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Everyday Examples of Situations to Apply Quadratic Equations
10 Ways Simultaneous Equations Can Be Used in Everyday Life
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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- Math is Fun: Real World Examples of Quadratic Equations
About the Author
Kevin Wandrei has written extensively on higher education. His work has been published with Kaplan, Textbooks.com, and Shmoop, Inc., among others. He is currently pursuing a Master of Public Administration at Cornell University.
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How are quadratic equations used in solving real-life problems and making decisions?
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.
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Home / Blog / Math / What Practical Applications do Quadratic Equations Have in Real Life?
What Practical Applications do Quadratic Equations Have in Real Life?
Quadratic Equations: What Are They?
A quadratic equation is “ any equation containing one term in which the unknown is squared and no term in which it is raised to a higher power. “ 1 The quadratic equation is most commonly written as ax² + bx + c = 0. The known numbers a, b, and c serve as the coefficients, while x denotes the unknown. 2 Quadratum, the Latin word for square, is the origin of the word quadratic. 3
Quadratic equations have many applications in daily life because they are crucial to human survival. 3 Quadratic equations must be used directly or indirectly in every field that involves calculating speed, area, or profit. 2, 4
Everyday Applications of Quadratic Equations
Construction workers use quadratic equations to calculate the area before starting a project. People also figure out the sizes of other things, like boxes and plots of land, using quadratic equations. For instance, the majority of buildings are square or rectangular. So, when constructing a rectangle, one side must cover twice as much space as the other sides. A quadratic equation will be created by calculating the area of the materials required to cover that area.
Quadratic equations are frequently used in sports and are now very helpful for gameplay and analysis. For instance, a football analyst must perform calculations to determine a team’s or athlete’s form. In these analyses, quadratic equation components can be found. Basketball players score by aiming for the exact distance and duration of the throw into the net. A quadratic equation can be used to determine the ball’s height. 2, 4
In order to determine business profit, quadratic equations are frequently used. You must solve a quadratic equation even when dealing with small products to figure out how many of them will be profitable. For instance, you may want to sell pillowcases and your financial goal is $50,000. For this, you must first figure out your average selling price, for example, $25 per case. To figure out how many cases you should sell to reach your goal if you make a profit of $10 per case, you must create a quadratic equation. 2
Quadratic equations are essential in our educational systems, something that we cannot ignore. Millions of students solve quadratic equations every day. You will probably encounter these calculations daily if you choose a career in math, physics, or computer science. Without sound knowledge of quadratic equations, learning certain aspects of science and math would be impossible.
Space science also makes use of some quadratic equation components. For instance, when installing satellite dishes. This is because it needs to be set up at specific angles to pick up signals effectively. 4
Students can understand different math concepts and learn about their importance both inside and outside of the classroom by using real-world examples and applications. On BYJU’S FutureSchool Blog , you can read about practical applications and examples of various such disciplines.
- Quadratic equation Definition & Meaning – Merriam-Webster . (n.d.). Retrieved September 6, 2022, from https://www.merriam-webster.com/dictionary/quadratic%20equation
- Everyday Examples of Situations to Apply Quadratic Equations . (n.d.). Retrieved September 6, 2022, from https://sciencing.com/everyday-examples-situations-apply-quadratic-equations-10200.html
- What Are Quadratic Equations? | Live Science . (n.d.). Retrieved September 6, 2022, from https://www.livescience.com/50411-quadratic-equations.html
- How Can I Use Quadratic Equations In Real Life? | by Jalal Mansoori | ILLUMINATION | Medium . (n.d.). Retrieved September 6, 2022, from https://medium.com/illumination/how-can-i-use-quadratic-equations-in-real-life-239a70eeb6d0
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Applying Quadratics to Real-Life Situations
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Quadratic equations lend themselves to modeling situations that happen in real life, such as the rise and fall of profits from selling goods, the decrease and increase in the amount of time it takes to run a mile based on your age, and so on.
The wonderful part of having something that can be modeled by a quadratic is that you can easily solve the equation when set equal to zero and predict the patterns in the function values.
The vertex and x -intercepts are especially useful. These intercepts tell you where numbers change from positive to negative or negative to positive, so you know, for instance, where the ground is located in a physics problem or when you’d start making a profit or losing money in a business venture.
The vertex tells you where you can find the absolute maximum or minimum cost, profit, speed, height, time, or whatever you’re modeling.
In 1972, you could buy a Mercury Comet for about $3,200. Cars can depreciate in value pretty quickly, but a 1972 Comet in pristine condition may be worth a lot of money to a collector today.
Let the value of one of these Comets be modeled by the quadratic function v ( t ) = 18.75 t 2 – 450 t + 3,200, where t is the number of years since 1972. When is the value of the function equal to 0 (what is an x -intercept), what was the car’s lowest value, and what was its value in 2010?
The car’s value never dropped to 0, the lowest value was $500, and the car was worth $13,175 in the year 2010. In this model, the y -intercept represents the initial value. When t = 0, the function is v (0) = 3,200, which corresponds to the purchase price.
