NCERT Solutions Class 10 Maths Chapter 4 for Exercise 4.2 - Quadratic Equations

NCERT Solutions Class 10 Maths Chapter 4 for Exercise 4.2 - Quadratic Equations

Updated on 02 Jun 2025, 02:04 PM IST

Quadratic equations are a polynomial equation which have the highest degree 2. Quadratic equations are used to model many scenarios, such as calculating the time or distance of a flight, optimising project cost or profit, etc. These scenarios or word problems are converted into quadratic equations. For solving these quadratic equations, there are three main methods: Factorising, Completing the square, and the Quadratic Formula. The factorising method is one of the above mentioned methods, and in this method, the middle term is split to determine the roots of the equation.

This Story also Contains

  1. Download Free PDF of NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2
  2. Assess NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2
  3. Topics Covered in Chapter 4, Quadratic Equation: Exercise 4.2
  4. NCERT Solutions Subject Wise
  5. Subject-Wise NCERT Exemplar Solutions
NCERT Solutions Class 10 Maths Chapter 4 for Exercise 4.2 - Quadratic Equations
NCERT Solutions for Exercise 4.2 Class 10 Maths Chapter 4 - Quadratic Equations

Class 10 NCERT Book maths exercise 4.2 comprises 6 straightforward questions with sub-questions. All these questions are based on the factorising method. This exercise 4.2 of class 10th also helps to determine the relation between roots of the equation and the nature of the equation. The NCERT solutions to these questions give a better understanding of solving the quadratic equation using the factorising method.

Download Free PDF of NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2

Assess NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2

Q1 (i) Find the roots of the following quadratic equations by factorisation: $x^2 - 3x - 10 =0$

Answer:

Given the quadratic equation: $x^2 - 3x - 10 =0$

Factorization gives, $x^2 - 5x+2x - 10 =0$

$\Rightarrow x^2 - 5x+2x - 10 =0$

$\Rightarrow x(x-5) +2(x-5) =0$

$\Rightarrow (x-5)(x+2) =0$

$\Rightarrow x= 5\ or\ -2$

Hence, the roots of the given quadratic equation are $5\ and\ -2$.

Q1 (ii) Find the roots of the following quadratic equations by factorisation: $2x^2 + x - 6 = 0$

Answer:

Given the quadratic equation: $2x^2 + x - 6 = 0$

Factorisation gives, $2x^2 +4x-3x - 6 = 0$

$\Rightarrow 2x(x+2) -3(x+2) =0$

$\Rightarrow (x+2)(2x-3) = 0$

$\Rightarrow x= -2\ or\ \frac{3}{2}$

Hence, the roots of the given quadratic equation are

$-2\ and\ \frac{3}{2}$

Q1 (iii) Find the roots of the following quadratic equations by factorisation:$\sqrt2x^2 + 7x + 5\sqrt2 = 0$

Answer:

Given the quadratic equation: $\sqrt2x^2 + 7x + 5\sqrt2 = 0$

Factorization gives, $\sqrt2x^2 + 5x+2x + 5\sqrt2 = 0$

$\Rightarrow x(\sqrt2 x +5) +\sqrt2 (\sqrt 2 x +5)= 0$

$\Rightarrow (\sqrt2 x +5)(x+\sqrt{2}) = 0$

$\Rightarrow x=\frac{-5}{\sqrt 2 }\ or\ -\sqrt 2$

Hence, the roots of the given quadratic equation are

$\frac{-5}{\sqrt 2 }\ and\ -\sqrt 2$

Q1 (iv) Find the roots of the following quadratic equations by factorisation:$2x^2 -x + \frac{1}{8} = 0$

Answer:

