CBSE Class 10 Mathematics: NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations (2025–2026)

For students preparing for CBSE Class 10 Mathematics in the 2025–2026 academic session, our comprehensive NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations guide provides conceptual clarity, rigorous step-by-step derivations, and standard solutions. Chapter 4 forms the algebraic cornerstone of secondary school mathematics. In the CBSE Mathematics curriculum, Quadratic Equations carries significant weight in board exam 10, appearing across multiple-choice questions (MCQs), short-answer derivations, and high-weightage word problems.

Key Concepts and Formulas in Quadratic Equations

A quadratic equation in the variable x is an algebraic equation of the second degree. The standard form of a quadratic equation is:

ax² + bx + c = 0

where a, b, and c are real numbers, and a ≠ 0. If a = 0, the equation reduces to a linear equation (bx + c = 0), which violates the definition of a quadratic polynomial.

1. Roots of a Quadratic Equation

A real number α is called a root (or solution) of the quadratic equation ax² + bx + c = 0 if it satisfies the equation, meaning a(α)² + b(α) + c = 0. A quadratic equation can have at most two real roots.

2. Methods for Finding Roots

Under the rationalised CBSE Class 10 syllabus, students must master two primary analytical techniques:

1. Factorisation Method (Splitting the Middle Term): We express the quadratic expression ax² + bx + c as the product of two linear factors (px + q)(rx + s) = 0 by identifying two numbers whose sum equals b and product equals ac. Setting each linear factor to zero yields the roots.
2. Quadratic Formula (Sridharacharya’s Method): For any quadratic equation ax² + bx + c = 0 where b² − 4ac ≥ 0, the roots are calculated directly using:
   
   x = [−b ± √(b² − 4ac)] / 2a

3. The Discriminant (D) and Nature of Roots

The term D = b² − 4ac is called the discriminant of the quadratic equation. It determines the nature of the roots without explicitly solving the equation:

• If D > 0: The equation has two distinct real roots: x = (−b + √D) / 2a and x = (−b − √D) / 2a.
• If D = 0: The equation has two equal real roots (coincident roots): x = −b / 2a.
• If D < 0: The equation has no real roots (roots are non-real complex numbers).

Real-World Applications

Quadratic equations are critical in modelling real-life situations involving parabolic motion, calculating rectangular areas in civil engineering, optimizing business revenue functions, and solving kinematics problems involving speed, distance, and time.

NCERT Exercise-Wise Overview

The NCERT textbook organizes Chapter 4 into targeted exercises designed to build proficiency progressively:

• Exercise 4.1: Focuses on checking whether given algebraic expressions form quadratic equations upon simplification, followed by translating real-life situations into standard quadratic equations.
• Exercise 4.2: Centers on solving quadratic equations using the method of factorisation, including equations with irrational/radical coefficients, and introductory application word problems.
• Exercise 4.3: Covers the discriminant test, identifying the nature of roots, applying the quadratic formula, and solving advanced word problems based on speed-stream mechanics, work-rate, and geometric dimensions.

Important CBSE Questions with Step-by-Step Solutions

Below are standard high-yield questions representative of the official CBSE question bank and previous years' board examinations.

Question 1: Nature of Roots (2 Marks)

Problem: Find the value(s) of k for which the quadratic equation (k − 12)x² + 2(k − 12)x + 2 = 0 has two equal real roots, given that k ≠ 12.

Solution:

1. Comparing the given equation with standard form Ax² + Bx + C = 0:
   A = (k − 12), B = 2(k − 12), C = 2.
2. For equal real roots, the discriminant must be zero (D = B² − 4AC = 0).
3. Substitute the coefficients:
   [2(k − 12)]² − 4(k − 12)(2) = 0
   4(k − 12)² − 8(k − 12) = 0
4. Factor out 4(k − 12):
   4(k − 12)[(k − 12) − 2] = 0
   4(k − 12)(k − 14) = 0
5. This implies k − 12 = 0 or k − 14 = 0 ⇒ k = 12 or k = 14.
6. Since the problem states k ≠ 12 (otherwise the coefficient of x² becomes zero), we obtain k = 14.

