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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.

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Ch 9: Quadratic Equations G) Quadratic Word Problems

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

  2. Definitions Projectile Motion: h = at2 + 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 x y − less than

  3. Example 1: Projectile Motion • The height of a model rocket that is fired into the air can be represented by the equation: h = -16t2 + 64t • What will be its maximum height? • How long will it stay in the air? y Find the vertex Find the time (t) when h = 0 -b 2a -(64) 2(-16) -64 -32 Vertex: x = = = = 2 height(in feet) Vertex y = -16(2)2 + 64(2) = 64 (2 seconds, 64 ft) x 10 20 30 40 50 60 70 80 90 100 time(in seconds) 0 0 (4 seconds, 0 ft) 2 64 1 2 3 4 5 6 7 8 9 10 11 4 0

  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 = -4.9t2 + 19.6t + 58.8 h Find the time (t) when h = 0 or Use the Quadratic Formula Solve by Graphing height(in meters) 10 20 30 40 50 60 70 80 90 100 t 1 2 3 4 5 6 7 8 9 10 11 time(in seconds) (6 seconds, 0 ft) Not Possible! Time can’t be negative t = -2 or t = 6

  5. Classwork 2) 1) • A model rocket is shot into the air and its path is approximated by the equation: h = -5t2 + 30t • When will it reach its highest point? • When will the rocket hit the ground? • An object is launched at 64 ft/s from a platform 80 ft high. • Note: h= -16t2 + 64t + 80 • What will be the objects maximum height? • When will it attain this height? h h height(in meters) height(in feet) 10 20 30 40 50 60 70 80 90 100 10 20 30 40 50 60 70 80 90 100 Vertex (2 seconds, 144 ft) t t (3 seconds, 45 ft) 1 2 3 4 5 6 7 8 9 10 11 1 2 3 4 5 6 7 8 9 10 11 time(in seconds) time(in seconds) (6 seconds, 0 ft)

  6. 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 x = y + 8 x – y = 8 Equation 2: Difference Solve for x (y + 8) x y = 65 Substitute y2 + 8y = 65 y2 + 8y − 65 = 0 Simplify = 5 or -13 Solve Plug in and solve for x x = 65 y x = 65 y 5 -13 x = 13 x = -5 5 & 13 and -13 and -5 Solution

  7. 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 x = y + 12 x – y = 12 Equation 2: Difference Solve for x (y + 12) x y = 640 Substitute y2 + 12y = 640 y2 + 12y − 640 = 0 Simplify = 20 or -32 Solve Plug in and solve for x x = 640 y x = 640 y 20 -32 x = 32 x = -20 20 & 32 and -32 and -20 Solution

  8. Classwork 3) 4) Find two numbers whose product is 36 and difference is 5. Find two numbers whose product is 48 and difference is 8. xy = 48 xy = 36 x – y = 8 x – y = 5 y2 + 8y − 48 = 0 y2 + 5y − 36 = 0 9 & 4 or -4 and -9 12 & 4 or -4 and -12

  9. Example 1: Dimensions You have 70 ft of material to fence in a rectangular garden that has an area of 150 ft2. What will be the dimensions of the fence? Let L = length Let w = width L w = 150 Equation 1: Area L = 35 − w 2L + 2w = 70 Equation 2: Perimeter Solve for L Substitute (35 − w) L w = 150 -w2 + 35w − 150 = 0 -w2 + 35w = 150 Simplify = 30 or 5 Solve Plug in and solve for L L  = 150 30 w L = 5 5 ft x 30 ft Solution

  10. Example 2: Dimensions The length of a rectangular garden is 143 ft less than the perimeter. The area of the rectangle is 2420 ft2. What are the dimensions of the rectangle? L = P − 143 L + 143 = P Let L = length Let w = width L w = 2420 Equation 1: Area P L + 143 L = 143 − 2w 2L + 2w = Equation 2: Perimeter Solve for L Substitute (143 − 2w) L w = 2420 -2w2 + 143w − 2420 = 0 -2w2 + 143w = 2420 Simplify = 44or 27.5 Solve L  = 2420 44 w or L  = 2420 27.5 w Plug in and solve for L L = 55 L = 88 55 ft x 44 ft 27.5 ft x 88 ft Solution

  11. Classwork 5) 6) The perimeter of a rectangle is 52 ft and its area is 168 ft2. What are the dimensions of the rectangle? The width of a rectangle is 46 ft less than 2 times its length. The area of the rectangle is 8580 ft2. What are the dimensions of the rectangle? L w = 8580 L w = 168 L  (2L – 46) = 8580 2L + 2w = 52 2L2 − 46L − 8580 = 0 -w2 + 26w − 168 = 0 110 ft x 78 ft or 156 ft x 55 ft 12 ft x 14 ft or 11 ft x 15 ft 3 11

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