NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.2

NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.2 focuses on the relationship between the zeroes and the coefficients of a polynomial. This is an important concept that helps students understand how the roots of a quadratic polynomial are connected to the numbers in the expression. The questions in this exercise require you to find the zeroes of a given quadratic polynomial and then verify the relationship using formulas.

NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.2 Solutions

Question 1
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients:
(i) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u
(v) t² – 15
(vi) 3x² – x – 4

Solution:
(i) x² – 2x – 8

Factorizing:
x² – 2x – 8 = (x – 4)(x + 2)
So, zeroes are x = 4 and x = –2

Sum of zeroes = 4 + (–2) = 2
Product of zeroes = 4 × (–2) = –8

Now, from the polynomial:
a = 1, b = –2, c = –8
–b/a = –(–2)/1 = 2
c/a = –8/1 = –8

Verified

(ii) 4s² – 4s + 1

Factorizing:
4s² – 4s + 1 = (2s – 1)²
Zeroes are s = 1/2 and s = 1/2

Sum = 1/2 + 1/2 = 1
Product = (1/2) × (1/2) = 1/4

Now, a = 4, b = –4, c = 1
–b/a = –(–4)/4 = 1
c/a = 1/4

Verified

(iii) 6x² – 3 – 7x

Rewriting:
6x² – 7x – 3

Factorizing:
6x² – 9x + 2x – 3 = 3x(2x – 3) + 1(2x – 3)
= (3x + 1)(2x – 3)

Zeroes are x = –1/3 and x = 3/2

Sum = –1/3 + 3/2 = (–2 + 9)/6 = 7/6
Product = (–1/3) × (3/2) = –1/2

Now, a = 6, b = –7, c = –3
–b/a = 7/6
c/a = –3/6 = –1/2

Verified

(iv) 4u² + 8u

Taking common factor:
= 4u(u + 2)

Zeroes are u = 0 and u = –2

Sum = 0 + (–2) = –2
Product = 0 × (–2) = 0

Now, a = 4, b = 8, c = 0
–b/a = –8/4 = –2
c/a = 0/4 = 0

Verified

(v) t² – 15

= t² – 0t – 15
Factorizing:
= (t – √15)(t + √15)

Zeroes: t = √15 and t = –√15

Sum = √15 + (–√15) = 0
Product = √15 × (–√15) = –15

Now, a = 1, b = 0, c = –15
–b/a = 0
c/a = –15

Verified

(vi) 3x² – x – 4

Factorizing:
3x² – 4x + 3x – 4 = x(3x – 4) + 1(3x – 4)
= (x + 1)(3x – 4)

Zeroes: x = –1 and x = 4/3

Sum = –1 + 4/3 = (–3 + 4)/3 = 1/3
Product = –1 × 4/3 = –4/3

Now, a = 3, b = –1, c = –4
–b/a = 1/3
c/a = –4/3

Verified

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

Solution:
If α and β are the zeroes (roots), and:

• Sum of roots = α + β = S
• Product of roots = α·β = P

Then the quadratic polynomial is:

x2–(Sum of roots)⋅x+(Product of roots)

(i) Sum = 1/4, Product = –1

Polynomial:
x² – (1/4)x – 1

(ii) Sum = √2, Product = 1/3

Polynomial:
x² – √2·x + 1/3

(iii) Sum = 0, Product = √5

Polynomial:
x² + √5
(Note: since sum = 0, middle term is missing)

(iv) Sum = 1, Product = 1

Polynomial:
x² – x + 1

(v) Sum = –1/4, Product = 1/4

Polynomial:
x² + (1/4)x + 1/4
(Note: –(–1/4) = +1/4)

(vi) Sum = 4, Product = 1

Polynomial:
x² – 4x + 1

You can access the official NCERT Solutions for Class 10 Mathematics on the NCERT website at the following link:

NCERT Class 10 Mathematics Solutions

This page will guide you to the textbook and solutions, as provided by the National Council of Educational Research and Training (NCERT).

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By solving NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.2, students gain a strong understanding of quadratic equations and how to determine their roots using factorization. Practicing from NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.2 helps build confidence in identifying the nature of roots and forming equations from given conditions.

These NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.2 are designed to strengthen your conceptual clarity and problem-solving approach. Whether you’re revising or preparing for exams, NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.2 serve as a reliable guide. Make sure to regularly practice NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.2 to boost your accuracy and performance in exams.