Question 2: Solving Quadratic Equation with Radical Coefficients (3 Marks)

Problem: Solve for x using factorisation: 4√3x² + 5x − 2√3 = 0.

Solution:

1. Here, a = 4√3, b = 5, and c = −2√3.
2. Product ac = (4√3) × (−2√3) = −8 × 3 = −24.
3. We need two numbers whose product is −24 and sum is +5. These numbers are +8 and −3.
4. Split the middle term:
   4√3x² + 8x − 3x − 2√3 = 0
5. Group the terms:
   4x(√3x + 2) − √3(√3x + 2) = 0
6. Factor out (√3x + 2):
   (4x − √3)(√3x + 2) = 0
7. Set each factor to zero:
   4x − √3 = 0 ⇒ x = √3 / 4
   √3x + 2 = 0 ⇒ x = −2 / √3 = −2√3 / 3

Question 3: Speed, Distance, and Time Word Problem (4/5 Marks)

Problem: A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Solution:

1. Let the speed of the stream be y km/h.
2. Speed of boat upstream = (18 − y) km/h (where y < 18).
   Speed of boat downstream = (18 + y) km/h.
3. Time taken for upstream journey: t₁ = 24 / (18 − y) hours.
   Time taken for downstream journey: t₂ = 24 / (18 + y) hours.
4. According to the given condition:
   t₁ − t₂ = 1
   [24 / (18 − y)] − [24 / (18 + y)] = 1
5. Simplify by taking the common denominator:
   24 [(18 + y) − (18 − y)] / [(18 − y)(18 + y)] = 1
   24 [2y] / (324 − y²) = 1
   48y = 324 − y²
6. Rearrange into standard form:
   y² + 48y − 324 = 0
7. Factorize by splitting the middle term (54 × −6 = −324, 54 − 6 = 48):
   y² + 54y − 6y − 324 = 0
   y(y + 54) − 6(y + 54) = 0
   (y − 6)(y + 54) = 0
8. Therefore, y = 6 or y = −54.
9. Since speed cannot be negative, we reject y = −54.
10. Final Answer: The speed of the stream is 6 km/h.

Common Mistakes to Avoid in Board Exam 10

• Sign Errors in the Quadratic Formula: In the formula x = [−b ± √(b² − 4ac)] / 2a, when b is already negative (e.g., b = −6), remember that −b becomes −(−6) = +6.
• Forgetting Both Roots in Square Root Simplification: When solving equations of the form x² = k, students frequently write x = √k instead of x = ±√k, losing half the marks.
• Not Checking Feasibility in Word Problems: In practical word problems involving dimensions (length, breadth), speed, age, or time, negative solutions must be explicitly stated as inadmissible with a brief justification.
• Ignoring the a ≠ 0 Condition: When an equation contains unknown coefficients in the second-degree term, always state the boundary condition a ≠ 0 before computing the discriminant.

How to Prepare for This Topic

1. Strengthen Factorisation Skills: Practice splitting the middle term with rational, irrational, and fractional coefficients until you can quickly identify factor pairs.
2. Master the Discriminant Criteria: Ensure rapid recall of conditions for real, distinct, equal, and non-real roots. Many 1-mark and 2-mark CBSE questions directly test discriminant relations.
3. Systematic Word Problem Conversion: Always follow a structured four-step protocol: define variables clearly, establish the relationship mathematically, reduce to standard quadratic form, and state final units clearly.
4. Solve Exemplar and Previous Years' Questions: Solve at least the last five years of CBSE Class 10 board papers to familiarize yourself with standard question patterns and step-marking schemes.

Where to Practice More

Consistent, exam-oriented practice is essential to score a perfect 100 in CBSE Mathematics. Visit Theorify QPTool (qptool.theorify.in) to access curated chapter-wise question banks, previous year board papers, AI-powered mock tests, and detailed CBSE marking scheme solutions for Chapter 4 Quadratic Equations.