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Last updated on October 30, 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.

Applications of Quadratics for US Students
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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 \(ax^2 + bx + c = 0\), 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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Tips and Tricks to Master Value of a Polynomial

Mastering the value of a polynomial means getting perfect at substituting numbers into expressions and simplifying them correctly. Here are some tips and tricks to master the value of a polynomial.
 

  1. Think of It as a number machine. Say to yourself that, “A polynomial is like a machine — I put in a number for 𝑥, and it gives me an output.” This helps visualize what’s happening when you substitute.
     

  2. Always use brackets when substituting. This avoids sign mistakes, one of the most common errors.
     

  3. Practice with different values for x. Like, x=0 (checks the constant term) x=1 (easy for mental math). If your answers make sense for all types of values, you’ve mastered the skill.
     

  4. Double-check with a calculator. After doing it by hand, use a calculator to confirm. Compare results to spot where you might’ve gone wrong, great for self-correction.
     
  5. Check your work by substitution. Plug the value back into the polynomial step-by-step and see if it makes sense. For example, results should increase for bigger 𝑥 if the leading coefficient is positive.
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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 \(h(t) = -4.9t^2 + 12t + 16\)

To find the ball when it hits the ground, find h(t) = 0

\(-4.9t^2 + 12t + 16 = 0\)

Using the quadratic formula to solve the equation, 

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\[1em] t = \frac{-12 \pm \sqrt{12^2 - 4(-4.9)(16)}}{2(-4.9)}\\[1em] t = \frac{-12 \pm \sqrt{144 - [4(-4.9)(16)]}}{-9.8}\\[1em] t = \frac{-12 \pm \sqrt{144 + 313.6}}{-9.8}\\[1em] t = \frac{-12 \pm \sqrt{457.6}}{-9.8}\\[1em] t = \frac{-12 \pm 21.4}{-9.8}\\[1em] t_1 = \frac{-12 + 21.4}{-9.8} = \frac{9.4}{-9.8} = -0.96\\[1em] t_2 = \frac{-12 - 21.4}{-9.8} = \frac{-33.4}{-9.8} = 3.41\\[1em] \boxed{t_1 = -0.96, \quad t_2 = 3.41} \)

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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

\(Height = -4.9t^2 - 5t + 50\)

When \(h(t) = 0, -4.9t^2 - 5t + 50 = 0\)

Solving the equation to find the time when the object hits the ground

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\[1em] t = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(-4.9)(50)}}{2(-4.9)}\\[1em] t = \frac{5 \pm \sqrt{25 - [-4(-4.9)(50)]}}{-9.8}\\[1em] t = \frac{5 \pm \sqrt{25 + 980}}{-9.8}\\[1em] t = \frac{5 \pm \sqrt{1005}}{-9.8} \)

The square root of, 1005 is 31.7

\(t = \frac{5 ± 31.7}{-9.8}\)

So, \(t = \frac{5 + 31.7}{-9.8} = \frac{36.7}{-9.8} = -3.745\)

\(t = \frac{5 - 31.7}{-9.8} = \frac{-26.7}{-9.8} = 2.724\)

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

\(x(x + 4) = 60\)

\(x^2 + 4x - 60 = 0\)

Solving the equation to find the length and width:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\[1em] x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-60)}}{2(1)}\\[1em] x = \frac{-4 \pm \sqrt{16 + 240}}{2}\\[1em] x = \frac{-4 \pm \sqrt{256}}{2}\\[1em] x = \frac{-4 \pm 16}{2}\\[1em] x_1 = \frac{-4 + 16}{2} = \frac{12}{2} = 6\\[1em] x_2 = \frac{-4 - 16}{2} = \frac{-20}{2} = -10\\[1em] \boxed{x = 6 \text{ or } x = -10} \)

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 \(x + 1\)

\(x(x + 1) = 132\)

\(x^2 + x - 132 = 0\)

Solving the quadratic equation using the quadratic formula:
 

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\[1em] x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-132)}}{2(1)}\\[1em] x = \frac{-1 \pm \sqrt{1 + 528}}{2}\\[1em] x = \frac{-1 \pm \sqrt{529}}{2}\\[1em] x = \frac{-1 \pm 23}{2}\\[1em] x_1 = \frac{-1 - 23}{2} = \frac{-24}{2} = -12\\[1em] x_2 = \frac{-1 + 23}{2} = \frac{22}{2} = 11\\[1em] \boxed{x = 11 \text{ or } x = -12} \)

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: \(y = -0.5x^2 + 6x\)

To find the top point of a curve, we use the vertex formula: \(x = \frac{-b}{2}a\)

Here,

\(a = -0.5\)

\(b = 6\)

So, \(x = \frac{-6}{2 }× (-0.5)\) 

\(\frac{-6}{2 × (-0.5)} = \frac {-6}{-1 }\)

= 6

So, the highest point happens at \(x = 6\).

Then substitute \(x = 6\) into the equation to find height:

\(y = -0.5(6)^2 + 6(6)\\[1em] y = -0.5(36) + 36\\[1em] y = -18 + 36\\[1em] y = 18 \)

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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: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

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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: \(ax^2 + bx + c = 0\), where a, b, and c are constants, and \(a ≠ 0\).

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6.What are quadratic equations, and why should my child learn them?

A quadratic equation is a math expression that involves a variable raised to the power of two. Children learn quadratics to understand how quantities change — it’s the foundation for analyzing curves, motion, and real-world patterns in physics, finance, and design.

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7.How can I help my child visualize quadratic applications?

Encourage visual learning:
 

Let your child experiment with changing coefficients 𝑎, 𝑏, 𝑐 and observe how the graph changes shape. Discuss how parabolas appear in nature, architecture, and sports. This brings the topic to life.

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8.My child says quadratics are “too abstract.” How can I make them engaging?

Turn abstract math into story-based or visual problems:

  • “If a ball takes 2 seconds to reach the ground after being thrown upward, how high did it go?”

  • “How can you maximize the area of a garden with 20m of fencing?” Relating quadratics to real outcomes (height, area, profit, time) sparks curiosity and meaning.

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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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