NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials Exercise 2.2

The NCERT Solutions Maths Class 10 for Chapter 2 Exercise 2.2 Polynomials are provided here in PDF format. These solutions are created by Maths experts who have reviewed them from time to time. These NCERT chapter-wise solutions help the students to study and prepare well for their CBSE exams. The Solutions of NCERT for Exercise 2.2 Chapter 2 – Polynomials are given here in a step-wise and easy-to-understand format. Students will surely be able to solve these easily once they go through these NCERT Solutions for Class 10 Maths.

All the important factors, like paying attention to NCERT guidelines while preparing these solutions, have been focused upon here. Exercise 2.2 NCERT Class 10 Maths Solutions explains the relationship between Zeroes and Coefficients of Polynomials.

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Access other exercise Solutions of Class 10 Maths Chapter 2 – Polynomials

Exercise 2.1 Solutions 1 Question
Exercise 2.3 Solutions 5 Questions (2 short, 3 long)
Exercise 2.4 Solutions 5 Questions (2 short, 3 long)

Access answers to NCERT Class 10 Maths Chapter 2 – Polynomials Exercise 2.2

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

Solutions:

(i) x²–2x –8

⇒x²– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)

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

Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x²)

Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x²)

(ii) 4s²–4s+1

⇒4s²–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)

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

Sum of zeroes = (½)+(1/2) = 1 = -(-4)/4 = -(Coefficient of s)/(Coefficient of s²)

Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s² )

(iii) 6x²–3–7x

⇒6x²–7x–3 = 6x² – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)

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

Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x²)

Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x² )

(iv) 4u²+8u

⇒ 4u(u+2)

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

Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u²)

Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u² )

(v) t²–15

⇒ t² = 15 or t = ±√15

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

Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t²)

Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t² )

(vi) 3x²–x–4

⇒ 3x²–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)

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

Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x²)

Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x² )

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

(i) 1/4 , -1

Solution:

From the formulas of sum and product of zeroes, we know,

Sum of zeroes = α+β

Product of zeroes = α β

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

Thus, 4x²–x–4 is the quadratic polynomial.

(ii)√2, 1/3

Solution:

Sum of zeroes = α + β =√2

Product of zeroes = α β = 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

Thus, 3x²-3√2x+1 is the quadratic polynomial.

(iii) 0, √5

Solution:

Given,

Sum of zeroes = α+β = 0

Product of zeroes = α β = √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

Thus, x²+√5 is the quadratic polynomial.

(iv) 1, 1

Solution:

Given,

Sum of zeroes = α+β = 1

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–x+1 = 0

Thus, x²–x+1 is the quadratic polynomial.

(v) -1/4, 1/4

Solution:

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

Thus, 4x²+x+1 is the quadratic polynomial.

(vi) 4, 1

Solution:

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

Thus, x²–4x+1 is the quadratic polynomial.

Exercise 2.2 of NCERT Solutions for Class 10 Maths Chapter 2 is the second exercise of Polynomials of Class 10 Maths. Polynomials are introduced in Class 9, and it is further discussed in detail in Class 10 by studying different cases of relationship between Zeroes and Coefficients of a Polynomial.

• Relationship between Zeroes and Coefficients of a Polynomial – It includes two questions with six different cases each.

Key Features of NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials Exercise 2.2 Page number 33

• These NCERT Class 10 Solutions help students solve and revise all questions of Exercise 2.1.
• Solving these NCERT Solutions will help learners score well in exams.
• These are the best study resources as they are prepared by Maths subject experts.
• It follows NCERT guidelines, which help in preparing the students accordingly.
• It contains all the important questions from the examination point of view.