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

NCERT Solutions Class 10 Maths Chapter 4- Quadratic Equations Exercise 4.2 contains the solutions to all the questions provided in the textbook on page number 76. The questions are provided with stepwise solutions for reference. The questions are solved by subject experts and therefore the answers provided are accurate.

NCERT Solutions Class 10 Maths has many such exercises along with the solutions on various topics provided in the textbook. The students can go through these for more practice. These might help the students score well in the examinations.

The solutions on this are provided in the form of a PDF. NCERT Solutions Class 10 Maths Chapter 4- Quadratic Equation is an important chapter from the examination perspective. The students should deal with this chapter thoroughly to score well.

NCERT Solutions are anyway the best guide for the examinations that carries all the solutions to the queries of the students. The students can refer these for further practice. These not only enhance their conceptual knowledge but also helps them to improve on their imperfections.

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Access Other Exercise Solutions of Class 10 Maths Chapter 4- Quadratic Equations

Exercise 4.1 Solutions– 2 Questions

Exercise 4.3 Solutions– 11 Questions

Exercise 4.4 Solutions– 5 Questions

Access Answers for Class 10 Maths NCERT Solutions for Chapter 4- Quadratic Equation Exercise 4.2

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

(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) √2 x2 + 7x + 5√2 = 0
(iv) 2x2 – x +1/8 = 0
(v) 100x2 – 20x + 1 = 0

Solutions:

(i) Given, x2 – 3x – 10 =0

Taking LHS,

=>x2 – 5x + 2x – 10

=>x(– 5) + 2(x – 5)

=>(x – 5)(x + 2)

The roots of this equation, x2 – 3x – 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, 2x2 + x – 6 = 0

Taking LHS,

=> 2x2 + 4x – 3x – 6

=> 2x(x + 2) – 3(x + 2)

=> (x + 2)(2x – 3)

The roots of this equation, 2x2 + x – 6=0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, x + 2 = 0 or 2x – 3 = 0

=> x = -2 or x = 3/2

(iii) √2 x2 + 7x + 5√2=0

Taking LHS,

=> √2 x+ 5x + 2x + 5√2 

=> x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(+ √2)

The roots of this equation, √2 x2 + 7x + 5√2=0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, √2x + 5 = 0 or x + √2 = 0

=> x = -5/√2 or x = -√2

(iv) 2x2 – x +1/8 = 0

Taking LHS,

=1/8 (16x2  – 8x + 1)

= 1/8 (16x2  – 4x -4x + 1)

= 1/8 (4x(4x  – 1) -1(4x – 1))

= 1/8 (4– 1)2

The roots of this equation, 2x2 – x + 1/8 = 0, are the values of x for which (4– 1)2= 0

Therefore, (4x – 1) = 0 or (4x – 1) = 0

⇒ x = 1/4 or x = 1/4

(v) Given, 100x2 – 20x + 1=0

Taking LHS,

= 100x2 – 10x – 10x + 1

= 10x(10x – 1) -1(10x – 1)

= (10x – 1)2

The roots of this equation, 100x2 – 20x + 1=0, are the values of x for which (10x – 1)2= 0

∴ (10x – 1) = 0 or (10x – 1) = 0

⇒x = 1/10 or x = 1/10  

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 have = x.

Therefore, number of marble Jivanti have = 45 – x

After losing 5 marbles each,

Number of marbles John have = x – 5

Number of marble Jivanti have = 45 – x – 5 = 40 – x

Given that the product of their marbles is 124.

∴ (– 5)(40 – x) = 124

⇒ x2 – 45x + 324 = 0

⇒ x2 – 36x – 9x + 324 = 0

⇒ x(x – 36) -9(x – 36) = 0

⇒ (x – 36)(x – 9) = 0

Thus, we can say,

x – 36 = 0 or x – 9 = 0

⇒ x = 36 or x = 9

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

(ii) Let us say, number of toys produced in a day be x.

Therefore, cost of production of each toy = Rs(55 – x)

Given, total cost of production of the toys = Rs 750

∴ x(55 – x) = 750

⇒ x2 – 55x + 750 = 0

⇒ x2 – 25x – 30x + 750 = 0

 x(x – 25) -30(x – 25) = 0

⇒ (x – 25)(x – 30) = 0

Thus, either x -25 = 0 or x – 30 = 0

⇒ x = 25 or x = 30

Hence, the number of toys produced in a day, will be either 25 or 30.

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.

Therefore, the product of two numbers

x(27 – x) = 182

⇒ x2 – 27x – 182 = 0

⇒ x2 – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x -14) = 0

Thus, either, x = -13 = 0 or x – 14 = 0

⇒ x = 13 or x = 14

Therefore, if first number = 13, then second number = 27 – 13 = 14

And if first number = 14, then second number = 27 – 14 = 13

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.

Therefore, as per the given questions,

x2 + (x + 1)2 = 365

⇒ xx+ 1 + 2x = 365

⇒ 2x2 + 2x – 364 = 0

⇒ x– 182 = 0

⇒ x+ 14x – 13x – 182 = 0

⇒ x(x + 14) -13(x + 14) = 0

⇒ (x + 14)(x – 13) = 0

Thus, either, x + 14 = 0 or x – 13 = 0,

⇒ x = – 14 or x = 13

since, the integers are positive, so x can be 13, only.

∴ 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:

Let us say, the base of the right triangle be x cm.

Given, the altitude of right triangle = (x – 7) cm

From Pythagoras theorem, we know,

Base2 + Altitude2 = Hypotenuse2

∴ x+ (x – 7)2 = 132

⇒ x+ x+ 49 – 14x = 169

⇒ 2x– 14x – 120 = 0

⇒ x– 7x – 60 = 0

⇒ x– 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 be x.

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

Given, total cost of production is Rs.90

∴ x(2x + 3) = 90

⇒ 2x+ 3x – 90 = 0

⇒ 2x+ 15x -12x – 90 = 0

⇒ x(2x + 15) -6(2x + 15) = 0

⇒ (2x + 15)(x – 6) = 0

Thus, either 2x + 15 = 0 or x – 6 = 0

⇒ 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 = Rs 15.


The exercise 4.2 for Class 10 Maths NCERT Solutions Chapter 4- Quadratic Equations contains a total of 6 Questions along with the solutions. 5 out of the 6 questions are short questions with bot very lengthy solutions. The sixth question requires a long solution based on the quadratic equation concepts. The students can refer to the NCERT Solutions Class 10 Maths for the required details related to Chapter 4- Quadratic Equations Exercise 4.2 .

Key Features of NCERT Solutions Class 10 Maths Chapter 4- Quadratic Equations Exercise 4.2

  • NCERT Solutions are created by subject experts.
  • The answers provided to the questions are accurate..
  • The stepwise solutions to the questions are provided in the NCERT Solutions Class 10 Maths Chapter 4- Quadratic Equations Exercise 4.2.
  • The questions are prepared from the examination perspective.

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