---
title: "NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations"
url: https://www.swavid.com/maths/class/10/chapter/quadratic-equations/ncert-solutions
dateModified: 2026-10-07T15:49:35+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

This chapter's questions cover checking whether given equations are quadratic, representing various real-life and mathematical situations as quadratic equations, finding the roots of quadratic equations using factorisation and the quadratic formula, determining the nature of roots using the discriminant, and solving related word problems.

Free PDF (10 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-4-quadratic-equations-fb40ded126.pdf

## EXERCISE 4.1

### Question 1

*3 marks · Short answer*

Check whether the following are quadratic equations :
(i) $(x + 1)^2 = 2(x - 3)$
(ii) $x^2 - 2x = (-2) (3 - x)$
(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) $x^2 + 3x + 1 = (x - 2)^2$
(vii) $(x + 2)^3 = 2x (x^2 - 1)$
(viii) $x^3 - 4x^2 - x + 1 = (x - 2)^3$

**Part (i)**

1. Given equation: $(x + 1)^2 = 2(x - 3)$
2. Expanding both sides gives $x^2 + 2x + 1 = 2x - 6$
3. Simplifying yields $x^2 + 7 = 0$, which is of the form $ax^2 + bx + c = 0$.

Answer (i): Yes, it is a quadratic equation.

**Part (ii)**

1. Given equation: $x^2 - 2x = (-2)(3 - x)$
2. Expanding the right side gives $x^2 - 2x = -6 + 2x$
3. Rearranging terms yields $x^2 - 4x + 6 = 0$, which is of the form $ax^2 + bx + c = 0$.

Answer (ii): Yes, it is a quadratic equation.

**Part (iii)**

1. Given equation: $(x - 2)(x + 1) = (x - 1)(x + 3)$
2. Expanding both sides gives $x^2 - x - 2 = x^2 + 2x - 3$
3. Cancelling $x^2$ and simplifying yields $-3x + 1 = 0$, which is not of the form $ax^2 + bx + c = 0$.

Answer (iii): No, it is not a quadratic equation.

**Part (iv)**

1. Given equation: $(x - 3)(2x + 1) = x(x + 5)$
2. Expanding both sides gives $2x^2 - 5x - 3 = x^2 + 5x$
3. Rearranging terms yields $x^2 - 10x - 3 = 0$, which is of the form $ax^2 + bx + c = 0$.

Answer (iv): Yes, it is a quadratic equation.

**Part (v)**

1. Given equation: $(2x - 1)(x - 3) = (x + 5)(x - 1)$
2. Expanding both sides gives $2x^2 - 7x + 3 = x^2 + 4x - 5$
3. Rearranging terms yields $x^2 - 11x + 8 = 0$, which is of the form $ax^2 + bx + c = 0$.

Answer (v): Yes, it is a quadratic equation.

**Part (vi)**

1. Given equation: $x^2 + 3x + 1 = (x - 2)^2$
2. Expanding the right side gives $x^2 + 3x + 1 = x^2 - 4x + 4$
3. Cancelling $x^2$ and simplifying yields $7x - 3 = 0$, which is not of the form $ax^2 + bx + c = 0$.

Answer (vi): No, it is not a quadratic equation.

**Part (vii)**

1. Given equation: $(x + 2)^3 = 2x(x^2 - 1)$
2. Expanding both sides gives $x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$
3. Rearranging terms yields $-x^3 + 6x^2 + 14x + 8 = 0$, which has degree 3.

Answer (vii): No, it is not a quadratic equation.

**Part (viii)**

1. Given equation: $x^3 - 4x^2 - x + 1 = (x - 2)^3$
2. Expanding the right side gives $x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$
3. Cancelling $x^3$ and rearranging terms yields $2x^2 - 13x + 9 = 0$, which is of the form $ax^2 + bx + c = 0$.

Answer (viii): Yes, it is a quadratic equation.

**Answer:** Refer to the parts for individual results.

> Common mistake: Failing to expand binomial cubes or products completely before comparing the degree of the resulting equation.

### Question 2

*3 marks · Short answer*

Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is $528\text{ m}^2$. 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 $480\text{ km}$ at a uniform speed. If the speed had been $8\text{ km/h}$ less, then it would have taken $3$ hours more to cover the same distance. We need to find the speed of the train.

