NCERT Solutions Class 10 Mathematics Chapter 4 Quadratic Equations

NCERT Solutions Class 10 Mathematics Chapter 4 Quadratic Equations provide comprehensive, step-by-step guidance for mastering second-degree polynomial equations and their real-world applications. As one of the core algebraic chapters in the CBSE Class 10 Mathematics syllabus, Chapter 4 carries significant weight in board examinations. Whether you are preparing for periodic assessments or the final Class 10 Board Exam, these NCERT Solutions step by step are designed to build conceptual clarity, strengthen analytical problem-solving skills, and help you score full marks.

This chapter bridges foundational algebra from Class 9 (polynomials and linear equations) with advanced algebraic modeling in Class 11 and 12. In this guide, you will find exhaustive solutions to all exercises, conceptual explanations of root-finding techniques, discriminant analysis, standard derivations, and whiteboard-style worked solutions to high-frequency CBSE board exam questions.

Chapter Overview

A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, and c are real numbers, and a ≠ 0. The condition a ≠ 0 is essential because if a = 0, the equation degenerates into a linear equation (bx + c = 0).

Quadratic equations arise naturally in diverse physical, geometric, and financial contexts—such as calculating projectile trajectories, determining the dimensions of geometric enclosures given their area, calculating relative speeds of upstream and downstream motion, and optimizing revenue in business mathematics.

Core Topics in Class 10 Quadratic Equations

• Standard Form & Identification: Expressing polynomial equations in the canonical form ax² + bx + c = 0 and verifying whether a given algebraic equation is quadratic.
• Solution by Factorisation (Splitting the Middle Term): Finding roots by resolving the quadratic polynomial into linear factors (px + q)(rx + s) = 0.
• The Quadratic Formula (Sridharacharya's Rule): Calculating roots directly using the formula x = (−b ± √(b² − 4ac)) / (2a).
• Nature of Roots & The Discriminant: Using D = b² − 4ac to classify roots into real and distinct, real and equal, or non-real (complex) roots without solving the equation.
• Applied Word Problems: Formulating and solving quadratic equations based on geometric dimensions, consecutive numbers, speed-time-distance relationships, and work rates.

Exercise Solutions: Step-by-Step NCERT Solutions

NCERT Exercise 4.1: Identification and Formulation

Question 1: Check whether the following are quadratic equations:

(i) (x + 1)² = 2(x − 3)

1. Given: (x + 1)² = 2(x − 3)
2. To Determine: Whether the equation is of the form ax² + bx + c = 0 with a ≠ 0.
3. Expansion & Simplification:
   LHS = x² + 2x + 1
   RHS = 2x − 6
   Equating LHS and RHS:
   x² + 2x + 1 = 2x − 6
   x² + 2x − 2x + 1 + 6 = 0
   x² + 7 = 0
   Here, x² + 0x + 7 = 0, where a = 1, b = 0, and c = 7.
4. Conclusion: Since the highest exponent of x is 2 and a ≠ 0, it is a quadratic equation.

(ii) x² − 2x = (−2)(3 − x)

1. Given: x² − 2x = (−2)(3 − x)
2. Simplification:
   x² − 2x = −6 + 2x
   x² − 2x − 2x + 6 = 0
   x² − 4x + 6 = 0
3. Conclusion: The equation is in standard form ax² + bx + c = 0 with a = 1 ≠ 0. Hence, it is a quadratic equation.

(iii) (x + 2)³ = 2x(x² − 1)

1. Given: (x + 2)³ = 2x(x² − 1)
2. Expansion:
   Using identity (a + b)³ = a³ + 3a²b + 3ab² + b³:
   LHS = x³ + 3(x²)(2) + 3(x)(2²) + 2³ = x³ + 6x² + 12x + 8
   RHS = 2x³ − 2x
   Equating LHS and RHS:
   x³ + 6x² + 12x + 8 = 2x³ − 2x
   x³ − 2x³ + 6x² + 12x + 2x + 8 = 0
   −x³ + 6x² + 14x + 8 = 0
3. Conclusion: The degree of the resulting polynomial equation is 3 (cubic). Therefore, it is NOT a quadratic equation.

