 |
VOOZH | about |
|
VOOZH | about |
Exercise 7.5 | Set 2 in Chapter 7 of RD Sharma's Class 8 Mathematics textbook continues the application of factorization techniques to solve equations and related problems. This set typically introduces more complex scenarios and word problems that require students to apply their factorization skills in various contexts.A quadratic equation ax² + bx + c = 0 has real and equal roots if its discriminant (b² - 4ac) is equal to zero. This condition can often be expressed as an equation to solve for unknown coefficients. It aims to further develop students' problem-solving abilities and deepen their understanding of the practical applications of factorization.
Solution:
50/x2 - 2x2/81
= 2(25/x2 - x2/81)
= 2[(5/x)2 - (x/9)2]
= 2(5/x - x/9) (5/x + x/9)
Solution:
256x5 - 81x
= x(256x4 - 81)
= x[(16x2)2 - (9)2]
= x(16x2 - 9) (16x2 + 9)
Solution:
a4 - (2b + c)4
= (a2)2 - [(2b + c)2]2
= [a2 - (2b + c)2] [a2 + (2b + c)2]
= [a + (2b + c)] [a - (2b + c)] [a2 + (2b + c)2]
= (a + 2b + c) (a - 2b - c) [a2 + (2b + c)2]
Solution:
(3x + 4y)4 - x4
= [(3x + 4y)2]2 - (x2)2
= [(3x + 4y)2 - x2] [(3x + 4y)2 + x2]
= [(3x + 4y) - x] [(3x + 4y) + x] [(3x + 4y)2 + x2]
= (2x + 4y) (4x + 4y) [(3x + 4y)2 + x2]
= 8(x + 2y) (x + y) [(3x + 4y)2 + x2]
Solution:
p2q2 - p4q4
= p2q2 (1 - p2q2)
= p2q2 (1 + pq) (1 - pq)
Solution:
3x3y - 243xy3
= 3xy (x2 - 81y2)
= 3xy [x2 - (9y)2]
= 3xy (x - 9y) (x + 9y)
Solution:
a4b4 - 16c4
= (a2b2)2 - (4c2)2
= (a2b2 - 4c2) (a2b2 + c2)
= [a2b2 - (2c)2] (a2b2 + c2)
= (ab - 2c) (ab + 2c) (a2b2 + c2)
Solution:
x4 - 625
= (x2)2 - (25)2
= (x2 - 25) (x2 + 25)
= (x2 - 52) (x2 + 25)
= (x - 5) (x + 5) (x2 + 25)
Solution:
x4 - 1
= (x2)2 - 12
= (x2 - 1)(x2 +1)
= (x + 1)(x - 1) (x2 + 1)
Solution:
49(a - b)2 - 25(a + b)2
= [7(a - b)]2 - [5 (a + b)]2
= [7(a - b) - 5 (a + b)] [7(a - b) + 5 (a + b)]
= (7a - 7b -5a - 5b) (7a - 7b + 5a + 5b)
= (2a - 12b) (12a - 2b)
= 4(a - 6b)(6a - b)
Solution:
x - y - x2 + y2
= (x - y) - (x2 - y2)
= (x - y) - (x - y)(x + y)
= (x - y) [1 - (x + y)]
= (x - y) (1 - x - y)
Solution:
16(2x - 1)2 - 25y2
= [4(2x - 1)]2 - (5y)2
= [4(2x - 1) - 5y] [4(2x - 1) + 5y]
= (8x - 4 - 5y) ( 8x - 4 + 5y)
Solution:
4(xy + 1)2 - 9( x- 1)2
= [2(xy + 1)]2 - [3(x - 1)]2
= [2(xy + 1) - 3(x - 1)] [2(xy + 1) + 3(x - 1)]
= (2xy + 2 - 3x + 3) (2xy + 2 + 3x - 3)
= (2xy - 3x + 5) (2xy + 3x - 1)
Solution:
(2x + 1)2 - 9x4
= (2x + 1)2 - (3x2 )2
= [(2x +1) - 3x2] [(2x + 1) + 3x2]
= (-3x2 + 2x + 1)(3x2 + 2x + 1)
= (-3x2 + 3x - x + 1) (3x2 + 2x + 1)
= [3x(1 - x) + (1 - x)] (3x2 + 2x + 1)
= (3x +1) (1 - x) (3x2 + 2x + 1)
Solution:
x4 - (2y - 3z)2
= (x2)2 - (2y - 3z)2
= [x2 - (2y - 3z)] [x2 + (2y - 3z)]
= (x2 - 2y + 3z) (x2 + 2y - 3z)
Solution:
a2 - b2 + a - b
= (a2 - b2) + (a - b)
= [(a - b) (a + b)] + (a - b)
= (a - b) (a + b + 1)
Solution:
16a4 - b4
= (4a2)2 - (b2)2
= (4a2 - b2) (4a2 + b2)
= [(2a)2 - b2] (4a2+ b2)
= (2a - b) (2a + b) (4a2 + b2)
Solution:
a4 - 16(b - c)4
= (a2)2 - [4(b - c)2]2
= [a2 - 4(b - c)2] [a2 + 4(b - c)2]
= {a2 - [2(b - c)]2}[a2 + 4(b - c)2]
= [a - 2(b - c)] [a + 2(b - c)] [a2 + 4(b - c)2]
= (a - 2b + 2c) (a + 2b - 2c) [a2 + 4(b - c)2]
Solution:
2a5 - 32a
= 2a(a5 - 1 )
= 2a [ (a2)2 - 12 ]
= 2a ( a2 - 1) (a2+ 1)
= 2a (a+1) (a-1) (a2+1)
Solution:
Answer:
a4b4 - 81c4
= (a2b2)2 - (9c2)2
= (a2b2 - 9c2) (a2b2 + 9c2)
= [(ab)2 - (3c)2 ] (a2b2 + 9c2)
= (ab - 3c) (ab + 3c) (a2b2 + 9c2)
Solution:
xy9 - x9y
= xy(y8 - x8)
= xy [(y4)2 - (x4)2]
= xy (y4 - x4)(y4 + x4)
= xy [(y2)2 - (x2)2] (y4 + x4)
= xy [(y2 - x2)(y2 + x2)(y4 + x4)
= xy (y - x) (y + x) (y2 + x2) (y4 + x4)
Solution:
x3 - x
= x(x2 - 1)
= x (x + 1)(x - 1)
Solution:
18a2x2 - 32
= 2(9a2x2 - 16)
= 2[(3ax)2 - 42]
= 2(3ax - 4) (3ax + 4)
Exercise 7.5 | Set 2 in Chapter 7 of RD Sharma's Class 8 Mathematics textbook builds upon the foundation laid in Set 1, focusing on more advanced applications of factorization in problem-solving. This set introduces a variety of word problems and real-life scenarios that require students to translate verbal descriptions into algebraic equations, which are then solved using factorization techniques. The problems cover a range of topics including geometry (areas and dimensions), number relationships, age problems, and more complex equation solving. Students are challenged to apply their factorization skills in conjunction with other mathematical concepts, such as consecutive integers, reciprocals, and the properties of quadratic equations. This exercise set aims to enhance students' ability to model real-world situations mathematically, reinforcing the practical relevance of factorization and algebraic problem-solving in everyday contexts.
