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Last updated on October 16, 2025

Applications of Quadratics

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Polynomials are algebraic expressions composed of variables, exponents, constants, and arithmetic operations. Quadratics are a type of polynomial where the highest degree of the variable is two. In this article, we will learn about the applications of quadratics in different fields.

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What are Quadratics?

A quadratic equation is a polynomial equation, which means the highest degree of the variable is 2. Its standard form of the quadratic equation is ax2 + bx + c = 0, where a ≠ 0.  
 

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What are the Applications of Quadratics in Different Fields?

Quadratic equations are used in everyday life to solve problems related to motion, area, finance, and design. In this section, we will learn how quadratic equations are used. 

 

 

Projectile Motion


When a stone skips across a pond, it follows a curved path that is known as projectile motion, which can be described using quadratic equations. By factoring in the initial speed and launch angle, engineers and scientists can predict the following:

  • The quadratic equation can be used to determine the maximum height a projectile reaches, which is essential for events like fireworks' success. They are also used to calculate the maximum range of a firefighter’s water cannon.  
  • In sports like the javelin throw, golf, and archery, the total distance traveled is calculated using a quadratic equation. 
  • In a defense system, the quadratic equation is used to predict the landing point of projectiles like missiles by modeling their curved trajectories under the influence of gravity.
     
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Optimizing Profit

In business, quadratic equations are used to identify the point at which profits are maximized. By representing the relationship between profit and the number of units sold in quadratic form. By analyzing this relationship through a quadratic equation, companies can determine: 
The optimal number of units to produce or sell to maximize profit
The most effective pricing strategy that balances production costs and consumer demand to achieve the highest possible revenue
To predict the height of a falling object, use quadratics to model motion under gravity. 

 

 

Designing

 
In fields like architecture, engineering, and technology, quadratic equations are used to create precise and effective designs. 
Satellite Dishes and Radio Telescopes
The parabolic design of satellite dishes and large telescopes is derived from the quadratic equations. This shape allows incoming signals, whether from satellites or distant galaxies, to a single receiver point, resulting in clearer images and clearer sound transmissions. 

 

 

Civil Engineering and Structural Design


In civil engineering, quadratic equations are used in the design and analysis of bridges, ramps, arches, and dams. It helps to ensure safety, functionality, and efficiency in infrastructure projects. 

  • Quadratic equations are used to model the curved and parabolic shapes in bridges and arches, which helps engineers to accurately calculate load-bearing capacity and to optimize stress distribution. 
  • To model the stable tunnel curves, engineers apply quadratic models to reduce stress on surrounding materials and improve long-term safety. 
  • When constructing the dams, engineers use quadratic principles to evenly distribute water pressure and prevent structural failure. 
     
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Solved Examples on Quadratics

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Problem 1

A ball is thrown from a height of 16m with a velocity of 12 m/s. The height is modeled by: h(t) = -4.9t2 + 12t + 16. When will the ball hit the ground?

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The ball hits the ground after approximately 3.41 seconds 
 

Explanation

 The height is given by h(t) = -4.9t2 + 12t + 16
To find the ball when it hits the ground, find h(t) = 0
-4.9t2 + 12t + 16 = 0
Using the quadratic formula to solve the equation, 
x = -b ± b2 - 4ac2a
So, t = -12 ± 122 - 4 × -4.9 × 162(-4.9) 
= -12 ± 144 - (-313.6)-9.8 
= -12 ± 457.6-9.8 
t = -12 ± 21.4-9.8 
t = -12 ± 21.4-9.8  = 9.4-9.8 = -0.96
t = -12 +- 21.4-9.8 = -33.4-9.8 = 3.41
 

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Problem 2

An object is dropped from a height of 50m with an initial downward velocity of 5 m/s. Its height after t seconds is: h(t) = -4.9t2 - 5t + 50. When does the object hit the ground?

