Gujarat Board GSEB Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.2

Question 1.

Find the zeroes of the following polynomial and verify the relationship between the zeroes and the coefficients

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

Solution:

1. p(x) = x^{2} – 2x – 8

= x^{2} – 4x + 2x – 8

[By splitting middle term]

= x(x – 4) + 2(x – 4)

= (x – 4) (x + 2)

For zeroes of p(x),

p(x) = 0

â‡’ (x – 4) (x + 2) = 0

â‡’ x – 4 = 0

or x + 2 = 0

â‡’ x = 4

or x = -2

â‡’ x = 4, – 2.

Zeroes of p(x) are 4 and -2.

â‡’ Î± = 4, Î² = -2

Verification of relationship between zeroes and coefficient:

Sum of the zeroes = 4 + (-2) = 2

â‡’ Î± + Î² = 2

and \(\frac{-b}{a}=\frac{-(\text { coefficient of } x)}{\text { coefficient of } x^{2}}\)

= \(\frac{-(-2)}{1}\) = 2

â‡’ Î± + Î² = \(\frac{-b}{a}\)

Product of the zeroes

Î±Î² = 4 Ã— (-2) = -8

and \(\frac{c}{a}=\frac{\text { constant term }}{\text { coefficient of } x^{2}}\)

= \(\frac{-8}{1}\) = -8

â‡’ Î±Î² = \(\frac{c}{a}\)

2. p(s) = 4s^{2} – 4s + 1

= 4s^{2} – 2s – 2s + 1

[By splitting middle term]

= 2s(2s – 1) – 1 (2s – 1)

(2s – 1) (2s – 1)

For zeroes of p(s),

P(s) = 0

â‡’ (2s – 1) (2s – 1) = 0

â‡’ 2s – 1 = 0

or 2s – 1 = 0

â‡’ s = \(\frac{1}{2}\)

or s = \(\frac{1}{2}\)

â‡’ s = \(\frac{1}{2}\), \(\frac{1}{2}\)

â‡’ zeroes of p(s) are \(\frac{1}{2}\), \(\frac{1}{2}\)

â‡’ Î± = \(\frac{1}{2}\), Î² = \(\frac{1}{2}\)

Verification of relationship between zeroes and coefficients:

Sum of the zeroes = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1

â‡’ Î± + Î² = 1

3. p(x) = 6x^{2} – 3 – 7x

p(x) = 6x^{2} – 7x – 3

= 6x^{2} – 9x + 2x – 3

= 3x(2x – 3) + 1(2x – 3)

= (2x – 3) (3x + 1)

For zeroes p(x) = 0

â‡’ (2x – 3) (3x + 1) = 0

â‡’ 2x – 3 = 0

or 3x + 1 = 0

â‡’ x = \(\frac{3}{2}\)

or x = \(\frac{-1}{3}\)

x = \(\frac{3}{2}\), \(\frac{-1}{3}\)

â‡’ Î± = \(\frac{3}{2}\), Î² = \(\frac{-1}{3}\)

Now, verification of relationship between zeroes and coefficients.

Sum of the zeroes

Also, product of the zeroes;

4. p(u) = 4u^{2} + 8 u

p(u) = 4u (u + 2)

For zeroes, p(u) = 0

â‡’ 4 u(u + 2) = 0

â‡’ u(u + 2) = 0

â‡’ u = 0

or u + 2 = 0

â‡’ u = 0

â‡’ u = -2

So, zeroes of p(u) are 0 and -2.

â‡’ Î± = 0, Î² = -2

Verification of relationship between zeroes and coefficients:

Sum of the zeroes,

Î± + Î² = 0 + (-2) = -2

5. t^{2} – 15

P(t) = t^{2} – 15

For zeroes p(t) = 0

t^{2} – 15 = 0

t^{2} – (\(\sqrt { 15 } \))^{2} = 0

(t – \(\sqrt { 15 } \)) (t + \(\sqrt { 15 } \)) = o

â‡’ t – \(\sqrt { 15 } \) = 0

or t + \(\sqrt { 15 } \) = 0

â‡’ t = \(\sqrt { 15 } \)

t = –\(\sqrt { 15 } \)

So, zeroes of p(t) are \(\sqrt { 15 } \) and –\(\sqrt { 15 } \)

â‡’ Î± = \(\sqrt { 15 } \), Î² = –\(\sqrt { 15 } \)

Verification of relationship between zeroes

and coefficients:

Sum of the zeroes,

Î± + Î² = \(\sqrt { 15 } \) + (-\(\sqrt { 15 } \)) = 0

6. p(x) = 3x^{2} – x – 4

p(x) = 3x^{2} – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4)

= (3x – 4) (x + 1)

For zeroes, p(x) = 0

â‡’ (3x – 4)(x + 1) = 0

â‡’ 3x – 4 = 0

or x + 1 = 0

â‡’ x = \(\frac{4}{3}\)

or x = -1

â‡’ Zeroes of p(x) are \(\frac{4}{3}\) and -1.

â‡’ Î± = \(\frac{4}{3}\), Î² = -1

Verification of relationship between zeroes and coefficients:

Now sum of the zeroes,

Question 2.

Find a quadratic 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. A quadratic polynomial, when the sum and product of its zeroes are given, is

p(x) = K [x^{2} – (sum of the zeroes) x + Product of zeroes]

where K is constant.

Now sum of the zeroes = \(\frac{1}{4}\)

Product of the zeroes = -1

âˆ´ Required quadratic polynomial is given

or K[4x^{2} – x – 4]

2. \(\sqrt { 2 } \), \(\frac{1}{3}\)

Sum of the zeroes, S = \(\sqrt { 2 } \)

Product of the zeroes, P = \(\frac{1}{3}\)

âˆ´ Required polynomial is

p(x) = K[x^{2} – Sx + P]

3. 0, \(\sqrt { 5 } \)

Sum of the zeroes S = 0

Product of the zeroes, P = \(\sqrt { 5 } \)

âˆ´ Required polynomial is

p(x) = K[x^{2} – Sx + P]

= x^{2} – 0x + \(\sqrt { 5 } \)

p(x) = x^{2} + \(\sqrt { 5 } \)

4. 1, 1

Sum of the polynomial, S = 1

Product of the polynomial, P = 1

âˆ´ Required polynomial is

p(x) = K[x^{2} – Sx + P]

= x^{2} – 1x + 1

p(x) = x^{2} – x + 1

5. \(\frac{-1}{4}\), \(\frac{1}{4}\)

Sum of the zeroes, S = \(\frac{-1}{4}\)

Product of the zeroes, P = \(\frac{1}{4}\)

âˆ´ Required polynomial is

p(x) = K[x^{2} – Sx + P]

6. 4, 1

Sum of the zeroes, S = 4

Product of the zeroes, P = 1

âˆ´ Required polynomial is

p(x) = K[x^{2} – Sx + P]

p(x) = x^{2} – 4x + 1