BSEB Class 10 Maths Chapter 2 Polynomials Ex 2.2 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 2 Polynomials Ex 2.2 Book Answers |
Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.2 Textbooks Solutions PDF
Bihar Board STD 10th Maths Chapter 2 Polynomials Ex 2.2 Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 10th Maths Chapter 2 Polynomials Ex 2.2 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.2 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 10th Maths Chapter 2 Polynomials Ex 2.2 Textbooks. These Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.2 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.Bihar Board Class 10th Maths Chapter 2 Polynomials Ex 2.2 Books Solutions
Board | BSEB |
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Format | DOC/PDF |
Class | 10th |
Subject | Maths Chapter 2 Polynomials Ex 2.2 |
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BSEB Class 10th Maths Chapter 2 Polynomials Ex 2.2 Textbooks Solutions with Answer PDF Download
Find below the list of all BSEB Class 10th Maths Chapter 2 Polynomials Ex 2.2 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:BSEB Bihar Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.2
Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
- x2 – 2x – 8
- 4s2 – 4s + 1
- 6x2 – 3 – 7x
- 4u2 + 8u
- t2 – 15
- 3x2 – x – 4
Solution:
1. We have:
x2 – 2x – 8 = x2 + 2 – 4x – 8
= x(x + 2) – 4(x + 2)
= (x + 2)(x – 4)
The value of x2 – 2x – 8 is 0, when the value of (x + 2)(x – 4) is 0, i.e., when x + 2 = 0 or x – 4 = 0, i.e., when x = – 2 or x = 4.
∴ The zeroes of x2 – 2x – 8 are – 2 and 4.
Therefore, sum of the zeroes = (- 2) + 4 = 2 = −(−2)1
and product of zeroes = (- 2)(4) = – 8 = −81
2. We have:
4s2 – 4s + 1 = 4s2 – 2s – 2s + 1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1)(2s – 1)
The value of 4s2 – 4s + 1 is 0, when the value of (2s – 1)(2s – 1) is 0, i.e., when 2s – 1 = 0 or 2s – 1 = 0, i.e.,
when s = 12 or s = 12.
∴ The zeroes of 4s2 – 4s + 1 are 12 and 12.
Therefore, sum of the zeroes = 12 + 12 = 1 = −(−4)4
3. We have:
6x2 – 3 – 7x = 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (3x + 1)(2x – 3)
The value of 6x2 – 3 – 7x is 0, when the value of (3x + 1)(2x – 3) is 0, i.e; when 3x + 1 = 0 or 2x – 3 = 0, i.e;
when x = – 13 or x = 32.
∴ The zeros of 6x2 – 3 – 7x are – 13 and 32.
Therefore, sum of the zeros = – 13 + 32 = 76 = −(−7)6
4. We have:
4u2 + 8u is 0, when the value of 4u(u + 2) is 0, i.e; when u = 0 or u + 2 = 0, i.e; when u = 0 or u = – 2.
∴ The zeroes of 4u2 + 8u are o and – 2.
Therefore, sum of the zeroes = 0 + (- 2) = – 2 = −84
and product of zeroes = (0)(- 2) = 0 = 04
5. We have:
t – 15 = (t – 15‾‾‾√)(t + 15‾‾‾√)
The value of t2 – 15 is 0, when the value of (t – 15‾‾‾√)(t + 15‾‾‾√) is 0, i.e; when t – 15‾‾‾√ = 0 or t + 15‾‾‾√ = 0,
i.e; when t = 15‾‾‾√ or t = – 15‾‾‾√.
∴ The zeroes of t2 – 15 are 15‾‾‾√ and – 15‾‾‾√.
Therefore, sum of the zeroes = 15‾‾‾√ + (- 15‾‾‾√) = 0
and product of the zeroes = (15‾‾‾√)(-15‾‾‾√)
= – 15 = −151
6. We have:
3x2 – x – 4 = 3x2 + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (x + 1)(3x – 4)
The value of 3x2 – x – 4 is 0, when the value of (x + 1)(3x – 4) is 0, i.e; when x + 1 = 0 or 3x – 4 = 0, i,e; when x = – 1 or x = 43.
∴ The zeroes of 3x2 – x – 4 are – 1 and 43.
Therefore, sum of the zeroes = – 1 + 43 = −3+43
= 13 = −(−1)3
and product of the zeroes = (- 1)(43) = – 43 = −43
Question 2.
Find a quadratie polynomial each with the given numbers as the sum and product of its zeroes respectively.
- 14, – 1
- 2‾√, 13
- 0, 5‾√
- 1, 1
- – 14, 14
- 4, 1
Solution:
1. Let the polynomial be ax2 + bx + c, and its zeroes be α and β. Then,
If a = 4, then b = – 1 and c = – 4.
∴ One quadratic polynomial which fits the given conditions is 4x2 – x – 4.
2. Let the polynomial be ax2 + bx + c, and its zeroes be α and β. Then,
If a = 3, then b = – 32‾√ and c = 1.
∴ One quadratic polynomial which fits the given conditions is 3x2 – 32𝑥‾‾‾√ + 1.
3. Let the polynomial be ax2 + bx + c, and its zeroes be α and β. Then,
If a = 1, then b = 0 and c = 5‾√.
∴ One quadratic polynomial which fits the given conditions is x2 – 0. x + 5‾√, i.e; x2 + 5‾√.
4. Let the polynomial be ax2 + bx + c, and its zeroes be α and β. Then,
If a = 1, then b = – 1 and c = 1.
∴ One quadratic polynomial which fits the given conditions is x2 – x + 1.
5. Let the polynomial be ax2 + bx + c and its zeroes be α and β. Then
If a = 4, then b = 1 and c = 1.
∴ One quadratic polynomial which fits the given conditions is 4x2 + x + 1.
6. Let the polynomial be ax2 + bx + c and its zeroes be α and β. Then,
If a = 1, then b = – 4, and c = 1.
∴ One quadratic polynomial which fits the given conditions is x2 – 4x + 1.
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