Exercise 2.2 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

Exercise 2.2 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

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Chapter 2. Polynomials

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Exercise 2.2 Class 10 maths Chapter 2. Polynomials

Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x² – 2x – 8

Solution:

Using Middle term splitting method:

       x² – 4x + 2x – 8 = 0

⇒ x(x – 4) + 2(x – 4) = 0

⇒  x(x – 4) + 2(x – 4) = 0

⇒ (x – 4) (x + 2) = 0

⇒ x – 4 = 0 x + 2 = 0

⇒ x = 4, x = – 2

Zeroes;  α = 4  β = – 2

Verifying the relationship between the zeroes and the coefficients.

a = 1, b = – 2, and c = – 8

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(ii) 4s² – 4s + 1

(iii) 6x² – 3 – 7x

(iv) 4u² + 8u

(v) t² – 15

(vi) 3x² – x – 4

Solution:​

(ii) 4s² – 4s + 1

Solution: 

4s²– 4s + 1 = 0

⇒ 4s² – 2s – 2s + 1 = 0

⇒ 2s (2s – 1) –1(2s – 1) = 0

⇒ (2s – 1) (2s – 1) = 0

⇒ 2s – 1= 0, 2s – 1= 0

⇒ 2s = 1, 2s = 1

Coefficients

a =  4, b = – 4, c = 1

Relationship between zeroes and coefficients

L.H.S = R.H.S

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L.H.S = R.H.S

Hence verified the relationship between coefficients and the zeroes in both cases.

(iii) 6x² – 3 – 7x

Solution:

⇒ 6x² – 3 – 7x = 0

⇒ 6x² – 7x – 3 = 0 (After rearranging the equation) 

⇒ 6x² – 9x + 2x – 3 = 0

⇒ 3x (2x – 3) +1 (2x – 3) = 0

⇒ (2x – 3) (3x + 1) = 0

⇒ 2x – 3 = 0,  3x + 1 = 0

⇒ 2x = 3, 3x = - 1

Hence verified the relationship between coefficients and the zeroes in both cases.

(iv) 4u² + 8u

Solution:  4u(u + 2) = 0

4u = 0, u + 2 = 0

u = 0,  u =  – 2

Zeroes: a = 0, b = – 2

Coefficients:

a = 4, b = 8, c = 0

Verifying relationship between zeroes and coefficients

⇒  0 = 0

⇒ L.H.S = R.H.S

Hence verified, the relationship between coefficients and zeroes in both cases.

(v) t² – 15

Solution:

t² – 15 = 0

t² = 15

t = ±√15

t = √15,    t = – √15

Zeroes: a = √15, b = – √15

Coefficients a = 1, b = 0, c = – 15

Verifying relationship between zeroes and Coefficients

Hence verified, the relationship between coefficients and zeroes in both cases.

(vi) 3x² – x – 4

Solution:

     3x² – x – 4 = 0

⇒ 3x² + 3x – 4x – 4 = 0

⇒ 3x (x + 1) – 4 (x + 1) = 0

⇒ (x + 1) (3x – 4) = 0

⇒ x  + 1 = 0, 3x – 4 = 0

⇒ x = –1 and 3x = 4

        L.H.S= R.H.S

Hence verified, the relationship between coefficients and zeroes in both cases.

Q2. . Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

 

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Hence, required quadratic polynomial is 4x² – x – 4

Hence, required quadratic polynomial is 3x² –3 x + 1

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Hence, required quadratic polynomial is x² + √5

Hence, required quadratic polynomial is x² – x + 1

Hence, required quadratic polynomial is 4X² + x + 1

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 Hence, required quadratic polynomial is x² – 4  + 1