**Part (i)**

1. Let the breadth of the plot be $x\text{ m}$.
2. Then the length of the plot is $(2x + 1)\text{ m}$.
3. Area = $\text{length} \times \text{breadth} = x(2x + 1) = 528$.
4. Simplifying gives $2x^2 + x - 528 = 0$.

Answer (i): $2x^2 + x - 528 = 0$

**Part (ii)**

1. Let the first positive integer be $x$.
2. Then the next consecutive positive integer is $x + 1$.
3. Their product is $x(x + 1) = 306$.
4. Simplifying gives $x^2 + x - 306 = 0$.

Answer (ii): $x^2 + x - 306 = 0$

**Part (iii)**

1. Let Rohan's present age be $x$ years.
2. Then Rohan's mother's present age is $(x + 26)$ years.
3. Three years from now, their ages will be $(x + 3)$ and $(x + 29)$ years respectively.
4. The product of their ages is $(x + 3)(x + 29) = 360$, which simplifies to $x^2 + 32x - 273 = 0$.

Answer (iii): $x^2 + 32x - 273 = 0$

**Part (iv)**

1. Let the uniform speed of the train be $x\text{ km/h}$.
2. Time taken to cover $480\text{ km}$ is $\frac{480}{x}\text{ hours}$.
3. If the speed is $(x - 8)\text{ km/h}$, the time taken is $\frac{480}{x - 8}\text{ hours}$.
4. According to the question, $\frac{480}{x - 8} - \frac{480}{x} = 3$, which simplifies to $x^2 - 8x - 1280 = 0$.

Answer (iv): $x^2 - 8x - 1280 = 0$

**Answer:** Refer to the parts for individual results.

> Common mistake: Formulating the algebraic expressions incorrectly or making errors in simplifying terms.

## EXERCISE 4.2

### Question 1

*3 marks · Short answer*

Find the roots of the following quadratic equations by factorisation:
(i) $x^2 - 3x - 10 = 0$
(ii) $2x^2 + x - 6 = 0$
(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$
(iv) $2x^2 - x + \frac{1}{8} = 0$
(v) $100x^2 - 20x + 1 = 0$

**Part (i)**

1. $x^2 - 3x - 10 = 0$
2. $x^2 - 5x + 2x - 10 = 0$
3. $x(x - 5) + 2(x - 5) = 0$
4. $(x - 5)(x + 2) = 0$
5. $x = 5 \text{ or } x = -2$

Answer (i): $x = 5, -2$

**Part (ii)**

1. $2x^2 + x - 6 = 0$
2. $2x^2 + 4x - 3x - 6 = 0$
3. $2x(x + 2) - 3(x + 2) = 0$
4. $(2x - 3)(x + 2) = 0$
5. $x = \frac{3}{2} \text{ or } x = -2$

Answer (ii): $x = \frac{3}{2}, -2$

**Part (iii)**

1. $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$
2. $\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0$
3. $x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0$
4. $(x + \sqrt{2})(\sqrt{2}x + 5) = 0$
5. $x = -\sqrt{2} \text{ or } x = -\frac{5}{\sqrt{2}}$

Answer (iii): $x = -\sqrt{2}, -\frac{5}{\sqrt{2}}$

**Part (iv)**

1. $2x^2 - x + \frac{1}{8} = 0$
2. $16x^2 - 8x + 1 = 0$
3. $(4x)^2 - 2(4x)(1) + (1)^2 = 0$
4. $(4x - 1)^2 = 0$
5. $x = \frac{1}{4}, \frac{1}{4}$

Answer (iv): $x = \frac{1}{4}, \frac{1}{4}$

**Part (v)**

1. $100x^2 - 20x + 1 = 0$
2. $(10x)^2 - 2(10x)(1) + (1)^2 = 0$
3. $(10x - 1)^2 = 0$
4. $(10x - 1)(10x - 1) = 0$
5. $x = \frac{1}{10}, \frac{1}{10}$

Answer (v): $x = \frac{1}{10}, \frac{1}{10}$

**Answer:** Roots of the given quadratic equations are found by factorisation.

> Common mistake: Sign errors while splitting the middle term.

