Class 9 Maths Chapter 4 Exercise 4.4 Solutions – Ganita Manjari
Welcome to Sid Classes, your trusted destination for accurate, reliable, and student-friendly academic resources! In this detailed study guide, we provide complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.4. Designed to reflect how top-scoring students write their answers in school examinations, this walkthrough makes splitting middle terms, algebraic identities, and product evaluations entirely clear.
🔑 Key Concepts Used in this Exercise:
- Splitting the Middle Term: Factoring quadratic polynomials of the form \(ax^2 + bx + c\) by finding two numbers whose sum is \(b\) and product is \(ac\).
- Algebraic Identities: Using \((a \pm b)^2 = a^2 \pm 2ab + b^2\), \((a+b)(a-b) = a^2 - b^2\), and \((a+b+c)^2\) for efficient mental calculations and polynomial factorization.
NCERT Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.4 Solutions
Question 1
Fill in the blanks to complete the following identities:
Solution:
(i) \(s^2 - 11s + 24 = ( \quad )( \quad )\)
Splitting the middle term \(-11s\) into two parts whose sum is \(-11\) and product is \(24\) (i.e., \(-8\) and \(-3\)):
\(= s^2 - 8s - 3s + 24\)
\(= s(s - 8) - 3(s - 8)\)
\(= \mathbf{(s - 8)(s - 3)}\)
(ii) \(( \quad )(x + 1) = (3x^2 - 4x - 7)\)
Factorizing the quadratic expression \(3x^2 - 4x - 7\):
Split the middle term \(-4x\) into \(-7x + 3x\) (since \(-7 \times 3 = -21 = 3 \times (-7)\)):
\(= 3x^2 - 7x + 3x - 7\)
\(= x(3x - 7) + 1(3x - 7)\)
\(= (3x - 7)(x + 1)\)
Therefore, the missing blank is \(\mathbf{(3x - 7)}\).
(iii) \(10x^2 - 11x - 6 = (2x - \quad )( \quad + 2)\)
Factorizing \(10x^2 - 11x - 6\) by splitting the middle term \(-11x\) into \(-15x + 4x\) (since \(-15 \times 4 = -60 = 10 \times (-6)\)):
\(= 10x^2 - 15x + 4x - 6\)
\(= 5x(2x - 3) + 2(2x - 3)\)
\(= (2x - 3)(5x + 2)\)
Matching with \((2x - \quad )( \quad + 2)\), the blanks are \(\mathbf{3}\) and \(\mathbf{5x}\), giving \(\mathbf{(2x - 3)(5x + 2)}\).
(iv) \(6x^2 + 7x + 2 = ( \quad )( \quad )\)
Splitting the middle term \(7x\) into \(4x + 3x\) (since \(4 \times 3 = 12 = 6 \times 2\)):
\(= 6x^2 + 4x + 3x + 2\)
\(= 2x(3x + 2) + 1(3x + 2)\)
\(= \mathbf{(2x + 1)(3x + 2)}\)
Question 2
Select and use the identity that will help you to find the following products without multiplying directly:
Solution:
(i) \((41)^2\)
\(= (40 + 1)^2\)
Using \((a + b)^2 = a^2 + 2ab + b^2\) with \(a = 40\) and \(b = 1\):
\(= (40)^2 + 2(40)(1) + (1)^2\)
\(= 1600 + 80 + 1 = \mathbf{1681}\)
(ii) \((27)^2\)
\(= (30 - 3)^2\)
Using \((a - b)^2 = a^2 - 2ab + b^2\) with \(a = 30\) and \(b = 3\):
\(= (30)^2 - 2(30)(3) + (3)^2\)
\(= 900 - 180 + 9 = \mathbf{729}\)
(iii) \((23 \times 17)\)
\(= (20 + 3)(20 - 3)\)
Using \((a + b)(a - b) = a^2 - b^2\) with \(a = 20\) and \(b = 3\):
\(= (20)^2 - (3)^2\)
\(= 400 - 9 = \mathbf{391}\)
(iv) \((135)^2\)
\(= (130 + 5)^2\)
Using \((a + b)^2 = a^2 + 2ab + b^2\) with \(a = 130\) and \(b = 5\):
\(= (130)^2 + 2(130)(5) + (5)^2\)
\(= 16900 + 1300 + 25 = \mathbf{18225}\)
(v) \((97)^2\)
\(= (100 - 3)^2\)
Using \((a - b)^2 = a^2 - 2ab + b^2\) with \(a = 100\) and \(b = 3\):
\(= (100)^2 - 2(100)(3) + (3)^2\)
\(= 10000 - 600 + 9 = \mathbf{9409}\)
(vi) \((18 \times 29)\)
\(= (20 - 2) \times 29 = 20(29) - 2(29)\)
\(= 580 - 58 = \mathbf{522}\)
(Alternatively, using standard algebraic distribution).
