NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations (NCERT 2026–27)

These Class 10 Maths Chapter 4 solutions cover Quadratic Equations from the latest NCERT textbook (Reprint 2026–27). Every question from Exercise 4.1, Exercise 4.2 and Exercise 4.3 is solved step by step — checking whether equations are quadratic, finding roots by factorisation, and using the discriminant to decide the nature of the roots — so you can master the chapter and revise it quickly before exams.

Class: 10 Subject: Mathematics Chapter: 4 Chapter name: Quadratic Equations Exercises: 4.1, 4.2, 4.3 Session: 2026–27
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Chapter 4 Overview

Chapter 4 of Class 10 Maths, Quadratic Equations, builds on the quadratic polynomials you met in Chapter 2. A quadratic equation is any equation that can be put in the standard form ax2 + bx + c = 0 with a ≠ 0. The chapter shows how such equations model everyday situations — areas of plots, ages, speeds and products of numbers — and teaches two main skills: finding the roots by factorisation (splitting the middle term), and using the discriminant b2 − 4ac together with the quadratic formula to decide whether an equation has two distinct, two equal, or no real roots, and to find those roots. The Class 10 Maths Chapter 4 solutions below work through every question in Exercises 4.1, 4.2 and 4.3 with full, exam-ready steps.

Key Concepts & Definitions

Quadratic equation: an equation of the form ax2 + bx + c = 0, where a, b, c are real numbers and a ≠ 0. This is the standard form (terms written in descending order of degree).

Root (solution): a real number α is a root of ax2 + bx + c = 0 if aα2 + bα + c = 0. The roots of the equation are exactly the zeroes of the polynomial ax2 + bx + c.

Number of roots: a quadratic equation has at most two roots.

Factorisation method: if ax2 + bx + c can be written as a product of two linear factors, the roots are found by setting each factor equal to zero (splitting the middle term).

Discriminant: the quantity D = b2 − 4ac decides the nature of the roots — it “discriminates” between the three cases below.

Nature of roots: two distinct real roots if D > 0; two equal (coincident) real roots if D = 0; no real roots if D < 0.

Important Formulas (Chapter 4)

Standard form: ax2 + bx + c = 0, a ≠ 0.

Quadratic formula: x = [ −b ± √(b2 − 4ac) ] / 2a, provided b2 − 4ac ≥ 0.

Discriminant: D = b2 − 4ac.

Nature of roots: D > 0 → two distinct real roots  •  D = 0 → two equal real roots, each −b/2a  •  D < 0 → no real roots.

Factorisation (split the middle term): find two numbers whose product is a×c and whose sum is b, split bx accordingly, then factor by grouping.

Exercise 4.1 Solutions

Questions are reproduced verbatim from the NCERT Mathematics textbook (Class 10); the worked solutions are original and verified.

1. Check whether the following are quadratic equations: (i) (x + 1)2 = 2(x − 3) (ii) x2 − 2x = (−2)(3 − x) (iii) (x − 2)(x + 1) = (x − 1)(x + 3) (iv) (x − 3)(2x + 1) = x(x + 5) (v) (2x − 1)(x − 3) = (x + 5)(x − 1) (vi) x2 + 3x + 1 = (x − 2)2 (vii) (x + 2)3 = 2x(x2 − 1) (viii) x3 − 4x2 − x + 1 = (x − 2)3

