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

Question 1. Find the roots of the following quadratic equations by factorization:

(i) x²– 3x – 10 = 0 

Solution:

Here, LHS = x²– 3x – 10 

= x² – 5x + 2x – 10

= x(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

Hence, x – 5 = 0 or x + 2 = 0

⇒ x = 5 or x = -2

(ii) 2x² + x – 6 = 0

Solution:

Here, LHS = 2x² + x – 6

= 2x² + 4x – 3x – 6

= 2x(x + 2) – 3(x + 2)

= (2x– 3)(x + 2)

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

(2x– 3)(x + 2) = 0

Hence, 2x– 3 = 0 or x + 2 = 0

⇒ x = 3/2 or x = –2

(iii) √2x² + 7x + 5√2 = 0

Solution:

Here, LHS = √2x² + 7x + 5√2

= √2x² + 5x + 2x + 5√2

= x(√2x + 5) + √2(√2x + 5)

= (√2x + 5) (x +√2)

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

(√2x + 5) (x +√2) = 0

Hence, √2x + 5 = 0 or x +√2 = 0

⇒ x = –5/√2 or x = –√2

(iv) 2x² – x + 1/8 = 0

Solution:

Here, LHS = 2x² – x + 1/8

= 1/8(16x² – 8x + 1)

= 1/8(16x² – 4x -4x + 1)

= 1/8(4x(4x-1) -1 (4x-1))

= 1/8 (4x-1) (4x-1) 

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

1/8 (4x-1) (4x-1)  = 0

(4x-1)² = 0

Hence, 4x-1 = 0 or 4x-1 = 0

⇒ x = 1/4 or x = 1/4

(v) 100x² – 20x + 1 = 0

Solution:

Here, LHS = 100x² – 20x + 1

= 100x² – 10x – 10x + 1

= 10x(10x – 1) – 1(10x – 1)

= (10x – 1) (10x – 1)

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

(10x – 1) (10x – 1) = 0

(10x – 1)² = 0

Hence, 10x – 1 = 0 or 10x – 1 = 0

⇒ x = 1/10 or x = 1/10

Question 2. Solve 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.

Solution:

Let's say, 

The number of marbles John have = x.

So, 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

Here, According to the given condition

(x – 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

Hence, 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) 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:

 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

So, 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

Hence, 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.

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

Solution:

Let's say, 

First number be x and the second number is 27 – x.

Therefore, the product of two numbers will be:

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

Hence, x – 13 = 0 or x – 14= 0

x = 13 or x = 14

Hence, 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.

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

Solution:

Let's say, 

Two consecutive positive integers be x and x + 1.

Here, According to the given condition,

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

Hence, x – 13 = 0 or x + 14= 0

x = 13 or x = – 14

As, here it is said positive integers, so x can be 13, only.

So,

x = 13

and, x + 1 = 13 + 1 = 14

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

Question 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's say, 

Base of the right triangle be x cm.

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

Base² + Altitude² = Hypotenuse²                       (Pythagoras theorem)

x² + (x – 7)² = 132

x² + x² + 49 – 14x = 169    (using identity (a-b)² = a² - 2ab + b²)

2x² – 14x – 120 = 0

x² – 7x – 60 = 0               (Dividing by 2)

x² – 12x + 5x – 60 = 0

x(x – 12) + 5(x – 12) = 0

(x – 12)(x + 5) = 0

Hence, x – 12 = 0 or x + 5= 0

x = 12 or x = – 5

As, here side will be a  positive integers, so x can be 12, only.

Therefore, the base of the given triangle is and,

the altitude of this triangle will be (12 – 7) cm =

Question 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's say,

Number of articles produced be x.

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

Here, According to the given condition

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

Hence, 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, 

So, x = 6.

Hence, number of articles produced = 6

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

Summary

Exercise 4.2 of Chapter 4 "Quadratic Equations" in Class 10 NCERT focuses on solving quadratic equations by factorization. This method involves rewriting the quadratic expression as a product of two linear factors. Students learn to identify common factors, use the split-middle term technique, and apply the difference of squares formula when appropriate. The exercise covers various types of quadratic equations, including those with rational coefficients and those requiring algebraic manipulation before factorization. Students practice solving equations and verifying their solutions, which helps reinforce their understanding of the factorization method and its application in finding roots of quadratic equations.