Class 9 Maths Chapter 4 Exercise 4.2 Solutions – Ganita Manjari
Welcome to Sid Classes, your trusted destination for accurate, reliable, and student-friendly academic resources! If you are working through polynomials and algebraic factorizations, you are in the right place. In this detailed study guide, we provide complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.2. Designed to reflect how top-scoring students write their answers in school examinations, this walkthrough makes complex algebraic expansions and identity applications entirely clear.
Our expert-written solutions focus on factoring quadratic expressions completely and evaluating numbers efficiently using the standard subtraction identity \((a - b)^2 = a^2 - 2ab + b^2\). Follow along to strengthen your problem-solving skills with absolute confidence.
🔑 Key Identities Used:
- Expansion Identity: \((a - b)^2 = a^2 - 2ab + b^2\)
- Factorization Identity: \(a^2 + 2ab + b^2 = (a + b)^2\)
- Common Factor Rule: Factoring out a rational coefficient (fractional constant) first can simplify awkward coefficients into standard perfect square trinomials.
NCERT Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.2 Solutions
Question 1
Factor completely:
Solution:
(i) \(9x^2 + 24xy + 16y^2\)
Expressing the terms in the form \(a^2 + 2ab + b^2\):
\(= (3x)^2 + 2(3x)(4y) + (4y)^2\)
Using the identity \(a^2 + 2ab + b^2 = (a + b)^2\), where \(a = 3x\) and \(b = 4y\):
\(= \mathbf{(3x + 4y)^2}\)
(ii) \(4s^2 + 20st + 25t^2\)
\(= (2s)^2 + 2(2s)(5t) + (5t)^2\)
Using the identity with \(a = 2s\) and \(b = 5t\):
\(= \mathbf{(2s + 5t)^2}\)
(iii) \(49x^2 + 28xy + 4y^2\)
\(= (7x)^2 + 2(7x)(2y) + (2y)^2\)
Using the identity with \(a = 7x\) and \(b = 2y\):
\(= \mathbf{(7x + 2y)^2}\)
(iv) \(64p^2 + \frac{32}{3}pq + \frac{4}{9}q^2\)
\(= (8p)^2 + 2(8p)\left(\frac{2}{3}q\right) + \left(\frac{2}{3}q\right)^2\)
Using the identity with \(a = 8p\) and \(b = \frac{2}{3}q\):
\(= \mathbf{\left(8p + \frac{2}{3}q\right)^2}\)
(v) \(3a^2 + 4ab + \frac{4}{3}b^2\)
Taking out \(\frac{1}{3}\) as a common factor:
\(= \frac{1}{3}(9a^2 + 12ab + 4b^2)\)
\(= \frac{1}{3}\left[(3a)^2 + 2(3a)(2b) + (2b)^2\right]\)
\(= \mathbf{\frac{1}{3}(3a + 2b)^2}\)
(vi) \(\frac{9}{5}s^2 + 6sv + 5v^2\)
Taking out \(\frac{1}{5}\) as a common factor:
\(= \frac{1}{5}(9s^2 + 30sv + 25v^2)\)
\(= \frac{1}{5}\left[(3s)^2 + 2(3s)(5v) + (5v)^2\right]\)
\(= \mathbf{\frac{1}{5}(3s + 5v)^2}\)
Question 2
Find the values of the following using the identity \((a - b)^2 = a^2 - 2ab + b^2\):
Solution:
(i) \((79)^2\)
Expressing \(79\) as a difference from a multiple of 10:
\(= (80 - 1)^2\)
Here, \(a = 80\) and \(b = 1\). Using \((a - b)^2 = a^2 - 2ab + b^2\):
\(= (80)^2 - 2(80)(1) + (1)^2\)
\(= 6400 - 160 + 1\)
\(= \mathbf{6241}\)
(ii) \((193)^2\)
Expressing \(193\) as a difference:
\(= (200 - 7)^2\)
Here, \(a = 200\) and \(b = 7\). Using the identity:
\(= (200)^2 - 2(200)(7) + (7)^2\)
\(= 40000 - 2800 + 49\)
\(= \mathbf{37249}\)
(iii) \((299)^2\)
Expressing \(299\) as a difference:
\(= (300 - 1)^2\)
Here, \(a = 300\) and \(b = 1\). Using the identity:
\(= (300)^2 - 2(300)(1) + (1)^2\)
\(= 90000 - 600 + 1\)
\(= \mathbf{89401}\)
Frequently Asked Questions (FAQs)
Q1: How do you factorize expressions like \(9x^2 + 24xy + 16y^2\)?
A: Recognize them as perfect square trinomials matching the identity \(a^2 + 2ab + b^2 = (a + b)^2\) by expressing the first and last terms as squares of monomials.
Q2: What should you do when coefficients are fractions, like in Question 1(v)?
A: Factor out the common denominator fraction (e.g., factoring out \(\frac{1}{3}\)) to transform the remaining expression into a clean integer-coefficient trinomial before applying the identity.
Q3: How do you evaluate numbers like \((79)^2\) using identities?
A: Rewrite the number as a subtraction from the nearest multiple of 10 (e.g., \(80 - 1\)) and apply the identity \((a - b)^2 = a^2 - 2ab + b^2\).
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