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The object hits the ground after approximately 2.73 seconds 
 

Explanation

Height = -4.9t2 - 5t + 50
When h(t) = 0, -4.9t2 - 5t + 50 = 0
Solving the equation to find the time when the object hits the ground
x = -b ± b2 - 4ac2a
t = -(-5)± (-5)2 - 4 × (-4.9) (50)2(-4.9)
t = 5 ± 25 - -9802(-4.9)
t = 5 ± 1005-9.8
The square root of, 1005 is 31.7
t = 5 ± 31.7-9.8
So, t = 5 + 31.7-9.8 = 36.7/-9.8 = -3.745
t = 5 - 31.7-9.8 = -26.7/-9.8 = 2.724
So, the object hit the ground after approximately 2.75 seconds. 
 

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Problem 3

The area of a rectangle is 60m2. Its length is 4m more than its width. Form an equation and use the quadratic formula to find the width.

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 The width is 6 meters, and the length is 10 meters
 

Explanation

 Let’s consider the width as x and the length as x + 4
Area of the rectangle = length x width
x(x + 4) = 60
x2 + 4x - 60 = 0
Solving the equation to find the length and width:
x = -b ± b2 - 4ac2a
x = -4 ± 42 - 4(1)(-60)2(1)
x = -4 ± 16 + 2402
x = -4 ± 2562
x = -4 ± 162
x = -4 + 162 = 12/2 = 6
x = -4 - 162 = -20/2 = -10
The width of the rectangle is 6 meters, as the width of a rectangle cannot be a negative value. 

Length = (x + 4)
Here, x = 6
Length = (6 + 4) = 10 meters


 

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Problem 4

The product of two consecutive integers is 132. Find the integers using the quadratic formula

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The numbers are 11 and 12
 

Explanation

The product of two consecutive integers is 132
Let’s consider the numbers as x and x + 1
x(x + 1) = 132
x2 + x - 132 = 0
Solving the quadratic equation using the quadratic formula: x = -b ± b2 - 4ac2a
x = -1 ± 12 - 4(1)(-132)2(1)
x = -1 ± 12 - 4(1)(-132)2(1)
x = -1 ± 1 + 5282
x = -1 ± 5292
x = -1 ± 232
x = -1 - 232 = -24/2 = -12
x = -1 + 232 = 22/2 = 11
Here, x = 11 
So, the numbers are 11 and 12
 

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Problem 5

The cross-section of an arch tunnel is y = -0.5x2 + 6x. Find the maximum height and the corresponding x values.

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 The maximum height is 18 units at x = 6
 

Explanation

 The given equation is: y = -0.5x2 + 6x
To find the top point of a curve, we use the vertex formula: x = -b/2a
Here, a = -0.5
b = 6
So, x = -6/2 × (-0.5) 
-62 × (-0.5) = -6-1 
= 6
So, the highest point happens at x = 6. 
Then substitute x = 6 into the equation to find height:
y = -0.5(6)2 + 6(6)
y = -0.5(36) + 36
   = -18 + 36
y = 18
 

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FAQs on Quadratics

1.What does a quadratic equation mean?

It refers to a mathematical expression where the highest exponent of the variable is 2.
 

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2.What are the methods to solve a quadratic equation?

To solve the quadratic equation, we use methods such as factoring, completing the square, using the quadratic formula, and graphing.
 

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3.What is the quadratic formula?

The quadratic formula used to find the roots of any quadratic equation is: x = -b ± b2 - 4ac2a
 

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4.Where are quadratic equations used in real life?

Quadratic equations are applied in areas such as projectile motion, business, engineering, architecture, and computer graphics, and the formula is used to find their solution. 
 

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5.What is the standard form of a quadratic equation?

The standard form of a quadratic equation is written as:ax2 + bx + c = 0, where a, b, and c are constants, and a ≠ 0. 
 

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Jaskaran Singh Saluja

About the Author

Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

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Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.

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