Question 2: Represent the following situations in the form of quadratic equations:

(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

• Let: Breadth of the rectangular plot = x metres.
• Then: Length of the plot = (2x + 1) metres.
• Formula: Area = Length × Breadth
• Substitution:
  x(2x + 1) = 528
  2x² + x = 528
  2x² + x − 528 = 0
• Answer: The required quadratic equation representing the situation is 2x² + x − 528 = 0.

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

• Let: Uniform speed of the train = x km/h.
• Total Distance: 480 km.
• Time taken at normal speed: t₁ = 480 / x hours.
• Reduced speed: (x − 8) km/h.
• Time taken at reduced speed: t₂ = 480 / (x − 8) hours.
• Given condition: t₂ − t₁ = 3
  [480 / (x − 8)] − [480 / x] = 3
  480 [(x − (x − 8)) / (x(x − 8))] = 3
  480 [8 / (x² − 8x)] = 3
  3840 = 3(x² − 8x)
  1280 = x² − 8x
  x² − 8x − 1280 = 0
• Answer: The required quadratic equation is x² − 8x − 1280 = 0.

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NCERT Exercise 4.2: Solving by Factorisation

Question 1: Find the roots of the following quadratic equations by factorisation:

(i) x² − 3x − 10 = 0

1. Step 1: Identify coefficients: a = 1, b = −3, c = −10. Product ac = −10, Sum b = −3.
2. Step 2: Find two integers whose product is −10 and sum is −3: these are −5 and +2.
3. Step 3: Split the middle term:
   x² − 5x + 2x − 10 = 0
4. Step 4: Factor by grouping:
   x(x − 5) + 2(x − 5) = 0
   (x − 5)(x + 2) = 0
5. Step 5: Equate each factor to zero:
   x − 5 = 0 ⇒ x = 5
   x + 2 = 0 ⇒ x = −2
6. Answer: The roots are x = 5 and x = −2.

(ii) √2 x² + 7x + 5√2 = 0

1. Step 1: Identify coefficients: a = √2, b = 7, c = 5√2.
   Product ac = √2 × 5√2 = 5 × 2 = 10.
   Sum b = 7.
2. Step 2: Two numbers with product 10 and sum 7 are 5 and 2.
3. Step 3: Split the middle term:
   √2 x² + 2x + 5x + 5√2 = 0
   √2 x² + (√2 × √2)x + 5x + 5√2 = 0
4. Step 4: Factor by grouping:
   √2 x(x + √2) + 5(x + √2) = 0
   (x + √2)(√2 x + 5) = 0
5. Step 5: Solve for x:
   x + √2 = 0 ⇒ x = −√2
   √2 x + 5 = 0 ⇒ x = −5 / √2 = −(5√2) / 2
6. Answer: The roots are x = −√2 and x = −5/√2.

Question 3: Find two numbers whose sum is 27 and product is 182.

1. Let: First number = x.
2. Then: Second number = (27 − x).
3. Given Product:
   x(27 − x) = 182
   27x − x² = 182
   x² − 27x + 182 = 0
4. Factorisation: Product = 182, Sum = −27. Factors are −13 and −14.
   x² − 13x − 14x + 182 = 0
   x(x − 13) − 14(x − 13) = 0
   (x − 13)(x − 14) = 0
   x = 13 or x = 14
5. Answer: The two numbers are 13 and 14.

Question 4: Find two consecutive positive integers, sum of whose squares is 365.

1. Let: Two consecutive positive integers be x and (x + 1), where x > 0.
2. According to Problem:
   x² + (x + 1)² = 365
   x² + (x² + 2x + 1) = 365
   2x² + 2x + 1 − 365 = 0
   2x² + 2x − 364 = 0
3. Dividing entire equation by 2:
   x² + x − 182 = 0
4. Factorising: Factors of −182 adding to +1 are +14 and −13.
   x² + 14x − 13x − 182 = 0
   x(x + 14) − 13(x + 14) = 0
   (x + 14)(x − 13) = 0
   x = 13 or x = −14
5. Discarding extraneous roots: Since x must be a positive integer, x ≠ −14. Therefore, x = 13.
6. Consecutive integers: x = 13 and x + 1 = 14.
7. Answer: The two consecutive positive integers are 13 and 14.

