# 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 step-by-step 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 to these exercises are provided in the form of a PDF. NCERT Solutions Class 10 Maths Chapter 4- Quadratic Equation covers an important chapter from the examination perspective. The students should deal with this chapter thoroughly to score well. NCERT Solutions are the best guide for the examinations that carries all the solutions to the queries of the students. The students can refer to these for further practice. They not only enhance their conceptual knowledge but also help them to overcome their imperfections.

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

### 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)Â x2Â â€“ 3xÂ â€“ 10 = 0
(ii) 2x2Â +Â xÂ â€“ 6 = 0
(iii) âˆš2Â x2Â + 7xÂ + 5âˆš2Â = 0
(iv) 2x2Â â€“Â xÂ +1/8 = 0
(v) 100x2Â â€“ 20xÂ + 1 = 0

Solutions:

(i) Given,Â x2Â â€“ 3xÂ â€“ 10 =0

Taking LHS,

=>x2Â – 5xÂ + 2xÂ â€“ 10

=>x(xÂ – 5)Â + 2(xÂ – 5)

=>(xÂ – 5)(xÂ +Â 2)

The roots of this equation, x2Â â€“ 3xÂ â€“ 10 = 0 are the values of x for whichÂ (xÂ – 5)(xÂ +Â 2) = 0

Therefore,Â xÂ – 5 = 0 orÂ xÂ + 2 = 0

=>Â xÂ = 5 orÂ xÂ = -2

(ii) Given, 2x2Â +Â xÂ â€“ 6 = 0

Taking LHS,

=> 2x2Â + 4xÂ – 3xÂ – 6

=> 2x(xÂ +Â 2) – 3(xÂ + 2)

=> (xÂ + 2)(2xÂ – 3)

The roots of this equation, 2x2Â +Â xÂ â€“ 6=0 are the values of x for whichÂ (xÂ + 2)(2xÂ – 3) = 0

Therefore,Â xÂ + 2Â = 0 orÂ 2xÂ – 3 = 0

=>Â xÂ = -2 orÂ xÂ = 3/2

(iii) âˆš2Â x2Â + 7xÂ + 5âˆš2=0

Taking LHS,

=> âˆš2Â x2Â + 5xÂ + 2xÂ + 5âˆš2

=>Â xÂ (âˆš2xÂ + 5)Â +Â âˆš2(âˆš2xÂ + 5)= (âˆš2xÂ + 5)(xÂ +Â âˆš2)

The roots of this equation, âˆš2Â x2Â + 7xÂ + 5âˆš2=0 are the values of x for whichÂ (âˆš2xÂ + 5)(xÂ +Â âˆš2)Â = 0

Therefore, âˆš2xÂ + 5Â = 0 orÂ xÂ +Â âˆš2Â = 0

=>Â xÂ = -5/âˆš2Â orÂ xÂ = -âˆš2

(iv) 2x2Â â€“Â xÂ +1/8 = 0

Taking LHS,

=1/8 (16x2 Â – 8xÂ + 1)

= 1/8 (16x2 Â – 4xÂ -4xÂ + 1)

= 1/8 (4x(4xÂ Â – 1) -1(4xÂ –Â 1))

= 1/8 (4xÂ – 1)2

The roots of this equation, 2x2Â â€“Â xÂ + 1/8 = 0, are the values of x for whichÂ (4xÂ – 1)2= 0

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

â‡’Â xÂ = 1/4 orÂ xÂ = 1/4

(v) Given, 100x2Â â€“ 20xÂ + 1=0

Taking LHS,

= 100x2Â â€“ 10xÂ – 10xÂ + 1

= 10x(10xÂ – 1) -1(10xÂ – 1)

= (10xÂ – 1)2

The roots of this equation, 100x2Â â€“ 20xÂ + 1=0, are the values of x for whichÂ (10xÂ – 1)2= 0

âˆ´ (10xÂ – 1) = 0 or (10xÂ – 1) = 0

â‡’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 wasÂ  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, number of marbles Jivanti has = 45 – x

After losing 5 marbles each,

Number of marbles John has = xÂ – 5

Number of marbles Jivanti has = 45 – xÂ – 5 = 40 –Â x

Given that the product of their marbles is 124.