SOLUTION A given equation is quadratic if, after simplifying, it has the form ax2 + bx + c = 0 with a ≠ 0. (i) LHS = x2 + 2x + 1; RHS = 2x − 6. So x2 + 2x + 1 = 2x − 6 ⇒ x2 + 7 = 0. Form ax2 + bx + c = 0 → Quadratic. (ii) LHS = x2 − 2x; RHS = (−2)(3 − x) = −6 + 2x. So x2 − 2x = −6 + 2x ⇒ x2 − 4x + 6 = 0 → Quadratic. (iii) LHS = x2 − x − 2; RHS = x2 + 2x − 3. So x2 − x − 2 = x2 + 2x − 3 ⇒ −3x + 1 = 0. The x2 terms cancel → Not quadratic (it is linear). (iv) LHS = 2x2 − 5x − 3; RHS = x2 + 5x. So 2x2 − 5x − 3 = x2 + 5x ⇒ x2 − 10x − 3 = 0 → Quadratic. (v) LHS = 2x2 − 7x + 3; RHS = x2 + 4x − 5. So 2x2 − 7x + 3 = x2 + 4x − 5 ⇒ x2 − 11x + 8 = 0 → Quadratic. (vi) RHS = x2 − 4x + 4. So x2 + 3x + 1 = x2 − 4x + 4 ⇒ 7x − 3 = 0. The x2 terms cancel → Not quadratic (linear). (vii) LHS = (x + 2)3 = x3 + 6x2 + 12x + 8; RHS = 2x3 − 2x. So x3 + 6x2 + 12x + 8 = 2x3 − 2x ⇒ −x3 + 6x2 + 14x + 8 = 0, i.e. x3 − 6x2 − 14x − 8 = 0. Degree 3 → Not quadratic. (viii) RHS = (x − 2)3 = x3 − 6x2 + 12x − 8. So x3 − 4x2 − x + 1 = x3 − 6x2 + 12x − 8 ⇒ 2x2 − 13x + 9 = 0. The x3 terms cancel → Quadratic.

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. (ii) The product of two consecutive positive integers is 306. We need to find the integers. (iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age. (iv) 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. We need to find the speed of the train.

SOLUTION (i) Let the breadth be x m. Then length = (2x + 1) m. Area = length × breadth = x(2x + 1) = 528 ⇒ 2x2 + x = 528. So 2x2 + x − 528 = 0. (ii) Let the smaller integer be x, so the next consecutive integer is (x + 1). Product = x(x + 1) = 306 ⇒ x2 + x = 306. So x2 + x − 306 = 0. (iii) Let Rohan’s present age be x years; his mother’s age = (x + 26) years. After 3 years their ages are (x + 3) and (x + 29). Product = (x + 3)(x + 29) = 360 ⇒ x2 + 32x + 87 = 360. So x2 + 32x − 273 = 0. (iv) Let the uniform speed be x km/h. Time at this speed = 480/x h; at speed (x − 8) km/h, time = 480/(x − 8) h. The slower trip takes 3 h more: 480/(x − 8) − 480/x = 3. Multiply through by x(x − 8): 480x − 480(x − 8) = 3x(x − 8) ⇒ 3840 = 3x2 − 24x. Dividing by 3: x2 − 8x − 1280 = 0.

Exercise 4.2 Solutions

1. Find the roots of the following quadratic equations by factorisation: (i) x2 − 3x − 10 = 0 (ii) 2x2 + x − 6 = 0 (iii) √2 x2 + 7x + 5√2 = 0 (iv) 2x2 − x + 1/8 = 0 (v) 100x2 − 20x + 1 = 0

SOLUTION (i) x2 − 3x − 10 = 0. Split −3x as −5x + 2x (since −5 × 2 = −10): x2 − 5x + 2x − 10 = 0 ⇒ x(x − 5) + 2(x − 5) = 0 ⇒ (x − 5)(x + 2) = 0. Roots: x = 5 or x = −2. (ii) 2x2 + x − 6 = 0. Here a×c = −12; split x as 4x − 3x: 2x2 + 4x − 3x − 6 = 0 ⇒ 2x(x + 2) − 3(x + 2) = 0 ⇒ (x + 2)(2x − 3) = 0. Roots: x = −2 or x = 3/2. (iii) √2 x2 + 7x + 5√2 = 0. Product a×c = √2 × 5√2 = 10; split 7x as 5x + 2x: √2 x2 + 5x + 2x + 5√2 = 0 ⇒ x(√2 x + 5) + √2(√2 x + 5) = 0 ⇒ (√2 x + 5)(x + √2) = 0. Roots: x = −5/√2 = −5√2/2 or x = −√2. (iv) 2x2 − x + 1/8 = 0. Multiply by 8: 16x2 − 8x + 1 = 0. This is (4x − 1)2 = 0, so 4x − 1 = 0 (repeated). Root: x = 1/4, 1/4 (equal roots). (v) 100x2 − 20x + 1 = 0 = (10x − 1)2 = 0, so 10x − 1 = 0 (repeated). Root: x = 1/10, 1/10 (equal roots).

