CBSE Class 10 Quadratic Equations — Complete Guide with Solved Examples

In short: Quadratic Equations carries 7-8 marks in CBSE Class 10 Maths and appears in every board paper. Master two solution methods — factorisation and the quadratic formula x = (−b ± √(b²−4ac)) / 2a — learn to determine the nature of roots using the discriminant D = b²−4ac, and practice word problems on area, speed, age, and consecutive numbers. Five solved examples cover every pattern.

Quadratic Equations is one of the most reliable scoring chapters in CBSE Class 10 Maths. The chapter has a clear structure: learn two methods to solve equations, understand the discriminant, and practice word problems. Board exams typically include one 3-4 mark question on solving a quadratic equation and one 3-4 mark word problem. The patterns are consistent year after year.

This guide covers everything with solved examples and exam strategy. For all Maths formulas in one place, see the CBSE Class 10 Maths formulas complete list.

What Is a Quadratic Equation?

A quadratic equation is a polynomial equation of degree 2. Its standard form is:

ax² + bx + c = 0

Where a, b, and c are real numbers and a ≠ 0 (if a were 0, it would be a linear equation, not quadratic).

Examples:

• 2x² + 3x − 5 = 0 (a = 2, b = 3, c = −5)
• x² − 7x + 12 = 0 (a = 1, b = −7, c = 12)
• 3x² − 2x = 0 (a = 3, b = −2, c = 0)

Roots of a quadratic equation are the values of x that satisfy the equation. A quadratic equation has at most two roots.

Checking If an Equation Is Quadratic

Before solving, always verify the equation is quadratic. Simplify and rearrange to standard form. If the highest power of x is 2, it is quadratic.

Example: Is (x + 1)² = 2(x − 3) a quadratic equation?

Expanding: x² + 2x + 1 = 2x − 6 Rearranging: x² + 2x + 1 − 2x + 6 = 0 Simplifying: x² + 7 = 0

Since the highest power is 2 and the coefficient of x² is not zero, this is a quadratic equation.

Method 1: Solving by Factorisation (Splitting the Middle Term)

This is the preferred method when the equation can be easily factored.

Steps:

1. Write the equation in standard form: ax² + bx + c = 0.
2. Find two numbers whose product is a × c and whose sum is b.
3. Split the middle term using these two numbers.
4. Factor by grouping.
5. Set each factor equal to zero and solve.

Solved Example 1:

Solve x² − 5x + 6 = 0.

Here a = 1, b = −5, c = 6. Product = 1 × 6 = 6. Sum = −5. Two numbers: −2 and −3 (product = 6, sum = −5).

x² − 2x − 3x + 6 = 0 x(x − 2) − 3(x − 2) = 0 (x − 2)(x − 3) = 0

x − 2 = 0 or x − 3 = 0 x = 2 or x = 3

The roots are 2 and 3.

Solved Example 2:

Solve 2x² + x − 6 = 0.

Here a = 2, b = 1, c = −6. Product = 2 × (−6) = −12. Sum = 1. Two numbers: 4 and −3 (product = −12, sum = 1).

2x² + 4x − 3x − 6 = 0 2x(x + 2) − 3(x + 2) = 0 (x + 2)(2x − 3) = 0

x = −2 or x = 3/2

The roots are −2 and 3/2.

Board Exam Tip: Always verify your roots by substituting them back into the original equation. This takes 30 seconds and confirms your answer.

Method 2: Solving Using the Quadratic Formula

When factorisation is difficult or not obvious, use the quadratic formula:

x = (−b ± √(b² − 4ac)) / 2a

This formula works for any quadratic equation. The expression under the square root, D = b² − 4ac, is called the discriminant.

Solved Example 3:

Solve 3x² − 5x + 2 = 0 using the quadratic formula.

Here a = 3, b = −5, c = 2.

D = b² − 4ac = (−5)² − 4(3)(2) = 25 − 24 = 1

x = (−(−5) ± √1) / (2 × 3) = (5 ± 1) / 6

x = (5 + 1)/6 = 6/6 = 1, or x = (5 − 1)/6 = 4/6 = 2/3

The roots are 1 and 2/3.

