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Last updated on October 16, 2025
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.
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.
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:
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.
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?
The ball hits the ground after approximately 3.41 seconds
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
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?
The object hits the ground after approximately 2.73 seconds
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.
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.
The width is 6 meters, and the length is 10 meters
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
The product of two consecutive integers is 132. Find the integers using the quadratic formula
The numbers are 11 and 12
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
The cross-section of an arch tunnel is y = -0.5x2 + 6x. Find the maximum height and the corresponding x values.
The maximum height is 18 units at x = 6
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
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.
: He loves to play the quiz with kids through algebra to make kids love it.