Question 5: The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

1. Let: Base of the right triangle = x cm.
2. Then: Altitude = (x − 7) cm.
3. By Pythagoras Theorem: (Base)² + (Altitude)² = (Hypotenuse)²
   x² + (x − 7)² = 13²
   x² + (x² − 14x + 49) = 169
   2x² − 14x + 49 − 169 = 0
   2x² − 14x − 120 = 0
   x² − 7x − 60 = 0
4. Factorising: Factors of −60 summing to −7 are −12 and +5.
   (x − 12)(x + 5) = 0
   x = 12 or x = −5
5. Evaluation: Side length cannot be negative, so x = 12 cm.
   Base = 12 cm, Altitude = 12 − 7 = 5 cm.
6. Answer: The other two sides are 5 cm and 12 cm.

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NCERT Exercise 4.3: Nature of Roots and Quadratic Formula

Question 1: Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) 2x² − 3x + 5 = 0

1. Identify coefficients: a = 2, b = −3, c = 5.
2. Calculate Discriminant (D):
   D = b² − 4ac
   D = (−3)² − 4(2)(5) = 9 − 40 = −31
3. Conclusion: Since D < 0, the equation has no real roots.

(ii) 3x² − 4√3 x + 4 = 0

1. Identify coefficients: a = 3, b = −4√3, c = 4.
2. Calculate Discriminant (D):
   D = (−4√3)² − 4(3)(4)
   D = (16 × 3) − 48 = 48 − 48 = 0
3. Nature of Roots: Since D = 0, the equation has two equal real roots.
4. Finding Roots:
   x = −b / (2a) = −(−4√3) / (2 × 3) = (4√3) / 6 = (2√3) / 3 = 2 / √3
5. Answer: The roots are 2/√3 and 2/√3.

Question 2: Find the values of k for each of the following quadratic equations, so that they have two equal roots:

(i) 2x² + kx + 3 = 0

1. Condition for two equal roots: D = b² − 4ac = 0
2. Coefficients: a = 2, b = k, c = 3.
3. Substitution:
   k² − 4(2)(3) = 0
   k² − 24 = 0
   k² = 24
   k = ±√24 = ± 2√6
4. Answer: The values of k are k = 2√6 and k = −2√6.

(ii) kx(x − 2) + 6 = 0

1. Standard Form: kx² − 2kx + 6 = 0
2. Coefficients: a = k, b = −2k, c = 6 (with condition k ≠ 0).
3. Condition D = 0:
   (−2k)² − 4(k)(6) = 0
   4k² − 24k = 0
   4k(k − 6) = 0
   k = 0 or k = 6
4. Constraint Check: If k = 0, the equation is no longer quadratic (coefficient of x² becomes 0). Therefore, k = 6.
5. Answer: k = 6.

Question 3: An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

1. Let: Average speed of passenger train = x km/h.
2. Then: Average speed of express train = (x + 11) km/h.
3. Distance: 132 km.
4. Time Equations:
   Time for passenger train = 132 / x hours.
   Time for express train = 132 / (x + 11) hours.
5. Formulating Equation:
   [132 / x] − [132 / (x + 11)] = 1
   132 [(x + 11 − x) / (x(x + 11))] = 1
   132 [11 / (x² + 11x)] = 1
   1452 = x² + 11x
   x² + 11x − 1452 = 0
6. Solving using Quadratic Formula:
   a = 1, b = 11, c = −1452
   D = (11)² − 4(1)(−1452) = 121 + 5808 = 5929
   √D = √5929 = 77
   x = (−11 ± 77) / (2 × 1)
   x = (−11 + 77) / 2 = 66 / 2 = 33
   x = (−11 − 77) / 2 = −88 / 2 = −44 (Rejected as speed cannot be negative)
7. Answer: The average speed of the passenger train is 33 km/h and that of the express train is 33 + 11 = 44 km/h.