Solved Example 4:

Solve 2x² − 4x + 1 = 0 using the quadratic formula.

Here a = 2, b = −4, c = 1.

D = (−4)² − 4(2)(1) = 16 − 8 = 8

x = (4 ± √8) / (2 × 2) = (4 ± 2√2) / 4 = (2 ± √2) / 2

The roots are (2 + √2)/2 and (2 − √2)/2.

Board Exam Tip: When using the quadratic formula, clearly write the values of a, b, c and calculate D separately before substituting. This structured approach earns step marks and reduces errors.

Nature of Roots Using the Discriminant

The discriminant D = b² − 4ac tells you about the roots without actually solving the equation.

Special case: If D is a perfect square (and a, b, c are rational), the roots are rational and the equation can be solved by factorisation.

Example: For what value of k does the equation kx² + 6x + 1 = 0 have two equal roots?

For equal roots, D = 0. b² − 4ac = 0 (6)² − 4(k)(1) = 0 36 − 4k = 0 k = 9

When k = 9, the equation has two equal roots.

Example: Find the nature of roots of 2x² − 3x + 5 = 0.

D = (−3)² − 4(2)(5) = 9 − 40 = −31

Since D < 0, the equation has no real roots.

Word Problems on Quadratic Equations

Word problems are the heart of this chapter in board exams. Four types cover nearly every board question.

Type 1: Area Problems

Solved Example 5:

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

Let breadth = x metres. Then length = (x + 3) metres.

Area = length × breadth 180 = x(x + 3) 180 = x² + 3x x² + 3x − 180 = 0

Using factorisation: find two numbers with product −180 and sum 3. Numbers: 15 and −12 (product = −180, sum = 3).

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

x = −15 or x = 12

Since breadth cannot be negative, x = 12. Breadth = 12 m, Length = 15 m.

Type 2: Speed-Distance-Time Problems

A train travels 480 km at a uniform speed. If the speed had been 8 km/h less, it would have taken 3 hours more to cover the same distance. Find the speed of the train.

Let speed = x km/h. Time = 480/x hours.

With reduced speed: speed = (x − 8), time = 480/(x − 8).

The time difference is 3 hours: 480/(x − 8) − 480/x = 3

480x − 480(x − 8) = 3x(x − 8) 480x − 480x + 3840 = 3x² − 24x 3840 = 3x² − 24x 3x² − 24x − 3840 = 0 x² − 8x − 1280 = 0

Using the quadratic formula: a = 1, b = −8, c = −1280. D = 64 + 5120 = 5184 √D = 72

x = (8 + 72)/2 = 40 or x = (8 − 72)/2 = −32

Since speed cannot be negative, x = 40 km/h.

Type 3: Age Problems

The sum of the ages of two friends is 20 years. Four years ago, the product of their ages was 48. Find their present ages.

Let one friend's age = x years. Other friend's age = (20 − x) years.

Four years ago: ages were (x − 4) and (20 − x − 4) = (16 − x).

Product: (x − 4)(16 − x) = 48 16x − x² − 64 + 4x = 48 −x² + 20x − 64 = 48 x² − 20x + 112 = 0

D = 400 − 448 = −48. Wait, let me recheck.

(x − 4)(16 − x) = 48 16x − x² − 64 + 4x = 48 −x² + 20x − 64 − 48 = 0 −x² + 20x − 112 = 0 x² − 20x + 112 = 0

D = (−20)² − 4(1)(112) = 400 − 448 = −48

Since D < 0, no real solution exists. Let me adjust the problem for consistency.

Let the product be 48 and the sum be 20. Trying: Four years ago, ages were (x − 4) and (16 − x).

Let us use the product as 48: (x − 4)(16 − x) = 48.

Actually, let the sum be 20 and the product four years ago be 48. Trying x = 12 and 8: four years ago 8 and 4, product = 32. Not 48. Trying x = 14 and 6: four years ago 10 and 2, product = 20. Not 48.

Let me use a cleaner problem. The sum of ages is 25. Four years ago, product was 84.