(vii) \((34 \times 43)\)
\(= (34) \times (40 + 3) = 34(40) + 34(3)\)
\(= 1360 + 102 = \mathbf{1462}\)
(viii) \((205)^2\)
\(= (200 + 5)^2\)
Using \((a + b)^2 = a^2 + 2ab + b^2\) with \(a = 200\) and \(b = 5\):
\(= (200)^2 + 2(200)(5) + (5)^2\)
\(= 40000 + 2000 + 25 = \mathbf{42025}\)
Question 3
Factor the following:
Solution:
(i) \(9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc\)
Using the trinomial identity \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\).
Notice the negative signs are in \(-6ab\) and \(-4bc\), which indicates that variable \(b\) is negative.
\(= (3a)^2 + (-b)^2 + (2c)^2 + 2(3a)(-b) + 2(-b)(2c) + 2(2c)(3a)\)
\(= \mathbf{(3a - b + 2c)^2}\)
(ii) \(16s^2 + 25t^2 - 40st\)
\(= (4s)^2 - 2(4s)(5t) + (5t)^2\)
Using \((a - b)^2 = a^2 - 2ab + b^2\):
\(= \mathbf{(4s - 5t)^2}\)
(iii) \(r^2 - r - 42\)
Splitting the middle term \(-r\) into \(-7r + 6r\) (since \(-7 \times 6 = -42\) and \(-7 + 6 = -1\)):
\(= r^2 - 7r + 6r - 42\)
\(= r(r - 7) + 6(r - 7)\)
\(= \mathbf{(r - 7)(r + 6)}\)
(iv) \(49g^2 + 14gh + h^2\)
\(= (7g)^2 + 2(7g)(h) + (h)^2\)
Using \((a + b)^2 = a^2 + 2ab + b^2\):
\(= \mathbf{(7g + h)^2}\)
(v) \(64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw\)
Using the trinomial identity \((a + b + c)^2\). Notice negative terms \(-176uv\) and \(-32uw\) indicate \(u\) is positive while \(v\) and \(w\) are negative (or vice versa; check \(+44vw\) which is \((-11v)(-2w) = +22\), so both \(v\) and \(w\) share the same sign opposite to \(u\)):
\(= (8u)^2 + (-11v)^2 + (-2w)^2 + 2(8u)(-11v) + 2(-11v)(-2w) + 2(-2w)(8u)\)
\(= \mathbf{(8u - 11v - 2w)^2}\)
Frequently Asked Questions (FAQs)
Q1: How do you split the middle term for quadratic expressions like \(s^2 - 11s + 24\)?
A: Find two numbers whose product equals the constant term (\(24\)) and whose sum equals the coefficient of the middle term (\(-11\)). Here, \(-8\) and \(-3\) fit perfectly.
Q2: How do we determine signs when factoring trinomial squares like \(9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc\)?
A: Identify which variable appears in all negative product terms. Since both \(-6ab\) and \(-4bc\) involve \(b\), \(b\) must be negative.
Q3: Can identities be used to multiply non-square numbers directly?
A: Yes, by expressing numbers as sums or differences from multiples of 10 or 100 (e.g., \(23 \times 17\) as \((20+3)(20-3)\)), you can apply standard algebraic identities quickly.
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