Ch 9 Quadratic Equations G Quadratic Word Problems

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

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 less than x − y

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

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? Find the time (t) when h = 0 Note: h = -4. 9 t 2 + 19. 6 t + 58. 8 or Solve by Graphing Use the Quadratic Formula 10 20 30 40 50 60 70 80 90 100 height (in meters) h (6 seconds, 0 ft) 1 2 3 4 5 6 7 8 9 10 11 time (in seconds) t t = -2 or Not Possible! Time can’t be negative t=6

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

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

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

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

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 = length Let w = width L w = 150 Equation 1: Area = 35 − w Equation 2: Perimeter 2 L + 2 w = 70 Solve for L L Substitute (35 − w) L w = 150 Simplify -w 2 + 35 w = 150 -w 2 + 35 w − 150 = 30 or 5 Solve Plug in and solve for L L 30 w = 150 L=5 5 ft x 30 ft Solution

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? L = P − 143 Let L = length Let w = width L + 143 = P L w = 2420 Equation 1: Area L + 143 L = 143 − 2 w Equation 2: Perimeter 2 L + 2 w = P Substitute (143 − 2 w) L w = 2420 Simplify -2 w 2 + 143 w = 2420 Solve for L -2 w 2 + 143 w − 2420 = 44 or 27. 5 L 44 w = 2420 or L 27. 5 w = 2420 L = 55 L = 88 27. 5 ft x 88 ft Solution 55 ft x 44 ft Plug in and solve for L

Classwork 5) The perimeter of a rectangle is 6) The width of a rectangle is 46 ft 52 ft and its area is 168 ft 2. What are the dimensions of the rectangle? L w = 168 2 L + 2 w = 52 -w 2 + 26 w − 168 = 0 12 ft x 14 ft or 3 11 ft x 1511 ft less than 2 times its length. The area of the rectangle is 8580 ft 2. What are the dimensions of the rectangle? L w = 8580 L (2 L – 46) = 8580 2 L 2 − 46 L − 8580 = 0 110 ft x 78 ft or 156 ft x 55 ft