Everyday Examples of Situations to Apply Quadratic Equations

Everyday Examples of Situations to Apply Quadratic Equations
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Quadratic equations are actually used in everyday life, as when calculating areas, determining a product's profit or formulating the speed of an object. Quadratic equations refer to equations with at least one squared variable, with the most standard form being ax² + bx + c = 0. The letter X represents an unknown, and a b and c being the coefficients representing known numbers and the letter a is not equal to zero.

Calculating Room Areas

People frequently need to calculate the area of rooms, boxes or plots of land. An example might involve building a rectangular box where one side must be twice the length of the other side. For example, if you have only 4 square feet of wood to use for the bottom of the box, with this information, you can create an equation for the area of the box using the ratio of the two sides. This means the area -- the length times the width -- in terms of x would equal x times 2x, or 2x^2. This equation must be less than or equal to four to successfully make a box using these constraints.

Figuring a Profit

Sometimes calculating a business profit requires using a quadratic function. If you want to sell something – even something as simple as lemonade – you need to decide how many items to produce so that you'll make a profit. Let's say, for example, that you're selling glasses of lemonade, and you want to make 12 glasses. You know, however, that you'll sell a different number of glasses depending on how you set your price. At $100 per glass, you're not likely to sell any, but at $0.01 per glass, you'll probably sell 12 glasses in less than a minute. So, to decide where to set your price, use P as a variable. You've estimated the demand for glasses of lemonade to be at 12 - P. Your revenue, therefore, will be the price times the number of glasses sold: P times 12 minus P, or 12P - P^2. Using however much your lemonade costs to produce, you can set this equation equal to that amount and choose a price from there.

Quadratics in Athletics

In athletic events that involve throwing objects like the shot put, balls or javelin, quadratic equations become highly useful. For example, you throw a ball into the air and have your friend catch it, but you want to give her the precise time it will take the ball to arrive. Use the velocity equation, which calculates the height of the ball based on a parabolic or quadratic equation. Begin by throwing the ball at 3 meters, where your hands are. Also assume that you can throw the ball upward at 14 meters per second, and that the earth's gravity is reducing the ball's speed at a rate of 5 meters per second squared. From this, we can calculate the height, h, using the variable t for time, in the form of h = 3 + 14t - 5t^2. If your friend's hands are also at 3 meters in height, how many seconds will it take the ball to reach her? To answer this, set the equation equal to 3 = h, and solve for t. The answer is approximately 2.8 seconds.

Finding a Speed

Quadratic equations are also useful in calculating speeds. Avid kayakers, for example, use quadratic equations to estimate their speed when going up and down a river. Assume a kayaker is going up a river, and the river moves at 2 km per hour. If he goes upstream against the current at 15 km, and the trip takes him 3 hours to go there and return, remember that time = distance divided by speed, let v = the kayak's speed relative to land, and let x = the kayak's speed in the water. While traveling upstream, the kayak's speed is v = x - 2 -- subtract 2 for the resistance from the river current-- and while going downstream, the kayak's speed is v = x + 2. The total time is equal to 3 hours, which is equal to the time going upstream plus the time going downstream, and both distances are 15km. Using our equations, we know that 3 hours = 15 / (x - 2) + 15 / (x + 2). Once this is expanded algebraically, we get 3x^2 - 30x -12 = 0. Solving for x, we know that the kayaker moved his kayak at a speed of 10.39 km per hour.

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