Applications of Quadratic Equations

Q: How do you know a word problem leads to a quadratic equation?

When the unknown appears in a product (area = length × breadth), as a denominator in two different fractions (time = distance/speed), or squared (Pythagoras), the resulting equation is quadratic.

Q: What if both roots are positive?

Both roots may be valid answers. For example, in 'Find a number whose sum with its reciprocal is 10/3', both x = 3 and x = 1/3 are correct. State both.

Q: When should I use factorisation vs the quadratic formula?

Try factorisation first — it is quicker. If the equation does not factor neatly, use the quadratic formula. The formula always works.

Q: How do I handle speed-distance problems?

Set up time = distance/speed for each scenario. The difference in times gives the equation. Cross-multiply to clear fractions, then simplify to standard quadratic form.

Q: What if the discriminant is negative?

A negative discriminant means no real solution exists. In application problems, this indicates the given conditions are impossible in real life.

Q: How do I verify my answer?

Substitute the answer back into the original word problem (not just the equation). Check that all conditions are satisfied and the answer makes practical sense.

Class 10Quadratic Equations

Applications of quadratic equations involve translating real-life problems into equations of the form ax² + bx + c = 0 and solving them. These problems cover areas such as geometry, consecutive numbers, speed-distance, profit-revenue, and age puzzles.

The quadratic equation arises naturally when a product of two related linear expressions equals a given value, or when area, speed, and time relationships are combined.

This is a high-scoring topic in Class 10 NCERT board exams, typically carrying 4-5 marks per question.

What is Applications of Quadratic Equations?

Definition: An application problem on quadratic equations is a word problem that, when translated into algebra, results in a polynomial equation of degree 2:

ax² + bx + c = 0, where a ≠ 0

Methods of solving:

Factorisation: Express ax² + bx + c as a product of two linear factors.
Quadratic Formula: x = (−b ± √(b² − 4ac)) / 2a
Completing the Square: Rewrite in the form (x + p)² = q.

Types and Properties

Common types of quadratic application problems:

Area Problems: Length and breadth related by a linear condition; area = product gives a quadratic.
Consecutive Number Problems: Product or sum of squares of consecutive integers.
Speed-Distance-Time Problems: When speed changes lead to a quadratic in time or speed.
Profit / Revenue Problems: Revenue = price × quantity, where one depends linearly on the other.
Age Problems: Product of ages or age differences leads to a quadratic.
Work Problems: Two workers with related rates lead to a quadratic in time.
Number Problems: A number and its reciprocal, or two parts of a number.

Solved Examples

Example 1: Area of a Rectangular Field

Problem: The length of a rectangular field is 3 m more than its breadth. If the area is 180 m², find its dimensions.

Solution:

Let: Breadth = x m, then Length = (x + 3) m

Equation:

x(x + 3) = 180
x² + 3x − 180 = 0
x² + 15x − 12x − 180 = 0
(x + 15)(x − 12) = 0
x = 12 or x = −15

Since breadth cannot be negative, x = 12.

Answer: Breadth = 12 m, Length = 15 m.

Example 2: Consecutive Numbers Problem

Problem: The product of two consecutive positive integers is 306. Find the integers.

Solution:

Let: The integers be x and (x + 1).

x(x + 1) = 306
x² + x − 306 = 0
x = (−1 + √(1 + 1224))/2 = (−1 + √1225)/2 = (−1 + 35)/2 = 17

Also x = (−1 − 35)/2 = −18 (rejected, since integers are positive).

Answer: The integers are 17 and 18.

Example 3: Speed-Time Problem

Problem: A train travels 480 km at a uniform speed. If the speed had been 8 km/h less, the train would have taken 3 hours more. Find the speed of the train.

Solution:

Let: Speed = x km/h. Time = 480/x hours.

With reduced speed: 480/(x − 8) = 480/x + 3

480/(x−8) − 480/x = 3
480[x − (x−8)] / [x(x−8)] = 3
480 × 8 / [x(x−8)] = 3
3840 = 3x(x − 8)
x² − 8x − 1280 = 0
(x − 40)(x + 32) = 0
x = 40 (rejecting −32)

Answer: The speed of the train is 40 km/h.

Example 4: Number and Its Reciprocal

Problem: The sum of a number and its reciprocal is 10/3. Find the number.

Solution:

Let: The number = x

x + 1/x = 10/3
(x² + 1)/x = 10/3
3x² + 3 = 10x
3x² − 10x + 3 = 0
3x² − 9x − x + 3 = 0
3x(x − 3) − 1(x − 3) = 0
(3x − 1)(x − 3) = 0
x = 3 or x = 1/3

Answer: The number is 3 or 1/3 (both are valid).

Example 5: Profit Problem

Problem: The cost price of an article is Rs. x. It is sold at a profit of x%. If the selling price is Rs. 75, find the cost price.

Solution:

Selling Price = Cost Price + Profit
75 = x + (x/100) × x = x + x²/100
7500 = 100x + x²
x² + 100x − 7500 = 0
(x + 150)(x − 50) = 0
x = 50 (rejecting −150)

Answer: The cost price is Rs. 50.

