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:

- x
^{2}– 2x – 8 - 4s
^{2}– 4s + 1 - 6x
^{2}– 3 – 7x - 4u
^{2}+ 8u - t
^{2}– 15 - 3x
^{2}– x – 4

Solution:

1. We have:

x^{2} – 2x – 8 = x^{2} + 2 – 4x – 8

= x(x + 2) – 4(x + 2)

= (x + 2)(x – 4)

The value of x^{2} – 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 x^{2} – 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:

4s^{2} – 4s + 1 = 4s^{2} – 2s – 2s + 1

= 2s(2s – 1) – 1(2s – 1)

= (2s – 1)(2s – 1)

The value of 4s^{2} – 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 4s^{2} – 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:

6x^{2} – 3 – 7x = 6x^{2} – 7x – 3

= 6x^{2} – 9x + 2x – 3

= 3x(2x – 3) + 1(2x – 3)

= (3x + 1)(2x – 3)

The value of 6x^{2} – 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 6x^{2} – 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:

4u^{2} + 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 4u^{2} + 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 t^{2} – 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 t^{2} – 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:

3x^{2} – x – 4 = 3x^{2} + 3x – 4x – 4

= 3x(x + 1) – 4(x + 1)

= (x + 1)(3x – 4)

The value of 3x^{2} – 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 3x^{2} – 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.

- \(\frac{1}{4}\), – 1
- \(\sqrt{2}\), \(\frac{1}{3}\)
- 0, \(\sqrt{5}\)
- 1, 1
- – \(\frac{1}{4}\), \(\frac{1}{4}\)
- 4, 1

Solution:

1. Let the polynomial be ax^{2} + 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 4x^{2} – x – 4.

2. Let the polynomial be ax^{2} + 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 3x^{2} – 3\(\sqrt{2x}\) + 1.

3. Let the polynomial be ax^{2} + 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 x^{2} – 0. x + \(\sqrt{5}\), i.e; x^{2} + \(\sqrt{5}\).

4. Let the polynomial be ax^{2} + 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 x^{2} – x + 1.

5. Let the polynomial be ax^{2} + 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 4x^{2} + x + 1.

6. Let the polynomial be ax^{2} + 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 x^{2} – 4x + 1.