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Chapter 6 of RD Sharma's Class 8 Mathematics textbook focuses on the Algebraic Expressions and Identities. This chapter is crucial for the building a strong foundation in algebra as it introduces students to the basics of the algebraic expressions how to manipulate them and various identities used in the solving complex problems. Exercise 6.6 specifically deals with the applying these concepts through a variety of problems designed to the reinforce understanding.
The Algebraic expressions are combinations of variables, constants and arithmetic operations. The Identities are equations that hold true for the all values of the variables involved. Together, they form the basis of the solving equations and simplifying expressions in algebra.
Chapter 6 Algebraic Expressions And Identities - Exercise 6.6 | Set 1
Solution:
Given in the question x – y = 7 and x y = 9
By squaring on both sides
(x – y)2 = 72
x2 + y2 – 2xy = 49
x2 + y2 – 2 (9) = 49 (since x y=9)
x2 + y2 – 18 = 49
x2 + y2 = 49 + 18
x2 + y2 =67
Solution:
Given in the question 3x + 5y = 11 and x y = 2
on squaring on both sides
(3x + 5y)2 = 112
(3x)2 + (5y)2 + 2(3x)(5y) = 121
9x2 + 25y2 + 2 (15xy) = 121 (given x y=2)
9x2 + 25y2 + 2(15(2)) = 121
9x2 + 25y2 + 60 = 121
9x2 + 25y2 = 121-60
9x2 + 25y2 = 61
We will using the formula (a + b)2 = a2 + b2 + 2ab
=(4x)2 + 2 (4x) (3) + 32
=(4x + 3)2
Putting when x = 7/4
=[4 (7/4) + 3]2
=(7 + 3)2
=100
We will using the formula (a + b)2 = a2 + b2 + 2ab
=(8x)2 + 2 (8x) (9y) + (9y)2 (8x + 9y)
Putting when x = 11 and y = 4/3
=[8 (11) + 9 (4/3)]2
=(88 + 12)2
=(100)2
=10000
We will using the formula (a + b)2 = a2 + b2 + 2ab
=(9x)2 + (4y)2 – 2 (9x) (4y)
=(9x – 4y)2
Putting x = 2/3 and y = 3/4
=[9 (2/3) – 4 (3/4)]2
=(6 – 3)2
=32
=9
Solution:
Given in the question x + 1/x = 9
squaring both sides
(x + 1/x)2 = (9)2
x2 + 2 × x × 1/x + (1/x)2 = 81
x2 + 2 + 1/x2 = 81
x2 + 1/x2 = 81 – 2
x2 + 1/x2 = 79
Now again when we square on both sides
(x2 + 1/x2)2 = (79)2
x4 + 2 × x2 × 1/x2 + (1/x2)2 = 6241
x4 + 2 + 1/x4 = 6241
x4 + 1/x4 = 6241- 2
x4 + 1/x4 = 6239
Solution:
Given in the question x + 1/x = 12
When squaring both sides
(x + 1/x)2 = (12)2
x2 + 2 × x × 1/x + (1/x)2 = 144
x2 + 2 + 1/x2 = 144
x2 + 1/x2 = 144 – 2
x2 + 1/x2 = 142
When subtracting 2 from both sides
x2 + 1/x2 – 2 × x × 1/x = 142 – 2
(x – 1/x)2 = 140
x – 1/x = √140
Solution:
