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Exercise 7.5 | Set 1 in Chapter 7 of RD Sharma's Class 8 Mathematics textbook typically focuses on the application of factorization techniques to solve equations. This exercise builds upon the factorization methods learned in previous sections and introduces students to the concept of using factorization as a problem-solving tool for equations. It aims to enhance students' algebraic thinking and prepare them for more advanced equation-solving techniques in higher mathematics.
Solution:
16x2 - 25y2
= (4x)2 - (5y)2
= (4x - 5y)(4x + 5y)
Solution:
27x2 - 12y2
= 3(9x2 - 4y2)
= 3[(3x)2 - (2y)2]
= 3(3x - 2y)(3x + 2y)
Solution:
144a2 - 289b2
= (12a)2 - (17b)2
= (12a - 17b)(12a + 17b)
Solution:
12m2 - 27
= 3(4m2 - 9)
= 3[(2m)2 - 32]
= 3(2m - 3)(2m + 3)
Solution:
125x2 - 45y2
= 5(25x2 - 9y2)
= 5[(5x)2 - (3y)2]
= 5(5x - 3y)(5x + 3y)
Solution:
144a2 - 169b2
= (12a)2 - (13b)2
= (12a - 13b)(12a + 13b)
Solution:
(2a - b)2 - 16c2
= (2a - b)2 - (4c)2
= [(2a - b) + 4c][(2a - b) - 4c]
= (2a - b + 4c)(2a - b - 4c)
Solution:
(x + 2y)2 - 4(2x - y)2
= (x + 2y)2 - [2(2x - y)]2
= [(x + 2y) - 2(2x - y)][(x + 2y) + 2(2x - y)]
= (x + 2y - 4x + 2y)(x + 2y + 4x - 2y)
= 5x(4y - 3x)
Solution:
3a5 - 48a3
= 3a3(a2 - 16)
= 3a3(a2 - (4)2)
= 3a3(a - 4)(a + 4)
Solution:
a4 - 16b4
= a4 - 24b4
= (a2)2 - (22b2)2
= (a2 - 22b2)(a2 + 22b2)
= [a2 - (2b)2](a2 + 4b2)
= [a2 - (2b)2](a2 + 4b2)
= (a - 2b)(a + 2b)(a2 + 4b2)
Solution:
x8 - 1
= (x4)2 - 12
= [(x2)2 - 12](x4 + 1)
= [(x2 - 1)(x2 + 1)](x4 + 1)
= (x - 1)(x + 1)(x2 + 1)(x4 + 1)
Solution:
64 - (a + 1)2
= (8)2 - (a + 1)2
= [8 - (a + 1)][8 + (a + 1)]
= (8 - a - 1)(8 + a + 1)
= (7-a)(9+a)
Solution:
36L2 - (m + n)2
= (6L)2 - (m + n)2
= [6L - (m + n)][6L + (m + n)]
= (6L - m - n)(6L + m + n)
Solution:
25x4y4 - 1
= (5x2y2)2 - 12
= (5x2y2 - 1)(5x2y2 + 1)
Solution:
a4 - 1/b4
= (a2)2 - (1/b2)2
= (a2 - 1/b2)(a2 + 1/b2)
= (a - 1/b)(a + 1/b)(a2 + 1/b2)
Solution:
x3 - 144x
= x(x2 - 144)
= x(x2 - 122)
= x(x - 12)(x + 12)
Solution:
(x - 4y)2 - 625
= (x - 4y)2 - 252
= [(x - 4y) - 25] [(x - 4y) + 25]
= (x - 4y - 25)(x - 4y + 25)
Solution:
9(a - b)2 - 100(x - y)2
= [3(a - b)]2 - [10 (x - y)]2
= [3(a - b) -10(x - y)] [3(a - b) + 10(x - y)]
= (3a - 3b - 10x + 10y) (3a - 3b + 10x - 10y)
Solution:
(3 + 2a)2 - 25a2
= (3 + 2a)2 - (5a)2
= (3 + 2a + 5a)(3 + 2a - 5a)
= (3 + 7a)(3 - 3a)
Solution:
(x + y)2 - (a - b)2
= [(x + y) + (a - b)] [(x + y) - (a - b)]
= (x + y + a - b) (x + y - a + b)
Solution:
1/16 x2y2 - 4/49y2z2
= (1/4xy)2 - (2/7 yz)2
= (1/4xy - 2/7yz) (1/4xy + 2/7yz)
= [y (1/4x - 2/7z)] [y(1/4x + 2/7z)]
= y2 (1/4x - 2/7z) (1/4x + 2/7z)
Solution:
75a3b2 - 108ab4
= 3ab2(25a2 - 36b2)
= 3ab2[(5a)2 - (6b)2]
= 3ab2[(5a - 6b)(5a + 6b)]
= 3ab2(5a - 6b)(5a + 6b)
Solution:
x5 - 16x3
= x3(x2 - 16)
= x3(x2 - 42)
= x3(x - 4)(x + 4)
Exercise 7.5 | Set 1 in Chapter 7 of RD Sharma's Class 8 Mathematics textbook focuses on applying factorization techniques to solve various types of equations. This exercise introduces students to the practical application of factorization in equation solving, covering quadratic equations, cubic equations, and even some higher-degree polynomial equations. Students learn to identify factors, use the zero-product property, and apply factorization to find solutions to equations. The problems in this set also include scenarios where students need to find unknown coefficients based on given conditions, understand the nature of roots, and apply factorization in the context of algebraic expressions and identities. This exercise serves as a crucial bridge between abstract factorization techniques and their practical applications in problem-solving, preparing students for more advanced algebraic concepts in higher grades.