### Question 2

*3 marks · Short answer*

Solve the problems given in Example 1.

**Part (i)**

1. The quadratic equation representing the problem is $x^2 - 45x + 324 = 0$.
2. $x^2 - 36x - 9x + 324 = 0$
3. $x(x - 36) - 9(x - 36) = 0$
4. $(x - 36)(x - 9) = 0$
5. Thus, $x = 36$ or $x = 9$. If John had 36 marbles, Jivanti had 9, and vice versa.

Answer (i): John and Jivanti had 36 and 9 marbles (or 9 and 36 marbles).

**Part (ii)**

1. The quadratic equation representing the problem is $x^2 - 55x + 750 = 0$.
2. $x^2 - 25x - 30x + 750 = 0$
3. $x(x - 25) - 30(x - 25) = 0$
4. $(x - 25)(x - 30) = 0$
5. Thus, $x = 25$ or $x = 30$.

Answer (ii): The number of toys produced on that day was 25 or 30.

**Answer:** Solutions to the problems given in Example 1.

> Common mistake: Not verifying if both values obtained satisfy the physical conditions of the problem.

### Question 3

*3 marks · Short answer*

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

**Solution**

1. Let the first number be $x$.
2. Then the second number is $27 - x$.
3. Their product is $x(27 - x) = 182$.
4. $27x - x^2 = 182$, which gives $x^2 - 27x + 182 = 0$.
5. $x^2 - 13x - 14x + 182 = 0$, so $(x - 13)(x - 14) = 0$.
6. Thus, $x = 13$ or $x = 14$.

**Answer:** The two numbers are 13 and 14.

> Common mistake: Errors in finding the factors of 182 that add up to -27.

### Question 4

*3 marks · Short answer*

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

**Solution**

1. Let the first consecutive positive integer be $x$.
2. Then the next consecutive positive integer is $x + 1$.
3. According to the given condition, $x^2 + (x + 1)^2 = 365$.
4. $x^2 + x^2 + 2x + 1 = 365$, which simplifies to $2x^2 + 2x - 364 = 0$ or $x^2 + x - 182 = 0$.
5. $x^2 + 14x - 13x - 182 = 0$, so $(x + 14)(x - 13) = 0$.
6. Since $x$ must be a positive integer, $x = 13$, and $x + 1 = 14$.

**Answer:** The two consecutive positive integers are 13 and 14.

> Common mistake: Ignoring the condition that the integers must be positive.

### Question 5

*3 marks · Short answer*

The altitude of a right triangle is $7\text{ cm}$ less than its base. If the hypotenuse is $13\text{ cm}$, find the other two sides.

**Solution**

1. Let the base of the right triangle be $x\text{ cm}$.
2. Then its altitude is $(x - 7)\text{ cm}$ and hypotenuse is given as $13\text{ cm}$.
3. By Pythagoras theorem, $x^2 + (x - 7)^2 = 13^2$.
4. $x^2 + x^2 - 14x + 49 = 169$, which simplifies to $2x^2 - 14x - 120 = 0$ or $x^2 - 7x - 60 = 0$.
5. $x^2 - 12x + 5x - 60 = 0$, so $(x - 12)(x + 5) = 0$.
6. Since length cannot be negative, $x = 12$, and the altitude is $12 - 7 = 5\text{ cm}$.

**Answer:** The other two sides are $5\text{ cm}$ and $12\text{ cm}$.

> Common mistake: Taking negative value for the base of the triangle.

### Question 6

*3 marks · Short answer*

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 $\text{`} 90$, find the number of articles produced and the cost of each article.

**Solution**

1. Let the number of articles produced on that day be $x$.
2. The cost of production of each article (in rupees) is $2x + 3$.
3. The total cost of production is given as $\text{`} 90$, so $x(2x + 3) = 90$.
4. Expanding and rearranging gives $2x^2 + 3x - 90 = 0$, which can be split as $2x^2 - 12x + 15x - 90 = 0$.
5. Factorising gives $(2x + 15)(x - 6) = 0$, so $x = 6$ (since the number of articles cannot be negative).
6. Thus, the number of articles produced is $6$ and the cost of each article is $2(6) + 3 = \text{`} 15$.

