In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.2. These solutions are based on new NCERT Syllabus.

## NCERT 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.

Solution:

(i) we have p(x) = x^{2} – 2x – 8

= x^{2} + 2x + – 4x – 8

= x(x + 2) – 4(x + 2)

= (x – 4)(x + 2)

For p(x) = 0, we must have (x – 4)(x + 2) = 0

Either x – 4 = 0 ⇒ x = 4

or x + 2 = 0 ⇒ x = -2

∴ The zeroes of x^{2} – 2x – 8 are 4 and -2

Now, sum of zeroes

Thus, the relationship between the zeroes and the coefficients in the polynomial x^{2} – 2x – 8 is verified.

(ii) We have p(s) = 4s^{2} – 4s + 1

= 4s^{2} – 2s – 2s + 1

= 2s(2s – 1) – l(2s – 1)

= (2s – 1)(2s – 1)

Thus, the relationship between the zeroes and coefficients in the polynomial 4s^{2} – 4s + 1 is verified.

(iii) We have p(x) = 6x^{2} – 3 – 7x

= 6x^{2} – 7x – 3

= 6x^{2} – 9x + 2x – 3

= 3x(2x – 3) + 1(2x – 3)

= (3x + 1)(2x – 3)

For p(x) = 0, we have,

Thus, the relationship between the zeroes and coefficients in the polynomial 6x^{2} – 3 – 7x is verified.

(iv) We have, p(u) = 4u^{2} + 8u = 4u(u + 2)

For p(u) = 0,

we have

Either 4u = 0 ⇒ u = 0

or u + 2 = 0 ⇒ u = -2

∴ The zeroes of 4u^{2} + 8u are 0 and – 2.

Now, 4u^{2} + 8u can be written as 4u^{2} + 8u + 0.

Thus, the relationship between the zeroes and coefficients in the polynomial 4u^{2} – 4s + 1 is verified.

(v) We have, p(t) = t^{2} – 15 = (t)^{2} – \((\sqrt{15})^{2}\)

Thus, the relationship between zeroes and the coefficients in the polynomial t^{2} – 15 is verified.

(vi) We have, p(x) = 3x^{2} – x – 4

= 3x^{2} + 3x – 4x – 4

= 3x(x + 1) – 4(x + 1)

= (x + 1)(3x – 4)

For p(x) = 0 we have,

Either (x + 1) = 0 ⇒ x = -1

Thus, the relationship between the zeroes and coefficients in the polynomial 3x^{2} – x – 4 is verified

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) Since, sum of zeroes, \((\alpha+\beta)=\frac{1}{4}\)

Product of zeroes, αβ = -1

∴ The required quadratic polynomial is

have samezeroes, therefore (4x^{2} – x – 4) is the required quadratic polynomial.

(ii) Since, sum of zeroes, \((\alpha+\beta)=\sqrt{2}\)

product of zeroes, αβ = \(\frac{1}{3}\)

∴ The required quadratic polynomial is

(iii) Since, sum of zeroes, (α + β) = 0

Product of zeroes, αβ = \(\sqrt{5}\)

∴ The required quadratic polynomial is

(iv) Since, sum of zeroes, (α + β) = 1

Product of zeroes, αβ = 1

∴ The required quadratic polynomial is

(v) Since, sum of zeroes, (α + β) = \(-\frac{1}{4}\)

Product of zeroes, αβ = \(\)[\frac{1}{4}/latex]

∴ The required quadratic polynomial is

same zeroes, therefore, the required quadratic polynomial is (4x^{2} + x + 1)

(vi) Since, sum of zeroes, (α + β) = 4 and

product of zeroes, αβ = 1

∴ The required quadratic polynomial is

x^{2} – (α + β)x + αβ = x^{2} – 4x + 1