Important Formulas and Theorems

Here is your comprehensive formula sheet for Class 10 Mathematics Chapter 4 Quadratic Equations:

Concept / Formula	Mathematical Expression	Significance / Condition
Standard Form	ax2 + bx + c = 0	a, b, c ∈ ℝ and a ≠ 0
The Discriminant (D)	D = b2 − 4ac	Determines the nature of roots without computing them
Two Distinct Real Roots	D > 0	Roots: x = (−b ± √D) / (2a)
Two Equal Real Roots	D = 0	Coincident roots: x = −b / (2a)
No Real Roots	D < 0	Roots are non-real complex conjugates
Quadratic Formula	x = (−b ± √(b2 − 4ac)) / (2a)	Direct algebraic method applicable to all quadratic equations
Sum and Product of Roots	α + β = −b/a αβ = c/a	Equation from roots: x2 − (α + β)x + αβ = 0

Derivation of the Quadratic Formula (Completing the Square Method)

To understand why the quadratic formula works, let us derive it from first principles:

1. Start with standard form: ax² + bx + c = 0
2. Divide throughout by a (since a ≠ 0):
   x² + (b/a)x + (c/a) = 0
3. Transpose the constant term to the RHS:
   x² + (b/a)x = −c/a
4. Add the square of half the coefficient of x, i.e., [b / (2a)]² = b² / (4a²) to both sides:
   x² + 2 × x × (b / 2a) + (b / 2a)² = [b² / (4a²)] − (c/a)
5. Factor the LHS as a perfect square and take a common denominator on the RHS:
   [x + (b / 2a)]² = (b² − 4ac) / (4a²)
6. Taking the square root on both sides:
   x + (b / 2a) = ± √(b² − 4ac) / (2a)
7. Subtracting b / (2a) from both sides:
   x = [−b ± √(b² − 4ac)] / (2a)

Common Mistakes and Tips

• Forgetting the Negative Sign in −b: If b is already negative (e.g., in x² − 6x + 8 = 0 where b = −6), students often write −6 instead of −(−6) = +6 in the quadratic formula.
• Sign Errors in Evaluating −4ac: When c is negative, −4ac becomes positive: −4(a)(−c) = +4ac. Always double-check your arithmetic when calculating the discriminant.
• Omitting the Non-Zero Condition (a ≠ 0): In parameter-finding questions such as kx² − 2kx + 6 = 0, students often retain k = 0 as a valid answer. Remember that k = 0 destroys the quadratic nature of the equation and must be discarded.
• Not Rejecting Inadmissible Roots in Real-Life Problems: In geometric and physical problems, physical quantities like distance, speed, breadth, and age cannot be negative. Clearly write a justification statement (e.g., "Speed cannot be negative, hence rejecting x = −44") to avoid losing method marks.
• Improper Splitting of Middle Term: Ensure that the product of the split terms equals exactly a × c (including signs) and their algebraic sum equals b.

Board Exam Relevance

In the CBSE Class 10 Mathematics Board Examination, Chapter 4 Quadratic Equations is part of the Algebra Unit, which carries a total of 20 marks. Quadratic Equations individually accounts for approximately 6 to 8 marks across different sections of the question paper:

• Section A (1 Mark - MCQs / Assertion-Reason): Questions testing the discriminant condition (D > 0, D = 0, D < 0) and identifying quadratic equations.
• Section B (2 Marks - Short Answer): Determining the value of an unknown constant k for which an equation has equal roots.
• Section C (3 Marks - Short Answer): Solving equations with radical coefficients or standard factorisation problems.
• Section D / Section E (4 & 5 Marks - Long Answer & Case Study Questions): Real-world situational problems involving speed-distance-time, upstream-downstream boats, taps filling cisterns, and rectangular construction geometry.

More NCERT Solutions and Practice

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