**Answer:** Number of articles produced = 6, Cost of each article = ₹15

> Common mistake: Taking the total cost as $x + 2x + 3$ instead of multiplying the number of articles by the cost per article.

## EXERCISE 4.3

### Question 1

*3 marks · Short answer*

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) $2x^2 - 3x + 5 = 0$
(ii) $3x^2 - 4\sqrt{3}x + 4 = 0$
(iii) $2x^2 - 6x + 3 = 0$

**Part (i)**

1. Here $a = 2, b = -3, c = 5$.
2. Discriminant $b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 - 40 = -31$.
3. Since $b^2 - 4ac < 0$, the given quadratic equation has no real roots.

Answer (i): No real roots

**Part (ii)**

1. Here $a = 3, b = -4\sqrt{3}, c = 4$.
2. Discriminant $b^2 - 4ac = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$.
3. Since $b^2 - 4ac = 0$, the equation has two equal real roots: $x = \frac{-b}{2a} = \frac{4\sqrt{3}}{2(3)} = \frac{2}{\sqrt{3}}$.

Answer (ii): $x = \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}$

**Part (iii)**

1. Here $a = 2, b = -6, c = 3$.
2. Discriminant $b^2 - 4ac = (-6)^2 - 4(2)(3) = 36 - 24 = 12 > 0$.
3. Since $b^2 - 4ac > 0$, the equation has two distinct real roots given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
4. Using the quadratic formula, $x = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}$.

Answer (iii): $x = \frac{3 + \sqrt{3}}{2}, \frac{3 - \sqrt{3}}{2}$

**Answer:** The nature of roots and their values are found for each equation.

> Common mistake: Forgetting to simplify the radical in the numerator or making sign errors while calculating the discriminant.

### Question 2

*3 marks · Short answer*

Find the values of $k$ for each of the following quadratic equations, so that they have two equal roots.
(i) $2x^2 + kx + 3 = 0$
(ii) $kx(x - 2) + 6 = 0$

**Part (i)**

1. The given equation is $2x^2 + kx + 3 = 0$. Here $a = 2, b = k, c = 3$.
2. For two equal roots, the discriminant must be zero, so $b^2 - 4ac = 0$.
3. Substituting the values, $k^2 - 4(2)(3) = 0$, which gives $k^2 = 24$.
4. Therefore, $k = \pm\sqrt{24} = \pm 2\sqrt{6}$.

Answer (i): $k = \pm 2\sqrt{6}$

**Part (ii)**

1. The given equation is $kx(x - 2) + 6 = 0$, which can be rewritten as $kx^2 - 2kx + 6 = 0$.
2. Here $a = k, b = -2k, c = 6$.
3. For two equal roots, $b^2 - 4ac = 0$, so $(-2k)^2 - 4(k)(6) = 0$.
4. This gives $4k^2 - 24k = 0$, or $4k(k - 6) = 0$. Since $k \neq 0$ for a quadratic equation, we get $k = 6$.

Answer (ii): $k = 6$

**Answer:** Values of $k$ are found using the condition $b^2 - 4ac = 0$.

> Common mistake: Forgetting that $k$ cannot be zero in the second part since the coefficient of $x^2$ is $k$.

### Question 3

*3 marks · Short answer*

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $800\text{ m}^2$? If so, find its length and breadth.

**Solution**

1. Let the breadth of the rectangular mango grove be $x\text{ m}$.
2. Then the length of the grove is $2x\text{ m}$.
3. The area of the rectangular grove is given as $800\text{ m}^2$, so $(2x)(x) = 800$.
4. Simplifying, $2x^2 = 800$, which gives $x^2 = 400$, or $x^2 - 400 = 0$.
5. Solving for $x$, we get $x = \pm 20$. Since breadth cannot be negative, $x = 20$.
6. Thus, breadth $= 20\text{ m}$ and length $= 2(20) = 40\text{ m}$.

**Answer:** Yes, it is possible. Length = $40\text{ m}$, breadth = $20\text{ m}$.

> Common mistake: Not discarding the negative value of $x$ when solving $x^2 = 400$.