Example 6: Work Problem

Problem: Two pipes together fill a tank in 6 hours. The larger pipe alone fills it 5 hours faster than the smaller one alone. Find the time taken by each pipe.

Solution:

Let: Smaller pipe takes x hours alone. Larger pipe takes (x − 5) hours.

1/x + 1/(x−5) = 1/6
[(x−5) + x] / [x(x−5)] = 1/6
6(2x − 5) = x(x − 5)
12x − 30 = x² − 5x
x² − 17x + 30 = 0
(x − 15)(x − 2) = 0
x = 15 or x = 2

If x = 2, larger pipe time = −3 (impossible). So x = 15.

Answer: Smaller pipe: 15 hours, Larger pipe: 10 hours.

Example 7: Hypotenuse Problem

Problem: The hypotenuse of a right triangle is 13 cm. If one side is 7 cm more than the other, find the two sides.

Solution:

Let: One side = x, Other side = (x + 7)

By Pythagoras: x² + (x + 7)² = 13²

x² + x² + 14x + 49 = 169
2x² + 14x − 120 = 0
x² + 7x − 60 = 0
(x + 12)(x − 5) = 0
x = 5 (rejecting −12)

Answer: The sides are 5 cm and 12 cm.

Example 8: Two Parts Problem

Problem: Divide 27 into two parts such that the sum of their reciprocals is 3/20.

Solution:

Let: One part = x, Other part = 27 − x

1/x + 1/(27−x) = 3/20
(27−x+x)/[x(27−x)] = 3/20
27/[x(27−x)] = 3/20
20 × 27 = 3x(27 − x)
540 = 81x − 3x²
3x² − 81x + 540 = 0
x² − 27x + 180 = 0
(x − 12)(x − 15) = 0
x = 12 or x = 15

Answer: The two parts are 12 and 15.

Real-World Applications

Quadratic equations model real-life situations in:

Architecture: Computing dimensions of rooms, gardens, paths given area constraints.
Physics: Projectile motion (h = ut − ½gt²), time of flight calculations.
Business: Revenue maximisation, break-even analysis.
Engineering: Pipe and tank flow rate calculations.
Sports: Trajectory of a ball, distance-time calculations.

Key Points to Remember

Always assign a single variable (x) and express all quantities in terms of it.
The equation is formed from the given condition (area = l × b, time = distance/speed, etc.).
After solving, always check both roots in the context of the problem.
Reject negative values for length, breadth, speed, age, and number of items.
Both roots may be valid in some problems (e.g., number and its reciprocal).
For speed-time problems, the equation usually comes from cross-multiplying after setting up a time difference.
For work problems, combine rates: 1/T₁ + 1/T₂ = 1/T (together).
Factorisation is faster but the quadratic formula always works.
Check the discriminant: D > 0 (two real roots), D = 0 (one repeated root), D < 0 (no real roots).
These problems carry 4-5 marks in CBSE board exams.

Practice Problems

The product of two consecutive even integers is 528. Find them.
A rectangular garden has an area of 400 m². Its length is 4 m more than twice its breadth. Find the dimensions.
A car travels 400 km at uniform speed. If it were 10 km/h faster, it would take 2 hours less. Find the speed.
The sum of the squares of two consecutive natural numbers is 365. Find the numbers.
The sum of the ages of a mother and daughter is 40. The product of their ages 3 years ago was 105. Find their present ages.
A takes 6 days more than B to complete a task. Together they finish it in 4 days. Find the time each takes alone.

Frequently Asked Questions

Q1. How do you know a word problem leads to a quadratic equation?

When the unknown appears in a product (area = length × breadth), as a denominator in two different fractions (time = distance/speed), or squared (Pythagoras), the resulting equation is quadratic.

Q2. What if both roots are positive?

Both roots may be valid answers. For example, in 'Find a number whose sum with its reciprocal is 10/3', both x = 3 and x = 1/3 are correct. State both.

Q3. When should I use factorisation vs the quadratic formula?

Try factorisation first — it is quicker. If the equation does not factor neatly, use the quadratic formula. The formula always works.

Q4. How do I handle speed-distance problems?

Set up time = distance/speed for each scenario. The difference in times gives the equation. Cross-multiply to clear fractions, then simplify to standard quadratic form.

Q5. What if the discriminant is negative?

A negative discriminant means no real solution exists. In application problems, this indicates the given conditions are impossible in real life.

Q6. How do I verify my answer?

Substitute the answer back into the original word problem (not just the equation). Check that all conditions are satisfied and the answer makes practical sense.

Applications of Quadratic Equations

What is Applications of Quadratic Equations?

Types and Properties

Solved Examples

Example 1: Area of a Rectangular Field

Example 2: Consecutive Numbers Problem

Example 3: Speed-Time Problem

Example 4: Number and Its Reciprocal

Example 5: Profit Problem

Example 6: Work Problem

Example 7: Hypotenuse Problem

Example 8: Two Parts Problem

Real-World Applications

Key Points to Remember

Practice Problems

Frequently Asked Questions

Q1. How do you know a word problem leads to a quadratic equation?

Q2. What if both roots are positive?

Q3. When should I use factorisation vs the quadratic formula?

Q4. How do I handle speed-distance problems?

Q5. What if the discriminant is negative?

Q6. How do I verify my answer?

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