# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

## BSEB Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Book Answers BSEB Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Book Answers

## Bihar Board Class 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Books Solutions

 Board BSEB Materials Textbook Solutions/Guide Format DOC/PDF Class 10th Subject Maths Chapter 4 Quadratic Equations Ex 4.2 Chapters All Provider Hsslive

2. Click on the Bihar Board Class 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Answers.
3. Look for your Bihar Board STD 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Textbooks PDF.

Find below the list of all BSEB Class 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

## BSEB Bihar Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x3 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) 2‾√x2 + 7x + 52‾√ = 0
(iv) 2x2 – x + 18 = 0
(v) 100x2 – 20x + 1 = 0
Solution:
(i) We have:
x2 – 3x – 10 = 0
or x2 – 5x + 2x – 10 = 0
or x(x – 5) + 2(x – 5) = 0
or (x – 5)(x + 2) = 0
i.e; x – 5 = 0 or x + 2 = 0
i.e; x = 5 or x = – 2
Thus, x = 5 and x = – 2 are two solutions of the equation x2 – 3x – 10 = 0. That is, roots of the equation are 5 and – 2.

(ii) We have:
2x2 + x – 6 = 0
or 2x2 + 4x – 3x – 6 = 0
or 2x(x + 2) – 3(x + 2) = 0
or (x + 2)(2x – 3) = 0
i.e; x + 2 = 0 or 2x – 3 = 0
i.e; x = – 2 or x = 32
Thus, – 2 and 32 are two roots of the equation 2x2 + x – 6 = 0.

(iii) We have:
2‾√x2 + 7x + 52‾√ = 0
or 2‾√x(x + 2‾√) + 5(x + 2‾√) = 0
or (x + 2‾√)(2‾√x + 5) = 0
i.e; x + 2‾√ = 0 or 2‾√x + 5 = 0
i.e; x = – 2‾√
or x = −52√=−52√2
Thus, – 2‾√ and −52√2 are two roots of the equation 2‾√x2 + 7x + 52‾√ = 0.

(iv) We have:
2x2 – x + 18 = 0
or 16x2 – 8x + 1 = 0
or 16x2 – 4x – 4x + 1 = 0
or 4x(4x – 1) – 1(4x – 1) = 0
or 4x – 1 = 0 or 4x – 1 = 0
i.e; x = 14 or x = 14
Thus, 14 and 14 are two equal roots of the equation 2x2 – x + 18 = 0

(v) We have:
100x2 – 20x + 1 = 0
or 100x2 – 10x – 10x + 1 = 0
or 10x(10x – 1) – 1(10x – 1) = 0
or (10x – 1)(10x – 1) = 0
i.e; 10x – 1 = 0 or 10x – 1 = 0
i.e; x = 110 or x = 110
Thus, 110 and 110 are two equal roots of the equation 100x2 – 20x = 1 = 0.

Question 2.
Solve the problems given in Example 1, i.e; to solve
(i) x2 – 45x + 324 = 0
(ii) x2 – 55x + 750 = 0
using factorisation method.
Solution:
(i) We have: x2 – 45x + 324 = 0
or x2 – 9x – 36x + 324 = 0
or x(x – 9) – 36(x – 9) = 0
or (x – 9)(x – 36) = 0
i.e; x – 9 = 0 or x – 36 = 0
i.e; x = 9 or x = 36
Thus, 9 and 36 are two roots of the equation x2 – 45x + 324 = 0.

(ii) We have:
x2 – 55x + 750 = 0
or x2 – 30x – 25x + 750 = 0
or x(x – 30) – 25(x – 30) = 0
or (x – 30)(x – 25) = 0
i.e; x – 30 = 0 or x – 25 = 0
i.e; x = 30 or x = 25
Thus, 30 and 25 are two roots of the equation.
x2 – 55x + 750 = 0.

Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let the required numbers be x and 27 – x. Then,
x(27 – x) = 182
or 27x – x2 = 182
or x2 – 27x + 182 = 0
or x2 – 13x – 14x + 182 = 0
or x(x – 13) – 14(x – 13) = 0
or (x – 13)(x – 14) = 0
i.e; x – 13 = 0 or x – 14 = 0
i.e; x = 13 or x = 14
Hence, the required numbers are 13 and 14.

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let the two consecutive positive integers be x and x + 1.
Then, x2 + (x + 1)2 = 365
or x2 + x2 + 2x + 1 = 365
or 2x2 + 2x – 364 = 0
or x2 + x – 182 = 0
or x(x + 14) – 13(x + 14) = 0
or (x + 14)(x – 13) = 0
i.e; x + 14 = 0 or x – 13 = 0
i.e; x = – 14 or x = 13
Since x, being a positive integer, cannot be negative. Therefore, x = 13.
Hence, the two consecutive positive integers are 13 and 14.

Question 5.
The altitude of a right traiangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:
Let the base of the right triangle be x cm. Then, its altitude is (x – 7) cm.
It is given that the hypotenuse = 13 cm
So, 𝑥2+(𝑥−7)2‾‾‾‾‾‾‾‾‾‾‾‾‾√ = 13
or x2 + (x2 – 14x + 49) = 169
or 2x2 – 14x – 120 = 0
or x2 – 7x – 60 = 0
or x2 – 12x + 5x – 60 = 0
or x(x – 12) + 5(x – 12) = 0
or (x – 12)(x + 5) = 0
i.e., x – 12 = 0 or x + 5 = 0
i.e., x = 12 or x = – 5
So, x = 12
[Since side of a triangle can never be negative]
∴ Length of the base = 12 cm
and length of the altitude
= (12 – 7) cm
= 5 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 Rs 90, find the number of articles produced and the cost of each article.
Solution:
Let the number of articles produced in a day = x
Then, cost of production of each article = Rs (2x + 3)
It is given that the total cost of production = Rs 90
∴ x × (2x + 3) = 90
or 2x2 + 3x – 90 = 0
or 2x2 – 12x + 15x – 90 = 0
or 2x(x – 6) + 15(x – 6) = 0
or (x – 6)(2x + 15) = 0
i.e; x – 6 = 0 or 2x + 15 = 0
i.e; x = 6 or x = −152
So, x = 6 [Since the number of articles produced cannot be negative]
Cost of each article = Rs (2 × 6 + 3) = Rs 15
Hence, the number of articles produced are 6 and the cost of each article is Rs 15.

## Bihar Board Class 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Textbooks for Exam Preparations

Bihar Board Class 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Textbook Solutions can be of great help in your Bihar Board Class 10th Maths Chapter 4 Quadratic Equations Ex 4.2 exam preparation. The BSEB STD 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Books State Board syllabus with maximum efficiency.

## FAQs Regarding Bihar Board Class 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Textbook Solutions

#### Can we get a Bihar Board Book PDF for all Classes?

Yes you can get Bihar Board Text Book PDF for all classes using the links provided in the above article.

## Important Terms

Bihar Board Class 10th Maths Chapter 4 Quadratic Equations Ex 4.2, BSEB Class 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Textbooks, Bihar Board Class 10th Maths Chapter 4 Quadratic Equations Ex 4.2, Bihar Board Class 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Textbook solutions, BSEB Class 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Textbooks Solutions, Bihar Board STD 10th Maths Chapter 4 Quadratic Equations Ex 4.2, BSEB STD 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Textbooks, Bihar Board STD 10th Maths Chapter 4 Quadratic Equations Ex 4.2, Bihar Board STD 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Textbook solutions, BSEB STD 10th Maths Chapter 4 Quadratic Equations Ex 4.2 Textbooks Solutions,
Share: