Bihar Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.2

Bihar Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.2 Textbook Questions and Answers.

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:

1. x² – 2x – 8
2. 4s² – 4s + 1
3. 6x² – 3 – 7x
4. 4u² + 8u
5. t² – 15
6. 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 = \(\frac{-(-2)}{1}\)

and product of zeroes = (- 2)(4) = – 8 = \(\frac{-8}{1}\)

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 = \(\frac{1}{2}\) or s = \(\frac{1}{2}\).
∴ The zeroes of 4s² – 4s + 1 are \(\frac{1}{2}\) and \(\frac{1}{2}\).
Therefore, sum of the zeroes = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1 = \(\frac{-(-4)}{4}\)

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 = – \(\frac{1}{3}\) or x = \(\frac{3}{2}\).
∴ The zeros of 6x² – 3 – 7x are – \(\frac{1}{3}\) and \(\frac{3}{2}\).
Therefore, sum of the zeros = – \(\frac{1}{3}\) + \(\frac{3}{2}\) = \(\frac{7}{6}\) = \(\frac{-(-7)}{6}\)

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 = \(\frac{-8}{4}\)

and product of zeroes = (0)(- 2) = 0 = \(\frac{0}{4}\)

5. We have:
t – 15 = (t – \(\sqrt{15}\))(t + \(\sqrt{15}\))
The value of t² – 15 is 0, when the value of (t – \(\sqrt{15}\))(t + \(\sqrt{15}\)) is 0, i.e; when t – \(\sqrt{15}\) = 0 or t + \(\sqrt{15}\) = 0,
i.e; when t = \(\sqrt{15}\) or t = – \(\sqrt{15}\).
∴ The zeroes of t² – 15 are \(\sqrt{15}\) and – \(\sqrt{15}\).
Therefore, sum of the zeroes = \(\sqrt{15}\) + (- \(\sqrt{15}\)) = 0

and product of the zeroes = (\(\sqrt{15}\))(-\(\sqrt{15}\))
= – 15 = \(\frac{-15}{1}\)

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 = \(\frac{4}{3}\).
∴ The zeroes of 3x² – x – 4 are – 1 and \(\frac{4}{3}\).
Therefore, sum of the zeroes = – 1 + \(\frac{4}{3}\) = \(\frac{-3+4}{3}\)
= \(\frac{1}{3}\) = \(\frac{-(-1)}{3}\)

and product of the zeroes = (- 1)(\(\frac{4}{3}\)) = – \(\frac{4}{3}\) = \(\frac{-4}{3}\)

Question 2.
Find a quadratie polynomial each with the given numbers as the sum and product of its zeroes respectively.

1. \(\frac{1}{4}\), – 1
2. \(\sqrt{2}\), \(\frac{1}{3}\)
3. 0, \(\sqrt{5}\)
4. 1, 1
5. – \(\frac{1}{4}\), \(\frac{1}{4}\)
6. 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 = – 3\(\sqrt{2}\) and c = 1.
∴ One quadratic polynomial which fits the given conditions is 3x² – 3\(\sqrt{2x}\) + 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 = \(\sqrt{5}\).
∴ One quadratic polynomial which fits the given conditions is x² – 0. x + \(\sqrt{5}\), i.e; x² + \(\sqrt{5}\).

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.