2x + 3y = 14 equation (1)
2x – 3y = 2 equation (2)
Now square both the equations and subtract equation (2) from equation (1)
(2x + 3y)2 – (2x – 3y)2 = (14)2 – (2)2
4x2 + 9y2 + 12xy – 4x2 – 9y2 + 12xy = 196 – 4
24 xy = 192
xy = 8
(i) x + y
(ii) x – y
(iii) x4 + y4
Solution:
(i) x + y
x2 + y2 = 29 (Given in the question)
x2 + y2 + 2xy – 2xy = 29
(x + y) 2 – 2 (2) = 29
(x + y) 2 = 29 + 4
x + y = ± √33
(ii) x – y
x2 + y2 = 29
x2 + y2 + 2xy – 2xy = 29
(x – y)2 + 2 (2) = 29
(x – y)2 + 4 = 29
(x – y)2 = 25
(x – y) = ± 5
(iii) x4 + y4
x2 + y2 = 29
Squaring both sides
(x2 + y2)2 = (29)2
x4 + y4 + 2x2y2 = 841
x4 + y4 + 2 (2)2 = 841
x4 + y4 = 841 – 8
= 833
(2x)2 – 2 (2x) (3) + 32 – 32 + 7
(2x – 3)2 – 9 + 7
(2x – 3)2 – 2
(2x)2 – 2 (2x) (5) + 52 – 52 + 20
(2x – 5)2 – 25 + 20
(2x – 5)2 – 5
(x2 – y2) (x2 + y2) (x4 + y4)
[(x2)2 – (y2)2] (x4 + y4)
(x4 – y4) (x4 – y4)
[(x4)2 – (y4)2]
x8 – y8
[(2x)2 – (1)2] (4x2 + 1) (16x4 + 1)
(4x2 – 1) (4x2 + 1) (16x4 + 1) 1
[(4x2)2 – (1)2] (16x4 + 1) 1
(16x4 – 1) (16x4 + 1) 1
[(16x4)2 – (1)2] 1
256x8 – 1
(7m)2 + (8n)2 – 2(7m)(8n) + (7m)2 + (8n)2 + 2(7m)(8n)
(7m)2 + (8n)2 – 112mn + (7m)2 + (8n)2 + 112mn
49m2 + 64n2 + 49m2 + 64n2
grouping the similar expression
98m2 + 64n2 + 64n2
98m2 + 128n2
on expansion
(2.5p)2 + (1.5q)2 – 2 (2.5p) (1.5q) – (1.5p)2 – (2.5q)2 + 2 (1.5p) (2.5q)
6.25p2 + 2.25q2 – 2.25p2 – 6.25q2
grouping the similar expression
4p2 – 6.25q2 + 2.25q2
4p2 – 4q2
4 (p2 – q2)
On expansion using (a + b)2 formula
(m2)2 – 2 (m2) (n2) (m) + (n2m)2 + 2m3n2
m4 – 2m3n2 + (n2m)2 + 2m3n2
m4+ n4m2 – 2m3n2 + 2m3n2
m4+ m2n4
LHS => (3x + 7)2 – 84x
We will using the formula (a + b)2 = a2 + b2 + 2ab
(3x)2 + (7)2 + 2 (3x) (7) – 84x
(3x)2 + (7)2 + 42x – 84x
(3x)2 + (7)2 – 42x
(3x)2 + (7)2 – 2 (3x) (7)
(3x – 7)2 = R.H.S
LHS => (9a – 5b)2 + 180ab
We will using the formula (a + b)2 = a2 + b2 + 2ab
(9a)2 + (5b)2 – 2 (9a) (5b) + 180ab
(9a)2 + (5b)2 – 90ab + 180ab
(9a)2 + (5b)2 + 9ab
(9a)2 + (5b)2 + 2 (9a) (5b)
(9a + 5b)2 = R.H.S
LHS => (4m/3 – 3n/4)2 + 2mn
(4m/3)2 + (3n/4)2 – 2mn + 2mn
(4m/3)2 + (3n/4)2
16/9m2 + 9/16n2 = R.H.S
LHS => (4pq + 3q)2 – (4pq – 3q)2
(4pq)2 + (3q)2 + 2 (4pq) (3q) – (4pq)2 – (3q)2 + 2(4pq)(3q)
24pq2 + 24pq2
48pq2 = RHS
LHS =>(a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
By using the identity (a – b) (a + b) = a2 – b2
(a2 – b2) + (b2 – c2) + (c2 – a2)
a2 – b2 + b2 – c2 + c2 – a2
0 = R.H.S
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