 13x 9x 9 22x 9 x 922 S x R x 922 2 The error is in the division by x 25 A solução correta é 5 x 25 x 25 5 x 25 0 x 25 0 x 25 3 1º method x16 1 5x2 5x2 x16 1 40x16 x16 1 39x 16 x 1639 2º method 16 x16 1 16 5x2 x 16 90x 16 39x 1639 x ou x 1634 9 Let x be the initial amount of money from the statement we have the following inequality x2 1920 2 x2 1620 x 3240 therefore she had an amount greater than or equal to 3290 2 Consider the story May June and July were running at the track May started first and ran at a steady pace of 1 mi every 11 min June started 5 min later than May and ran at a steady pace of 1 mi every 9 min July started 2 min after June and ran at a steady pace running the first lap 14 mi in 15 min She maintained this steady pace for 3 more laps and then slowed down to 1 lap every 3 min a Sketch May June and Julys distanceversustime graphs on a coordinate plane b Create linear equations that represent each girls mileage in terms of time in minutes You will need two equations for July since her pace changes after 4 laps 1 mi Equations for May June and July are shown below Notice that July has two equations since her speed changes after her first mile which occurs 13 min after May starts running May d 111 t June d 19 t 5 July d 16 t 7 t 13 and d 112 t 13 1 t 13 c Who was the first person to run 3 mi June at time 32 min d Did June and July pass May on the track If they did when and at what mileage Assuming that they started at the same place June passes May at time 275 min at the 25 mi marker July does not pass May e Did July pass June on the track If she did when and at what mileage July passes June at time 11 min at the 23 mi marker Exit Ticket Sample Solutions Maya and Earl live at opposite ends of the hallway in their apartment building Their doors are 50 ft apart Each person starts at his or her own door and walks at a steady pace toward the other They stop walking when they meet Suppose Maya walks at a constant rate of 3 ft every second Earl walks at a constant rate of 4 ft every second 1 Graph both peoples distance from Mayas door versus time in seconds Graphs should be scaled and labeled appropriately Mayas graph should start at 0 0 and have a slope of 3 and Earls graph should start at 0 50 and have a slope of 4 2 According to your graphs approximately how far will they be from Mayas door when they meet The graphs intersect at approximately 7 sec at a distance of about 21 ft from Mayas door 3 Suppose two cars are travelling north along a road Car 1 travels at a constant speed of 50 mph for two hours then speeds up and drives at a constant speed of 100 mph for the next hour The car breaks down and the driver has to stop and work on it for two hours When he gets it running again he continues driving recklessly at a constant speed of 100 mph Car 2 starts at the same time that Car 1 starts but Car 2 starts 100 mi farther north than Car 1 and travels at a constant speed of 25 mph throughout the trip a Sketch the distanceversustime graphs for Car 1 and Car 2 on a coordinate plane Start with time 0 and measure time in hours A graph is shown below that approximates the two cars traveling north b Approximately when do the cars pass each other The cars pass after about 212 hr after 4 hr and after about 512 hr c Tell the entire story of the graph from the point of view of Car 2 What does the driver of Car 2 see along the way and when The driver of Car 2 is carefully driving along at 25 mph and he sees Car 1 pass him at 100 mph after about 212 hr About 112 hr later he sees Car 1 broken down along the road After about another 112 hr Car 1 whizzes past again d Create linear equations representing each cars distance in terms of time in hours Note that you will need four equations for Car 1 and only one for Car 2 Use these equations to find the exact coordinates of when the cars meet Using the variables d for distance in miles and t for time in hours Equation for Car 2 d 25t 100 Equations for Car 1 d 50t 0 t 2 d 100t 2 100 100t 1 2 t 3 d 200 3 t 5 d 100t 5 200 100t 3 5 t Intersection points First solving 100t 1 25t 100 gives 20075 2520075 100 27 1667 Second solving 200 25t 100 gives 4 200 and Third solving 100t 3 25t 100 gives 40075 2540075 100 53 2333 Suppose that in Problem 3 above Car 1 travels at the constant speed of 25 mph the entire time Sketch the distanceversustime graphs for the two cars on a graph below Do the cars ever pass each other What is the linear equation for Car 1 in this case A sample graph is shown below Car 1 never overtakes Car 2 and they are 100 mi apart the entire time The equation for Car 1 is y 25t 5 Generate six distinct random whole numbers between 2 and 9 inclusive and fill in the blanks below with the numbers in the order in which they were generated A 0 B C 10 D 0 E 10 Link to a random number generator httpwwwmathgoodiescomcalculatorsrandomnocustomhtml a On a coordinate plane plot points A B and C Draw line segments from point A to point B and from point B to point C b On the same coordinate plane plot points D and E and draw a line segment from point D to point E c Write a graphing story that describes what is happening in this graph Include a title x and yaxis labels and scales on your graph that correspond to your story Answers will vary depending on the random points generated EUREKA MATH Lesson 5 Two Graphing Stories 61 This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eureka mathorg ALG IM1TEB1130052015 6 The following graph shows the revenue or income a company makes from designer coffee mugs and the total cost including overhead maintenance of machines etc that the company spends to make the coffee mugs a How are revenue and total cost related to the number of units of coffee mugs produced Definition Profit Revenue Cost Revenue is the income from the sales and is directly proportional to the number of coffee mugs actually sold it does not depend on the units of coffee mugs produced Total cost is the sum of the fixed costs overhead maintaining the machines rent etc plus the production costs associated with the number of coffee mugs produced it does not depend on the number of coffee mugs sold b What is the meaning of the point 0 4000 on the total cost line The overhead costs the costs incurred regardless of whether 0 or 1000 coffee mugs are made or sold is 4000 c What are the coordinates of the intersection point What is the meaning of this point in this situation 500 6000 The revenue 6000 from selling 500 coffee mugs is equal to the total cost 6000 of producing 500 coffee mugs This is known as the breakeven point When Revenue Cost the Profit is 0 After this point the more coffee mugs sold the more the positive profit before this point the company loses money d Create linear equations for revenue and total cost in terms of units produced and sold Verify the coordinates of the intersection point If u is a whole number for the number of coffee mugs produced and sold C is the total cost to produce u mugs and R is the total revenue when u mugs are sold then C 4000 4u R 12u When u 500 both C 4000 4 500 6000 and R 12 500 6000 e Profit for selling 1000 units is equal to revenue generated by selling 1000 units minus the total cost of making 1000 units What is the companys profit if 1000 units are produced and sold Profit Revenue Total Cost Hence P R C 12 1000 4000 4 1000 12000 8000 4000 The companys profit is 4000 EUREKA MATH 62 Lesson 5 Two Graphing Stories This work is derived from Eureka Math and licensed by Great Minds 2015 Great Minds eurekamathorg ALG IM1TEB1130052015