### Question 4

*3 marks · Short answer*

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**

1. Let the present age of one friend be $x$ years.
2. Then the present age of the other friend is $(20 - x)$ years.
3. Four years ago, their ages were $(x - 4)$ years and $(20 - x - 4) = (16 - x)$ years respectively.
4. According to the question, the product of their ages 4 years ago was 48, so $(x - 4)(16 - x) = 48$.
5. Expanding and simplifying, $16x - x^2 - 64 + 4x = 48$, which gives $-x^2 + 20x - 112 = 0$, or $x^2 - 20x + 112 = 0$.
6. Checking the discriminant, $b^2 - 4ac = (-20)^2 - 4(1)(112) = 400 - 448 = -48 < 0$.
7. Since the discriminant is negative, there are no real roots, meaning this situation is not possible.

**Answer:** Not possible

> Common mistake: Jumping to conclusions without calculating the discriminant to check the existence of real roots.

### Question 5

*3 marks · Short answer*

Is it possible to design a rectangular park of perimeter $80\text{ m}$ and area $400\text{ m}^2$? If so, find its length and breadth.

**Solution**

1. Let the breadth of the rectangular park be $x\text{ m}$.
2. The perimeter of the park is given as $80\text{ m}$, so $2(\text{length} + \text{breadth}) = 80$, which means $\text{length} + \text{breadth} = 40$.
3. Thus, the length of the park is $(40 - x)\text{ m}$.
4. The area of the park is $400\text{ m}^2$, so $x(40 - x) = 400$.
5. Simplifying, $40x - x^2 = 400$, which gives $x^2 - 40x + 400 = 0$.
6. Solving this quadratic equation using factorisation or formula, $(x - 20)^2 = 0$, which gives $x = 20$.
7. Thus, breadth $= 20\text{ m}$ and length $= 40 - 20 = 20\text{ m}$. The park is a square of side $20\text{ m}$.

**Answer:** Yes, it is possible. Length = $20\text{ m}$, breadth = $20\text{ m}$.

> Common mistake: Assuming a rectangle cannot have equal sides (a square is a special case of a rectangle).

## Frequently asked questions

### How many exercises and questions are there in NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations for the 2026-27 session?

This chapter has a total of 3 exercises with 13 short answer questions in the NCERT textbook. Exercise 4.1 has 2 questions, Exercise 4.2 has 6 questions, and Exercise 4.3 has 5 questions. You can find SwaVid's free PDF and step-by-step solutions for all of them on this page.

### What concepts and topics do the questions in this chapter cover?

The questions cover checking and representing quadratic equations, factorisation methods, and using the quadratic formula. Other key topics include the nature of roots, the condition for equal roots, and various word problems related to ages, perimeter, area, and the Pythagoras theorem.

### Which question type is considered the hardest in this chapter and how should I approach it?

Word problems leading to quadratic equations, especially those involving ages, perimeter and area, are often found to be the toughest. To approach them, first define the unknown variable clearly based on the given statement, translate the conditions into a mathematical equation of the form $ax^2 + bx + c = 0$, and then solve it using factorisation or the quadratic formula.

### How should I write my answers to score full marks in board exams?

To secure full marks, always write down the given information and let statements clearly before forming the equation. Show every intermediate calculation step clearly, especially when applying the quadratic formula or factorising. SwaVid's step-by-step solutions on this page demonstrate the exact presentation format recommended for exams.

### Is the free PDF for these NCERT solutions available for download?

Yes, the complete set of solutions for Class 10 Maths Chapter 4 is available as a free PDF on this page. You can easily access and download SwaVid's detailed step-by-step solutions to practice offline and prepare thoroughly for your exams.

## Related pages

- [Exercise 4.1 solutions](https://www.swavid.com/maths/class/10/chapter/quadratic-equations/ncert-solutions/exercise-4-1)
- [Exercise 4.2 solutions](https://www.swavid.com/maths/class/10/chapter/quadratic-equations/ncert-solutions/exercise-4-2)
- [Exercise 4.3 solutions](https://www.swavid.com/maths/class/10/chapter/quadratic-equations/ncert-solutions/exercise-4-3)
- [Quadratic Equations: CBSE previous year questions](https://www.swavid.com/cbse/class-10/maths/pyq/quadratic-equations)
- [Class 10 Maths chapters](https://www.swavid.com/maths/class/10)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student