Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2 NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2 focus on solving quadratic equations using the factorisation method. This exercise helps in breaking down equations into simpler factors to determine their roots clearly and systematically. It forms an important part of the CBSE Class 10 Maths syllabus and strengthens core algebraic concepts.

The step-by-step solutions explain how to identify suitable factors using given coefficients and solve equations accurately. Regular practice of these questions improves problem-solving skills and builds a strong base for understanding other methods like completing the square and the quadratic formula.

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2

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

(i) x 2 – 3x – 10 = 0

(ii) 2x  2 + x – 6 = 0

(iii) √2 x 2 + 7x + 5√2 = 0

(iv) 2x  2 – x +1/8 = 0

(v) 100x  2 – 20x + 1 = 0

Solutions:

(i) Given, x 2 – 3 x – 10 =0 

Taking LHS, 

=> x 2 – 5 x + 2 x – 10 

=> x ( x – 5) + 2( x – 5) 

=>( x – 5)( x + 2) 

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

Taking LHS, => 2 x 2 + 4 x – 3 x – 6 

=> 2 x ( x + 2) – 3( x + 2) 

=> ( x + 2)(2 x – 3) 

The roots of this equation, 2 x 2 + x – 6=0 are the values of x for which ( x + 2)(2 x – 3) = 0 

Therefore, x + 2 = 0 or 2 x – 3 = 0 

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

(iii) √2  x 2 + 7 x + 5√2=0 

Taking LHS, => √2 x 2 + 5 x + 2 x + 5√2 

=> x (√2 x + 5) + √2(√2 x + 5)

= (√2 x + 5)( x + √2) 

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

(√2 x + 5)( x + √2) = 0 

Therefore, √2 x + 5 = 0 

or 

x + √2 = 0 

=> x = -5/√2 or x = -√2

(iv) 2  x 2 – x +1/8 = 0 

Taking LHS, =1/8 (16 x 2 – 8 x + 1) 

= 1/8 (16 x 2 – 4 x -4 x + 1) 

= 1/8 (4 x (4 x – 1) -1(4 x – 1)) 

= 1/8 (4 x – 1) 2 

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

(4 x – 1) 2 = 0 

Therefore, (4 x – 1) = 0 

or 

(4 x – 1) = 0 

⇒ x = 1/4 or x = 1/4

(v) Given, 100x  2 – 20x + 1=0 

Taking LHS, = 100x 2 – 10x – 10x + 1 

= 10x(10x – 1) -1(10x – 1) 

= (10x – 1) 2 

The roots of this equation, 100x 2 – 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 has = x. Therefore, the number of marbles Jivanti has = 45 – x. 

After losing 5 marbles each, Number of marbles John has =  x – 5 Number of marbles Jivanti has = 45 – x – 5 = 40 – x

Given that the product of their marbles is 124. 

∴ ( x – 5)(40 – x ) = 124

⇒ x 2 – 45 x + 324 = 0 

⇒ x 2 – 36 x – 9 x + 324 = 0 

⇒ x ( x – 36) -9( x – 36) = 0 

⇒ ( x – 36)( x – 9) = 0 

hus, 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, the 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 

⇒ x 2 – 55 x + 750 = 0 

⇒ x 2 – 25 x – 30 x + 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, the first number is x and the second number is 27–x.

Therefore, the product of two numbers x(27 – x) = 182 

⇒ x 2 – 27x – 182 = 0 

⇒ x 2 – 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, 

x 2 + ( x + 1) 2 = 365 

⇒ x 2 + x 2 + 1 + 2 x = 365 

⇒ 2 x 2 + 2x – 364 = 0 

⇒ x 2 + x – 182 = 0 

⇒ x 2 + 14 x – 13 x – 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 is x cm.

Given, the altitude of right triangle = (x – 7) cm From Pythagoras theorem, we know, Base 2 + Altitude 2 = Hypotenuse 2 

∴ x 2 + (x – 7) 2 = 13 2 

⇒ x 2 + x 2 + 49 – 14x = 169 

⇒ 2x 2 – 14x – 120 = 0 

⇒ x 2 – 7x – 60 = 0 

⇒ x 2 – 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 is x.

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

Given, total cost of production is Rs. 90 

∴  x (2 x + 3) = 90 

⇒ 2 x 2 + 3 x – 90 = 0 

⇒ 2 x 2 + 15 x -12 x – 90 = 0 

⇒ x (2 x + 15) -6(2 x + 15) = 0 

⇒ (2 x + 15)( x – 6) = 0

 Thus, either 2  x + 15 = 0 or x – 6 = 0 ⇒ x = -15/2 or x = 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 × 6 + 3 = Rs. 15.

How to Score Better in Class 10 Maths Exam?

Scoring well in Class 10 Maths requires clear concepts, regular practice, and a focus on accuracy and answer presentation. To score better, you should:

• Build Strong Concepts:

Focus on understanding concepts in Class 10 Maths instead of memorising steps, as this helps in solving application-based questions.

• Work on Weak Areas:

Focus on difficult topics from the Class 10 Maths syllabus instead of skipping them to avoid losing marks.

• Revise Formulas Daily:

Regular revision of PW Class 10 Maths MIQs helps avoid calculation mistakes in exams.

• Practise Regularly:

Solve all CBSE Class 10 NCERT questions multiple times to strengthen your basics and improve accuracy.

• Solve Previous Year Papers:

Practising CBSE Class 10 Maths previous year questions (PYQs) helps you understand question patterns and important topics.

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