NCERT Solutions Class 10 Maths Chapter 4  Quadratic Equations Exercise 4.2

Class 10 NCERT Solutions Maths Chapter 4 Quadratic Equations Exercise 4.2 will help students grasp the fundamentals of quadratic equations. The exercise will reiterate the method of factorisation, which you may have studied in previous classes, for solving these quadratic equations. The method is the most basic and widely used technique for tackling these questions.

Here, you will be able to find some important concepts of this method in brief, along with the NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2 PDF.

1.0Download NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2: Free PDF

2.0Introduction to Quadratic Equations: 

Before getting into the realm of the method of factorisation, let’s quickly recall the definition of a quadratic equation itself. A mathematical equation with the highest degree of its variable, mostly 2, is known as a quadratic equation, hence named “quadratic”, which basically means two. A quadratic equation must always be written in its standard form, that is: 

$a{x}^2+bx+c=0$

Here, a and b are the coefficients of a single variable x, while c is the constant where a0. Quadratic equations are the fundamental concept of algebra in mathematics, with a variety of real-life applications in other fields, too.

3.0Key Concepts of Exercise 4.2: Overview 

Exercise 4.2 includes questions related to the concepts of factorisation of quadratic equations to calculate the roots of such equations. Let’s understand some key concepts that will help you solve the exercise with ease: 

Understanding Roots of Quadratic Equations:

Roots of quadratic equations, also known as zeroes, are the values of the single variable present in these equations, making the whole equation equal to 0. In general, a quadratic equation contains two roots. However, the roots of quadratic equations can be classified into different types that are: 

• Two different real roots
• Two equal real roots
• No roots or complex roots

To further understand the zeroes of the quadratic equations, consider an equation $a{x}^2+bx+c=0$ put x=$\alpha$ , such that “$\alpha$” is a real number. If, after the substitution, the value comes out to be 0, then “$\alpha$” is the root of the given equation. 

Solution of a Quadratic Equation by Factorisation: 

The method of factorisation involves factoring a quadratic polynomial into the product of two binomials and then solving for the values of x that make the equation zero. The method can be used by following the below-mentioned approach: 

• State with Standard form: Before starting the solution procedure, write the equation in its standard form, that is $a{x}^2+bx+c=0$
• Multiply: Multiply the coefficient of x² that is a with the constant term that is c to get the product ac. 
• Split the middle term: Now split the middle term ‘b’ into two numbers so that after adding or subtracting, they come out to be equal to it. Also, after multiplying these two numbers, they equal to ac. Understand it like this: consider the two numbers to be m and n, then it must be: 

$m\pm n=b$

• mn=ac
• Group and factor the term: After splitting the middle term, group the terms in pairs with a common factor in them. Factor out this common term from each group to form two binomials. In these two binomials, one will be common in both; factor out that one binomial. 
• Solve for x: Once we have the product of two factored binomials, find the value of x. This can be done by keeping them equal to zero and then solving them using proper mathematical operations. 

4.0Exercise 4.2: Overview 

• Exercise 4.2 involves various numerical questions of quadratic equations to be solved with factorisation. 
• The exercise also includes different types of real-life word problems. These questions require forming the equation according to the question and then solving for the value of the unknown variable. 

Mastering the factorisation technique will help you solve quadratic equations with ease, and to further develop your knowledge, visit the NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2 for further practice and learning.

5.0NCERT Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2 : Detailed Solutions

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

(i) x² - 3x - 10 = 0

(ii) 2x² + x - 6 = 0

(iii) √2x² + 7x + 5√2 = 0

(iv) 2x² - x + 1/8 = 0

(v) 100x² - 20x + 1 = 0

Solution:

(i) x² - 3x - 10 = 0

=> x² - 5x + 2x - 10 = 0

=> x(x - 5) + 2(x - 5) = 0

=> (x + 2)(x - 5) = 0

=> x + 2 = 0 or x - 5 = 0

=> x = -2 or x = 5

Hence, the roots are -2 and 5.

(ii) 2x² + x - 6 = 0

=> 2x² + 4x - 3x - 6 = 0

=> 2x(x + 2) - 3(x + 2) = 0

=> (x + 2)(2x - 3) = 0

=> x + 2 = 0 or 2x - 3 = 0

=> x = -2 or x = 3/2

(v) 100x² - 20x + 1 = 0

=> (10x)² - 2 × 10x × 1 + 1² = 0

=> (10x - 1)² = 0

=> Both roots are given by 10x - 1 = 0, i.e. x = 1/10.