2. Solve the problems given in Example 1.

SOLUTION Problem (i) — John and Jivanti’s marbles. The situation gives x2 − 45x + 324 = 0, where x is the number of marbles John started with. Split −45x as −36x − 9x (since −36 × −9 = 324): x2 − 36x − 9x + 324 = 0 ⇒ x(x − 36) − 9(x − 36) = 0 ⇒ (x − 36)(x − 9) = 0. So x = 36 or x = 9. If John had 36, Jivanti had 45 − 36 = 9; if John had 9, Jivanti had 36. Either way the two started with 36 and 9 marbles. (Check: (36 − 5)(9 − 5) = 31 × 4 = 124. ✓) Problem (ii) — cottage industry toys. The situation gives x2 − 55x + 750 = 0, where x is the number of toys. Split −55x as −30x − 25x (since −30 × −25 = 750): x2 − 30x − 25x + 750 = 0 ⇒ x(x − 30) − 25(x − 30) = 0 ⇒ (x − 30)(x − 25) = 0. So x = 30 or x = 25. Both are valid, so the number of toys produced that day is 25 or 30. (Check: 25 × (55 − 25) = 25 × 30 = 750, and 30 × (55 − 30) = 30 × 25 = 750. ✓)

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

SOLUTION Let one number be x; then the other is (27 − x). Product: x(27 − x) = 182 ⇒ 27x − x2 = 182 ⇒ x2 − 27x + 182 = 0. Split −27x as −13x − 14x (since −13 × −14 = 182): x2 − 13x − 14x + 182 = 0 ⇒ x(x − 13) − 14(x − 13) = 0 ⇒ (x − 13)(x − 14) = 0. So x = 13 or x = 14. ∴ the two numbers are 13 and 14. (Check: 13 + 14 = 27 and 13 × 14 = 182. ✓)

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

SOLUTION Let the integers be x and (x + 1). Then x2 + (x + 1)2 = 365 ⇒ x2 + x2 + 2x + 1 = 365 ⇒ 2x2 + 2x − 364 = 0 ⇒ x2 + x − 182 = 0. Split x as 14x − 13x (since 14 × −13 = −182): x2 + 14x − 13x − 182 = 0 ⇒ x(x + 14) − 13(x + 14) = 0 ⇒ (x + 14)(x − 13) = 0. So x = 13 or x = −14. Since the integers must be positive, x = 13. The integers are 13 and 14. (Check: 132 + 142 = 169 + 196 = 365. ✓)

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.

SOLUTION Let the base be x cm; then the altitude = (x − 7) cm. By the Pythagoras theorem: base2 + altitude2 = hypotenuse2. x2 + (x − 7)2 = 132 ⇒ x2 + x2 − 14x + 49 = 169 ⇒ 2x2 − 14x − 120 = 0 ⇒ x2 − 7x − 60 = 0. Split −7x as −12x + 5x (since −12 × 5 = −60): x2 − 12x + 5x − 60 = 0 ⇒ x(x − 12) + 5(x − 12) = 0 ⇒ (x − 12)(x + 5) = 0. So x = 12 or x = −5. A length cannot be negative, so x = 12. Base = 12 cm, altitude = 12 − 7 = 5 cm. (Check: 122 + 52 = 144 + 25 = 169 = 132. ✓)

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.

SOLUTION Let the number of articles produced be x. Cost of each article = (2x + 3) rupees. Total cost = number × cost per article = x(2x + 3) = 90. 2x2 + 3x − 90 = 0. Here a×c = −180; split 3x as 15x − 12x: 2x2 + 15x − 12x − 90 = 0 ⇒ x(2x + 15) − 6(2x + 15) = 0 ⇒ (2x + 15)(x − 6) = 0. So x = 6 or x = −15/2. The number of articles cannot be negative or fractional, so x = 6. Number of articles = 6; cost of each article = 2(6) + 3 = ₹ 15. (Check: 6 × 15 = ₹ 90. ✓)

Exercise 4.3 Solutions

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: (i) 2x2 − 3x + 5 = 0 (ii) 3x2 − 4√3 x + 4 = 0 (iii) 2x2 − 6x + 3 = 0

SOLUTION Use the discriminant D = b2 − 4ac. (i) a = 2, b = −3, c = 5. D = (−3)2 − 4(2)(5) = 9 − 40 = −31 < 0. So the equation has no real roots. (ii) a = 3, b = −4√3, c = 4. D = (−4√3)2 − 4(3)(4) = 48 − 48 = 0. So it has two equal real roots. Root = −b/2a = 4√3 / (2×3) = 4√3/6 = 2√3/3 = 2/√3 (repeated). (iii) a = 2, b = −6, c = 3. D = (−6)2 − 4(2)(3) = 36 − 24 = 12 > 0. So it has two distinct real roots. By the quadratic formula: x = [6 ± √12] / 4 = [6 ± 2√3] / 4 = (3 ± √3)/2. Roots: x = (3 + √3)/2 or x = (3 − √3)/2.