Find the x -intercepts by solving 18.75 t 2 –450 t + 3,200 = 0. Using the quadratic formula (you could try factoring, but it’s a bit of a challenge and, as it turns out, the equation doesn’t factor), you get –37,500 under the radical in the formula. You can’t get a real-number solution, so the graph has no x -intercept. The value of the Comet doesn’t ever get down to 0.
Find the lowest value by determining the vertex. Using the formula,
This coordinate tells you that 12 years from the beginning (1984 — add 12 to 1972), the value of the Comet is at its lowest. Replace the t ’s in the formula with 12, and you get v (12) = 18.75(12) 2 – 450(12) + 3,200 = 500.
The Comet was worth $500 in 1984. To find the value of the car in 2010, you let t = 38, because the year 2010 is 38 years after 1972. The value of the car in 2010 is v (38) = 18.75(38) 2 – 450(38) + 3,200 = $13,175.
The height of a ball t seconds after it’s thrown into the air from the top of a building can be modeled by h ( t ) = –16 t 2 + 48 t + 64, where h ( t ) is height in feet. How high is the building, how high does the ball rise before starting to drop downward, and after how many seconds does the ball hit the ground?
The profit function telling Georgio how much money he will net for producing and selling x specialty umbrellas is given by P ( x ) = –0.00405 x 2 + 8.15 x – 100.
What is Georgio’s loss if he doesn’t sell any of the umbrellas he produces, how many umbrellas does he have to sell to break even, and how many does he have to sell to earn the greatest possible profit?
Chip ran through a maze in less than a minute the first time he tried. His times got better for a while with each new try, but then his times got worse (he took longer) due to fatigue.
The amount of time Chip took to run through the maze on the a th try can be modeled by T ( a ) = 0.5 a 2 – 9 a + 48.5. How long did Chip take to run the maze the first time, and what was his best time?
A highway underpass is parabolic in shape. If the curve of the underpass can be modeled by h ( x ) = 50 – 0.02 x 2 , where x and h ( x ) are in feet, then how high is the highest point of the underpass, and how wide is it?
Following are answers to the practice questions:
The building is 64 feet tall, the ball peaks at 100 feet, and it takes 4 seconds to hit the ground.
The ball is thrown from the top of the building, so you want the height of the ball when t = 0. This number is the initial t value (the y -intercept). When t = 0, h = 64, so the building is 64 feet high.
The ball is at its highest at the vertex of the parabola. Calculating the t value, you get that the vertex occurs where t = 1.5 seconds. Substituting t = 1.5 into the formula, you get that h = 100 feet.
The ball hits the ground when h = 0. Solving –16 t 2 + 48 t + 64 = 0, you factor to get –16( t – 4)( t + 1) = 0. The solution t = 4 tells you when the ball hits the ground.
The t = –1 represents going backward in time, or in this case, where the ball would have started if it had been launched from the ground — not the top of a building.
Georgio loses $100 (earns –$100) if he sells 0, needs to sell 13 to break even, and can maximize profits if he sells 1,006 umbrellas.
If Georgio sells no umbrellas, then x = 0, and he makes a negative profit (loss) of $100. The break-even point comes when the profit changes from negative to positive, at an x -intercept. Using the quadratic formula, you get two intercepts: at x = 2,000 and x is approximately 12.35.
The first (smaller) x -intercept is where the function changes from negative to positive. The second is where the profit becomes a loss again (too many umbrellas, too much overtime?). So, 13 umbrellas would yield a positive profit — he’d break even (have zero profit).
The maximum profit occurs at the vertex. Using the formula for the x -value of the vertex, you get that x is approximately 1,006.17. Substituting 1,006 into the formula, you get 4,000.1542; then substituting 1,007 into the formula, you get 4,000.15155.
You see that Georgio gets slightly more profit with 1,006 umbrellas, but that fraction of a cent doesn’t mean much. He’d still make about $4,000.
Chip took 40 seconds the first time; his best time was 8 seconds.
Because the variable a represents the number of the attempt, find T (1) for the time of the first attempt. T (1) = 40 seconds. The best (minimum) time is at the vertex. Solving for the a value (which is the number of the attempt),
He had the best time on the ninth attempt, and T (9) = 8.
The underpass is 50 feet high and 100 feet wide.
The highest point occurs at the vertex:
The x -coordinate of the vertex is 0, so the vertex is also the y -intercept, at (0, 50). The two x -intercepts represent the endpoints of the width of the overpass. Setting 50 – 0.02 x 2 equal to 0, you solve for x and get x = 50, –50. These two points are 100 units apart — the width of the underpass.
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Answer to Question #85317 in Math for Janet
Quadratic equations lend themselves to modeling situations that happen in real life and in decision making, such as the rise and fall of profits from selling goods, the decrease and increase in the amount of time it takes to run a mile based on your age, and so on.
The wonderful part of having something that can be modeled by a quadratic is that you can easily solve the equation when set equal to zero and predict the patterns in the function values.
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