âˆ´ (xÂ – 5)(40 –Â x) = 124

â‡’Â x2Â â€“ 45xÂ + 324 = 0

â‡’Â x2Â â€“ 36xÂ – 9xÂ + 324 = 0

â‡’Â x(xÂ – 36) -9(xÂ – 36) = 0

â‡’ (xÂ – 36)(xÂ – 9) = 0

Thus, we can say,

xÂ – 36 = 0 orÂ xÂ – 9 = 0

â‡’Â xÂ = 36 orÂ xÂ = 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 beÂ x.

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

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

âˆ´Â x(55 –Â x) = 750

â‡’Â x2Â â€“Â 55xÂ + 750 = 0

â‡’Â x2Â â€“ 25xÂ –Â 30xÂ + 750 = 0

â‡’Â x(xÂ – 25) -30(xÂ – 25) = 0

â‡’ (xÂ – 25)(xÂ – 30) = 0

Thus, either xÂ -25 = 0 orÂ xÂ – 30 = 0

â‡’Â xÂ = 25 orÂ xÂ = 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 beÂ xÂ and the second number is 27 –Â x.

Therefore, the product of two numbers

x(27 –Â x) = 182

â‡’Â x2Â â€“ 27xÂ – 182 = 0

â‡’Â x2Â â€“ 13xÂ – 14xÂ + 182 = 0

â‡’Â x(xÂ – 13) -14(xÂ – 13) = 0

â‡’ (xÂ – 13)(xÂ -14) = 0

Thus, either, xÂ = -13Â = 0 orÂ xÂ – 14 = 0

â‡’Â xÂ = 13 orÂ xÂ = 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 beÂ xÂ andÂ xÂ + 1.

Therefore, as per the given questions,

x2Â + (xÂ + 1)2Â = 365

â‡’Â x2Â +Â x2Â + 1 + 2xÂ = 365

â‡’ 2x2Â + 2x – 364 = 0

â‡’Â x2Â +Â xÂ – 182 = 0

â‡’Â x2Â + 14xÂ – 13xÂ – 182 = 0

â‡’Â x(xÂ +Â 14) -13(xÂ +Â 14) = 0

â‡’ (xÂ + 14)(xÂ – 13) = 0

Thus, either, xÂ + 14 = 0 orÂ xÂ – 13 = 0,

â‡’Â xÂ = – 14 orÂ xÂ = 13

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

âˆ´Â xÂ + 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 isÂ  x cm.

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

From Pythagoras theorem, we know,

Base2Â + Altitude2Â = Hypotenuse2

âˆ´Â x2Â + (xÂ – 7)2Â = 132

â‡’Â x2Â +Â x2Â + 49 – 14xÂ = 169

â‡’ 2x2Â – 14xÂ – 120 = 0

â‡’Â x2Â – 7xÂ – 60 = 0

â‡’Â x2Â – 12xÂ + 5xÂ – 60 = 0

â‡’Â x(xÂ – 12) + 5(xÂ – 12) = 0

â‡’ (xÂ – 12)(xÂ + 5) = 0

Thus, eitherÂ xÂ – 12 = 0 orÂ xÂ + 5 = 0,

â‡’Â xÂ = 12 orÂ xÂ = – 5

Since sides cannot be negative,Â xÂ 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.

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

Given, total cost of production is Rs. 90

âˆ´Â x(2xÂ + 3) = 90

â‡’ 2x2Â + 3xÂ – 90 = 0

â‡’ 2x2Â + 15xÂ -12xÂ – 90 = 0

â‡’Â x(2xÂ + 15) -6(2xÂ + 15) = 0

â‡’ (2xÂ + 15)(xÂ – 6) = 0

Thus, either 2xÂ + 15 = 0 orÂ xÂ – 6 = 0

â‡’Â xÂ = -15/2 orÂ xÂ = 6

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

Hence, number of articles produced = 6

Cost of each article = 2 Ã— 6 + 3 = Rs. 15.

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 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 Chapter 4 – Quadratic Equations for the required details related to 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 step-by-step 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.