Class 10 Maths Chapter 2 Polynomials Exercise 2.2 NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2 focus on explaining the relationship between the Zeros of a Polynomial and its coefficients. The solutions cover linear, quadratic, and cubic polynomials with clear step-by-step methods that make each concept easier to understand. 

These NCERT Solutions are designed according to the CBSE syllabus. It will help you prepare effectively for board exams. Using these solutions, you can verify answers and gain confidence in scoring well in the polynomial concepts. 

NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2

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

Solutions:

(i) x 2 –2x –8

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

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

Sum of zeroes = 4–2 = 2 = -(-2)/1 = 

-(Coefficient of x)/(Coefficient of x 2 ) 

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

= (Constant term)/(Coefficient of x 2 )

(ii) 4s 2 –4s+1

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

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

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

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

(iii) 6x 2 –3–7x

⇒6x 2 –7x–3 = 6x 2 – 9x + 2x – 3 

= 3x(2x – 3) +1(2x – 3) 

= (3x+1)(2x-3) 

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

Sum of zeroes = -(1/3)+(3/2) = (7/6) 

= -(Coefficient of x)/(Coefficient of x 2 ) 

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

= (Constant term) /(Coefficient of x 2 )

(iv) 4u 2 +8u

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

Sum of zeroes = 0+(-2) = -2 = -(8/4) 

= -(Coefficient of u)/(Coefficient of u 2 ) 

Product of zeroes = 0×-2 = 0 = 0/4 

= (Constant term)/(Coefficient of u 2 )

(v) t 2 –15

⇒ t 2 = 15 or t = ±√15 

Therefore, zeroes of the polynomial equation t 2 –15 are (√15, -√15) 

Sum of zeroes =√15+(-√15) = 0= -(0/1)

= -(Coefficient of t) / (Coefficient of t 2 ) Product of zeroes 

= √15×(-√15) = -15 = -15/1 

= (Constant term) / (Coefficient of t 2 )

(vi) 3x 2 –x–4

⇒ 3x 2 –4x+3x–4 = x(3x-4)+1(3x-4) 

= (3x – 4)(x + 1) 

Therefore, zeroes of the polynomial equation 3x 2 – x – 4 are (4/3, -1) 

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

= -(-1/3) = -(Coefficient of x) / (Coefficient of x 2 ) 

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

= (Constant term) /(Coefficient of x 2 )

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

x 2 –(α+β)x +αβ = 0

x 2 –(1/4)x +(-1) = 0

4x 2 –x-4 = 0

Thus, 4x 2 –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

x 2 –(α+β)x +αβ = 0

x 2 –( √2 )x + (1/3) = 0

3x 2 -3√2x+1 = 0

Thus, 3x 2 -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

x 2 –(α+β)x +αβ = 0

x 2 –(0)x + √5 = 0

Thus, x 2 +√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

x 2 –(α+β)x +αβ = 0

x 2 –x+1 = 0

Thus, x 2 –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

x 2 –(α+β)x +αβ = 0

x 2 –(-1/4)x +(1/4) = 0

4x 2 +x+1 = 0

Thus, 4x 2 +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

x 2 –(α+β)x+αβ = 0

x 2 –4x+1 = 0

Thus, x 2 –4x+1 is the quadratic polynomial.

How to Score Better in Class 10 Maths Exam?

Scoring well in Class 10 Maths requires clear concepts, regular practice, and a focus on accuracy and answer presentation. To score better, you should:

• Build Strong Concepts:

Focus on understanding concepts in Class 10 Maths instead of memorising steps, as this helps in solving application-based questions.

• Practise Regularly:

Solve all CBSE Class 10 NCERT questions multiple times to strengthen your basics and improve accuracy.

• Revise Formulas Daily:

Regular revision of PW Class 10 Maths MIQs helps avoid calculation mistakes in exams.

• Solve Previous Year Papers:

Practising CBSE Class 10 Maths previous year questions (PYQs) helps you understand question patterns and important topics.

• Attempt Sample Papers:

Solving PW Class 10 Maths sample papers improves time management and gives you exam-like practice.

• Work on Weak Areas:

Focus on difficult topics from the Class 10 Maths syllabus instead of skipping them to avoid losing marks.

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