Given the quadratic equation: $2x^2 -x + \frac{1}{8} = 0$

Solving the quadratic equations, we get

$16x^2-8x+1 = 0$

Factorization gives, $\Rightarrow 16x^2-4x-4x+1 = 0$

$\Rightarrow 4x(4x-1)-1(4x-1) = 0$

$\Rightarrow (4x-1)(4x-1) = 0$

$\Rightarrow x=\frac{1}{4}\ or\ \frac{1}{4}$

Hence, the roots of the given quadratic equation are

$\frac{1}{4}\ and\ \frac{1}{4}$

Q1 (v) Find the roots of the following quadratic equations by factorisation: $100x^2 -20x +1 = 0$

Answer:

Given the quadratic equation: $100x^2 -20x +1 = 0$

Factorization gives, $100x^2 -10x-10x +1 = 0$

$\Rightarrow 10x(10x-1)-10(10x-1) = 0$

$\Rightarrow (10x-1)(10x-1) = 0$

$\Rightarrow x=\frac{1}{10}\ or\ \frac{1}{10}$

Hence, the roots of the given quadratic equation are

$\frac{1}{10}\ and\ \frac{1}{10}$ .

Q2 Solve the problems given in Example 1. (i) $x^2-45x+324 = 0$ (ii) $x^2-55x+750 = 0$

Answer:

From Example 1, we get:

Equations:

(i) $x^2-45x+324 = 0$

Solving by the factorisation method:

Given the quadratic equation: $x^2-45x+324 = 0$

Factorization gives, $x^2-36x-9x+324 = 0$

$\Rightarrow x(x-36) - 9(x-36) = 0$

$\Rightarrow (x-9)(x-36) = 0$

$\Rightarrow x=9\ or\ 36$

Hence, the roots of the given quadratic equation are $x=9\ and \ 36$.

Therefore, John and Jivanti have 36 and 9 marbles, respectively, in the beginning.

(ii) $x^2-55x+750 = 0$

Solving by the factorisation method:

Given the quadratic equation: $x^2-55x+750 = 0$

Factorization gives, $x^2-30x-25x+750 = 0$

$\Rightarrow x(x-30) -25(x-30) = 0$

$\Rightarrow (x-25)(x-30) = 0$

$\Rightarrow x=25\ or\ 30$

Hence, the roots of the given quadratic equation are $x=25\ and \ 30$.

Therefore, the number of toys on that day was $30\ or\ 25.$

Q3 Find two numbers whose sum is 27 and the product is 182.

Answer:

Let two numbers be x and y .

Then, their sum will be equal to 27, and the product equals 182.

$x+y = 27$ ..........(1)

$xy =182$ ...........(2)

From equation (2) we have:

$y = \frac{182}{x}$

Then, putting the value of y in equation (1), we get

$x+\frac{182}{x} = 27$

Solving this equation:

$\Rightarrow x^2-27x+182 = 0$

$\Rightarrow x^2-13x-14x+182 = 0$

$\Rightarrow x(x-13)-14(x-13) = 0$

$\Rightarrow (x-14)(x-13) = 0$

$\Rightarrow x = 13\ or\ 14$

Hence, the two required numbers are $13\ and \ 14$.

Q4 Find two consecutive positive integers, the sum of whose squares is 365.

Answer:

Let the two consecutive integers be $'x'\ and\ 'x+1'.$

Then the sum of the squares is 365.

. $x^2+ (x+1)^2 = 365$

$\Rightarrow x^2+x^2+1+2x = 365$

$\Rightarrow x^2+x-182 = 0$

$\Rightarrow x^2 - 13x+14x+182 = 0$

$\Rightarrow x(x-13)+14(x-13) = 0$

$\Rightarrow (x-13)(x-14) = 0$

$\Rightarrow x =13\ or\ 14$

Hence, the two consecutive integers are $13\ and\ 14$.

Q5 The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Answer:

Let the length of the base of the triangle be $b\ cm$.

Then, the altitude length will be: $b-7\ cm$.

Given if hypotenuse is $13\ cm$ .