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 (ii) kx(x − 2) + 6 = 0

SOLUTION Two equal roots occur when the discriminant D = b2 − 4ac = 0. (i) a = 2, b = k, c = 3. D = k2 − 4(2)(3) = k2 − 24 = 0 ⇒ k2 = 24 ⇒ k = ±2√6 (i.e. k = 2√6 or −2√6). (ii) Expand: kx2 − 2kx + 6 = 0, so a = k, b = −2k, c = 6. D = (−2k)2 − 4(k)(6) = 4k2 − 24k = 0 ⇒ 4k(k − 6) = 0 ⇒ k = 0 or k = 6. If k = 0 the equation is not quadratic (the x2 term vanishes), so we reject it. Hence k = 6.

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

SOLUTION Let the breadth be x m; then length = 2x m. Area = length × breadth = 2x × x = 2x2 = 800 ⇒ x2 = 400. Written as 2x2 − 800 = 0, the discriminant D = 02 − 4(2)(−800) = 6400 > 0, so real roots exist — the design is possible. x2 = 400 ⇒ x = 20 (taking the positive value, since x is a length). Breadth = 20 m, length = 2 × 20 = 40 m.

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

SOLUTION Let one friend’s present age be x years; the other’s is (20 − x) years. Four years ago their ages were (x − 4) and (16 − x). Their product was 48: (x − 4)(16 − x) = 48. 16x − x2 − 64 + 4x = 48 ⇒ −x2 + 20x − 64 = 48 ⇒ x2 − 20x + 112 = 0. Discriminant D = (−20)2 − 4(1)(112) = 400 − 448 = −48 < 0. There are no real roots, so the given situation is not possible.

5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.

SOLUTION Let the length be x m and the breadth be y m. Perimeter = 2(x + y) = 80 ⇒ x + y = 40, so y = 40 − x. Area = xy = 400. Substitute: x(40 − x) = 400 ⇒ 40x − x2 = 400 ⇒ x2 − 40x + 400 = 0. Discriminant D = (−40)2 − 4(1)(400) = 1600 − 1600 = 0, so equal real roots exist — the design is possible. x = −b/2a = 40/2 = 20. So length = 20 m and breadth = 40 − 20 = 20 m. The park is a square of side 20 m (length = breadth = 20 m).

Common Mistakes to Avoid

Watch out for these

  • Calling an equation quadratic before simplifying — always bring it to standard form first; the x2 terms may cancel (Ex 4.1 parts iii, vi) or a cubic may reduce to a quadratic (part viii).
  • Forgetting to reject invalid roots in word problems — lengths, ages and counts cannot be negative or fractional.
  • Sign errors when splitting the middle term — the two numbers must multiply to a×c and add to b.
  • Confusing the discriminant cases: D > 0 gives two distinct roots, D = 0 two equal roots, D < 0 no real roots.
  • Accepting k = 0 in “equal roots” problems — if it makes a = 0 the equation is no longer quadratic and must be rejected.
  • Mis-applying the quadratic formula — remember it is x = (−b ± √D)/2a, with the whole −b on top.

Practice MCQs & Assertion–Reason

1. Which of the following is a quadratic equation?

(a) x2 + 1/x = 5    (b) x3 − 2x = 0    (c) (x − 1)2 = x2 + 3    (d) 2x2 − 3x + 1 = 0

