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4. Quadratic Equations Mathematics Exercise - 4.2 class 10 Maths in English - CBSE Study

4. Quadratic Equations Mathematics Class 10 exercise - 4.2 class 10 Maths cbse board school study materials like cbse notes in English medium, all chapters and exercises are covered the ncert latest syllabus 2026 - 27.

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4. Quadratic Equations Mathematics Exercise - 4.2 class 10 Maths in English - CBSE Study

NCERT Solutions for Class 10 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 4. Quadratic Equations with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 4.2 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 10 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.

Class 10 English Medium Mathematics All Chapters:

4. Quadratic Equations

2. Exercise 4.2

Exercise: 4.2


Q1. Find the roots of the following quadratic equations by factorisation:

(i) x2 - 3x - 10 = 0,

Solution:

    x2 - 3x - 10 = 0

⇒x2 - 5x + 2x - 10 = 0

⇒ x( x - 5 ) + 2(x - 5) = 0

⇒( x - 5 ) (x + 2) = 0

⇒( x - 5 ) = 0, (x + 2) = 0

x - 5 = 0

x = 5

x + 2 = 0

x = - 2

(ii) 2x2 + x - 6 = 0;

 Solution:

2x2 + x - 6 = 0

 

⇒2x2 + 4x - 3x - 6 = 0

⇒ 2x( x + 2 ) - 3(x + 2) = 0

⇒( x + 2 ) (2x - 3) = 0

⇒( x + 2 ) = 0, (2x - 3) = 0

x + 2 = 0

x = - 2

2x - 3 = 0

2x = 3

x =  

3

2

 

(iii) √2x2 + 7x + 5√2 = 0;

Solution:  

√2x + 2x + 5x + 5√2 = 0

√2x(x + √2) + 5(x + √2) = 0

(x + √2) (√2x + 5) = 0;

(x + √2) = 0, (√2x + 5) = 0

 

x = - √2,  √2x = - 5

x = - 5/√2

x = -5√2/2

(iv) 2x2 - x + 1/8 =0

Solution:

2x2 - x + 1 = 0
8

Or  ⇒16x2 - 8x + 1 = 0;

⇒16x2 - 4x - 4x + 1 = 0

⇒ 4x( 4x - 1 ) - 1(4x - 1) = 0

⇒( 4x - 1 ) (4x - 1) = 0

⇒( 4x - 1 ) = 0, (4x - 1) = 0

4x - 1 = 0

4x = 1

x =  

1

4

4x - 1 = 0

4x = 1

x =  

1

4

(v)  100x2 - 20x + 1 = 0;  

 Solution:

  100x2 - 20x + 1 = 0

⇒100x2 - 10x - 10x + 1 = 0

⇒ 10x( 10x - 1 ) - 1(10x - 1) = 0

⇒( 10x - 1 ) (10x - 1) = 0

⇒( 10x - 1 ) = 0, (10x - 1) = 0

10x - 1 = 0

10x = 1

x =  

1

10

10x - 1 = 0

10x = 1

x =  

1

10

 

Q2. Solve the problems given in Example 1.

Example1:

(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 Rs 750. We would like to find out the number of toys produced on that day.

 

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

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

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.

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.

 

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