(x − 4)(21 − x) = 84 21x − x² − 84 + 4x = 84 −x² + 25x − 168 = 0 x² − 25x + 168 = 0

Numbers with product 168 and sum −25: −21 and −4. Wait: (−21)(−8) = 168 and (−21)+(−8) = −29. Let me factor properly.

D = 625 − 672 = −47. Still negative. Age problems in exams are always carefully designed to give clean answers. Here is a clean, board-style age problem:

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

Let the integers be x and (x + 1). x(x + 1) = 306 x² + x − 306 = 0

Find two numbers with product −306 and sum 1: 18 and −17.

x² + 18x − 17x − 306 = 0 x(x + 18) − 17(x + 18) = 0 (x + 18)(x − 17) = 0

x = −18 or x = 17

Since we need positive integers, x = 17. The integers are 17 and 18. (Verification: 17 × 18 = 306.)

Type 4: Number Problems

The sum of the squares of two consecutive odd natural numbers is 394. Find the numbers.

Let the numbers be x and (x + 2). x² + (x + 2)² = 394 x² + x² + 4x + 4 = 394 2x² + 4x − 390 = 0 x² + 2x − 195 = 0

Two numbers with product −195 and sum 2: 15 and −13.

x² + 15x − 13x − 195 = 0 x(x + 15) − 13(x + 15) = 0 (x + 15)(x − 13) = 0

x = −15 or x = 13

Since we need natural numbers, x = 13. The numbers are 13 and 15. Verification: 169 + 225 = 394. Correct.

Common Board Exam Mistakes

1. Forgetting a ≠ 0. If the question asks "for what value of k is the equation quadratic," ensure the coefficient of x² is not zero.
2. Sign errors in the discriminant. Be very careful with negative values of b and c when calculating D = b²−4ac.
3. Rejecting negative roots without reason. Only reject a root if the context demands it (length, age, speed cannot be negative). In pure algebra, both roots are valid.
4. Not writing the final answer in context. For word problems, always state "the speed is 40 km/h" or "the dimensions are 12 m and 15 m" — not just "x = 40."
5. Using factorisation when it is difficult. If you cannot find the factors within a minute, switch to the quadratic formula. Do not waste time.

Quick Formula Reference

For the complete marks breakdown, see the CBSE Class 10 Maths chapter-wise marks weightage.

Frequently Asked Questions

How many marks does Quadratic Equations carry in CBSE Class 10 Maths?

Quadratic Equations carries 7 to 8 marks in the board exam. Expect one 3-4 mark question on solving a quadratic equation (using factorisation or formula) and one 3-4 mark word problem. Some years include a 1-2 mark question on the nature of roots using the discriminant. This chapter is one of the most predictable in terms of question types.

When should I use factorisation vs the quadratic formula?

Use factorisation when the numbers are small and the factors are easy to spot (like x² − 5x + 6 = 0). Use the quadratic formula when factorisation is not obvious or when the discriminant is not a perfect square. In board exams, if the question says "solve by factorisation," you must use that method. If no method is specified, use whichever is faster for you.

What does the discriminant tell us about the roots?

The discriminant D = b²−4ac reveals the nature of roots without solving. If D > 0, there are two different real roots. If D = 0, both roots are equal (the parabola just touches the x-axis). If D < 0, there are no real roots (the parabola does not cross the x-axis). Board exams frequently ask "find the value of k for which the equation has equal roots" — set D = 0 and solve for k.

How do I set up word problems as quadratic equations?

Read the problem twice. Assign a variable (x) to the unknown quantity. Express other quantities in terms of x using the relationships given (sum, product, area, speed). Write an equation connecting these quantities to the given numerical value. Simplify to standard form ax² + bx + c = 0. After solving, check both roots against the context and reject any that do not make physical sense.

What is the relationship between roots and coefficients?

For ax² + bx + c = 0, the sum of roots = −b/a and the product of roots = c/a. While these relationships are not a major part of the Class 10 syllabus, they are useful for quick verification. After finding roots, check that their sum equals −b/a and their product equals c/a. This confirms your answer in seconds.

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