These NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

## Exercise 2.2

Question 1.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x^{2} – 2x – 8

(ii) 4s^{2} – 4s + 1

(iii) 6x^{2} – 3 – 7x

(iv) 4u^{2} + 8u

(v) t^{2} -15

(vi) 3x^{2} – x – 4

Solution:

(i) x² – 2x – 8

⇒ x² – 4x + 2x – 8

⇒ x(x – 4) + 2(x – 4)

⇒ (x – 4) (x + 2)

⇒ x – 4 = 0 or x + 2 = 0

⇒ x = 4 or x= -2

Verification:

(ii) 4s^{2} – 4s + 1

⇒ 4s² – 2s – 2s + 1

⇒ 2s(2s – 1) – 1(2s – 1)

⇒ (2s – 1) (2s – 1)

2s – 1 = 0 or 2s – 1 = 0

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

Verification:

(iii) 6x^{2} – 3 – 7x

⇒ 6x² – 7x + 2x – 3

⇒ 3x(2x – 3) + 1(2x – 3)

⇒ (2x – 3) (3x + 1)

2x – 3 = 0 or 3x + 1 = 0

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

Verification:

(iv) 4u^{2} + 8u

⇒ 4u(u + 2)

⇒ u – 4 = 0 or u + 2 = 0

⇒ u = 4 or u= – 2

Verification:

(v) t^{2} -15

⇒ t² = 15

⇒ t = ± \(\sqrt{15}\)

t = \(\sqrt{15}\) or t = – \(\sqrt{15}\)

Verification:

(vi) 3x^{2} – x – 4

⇒ 3x² – 4x + 3x – 4

⇒ x(3x – 4) + 1(3x – 4)

⇒ (3x – 4) (x + 1)

3x – 4 = 0 or x + 1 = 0

⇒ x = \(\frac { 4 }{ 3 }\) or x = – 1

Verification:

Question 2.

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

(i) \(\frac { 1 }{ 4 }\), – 1

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

(iii) 0, \(\sqrt{5}\)

(iv) 1, 1

(v) \(\frac { -1 }{ 4 }\), \(\frac { 1 }{ 4 }\)

(vi) 4, 1

Solution:

(i) Let the polynomial be ax² + bx + c and its zeroes be α and ß.

Then α + β = \(\frac { 1 }{ 4 }\) = \(\frac { -b }{ a }\)

and α.β = – 1 = \(\frac { c }{ a }\)

If a = 4, then b = – 1 and c = – 4

So, one quadratic polynomial which fits the given condition is 4x² – x – 4.

(ii) Let the polynomial be ax² + bx + c and its zeroes be α and ß.

Then α + β = \(\sqrt{2}\) = \(\frac { -b }{ a }\)

and α.β = \(\frac { 1 }{ 3 }\) = \(\frac { c }{ a }\)

If a = 4, then b = – \(\sqrt{2}\) and c = \(\frac { 1 }{ 3 }\)

So, one quadratic polynomial which fits the given condition is x² – \(\sqrt{2x}\) + \(\frac { 1 }{ 3 }\) or 3x² – 3\(\sqrt{2x}\) + 1

(iii) Let the polynomial be ax² + bx + c and its zeroes be α and ß.

Then α + β = 0 = \(\frac { -b }{ a }\)

and α.β = \(\sqrt{5}\) = \(\frac { c }{ a }\)

If a = 1, then b = 0 and c = \(\sqrt{5}\)

So, one quadratic polynomial which fits the given condition is x² – \(\sqrt{5}\)

(iv) Let the polynomial be ax² + bx + c and its zeroes be α and ß.

Then α + β = 1 = \(\frac { -b }{ a }\)

and α.β = 1 = \(\frac { c }{ a }\)

If a = 1, then b = – 1 and c = 1

Now, put the values of a, b and c in equation ax² + bx + c

So, one quadratic polynomial which fits the given condition is x² – \(\sqrt{5}\)

1x² – 1x 1 = 0

or x² – x + 1 = 0

(v) Let the polynomial be ax² + bx + c and its zeroes be α and ß.

Then α + β = –\(\frac { 1 }{ 4 }\) = \(\frac { -b }{ a }\)

and α.β = \(\frac { 1 }{ 4 }\) = \(\frac { c }{ a }\)

If a = 1, then b = 1 and c = 1

Now, put the values of a, b and c in equation ax² + bx + c

So, one quadratic polynomial which fits the given condition is 4x² + x + 1 = 0

(vi) Let the polynomial be ax² + bx + c and its zeroes be α and ß.

Then α + β = 4 = \(\frac { -b }{ a }\)

and α.β = 1 = \(\frac { c }{ a }\)

If a = 1, then b = – 4 and c = 1

So, one quadratic polynomial which fits the given condition is x² – 4x + 1 = 0