Applying the Pythagoras theorem, we get

$Hypotenuse^2 = Perpendicular^2 + Base^2$

So, $(13)^2 = (b-7)^2 +b^2$

$\Rightarrow 169 = 2b^2+49-14b$

$\Rightarrow 2b^2-14b-120 = 0$ Or $b^2-7b-60 = 0$

$\Rightarrow b^2-12b+5b-60 = 0$

$\Rightarrow b(b-12) + 5(b-12) = 0$

$\Rightarrow (b-12)(b+5) = 0$

$\Rightarrow b= 12\ or\ -5$

But the length of the base cannot be negative.

Hence, the base length will be $12\ cm$.

Therefore, we have

Altitude length $= 12cm -7cm = 5cm$ and Base length $= 12\ cm$

Q6 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.

Answer:

Let the number of articles produced in a day $= x$

The cost of production of each article will be $=2x+3$

Given that the total production on that day was $Rs.90$.

Hence, we have the equation;

$x(2x+3) = 90$

$2x^2+3x-90 = 0$

$\Rightarrow 2x^2+15x-12x-90 = 0$

$\Rightarrow x(2x+15) - 6(2x+15) = 0$

$\Rightarrow (2x+15)(x-6) = 0$

$\Rightarrow x =-\frac{15}{2}\ or\ 6$

But, x cannot be negative as it is the number of articles.

Therefore, $x=6$ and the cost of each article $= 2x+3 = 2(6)+3 = 15$

Hence, the number of articles is 6 and the cost of each article is Rs 15.

Also Read-

Topics Covered in Chapter 4, Quadratic Equation: Exercise 4.2

  1. Factorisation: In this method, the quadratic equation is broken down into the product of two linear factors by splitting the middle term of the quadratic equation.
  2. Roots and Zeros: This exercise defines the relation between the roots of the quadratic equation and with zeros of the corresponding polynomial.
  3. Standard Form of a Quadratic Equation: The standard form of a quadratic equation, which is ax² + bx + c = 0, where 'a is not equal to 0, and this exercise deals with the standard form of the quadratic equation.
  4. Problem Solving: In this exercise, the word problem is converted into a quadratic equation, and the equation is solved by applying the factorisation method.

Also see-

Subject-Wise NCERT Exemplar Solutions

Students must check the NCERT exemplar solutions for Class 10 Maths and Science given below:

Frequently Asked Questions (FAQs)

Q: What is the general form of the quadratic equation?
A:

The common shape of the quadratic condition is ax62+bx+c=0 where a, b, c are real numbers. 

Q: Whether the roots of the condition and the zeroes of the condition are the same ?
A:

Yes, the roots of the equation and the zeroes of the equation are the same. 

Q: Factorize the polynomial a^2-10a+24 .
A:

a^2-10a+24=a2-4a-6a+24 

=a(a-4)-6(a-4) 

=(a-4)(a-6)  

Q: What is the vital condition for understanding the quadratic conditions by utilizing the strategy of factorization?
A:

In the strategy of factorization, the product of 1st and final terms of a given condition must be broken even with the product of 2nd and 3rd terms of the same given condition. 

Q: What is the splitting of the middle term?
A:

Splitting of the middle term is nothing but we have to rewrite the middle term of the quadratic expression as the sum or difference of the two terms, that is we have to split the middle term into two parts in terms of sum or difference of the terms. 

Q: What is the sum product form according to NCERT solutions for Class 10 Maths chapter 4 exercise 4.2 ?
A:

The sum-product form is nothing but in the equation ax^2+bx+c=0 , the product of the middle term after splitting must be equal to a×c and the sum must be equal to b.

Q: How numerous questions and what sort of questions are secured inside the NCERT solutions for Class 10 Maths chapter 4 exercise 4.2 ?
A:

NCERT solutions for Class 10 Maths chapter 4 exercise 4.2 comprises of six questions which are based on the factorization strategy.

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