Quadratic Equations Class 10 Exercise 4.2 NCERT Solutions PDF
Neha Tanna23 Nov, 2025
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NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2: Chapter 4 of Class 10 Maths, Quadratic Equations, introduces students to various methods of solving quadratic equations. NCERT Solutions class 10 maths Exercise 4.2 focuses on the factorisation method, where equations are expressed in factored form to find their roots. This approach is simple, effective, and strengthens the algebraic skills students have learned in earlier chapters.
Exercise 4.2 provides clear explanations and ample practice to help students understand how coefficients relate to factors. Mastering this exercise is crucial for building strong problem-solving skills and preparing learners for advanced methods like the quadratic formula and completing the square. These solutions are especially helpful for students searching for quadratic equation class 10 exercise 4.2.
Understanding Chapter 4- Quadratic Equations Through Factorisation
Exercise 4.2 of Class 10 Maths Chapter 4 focuses on solving quadratic equations using the factorisation method, helping students break equations into linear factors to find their roots easily.NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2 Quadratic Equations
Below is the NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2 Quadratic Equations -1. Find the roots of the following quadratic equations by factorisation:
(i) x 2 – 3x – 10 = 0
(ii) 2x 2 + x – 6 = 0
(iii) √2 x 2 + 7x + 5√2 = 0
(iv) 2x 2 – x +1/8 = 0
(v) 100x 2 – 20x + 1 = 0
Solutions:
(i) Given, x 2 – 3 x – 10 =0 Taking LHS, => x 2 – 5 x + 2 x – 10 => x ( x – 5) + 2( x – 5) =>( x – 5)( x + 2) The roots of this equation, x 2 – 3 x – 10 = 0 are the values of x for which ( x – 5)( x + 2) = 0 Therefore, x – 5 = 0 or x + 2 = 0 => x = 5 or x = -2(ii) Given, 2 x 2 + x – 6 = 0 Taking LHS, => 2 x 2 + 4 x – 3 x – 6 => 2 x ( x + 2) – 3( x + 2) => ( x + 2)(2 x – 3) The roots of this equation, 2 x 2 + x – 6=0 are the values of x for which ( x + 2)(2 x – 3) = 0 Therefore, x + 2 = 0 or 2 x – 3 = 0 => x = -2 or x = 3/2
(iii) √2 x 2 + 7 x + 5√2=0 Taking LHS, => √2 x 2 + 5 x + 2 x + 5√2 => x (√2 x + 5) + √2(√2 x + 5)= (√2 x + 5)( x + √2) The roots of this equation, √2 x 2 + 7 x + 5√2=0 are the values of x for which (√2 x + 5)( x + √2) = 0 Therefore, √2 x + 5 = 0 or x + √2 = 0 => x = -5/√2 or x = -√2
(iv) 2 x 2 – x +1/8 = 0 Taking LHS, =1/8 (16 x 2 – 8 x + 1) = 1/8 (16 x 2 – 4 x -4 x + 1) = 1/8 (4 x (4 x – 1) -1(4 x – 1)) = 1/8 (4 x – 1) 2 The roots of this equation, 2 x 2 – x + 1/8 = 0, are the values of x for which (4 x – 1) 2 = 0 Therefore, (4 x – 1) = 0 or (4 x – 1) = 0 ⇒ x = 1/4 or x = 1/4
2. Solve the problems given in Example 1.
Represent the following situations mathematically:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was 750. We would like to find out the number of toys produced on that day.
Solutions:
(i) Let us say the number of marbles John has = x. Therefore, the number of marbles Jivanti has = 45 – x.Therefore, If, John’s marbles = 36, Then, Jivanti’s marbles = 45 – 36 = 9 And if John’s marbles = 9, Then, Jivanti’s marbles = 45 – 9 = 36
3. Find two numbers whose sum is 27 and product is 182.
Solution:
Let us say, first number be x and the second number is 27–x.Hence, the numbers are 13 and 14.
4. Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let us say the two consecutive positive integers be x and x + 1.5. 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 us say the base of the right triangle is x cm.Given, the altitude of right triangle = (x – 7) cm From Pythagoras theorem, we know, Base 2 + Altitude 2 = Hypotenuse 2 ∴ x 2 + (x – 7) 2 = 13 2 ⇒ x 2 + x 2 + 49 – 14x = 169 ⇒ 2x 2 – 14x – 120 = 0 ⇒ x 2 – 7x – 60 = 0 ⇒ x 2 – 12x + 5x – 60 = 0 ⇒ x(x – 12) + 5(x – 12) = 0 ⇒ (x – 12)(x + 5) = 0
Thus, either x – 12 = 0 or x + 5 = 0, ⇒ x = 12 or x = – 5
Since sides cannot be negative, x can only be 12. Therefore, the base of the given triangle is 12 cm, and the altitude of this triangle will be (12 – 7) cm = 5 cm.
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 Rs. 90, find the number of articles produced and the cost of each article.
Solution:
Let us say the number of articles produced is x.NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2 PDF
Chapter 4 of Class 10 Maths explains how to solve quadratic equations, and Exercise 4.2 focuses on the factorisation method to find roots easily. We’ve provided a clear, step-by-step solution PDF to help students understand the method better. These solutions support quadratic equation class 10 exercise 4.2 preparation and follow the CBSE Class 10 Maths syllabus.
NCERT solutions Class 10 Maths PDF
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