1. Solving Quadratic Word Problems by Factoring March 11, 2014 SWBAT Solve Quadratic Word Problems by Factoring

2. WARM – UP 9 cm 15 cm 1. Solve for the missing side. 2.Find two consecutive odd integers such that twice the larger is seven less than three times the smaller. a 2 + b 2 =c 2 9 2 + x 2 = 15 2 81 + x 2 = 225 x 2 = 144 x = 12 Let x = 1 st integer x + 2 = 2 nd odd integer 2(x + 2) = 3x – 7 2x + 4 = 3x – 7 11 = x 13 = x + 2

3. SWBAT Solve Quadratic Word Problems by Factoring STEPS IN SOLVING WORD PROBLEMS 1.Define the variables that you want to find with let statements. 2.Create equation(s) that express the information given in the problem’s scenario. 3. Solve using algebraic methods. 4. Consider if your answer(s) is/are reasonable. 5. Label your solution(s) appropriately. 6.Check your answer(s) with the conditions given in the problem.

4. SWBAT Solve Quadratic Word Problems by Factoring Example 1 The product of two consecutive even integers is 48. Find the integers. Let x = the smaller integer x + 2 = the larger integer x(x + 2) = 48 x 2 + 2x = 48 x 2 + 2x – 48 = 0 (x + 8)(x – 6) = 0 x + 8 = 0 x – 6 = 0 x = -8 x = 6 x + 2 = -6 x + 2 = 8 Solutions: {-8, -6} & {6, 8}

5. SWBAT Solve Quadratic Word Problems by Factoring Example 2 Find three consecutive positive odd integers such that the product of the first and the third is 4 less than 7 times the second. Let x = 1 st positive consecutive integer x + 2 = 2 nd positive consecutive integer x + 4 = 3 rd positive consecutive integer x(x + 4) = 7(x + 2) – 4 x 2 + 4x = 7x + 14 – 4 x 2 + 4x = 7x + 10 x 2 - 3x – 10 = 0 (x – 5) (x + 2) = 0 x – 5 = 0 x + 2 = 0 x = 5 x = -2 REJECT x + 2 = 7 x + 4 = 9 {5, 7, 9}

6. SWBAT Solve Quadratic Word Problems by Factoring Practice: Find three consecutive positive integers such that the product of the first two is 22 less than 11 times the third. Let x = 1 st integer x + 1 = 2 nd integer x + 2 = 3 rd integer x(x + 1) = 11(x + 2) – 22 x 2 + x = 11x + 22 – 22 x 2 – 10x = 0 x(x – 10) = 0 x = 0 x – 10 = 0 x = 10 x + 1 = 11 x + 2 = 12 {10, 11, 12}

7. SWBAT Solve Quadratic Word Problems by Factoring 2.The product of two consecutive odd integers is equal to 30 more than the first. Find the integers. Let x = 1 st integer x + 2 = 2 nd odd integer x(x + 2) = x + 30 x 2 + 2x = x + 30 x 2 + x – 30 = 0 (x + 6)(x – 5) = 0 x + 6 = 0 x – 5 = 0 x = -6 x = 5 x + 2 = 7 {5, 7}

8. SWBAT Solve Quadratic Word Problems by Factoring Example 3 An object is moving in a straight line. It initially travels at a speed of 9 meters per second, and it speeds up at a constant rate of 2 meters per second each second. Under such conditions, the distance d, in meters, that the object travels is given by the equation d = t 2 + 9t, where t is in seconds. According to this equation, how long will it take the object to travel 22 meters? d = t 2 + 9t 22 = t 2 + 9t t 2 + 9t – 22 = 0 (t + 11)(t – 2) = 0 t + 11 = 0 t – 2 = 0 t = -11 REJECT t = 2 seconds

9. SWBAT Solve Quadratic Word Problems by Factoring Practice: An object is moving in a straight line. It initially travels at a speed of 6 meters per second, and it speeds up at a constant acceleration of 4 meters per second each second. The distance d in meters, that this object travels is given by the equation d = 2t 2 + 6t, where t is in seconds. According to this equation, how long will it take the object to travel 108 meters? 2t 2 + 6t = 108 2t 2 + 6t – 108 = 0 2(t 2 + 3t – 54) = 0 2(t + 9)(t – 6) = 0 t + 9 = 0 t – 6 = 0 t = -9 t = 6 sec.

10. SWBAT Solve Quadratic Word Problems by Factoring Example 4 One leg of a right triangle is one inch shorter than the other leg. If the hypotenuse is 5 inches, find the length of the shorter leg. x x - 1 5 in. a 2 + b 2 =c 2 x 2 + (x – 1) 2 = 5 2 x 2 + x 2 - 2x +1 = 25 2x 2 - 2x – 24 = 0 2(x 2 - x – 12) = 0 2(x - 4)(x + 3) = 0 x - 4 = 0x + 3 = 0 x = 4 x = - 3 x – 1 = 3

11. SWBAT Solve Quadratic Word Problems by Factoring Practice: The longer leg of a right triangle is two inches more than twice the length of the shorter leg. The hypotenuse is two inches less than three times the length of the shorter leg. Find the length of the hypotenuse. x 2x + 2 3x - 2 (x) 2 +(2x + 2) 2 = (3x – 2) 2 x 2 + 4x 2 + 8x + 4 = 9x 2 – 12x + 4 5x 2 + 8x + 4 = 9x 2 – 12x + 4 4x 2 – 20x = 0 4x(x – 5) = 0 4x = 0 x – 5 = 0 x = 0 x = 5

12. SWBAT Solve Quadratic Word Problems by Factoring 5. A square is altered so that one dimension is increased by 4, while the other dimension is decreased by 2. The area of the resulting rectangle is 55. Find the area of the original square. x x + 4 x - 2 A = 55 (x – 2)(x + 4) = 55 x 2 + 2x – 63 = 0 (x + 9)(x – 7) = 0 x = -9 x = 7 7 2 = 49

13. SWBAT Solve Quadratic Word Problems by Factoring Summary 1.Define the variables that you want to find with let statements. 2. Translate the problem into a mathematical equation. 3. Get all terms on the same side. 4. Factor 5.Set each factor equal to 0 and solve for x. 6.Check your answer(s) with the conditions given in the problem.

14. SWBAT Solve Quadratic Word Problems by Factoring Exit Ticket The square of a number exceeds 5 times the number by 24. Find the number(s).

15. SWBAT Solve Quadratic Word Problems by Factoring