1. Introduction
Quadratic equations are not just algebraic expressions—they appear in many real-life situations. Whenever there is an area relationship, a product-sum relation, motion under gravity, or optimization of quantity, quadratic equations naturally arise.
In this topic, we explore how to form quadratic equations from word problems and solve them step-by-step.
2. How Quadratic Equations Arise in Real Life
Quadratic equations appear whenever:
- A variable is multiplied by itself (square relationships)
- Two quantities change proportionally
- The motion involves acceleration
- Areas, dimensions, or geometry constraints are given
3. Steps to Form a Quadratic Equation from a Word Problem
Most application problems follow a simple structure:
- Identify the unknown quantity and represent it using a variable.
- Translate the problem into mathematical relationships.
- Write the equation based on the relationships.
- Rearrange into the standard quadratic form: \(ax^2 + bx + c = 0\).
- Solve using any appropriate method.
- Check which solution is meaningful (some roots may not make sense in real life).
4. Types of Application Problems
Quadratic equations are commonly used in several categories of word problems.
4.1. 1. Number Problems
These involve relationships between numbers, often using product, sum, or difference.
Example situation: A number and its square differ by a given value.
4.2. 2. Geometry Problems
Quadratic equations frequently arise in areas, rectangles, triangles, and diagonal relationships.
Example: Length × Breadth = Area.
4.3. 3. Motion Problems
4.4. 4. Age and Money Problems
Age differences or investment conditions often form quadratic equations.
5. Worked Examples
These examples show how to convert real situations into quadratic equations.
5.1. Example 1: Geometry – Rectangle Dimensions
The area of a rectangle is 96 cm². The length is 4 cm more than its breadth. Find the length and breadth.
Let breadth be \(x\). Length = \(x + 4\).
Area: \(x(x + 4) = 96\)
\(x^2 + 4x - 96 = 0\)
Solving gives: \(x = 8\) (positive value)
Length = 12 cm.
5.2. Example 2: Number Relationship
The product of two consecutive positive integers is 72. Find the numbers.
Let the smaller number be \(x\). Then the next integer is \(x + 1\).
Equation: \(x(x + 1) = 72\)
\(x^2 + x - 72 = 0\)
Solution gives: \(x = 8\).
Numbers are 8 and 9.
5.3. Example 3: Motion
A car travels 300 km at a speed of \(x\) km/h. If the speed were increased by 20 km/h, the travel time would decrease by 1 hour. Find the speed.
Time relation:
\(\dfrac{300}{x} - \dfrac{300}{x+20} = 1\)
Multiply and rearrange to form a quadratic: \(x^2 + 20x - 6000 = 0\)
Solving gives: \(x = 60\) km/h.
5.4. Example 4: Profit and Loss (Money Problem)
A shopkeeper sells an item for ₹x. If he had sold it for ₹10 more, his profit would have doubled. Cost price is ₹40. Find the selling price.
Equation: \(x - 40 = 2(x + 10 - 40)\)
Simplify → \(x^2 - 20x - 400 = 0\)
Solution: \(x = 40\) or \(x = 20\). Only positive profit makes sense → \(x = 40\).
6. Choosing the Correct Root
Many quadratic equations give two solutions, but only one may fit the real-world situation.
For example, negative values are not acceptable when finding:
- Length
- Speed
- Age
- Distance
- Money
Always check which root is practically meaningful.
7. Common Mistakes
- Incorrectly forming the equation from the word problem.
- Using the wrong variable or not defining variables clearly.
- Missing units or ignoring real-life constraints.
- Choosing an invalid root (e.g., negative length).
- Arithmetic mistakes while simplifying.