2. The roots of x2 − 3x − 10 = 0 are:

(a) 5, −2    (b) −5, 2    (c) 5, 2    (d) −5, −2

3. The discriminant of 2x2 − 4x + 3 = 0 is:

(a) 8    (b) −8    (c) 40    (d) −40

4. A quadratic equation ax2 + bx + c = 0 has two equal real roots when:

(a) b2 − 4ac > 0    (b) b2 − 4ac = 0    (c) b2 − 4ac < 0    (d) a = 0

5. The maximum number of roots a quadratic equation can have is:

(a) 1    (b) 2    (c) 3    (d) 4

6. The roots of 100x2 − 20x + 1 = 0 are:

(a) 1/10, 1/10    (b) 10, 10    (c) 1/10, −1/10    (d) 1/5, 1/5

7. For 2x2 + kx + 3 = 0 to have two equal roots, k equals:

(a) ±6    (b) ±2√6    (c) ±24    (d) ±√6

8. If the product of two consecutive positive integers is 306, the smaller integer is:

(a) 16    (b) 17    (c) 18    (d) 15

9. The nature of the roots of 3x2 − 4√3 x + 4 = 0 is:

(a) two distinct real roots    (b) two equal real roots    (c) no real roots    (d) cannot be determined

10. By the quadratic formula, the roots of ax2 + bx + c = 0 are given by:

(a) (b ± √(b2 − 4ac))/2a    (b) (−b ± √(b2 − 4ac))/2a    (c) (−b ± √(b2 + 4ac))/2a    (d) (−b ± √(4ac − b2))/2a

Answer key: 1-(d), 2-(a), 3-(b), 4-(b), 5-(b), 6-(a), 7-(b), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The equation (x − 2)(x + 1) = (x − 1)(x + 3) is not a quadratic equation.

Reason: On simplifying, the x2 terms cancel and the equation reduces to a linear equation.

A-R 2. Assertion: The equation 2x2 − 3x + 5 = 0 has no real roots.

Reason: Its discriminant b2 − 4ac = −31, which is negative.

A-R 3. Assertion: A quadratic equation can have at most two roots.

Reason: A quadratic polynomial can have at most two zeroes.

A-R 4. Assertion: For kx(x − 2) + 6 = 0 to have equal roots, the only acceptable value is k = 6.

Reason: The discriminant gives k = 0 or k = 6, and k = 0 makes the equation non-quadratic.

A-R 5. Assertion: A rectangular park of perimeter 80 m and area 400 m2 can be designed.

Reason: The resulting equation x2 − 40x + 400 = 0 has discriminant zero, so real (equal) roots exist.

Answer key: 1-(A), 2-(A), 3-(A), 4-(A), 5-(A).

Quick Revision Summary

  • A quadratic equation has the standard form ax2 + bx + c = 0 with a ≠ 0; always simplify before deciding if an equation is quadratic.
  • A real number is a root if it makes the equation zero; the roots are the zeroes of ax2 + bx + c, and there are at most two.
  • Factorisation: split the middle term using two numbers whose product is a×c and sum is b, then factor by grouping and set each factor to zero.
  • Quadratic formula: x = (−b ± √(b2 − 4ac))/2a, valid when b2 − 4ac ≥ 0.
  • Discriminant D = b2 − 4ac: D > 0 → two distinct real roots; D = 0 → two equal real roots; D < 0 → no real roots.
  • In word problems, form the equation carefully and reject any root that is negative, fractional or otherwise impossible for the quantity.

How to score full marks in this chapter

Write the equation in standard form first and clearly state a, b and c before using the formula or discriminant. For factorisation, show the middle-term split and the grouping so each step earns its mark. In application questions, always define the variable, justify which root you keep, and verify the answer by substituting back. Compute the discriminant whenever a question asks about the “nature” of roots or whether a design/situation is “possible” — the sign of b2 − 4ac is the whole answer.

Frequently Asked Questions

What is Class 10 Maths Chapter 4 about?

Chapter 4, Quadratic Equations, covers the standard form ax2 + bx + c = 0, finding roots by factorisation (splitting the middle term), the quadratic formula, and using the discriminant b2 − 4ac to determine the nature of the roots, with many real-life word problems.

How many exercises are there in Class 10 Maths Chapter 4?

There are three exercises — Exercise 4.1 (checking and forming quadratic equations), Exercise 4.2 (finding roots by factorisation) and Exercise 4.3 (nature of roots using the discriminant) — all solved step by step on this page.

What is the discriminant and why is it important?

The discriminant is D = b2 − 4ac. Its sign tells you the nature of the roots without solving: D > 0 means two distinct real roots, D = 0 means two equal real roots, and D < 0 means there are no real roots.

Are these Class 10 Maths Chapter 4 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for Class 10 (2026–27 session), with every answer worked out and verified against the book.

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