Lesson 8.2 Exercises, pages

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1 Lesson 8. Eercises, pages 38 A Students should verif the solutions to all equations.. Which values of are not roots of each equation? a) ƒ - 3 ƒ = 7 = 5 or =- Use mental math. 5: L.S. 7 R.S. 7 : L.S. 7 R.S. 7 So, both 5 and are roots. b) ƒ - + ƒ = 8 =-0.5 or = 3 Use mental math. 0.5: L.S. 8 R.S. 8 3: L.S. R.S. 8 So, 3 is not a root of the equation. c) ƒ ƒ = 8 = or =-0.75 Use a calculator. : L.S. 8 R.S : L.S..815 R.S. 8 So, 0.75 is not a root of the equation. 5. Use the graphs to determine the approimate solutions of each equation. Where necessar, give the solutions to the nearest tenth. a) ƒ ƒ = 9 b) ƒ - ƒ = The line 9 intersects The line 1 intersects 3 3 at points: at points, (, 9) and (, 9). So, the which appear to be: solutions are and. ( 1.7, 1), ( 1, 1), (1, 1), and (1.7, 1). So, the solutions are 1.7, 1, 1, and Solving Absolute Value Equations Solutions DO NOT COPY. P

2 B. Solve b graphing. a) 5 = ƒ ƒ b) = ƒ 5 - ƒ To graph 1, To graph 5, graph graph 1, then 5, then reflect, in the reflect, in the -ais, the part -ais, the part of the graph that of the graph that is below the is below the -ais. The line -ais. The line 5 intersects does not intersect 1 at ( 1, 5) and 5. So, the (1.5, 5). So, the solutions are equation has no solution. 1 and 1.5. c) ƒ - 3 ƒ = 9 d) ƒ + 8 ƒ = To graph 3, To graph 8, graph graph 3, then 8, then reflect, in the reflect, in the -ais, the -ais, the part of the graph that part of the graph that is below the -ais. The line is below the -ais. 0 intersects 8 The line 9 intersects at (, 0). So, the solution is 3 at ( 3, 9). and (, 9). So, the solutions are 3 and. P DO NOT COPY. 8. Solving Absolute Value Equations Solutions 11

3 7. Solve b graphing. Where necessar, give the solutions to the nearest tenth. a) ƒ ƒ = b) = ƒ ƒ Enter 1 and Enter 3 and in the graphing calculator. in the graphing calculator. The line appears to The line appears to intersect intersect 1 at 3 at points: points: ( , ) and ( ,),( 1.5, ), (0, ), (.1..., ). and ( ,).So,the So, the equation has solutions: equation has solutions: 0. and...9, 1.5, 0, and 1.. c) = ƒ ƒ d) Enter 7 3 and in the graphing calculator. The line appears to intersect 7 3 at points: ( , ) and ( , ). So, the equation has solutions: 3.9 and 0.. ƒ ƒ = Enter and in the graphing calculator. The line does not intersect. So, the equation has no solution. 8. Use algebra to solve each equation. a) ƒ + 3 ƒ = 3 if 3» 0 that is, if» 3 When» 3 : » 3, so this root ( 3) if 3<0 that is, if < 3 When < 3 : ( 3) < 3, so this root The solutions are 1 and Solving Absolute Value Equations Solutions DO NOT COPY. P

4 b) 3 = ƒ + 5 ƒ 5 3 if 5» 0 that is, if» 5 ( 5) if 5<0 that is, if < 5 When» 5 : » 5, so this root The solutions are 1and. When < 5 : ( 5) < 5, so this root c) = ƒ ƒ When 5» 0: When 5<0: 5 ( 5) ( 1)( 3) or 3 _ () (1)(7) (1) _ 1 This is not a real number. So, 1and 3are the solutions. d) ƒ ƒ = 5 When 5» 0: When 5<0: 5 5 ( 5) 5 0 ( ) 0 0 or _ ( ) (1)(10) (1) _ This is not a real number. So, 0 and are the solutions. P DO NOT COPY. 8. Solving Absolute Value Equations Solutions 13

