1. Ch 9: Quadratic Equations G) Quadratic Word Problems Objective: To solve word problems using various methods for solving quadratic equations.

2. Definitions Projectile Motion: h = at 2 + vt + s The path of an object that is thrown, shot, or dropped. h =height, t = time, v =velocity, s = initial height Area of a Rectangle: Length  width Perimeter of a Rectangle: 2  Length + 2  width Product: multiplication Sum: addition Difference: subtraction Less than: subtraction & switch order − xy lessthan

3. x y time (in seconds) height (in feet) Example 1: Projectile Motion The height of a model rocket that is fired into the air can be represented by the equation: h = -16t 2 + 64t a)What will be its maximum height? b)How long will it stay in the air? Vertex: x = -b 2a = -(64) 2(-16) = -64 -32 = 2 y = -16(2) 2 + 64(2)= 64 xy Left Vertex Right 264 00 40 10 20 30 40 50 60 70 80 90 100 1 2 3 4 5 6 7 8 9 10 11 Find the vertex Find the time (t) when h = 0 Vertex (2 seconds, 64 ft) (4 seconds, 0 ft)

4. Example 2: Projectile Motion An object is launched at 19.6 m/s from a 58.8 meter-tall platform. When does the object strike the ground? Note: h = -4.9t 2 + 19.6t + 58.8 Find the time (t) when h = 0 t h time (in seconds) height (in meters) 10 20 30 40 50 60 70 80 90 100 1 2 3 4 5 6 7 8 9 10 11 (6 seconds, 0 ft) Solve by Graphing or Use the Quadratic Formula t = -2 or t = 6 Not Possible! Time can’t be negative

5. t h time (in seconds) height (in feet) 10 20 30 40 50 60 70 80 90 100 1 2 3 4 5 6 7 8 9 10 11 Classwork A model rocket is shot into the air and its path is approximated by the equation: h = -5t 2 + 30t a)When will it reach its highest point? b)When will the rocket hit the ground? 2) t h time (in seconds) height (in meters) 10 20 30 40 50 60 70 80 90 100 1 2 3 4 5 6 7 8 9 10 11 1) An object is launched at 64 ft/s from a platform 80 ft high. Note: h= -16t 2 + 64t + 80 a)What will be the objects maximum height? b)When will it attain this height? (2 seconds, 144 ft) Vertex (3 seconds, 45 ft) (6 seconds, 0 ft)

6. Example 1: Integers Find two numbers whose product is 65 and difference is 8. Let x = one of the numbersLet y = the other number Equation 1: Product Equation 2: Difference x  y = 65 x – y = 8 x = y + 8  y = 65 Substitute x Solve for x (y + 8) Simplify y 2 + 8y = 65y 2 + 8y − 65 = 0 Solve = 5 or -13 Plug in and solve for x x  = 65 y 5 x = 13 x  = 65 y -13 x = -5 Solution 5 & 13 and -13 and -5

7. Example 2: Integers Find two numbers whose product is 640 and difference is 12. Let x = one of the numbersLet y = the other number Equation 1: Product Equation 2: Difference x  y = 640 x – y = 12 x = y + 12  y = 640 Substitute x Solve for x (y + 12) Simplify y 2 + 12y = 640y 2 + 12y − 640 = 0 Solve = 20 or -32 Plug in and solve for x x  = 640 y 20 x = 32 x  = 640 y -32 x = -20 Solution 20 & 32 and -32 and -20

8. Find two numbers whose product is 36 and difference is 5. 4)3) Find two numbers whose product is 48 and difference is 8. 9 & 4 or -4 and -9 12 & 4 or -4 and -12 x  y = 36 x – y = 5 x  y = 48 x – y = 8 y 2 + 5y − 36 = 0 y 2 + 8y − 48 = 0 Classwork

9. Example 1: Dimensions You have 70 ft of material to fence in a rectangular garden that has an area of 150 ft 2. What will be the dimensions of the fence? Let L = lengthLet w = width Equation 1: Area Equation 2: Perimeter L  w = 150 2L + 2w = 70 L = 35 − w  w = 150 Substitute L Solve for L (35 − w) Simplify -w 2 + 35w = 150 -w 2 + 35w − 150 = 0 Solve = 30 or 5 Plug in and solve for L L  = 150 w30 L = 5 Solution 5 ft x 30 ft

10. Example 2: Dimensions The length of a rectangular garden is 143 ft less than the perimeter. The area of the rectangle is 2420 ft 2. What are the dimensions of the rectangle? Let L = lengthLet w = width Equation 1: Area Equation 2: Perimeter L  w = 2420 2L + 2w = L = 143 − 2w  w = 2420 Substitute L Solve for L (143 − 2w) Simplify -2w 2 + 143w = 2420 -2w 2 + 143w − 2420 = 0 Solve = 44 or 27.5 Plug in and solve for L L  = 2420 w44 L = 55 Solution 55 ft x 44 ft L = P − 143 L + 143 L + 143 = P or L  = 2420 w27.5 L = 88 27.5 ft x 88 ft P

11. The perimeter of a rectangle is 52 ft and its area is 168 ft 2. What are the dimensions of the rectangle? 6)5) The width of a rectangle is 46 ft less than 2 times its length. The area of the rectangle is 8580 ft 2. What are the dimensions of the rectangle? 12 ft x 14 ft or 11 ft x 15 ft 110 ft x 78 ft or 156 ft x 55 ft 3 11 L  w = 168 2L + 2w = 52 L  w = 8580 L  (2L – 46) = 8580 -w 2 + 26w − 168 = 0 2L 2 − 46L − 8580 = 0 Classwork