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Quadratic Equations Class 10 Exercise 4.1 NCERT Solutions PDF

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 offer clear, step-by-step explanations to help students understand and identify quadratic equations. These solutions are ideal for quick revision and effective exam preparation.
authorImageNeha Tanna23 Nov, 2025
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NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1: Chapter 4 of Class 10 Maths introduces students to Quadratic Equations, and Exercise 4.1 helps them identify equations in the standard quadratic form. Through simplifying and analysing expressions, learners understand whether an equation is truly quadratic.

The NCERT solutions make concepts easy with step-by-step explanations. Regular practice of the problems builds a strong base for solving quadratic equations in later exercises.

Class 10 Maths Chapter 4 Exercise 4.1 Quadratic Equations Solutions

Below are the NCERT solutions, for Class 10 Maths Chapter 4 Exercise 4.1 – Quadratic Equations. These solutions help students clearly identify and understand quadratic expressions, forming a strong base for the chapter. They are especially useful for learners searching for quadratic equation class 10 exercise 4.1.

1. 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

Solutions:

(i) Given, (x + 1) 2 = 2(x – 3) By using the formula for (a+b) 2 = a 2 +2ab+b 2 ⇒ x 2 + 2x + 1 = 2x – 6 ⇒ x 2 + 7 = 0 The above equation is in the form of ax  2 + bx + c = 0. Therefore, the given equation is a quadratic equation.

(ii) Given, x  2 – 2x = (–2) (3 – x) ⇒ x 2 – 2x = -6 + 2x ⇒ x 2 – 4x + 6 = 0 The above equation is in the form of ax 2 + bx + c = 0. Therefore, the given equation is a quadratic equation.

(iii) Given, (x – 2)(x + 1) = (x – 1)(x + 3) By multiplication ⇒ x2 – x – 2 = x 2 + 2x – 3 ⇒ 3x – 1 = 0 The above equation is not in the form of ax 2 + bx + c = 0. Therefore, the given equation is not a quadratic equation.

(iv) Given, (x – 3)(2x +1) = x(x + 5) By multiplication ⇒ 2x2 – 5x – 3 = x 2 + 5x ⇒  x 2 – 10x – 3 = 0 The above equation is in the form of ax 2 + bx + c = 0. Therefore, the given equation is a quadratic equation.
 
(v) Given, (2x – 1)(x – 3) = (x + 5)(x – 1) By multiplication ⇒ 2x2 – 7x + 3 = x 2 + 4x – 5 ⇒ x 2 – 11x + 8 = 0 The above equation is in the form of ax 2 + bx + c = 0. Therefore, the given equation is a quadratic equation.

(vi) Given, x  2 + 3x + 1 = (x – 2) 2 By using the formula for (a-b) 2 =a 2 -2ab+b 2 ⇒ x 2 + 3x + 1 = x 2 + 4 – 4x ⇒ 7x – 3 = 0 The above equation is not in the form of ax 2 + bx + c = 0. Therefore, the given equation is not a quadratic equation.

(vii) Given, (x + 2)  3 = 2x(x 2 – 1) By using the formula for (a+b) 3 = a 3 +b 3 +3ab(a+b) ⇒ x 3 + 8 + x 2 + 12x = 2x 3 – 2x ⇒ x 3 + 14x – 6x 2 – 8 = 0 The above equation is not in the form of ax 2 + bx + c = 0. Therefore, the given equation is not a quadratic equation.

(viii) Given, x 3 – 4x 2 – x + 1 = (x – 2) 3 By using the formula for (a-b) 3 = a 3 -b 3 -3ab(a-b) ⇒  x 3 – 4x 2 – x + 1 = x 3 – 8 – 6x 2 + 12x ⇒ 2x 2 – 13x + 9 = 0 The above equation is in the form of ax 2 + bx + c = 0. Therefore, the given equation is a quadratic equation.


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

(i) The area of a rectangular plot is 528 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 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken

Solutions:

(i) Let us consider, The breadth of the rectangular plot = x m Thus, the length of the plot = (2x + 1) m
As we know, Area of rectangle = length × breadth = 528 m  2 Putting the value of the length and breadth of the plot in the formula, we get (2x + 1) × x = 528 ⇒ 2x 2 + x =528 ⇒ 2x 2 + x – 528 = 0
Therefore, the length and breadth of the plot satisfy the quadratic equation, 2x  2 + x – 528 = 0, which is the required representation of the problem mathematically.
 
(ii) Let us consider, The first integer number = x Thus, the next consecutive positive integer will be = x + 1 Product of two consecutive integers = x × (x +1) = 306 ⇒ x 2 + x = 306 ⇒ x 2 + x – 306 = 0 
Therefore, the two integers, x and x+1, satisfy the quadratic equation, x2 + x – 306 = 0, which is the required representation of the problem mathematically.

(iii) Let us consider, Age of Rohan’s = x  years Therefore, as per the given question, Rohan’s mother’s age = x + 26 After 3 years, Age of Rohan’s = x + 3 Age of Rohan’s mother will be = x + 26 + 3 = x + 29
The product of their ages after 3 years will be equal to 360, such that (x + 3)(x + 29) = 360 ⇒ x  2 + 29x + 3x + 87 = 360 ⇒ x 2 + 32x + 87 – 360 = 0 ⇒ x 2 + 32x – 273 = 0

Therefore, the age of Rohan and his mother satisfies the quadratic equation, x  2 + 32x – 273 = 0, which is the required representation of the problem mathematically.

(iv) Let us consider the speed of the train =  x km/h. And the Time taken to travel 480 km = 480/x km/hr. As per second condition, the speed of train = ( x – 8) km/h Also given, the train will take 3 hours to cover the same distance. 

Therefore, time taken to travel 480 km = (480/x)+3 km/h As we know, Speed × Time = Distance Therefore, (  x – 8)(480/ x) + 3 = 480 ⇒ 480 + 3 x – 3840/ x – 24 = 480 ⇒ 3 x – 3840/ x = 24 ⇒ x 2 – 8 x – 1280 = 0

Therefore, the speed of the train satisfies the quadratic equation, x 2 – 8 x – 1280 = 0, which is the required representation of the problem mathematically.

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 PDF

Chapter 4 of Class 10 Maths, Quadratic Equations, introduces students to identifying and verifying quadratic expressions. Exercise 4.1 lays the foundation for understanding how quadratic equations are formed. To support easy learning, we’ve provided a detailed solution PDF with clear, step-by-step explanations. Check the Link given below; 

NCERT solutions Class 10 Maths PDF

Check More Related Chapters

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 FAQs

What does Exercise 4.1 of Class 10 Quadratic Equations cover?

Exercise 4.1 explains how to identify quadratic equations and check whether a given equation is quadratic or not.

Why is Exercise 4.1 important for Class 10 students?

It builds the foundation for solving quadratic equations using different methods in later exercises.

Are the NCERT Solutions for Exercise 4.1 helpful for board exam preparation?

Yes, these solutions help students understand concepts clearly and solve similar questions easily in the exam.

What types of questions are included in Exercise 4.1?

The exercise mainly includes problems where students determine if an equation is quadratic and rewrite equations in standard form.
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