Hence, the roots are 1/10, 1/10.

2. Solve the following problems :

(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.

Solution:

(i) Let the number of marbles John had be x. Then the number of marbles Jivanti had =45−x

The number of marbles left with John, when he lost 5 marbles =x−5 The number of marbles left with Jivanti, when she lost 5 marbles =45−x−5=40−x

Therefore, their product =(x−5)(40−x)

=40x−x2−200+5x

=−x2+45x−200

So, −x2+45x−200=124

(Given that product =124 )

⇒−x2+45x−324=0

⇒x2−45x+324=0

⇒x2−36x−9x+324=0

⇒x(x−36)−9(x−36)=0

⇒(x−9)(x−36)=0

⇒x−9=0 or x−36=0

⇒x=9 or x=36

⇒45−x=36 or 45−x=9

Therefore, the number of marbles John and Jivanti had 9 and 36 respectively or 36 and 9 respectively.

(ii) Let the number of toys produced be x .

∴ Cost of production of each toy =₹(55−x) It is given that, total cost of production of the toys =₹750

∴x(55−x)=750

⇒x2−55x+750=0

⇒x2−30x−25x+750=0

⇒x(x−30)−25(x−30)=0

⇒(x−25)(x−30)=0

⇒x−25=0 or x−30=0

⇒x=25 or x=30

⇒55−x=30 or 55−x=25

Therefore, the number of toys produced are either 25 or 30 .

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

Solution:

Let one number be x , then second number

= 27−x

Given that x×(27−x)=182

⇒27x−x2=182

⇒x2−27x+182=0

⇒x2−14x−13x+182=0

⇒x(x−14)−13(x−14)=0

⇒(x−13)(x−14)=0

Either x−13=0 or x−14=0,

i.e. x=13 or 14

⇒27−x=14 or 13

Hence, the two marbles are 13 and 14.

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

Solution:

Let the consecutive positive integers be x and (x+1).

Given that x² + (x+1)² = 365

=> x² + x² + 1 + 2x = 365

=> 2x² + 2x - 364 = 0

=> x² + x - 182 = 0

=> x² + 14x - 13x - 182 = 0

=> x(x+14) - 13(x+14) = 0

=> (x+14)(x-13) = 0

Either x+14 = 0 or x-13 = 0,

i.e. x = -14 or x = 13

Since the integers are positive, x can only be 13.

Therefore, x+1 = 13+1 = 14

Therefore, two consecutive positive integers will be 13 and 14.

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:

In triangle ABC, base BC = x cm, altitude AC = (x - 7) cm, angle ACB = 90 degrees, and AB = 13 cm.

By Pythagoras theorem, we have:

BC² + AC² = AB²

=> x² + (x - 7)² = 13²

=> x² + x² - 14x + 49 = 169

=> 2x² - 14x - 120 = 0

=> x² - 7x - 60 = 0

=> x² - 12x + 5x - 60 = 0

=> x(x - 12) + 5(x - 12) = 0

=> (x + 5)(x - 12) = 0

i.e., x = -5 or x = 12

We reject x = -5, because length cannot be negative.

Therefore, x = 12.

Therefore, BC = 12 cm and AC = 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 ₹ 90, find the number of articles produced and the cost of each article.

Solution:

Let the number of articles produced be x.

Therefore, cost of production of each article = ₹(2x + 3)

It is given that the total cost of production is ₹ 90.

Therefore, x(2x + 3) = 90

=> 2x² + 3x - 90 = 0

=> 2x² + 15x - 12x - 90 = 0

=> x(2x + 15) - 6(2x + 15) = 0

=> (2x + 15)(x - 6) = 0

Either 2x + 15 = 0 or x - 6 = 0,

i.e., x = -15/2 or x = 6

As the number of articles produced can only be a positive integer, therefore, x can only be 6.

Hence, number of articles produced = 6

Cost of each article = 2 × 6 + 3 = ₹ 15.

6.0Benefits of studying NCERT Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2

1. Helps solve common board exam questions confidently and accurately, especially when factoring isn’t possible.
2. Strengthens step-by-step logical thinking and accuracy in applying mathematical formulas.
3. Builds essential knowledge for higher-level algebra and real-life applications in physics and economics.