5 9. For which values of c does the equation ƒ 3 + ƒ = c have: a) solutions? This is the graph of 3. For the equation to have solutions, the line c must intersect the graph of 3 at points; that is, c> b) 1 solution? For the equation to have 1 solution, the line c must intersect the graph of 3 at 1 point; that is, c 0. c) no solution? For the equation to have no solution, the line c must not intersect the graph of 3 ; that is, c< A manufacturer rejects 75-g boes of crackers when the actual mass of the bo differs from the stated mass b more than 3.5 g. a) Write an absolute value equation that can be used to determine the greatest and least masses that are acceptable. Let m grams represent the mass of a bo of crackers. So, an equation is: m b) Solve the equation. What is the least mass that is acceptable? What is the greatest mass? When m 75» 0: When m 75<0: m (m 75) 3.5 m 78.5 m m 71.5 So, the least acceptable mass is 71.5 g and the greatest mass is 78.5 g Solving Absolute Value Equations Solutions DO NOT COPY. P

6 11. Use algebra to solve each equation. a) ƒ + 1 ƒ = (7 ) 1 1 (7 ) if 1» 0 that is, if» 1 ( 1) 1 (7 ) if 1<0 that is, if < 1 When» 1 : When < 1 : 1 1 (7 ) ( 1) 1 (7 ) » 1, so this root 9 < 1, so this root 5 The solutions are 5 and b) = ƒ - 11 ƒ When 11» 0: When 11<0: ( 11) ( 11)( 1) or 1 ( 11)( 1) 0 11 or 1 So, 1, 1, and 11 are the solutions. c) ` 1-3 ` = ( ) if » 0 if <0 that is, if» 1.5 that is, if <1.5 When» 1.5: When <1.5: ( ) » 1.5, so this root.5.5<1.5, so this root The solutions are.5 and 9.5. P DO NOT COPY. 8. Solving Absolute Value Equations Solutions 15

7 d) = ƒ + ƒ When» 0: When <0: 3 18 ( ) 3 18 ( ) ( )( 3) ( )( 3) 0 or 3 or 3 So,, 3, and 3 are the solutions. 1. A student solved the equation ƒ ƒ - = -3 and reasoned that since the absolute value of an epression cannot be negative, the equation has no solution. Is the student correct? Eplain. If the student is not correct, describe the error the student made and solve the equation. The student is incorrect. The student should have simplified the equation first b adding to both sides. Then the equation becomes Correct solution: ( 3 1) ( 1)( ) ( 3) 0 1 or 0 or 3 So,, 3, 0, and 1 are the solutions. 13. A car is travelling toward the British Columbia-Alberta border. The car is 150 km from the border and is travelling at an average speed of 100 km/h. a) Write an absolute value equation to represent the distance, d kilometres, of the car from the border after t hours. After 1 h, the car has travelled 100 km. After t hours, the car has travelled 100t kilometres. So, the distance from the border after t hours is represented b the equation: d t 1 8. Solving Absolute Value Equations Solutions DO NOT COPY. P

8 b) Determine when the car is 5 km from the border. Eplain our strateg. Substitute: d t When t» 0: When t<0: t 5 ( t) 100t t t t 175 t 1 3 So, the car is 5 km from the border after 1 1 h and after 1 3 h. The car can be 5 km from the border on the Alberta side or 5 km from the border on the British Columbia side. C 1. The function = f () is linear. The line = intersects = ƒ f () ƒ at = and = 0. The line = 3 intersects = ƒ f () ƒ at = 1 and = 3. What is an equation for the function = f ()? f() 0 f() passes through the points (0, ), (, ), (1, 3), and (3, 3). Plot the points. The points are smmetrical about the line, so plot a point at (, 0). Join the points from (, 0) to (0, ) and from (, 0) to (, ) with straight lines. An equation for the right branch of the graph has the form m b. m 3 3 m 3 Use: 3 b Substitute: and 3() b b So, an equation for the function f() is 3. P DO NOT COPY. 8. Solving Absolute Value Equations Solutions 17

9 15. A student used this graph to solve an absolute value equation. What might the equation have been? Eplain our strateg The line has -intercept 3 and -intercept 3. An equation of the line has the form m 3. Use the point ( 3, 0). Substitute: 3and 0 0 m( 3) 3 m 1 So, an equation of the line is: 3 Assume the middle piece of the graph was reflected in the -ais. So, the graph of the quadratic function opens up and has verte (0, 9). So, the equation has the form a 9. An -intercept of the graph is 3, so use the point (3, 0). Substitute 3 and 0. 0 a(3) 9 0 9a 9 a 1 An equation for the quadratic function is 9, and an equation for the absolute value function is 9. So, the student might have used the graph to solve the equation: Solving Absolute Value Equations Solutions DO NOT COPY. P

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