Write 2 word problems that illustrate a Quadratic Equation with a solution.
Answer (1)
Answer:Here are two word problems that illustrate a quadratic equation with a solution:Problem 1:A rectangular garden has a length that is 5 meters more than its width. If the area of the garden is 84 square meters, what are the dimensions of the garden?Solution:Let w be the width of the garden in meters.Then the length of the garden is w+5 meters.The area of a rectangle is given by length \times width.So, w(w+5) = 84.w^2 + 5w = 84.w^2 + 5w - 84 = 0.We can solve this quadratic equation by factoring. We need two numbers that multiply to -84 and add to 5. These numbers are 12 and -7.(w+12)(w-7) = 0.This gives two possible solutions for w: w = -12 or w = 7.Since width cannot be negative, we take w = 7 meters.The length is w+5 = 7+5 = 12 meters.Therefore, the dimensions of the garden are 7 meters by 12 meters.Problem 2:A projectile is launched vertically upwards from the ground with an initial velocity of 30 meters per second. The height h (in meters) of the projectile above the ground after t seconds is given by the equation h = 30t - 5t^2. After how many seconds will the projectile be at a height of 40 meters?Solution:We are given the equation for the height: h = 30t - 5t^2.We want to find t when h = 40.So, we set the equation to 40:40 = 30t - 5t^2.Rearrange the equation into standard quadratic form (at^2 + bt + c = 0):5t^2 - 30t + 40 = 0.Divide the entire equation by 5 to simplify:t^2 - 6t + 8 = 0.We can solve this quadratic equation by factoring. We need two numbers that multiply to 8 and add to -6. These numbers are -2 and -4.(t-2)(t-4) = 0.This gives two possible solutions for t: t = 2 or t = 4.Therefore, the projectile will be at a height of 40 meters after 2 seconds (on its way up) and again after 4 seconds (on its way down)
