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Class 10 NCERT Solutions – Chapter 2 Polynomials – Exercise 2.2

  • Last Updated : 25 Feb, 2021
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Question 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x2 – 2x – 8

x2 – 2x – 8 = x2 – 4x + 2x – 8 

= x (x – 4) + 2(x – 4) 

= (x – 4) (x + 2)

Therefore, zeroes of equation x2 – 2x – 8 are (4, -2)



Sum of zeroes is equal to [4 – 2]= 2 = -(-2)/1 

i.e. = -(Coefficient of x) / (Coefficient of x2)

Product of zeroes is equal to 4 × (-2) = -8 =-(8)/1  

i.e.= (Constant term) / (Coefficient of x2)

(ii) 4s2 – 4s + 1

4s2 – 4s + 1 = 4s2 – 2s – 2s +1 

= 2s(2s – 1) – 1(2s – 1) 

= (2s – 1) (2s – 1)

Therefore, zeroes of  equation 4s2 – 4s +1 are (1/2, 1/2)

Sum of zeroes is equal to [(1/2) + (1/2)] = 1 = -4/4 

i.e.= -(Coefficient of s) / (Coefficient of s2)

Product of zeros is equal to [(1/2) × (1/2)] = 1/4 

i.e.= (Constant term) / (Coefficient of s2 )

(iii) 6x2 – 3 – 7x

6x2 – 3 – 7x = 6x2 – 7x – 3 

= 6x2 – 9x + 2x – 3 

= 3x(2x – 3) + 1(2x – 3) 

= (3x + 1) (2x – 3)



Therefore, zeroes of equation 6x2 – 3 – 7x are (-1/3, 3/2)

Sum of zeroes is equal to -(1/3) + (3/2) = (7/6) 

i.e.= -(Coefficient of x) / (Coefficient of x2)

Product of zeroes is equal to -(1/3) × (3/2) = -(3/6) 

i.e.= (Constant term) / (Coefficient of x2 )

(iv) 4u2 + 8u

4u2 + 8u = 4u(u + 2)

Therefore, zeroes of equation 4u2 + 8u are (0, -2).

Sum of zeroes is equal to [0 + (-2)] = -2 = -(8/4)

i.e. = -(Coefficient of u) / (Coefficient of u2)

Product of zeroes is equal to 0 × -2 = 0 = 0/4 

i.e. = (Constant term) / (Coefficient of u2 )

(v) t2 – 15

 t2 – 15

⇒ t2 = 15 or t = ±√15

Therefore, zeroes of equation t2 – 15 are (√15, -√15)

Sum of zeroes is equal to [√15 + (-√15)] = 0 = -(0/1) 

i.e.= -(Coefficient of t) / (Coefficient of t2)

Product of zeroes is equal to √15 × (-√15) = -15 = -15/1 

i.e. = (Constant term) / (Coefficient of t2 )



(vi) 3x2 – x – 4

3x2 – x – 4 = 3x2 – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4) 

= (3x – 4) (x + 1)

Therefore, zeroes of equation 3x2 – x – 4 are (4/3, -1)

Sum of zeroes is equal to (4/3) + (-1) = (1/3) = -(-1/3) 

i.e. = -(Coefficient of x) / (Coefficient of x2)

Product of zeroes is equal to (4/3) × (-1) = (-4/3) 

i.e. = (Constant term) / (Coefficient of x2)

Question 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4, -1

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 1/4

Product of zeroes = α β = -1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2 – (α + β)x + αβ = 0

x2 – (1/4)x +(-1) = 0

4x2 – x – 4 = 0

∴ 4x2 – x – 4 is the quadratic polynomial.



(ii) √2, 1/3

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given Sum of zeroes = α + β =√2

Product of zeroes = αβ = 1/3

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:

x2 – (α + β)x + αβ = 0

x2 – (√2)x + (1/3) = 0

3x2 – 3√2x + 1 = 0

∴ 3x2 – 3√2x + 1 is the quadratic polynomial.

(iii) 0, √5

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 0

Product of zeroes = αβ = √5

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2 – (α + β)x + αβ = 0

x2 – (0)x + √5 = 0

∴ x2 + √5 is the quadratic polynomial.

(iv) 1, 1

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 1

Product of zeroes = αβ = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2 – (α + β)x + αβ = 0

x2 – x + 1 = 0

∴ x2 – x + 1 is the quadratic polynomial.

(v) -1/4, 1/4

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = -1/4

Product of zeroes = α β = 1/4

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2 – (α + β)x + αβ = 0

x2 – (-1/4)x + (1/4) = 0



4x2 + x + 1 = 0

∴ 4x2 + x + 1 is the quadratic polynomial.

(vi) 4, 1

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 4

Product of zeroes = αβ = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:

x2 – (α + β)x + αβ = 0

x2 – 4x + 1 = 0

∴ x2 – 4x + 1 is the quadratic polynomial.

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