Class 8 RD Sharma Solutions Chapter 6 Algebraic Expressions And Identities Exercise 6.6 | Set 2 (original) (raw)
Last Updated : 23 Jul, 2025
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.
Algebraic Expressions And Identities
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
**Question 11. If x – y = 7 and xy = 9, find the value of x 2 +y 2
**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
**Question 12. If 3x + 5y = 11 and x y = 2, find the value of 9x 2 + 25y 2
**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
**Question 13. Find the values of the following expressions:
****(i) 16x** 2 + 24x + 9 when x = 7/4
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
****(ii) 64x** 2 + 81y 2 + 144xy when x = 11 and y = 4/3
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
****(iii) 81x** 2 + 16y 2 – 72xy when x = 2/3 and y = ¾
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
**Question 14. If x + 1/x = 9 find the value of x4 + 1/ x4.
**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
**Question 15. If x + 1/x = 12 find the value of x – 1/x.
**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
**Question 16. If 2x + 3y = 14 and 2x – 3y = 2, find value of xy. [Hint: Use (2x+3y) 2 – (2x-3y) 2 = 24xy]
**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
**Question 17. If x 2 + y 2 = 29 and xy = 2, find the value of
****(i) x + y**
****(ii) x – y**
****(iii) x** 4 + y 4
**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
**Question 18. What must be added each of the following expression to make it a whole square?
****(i) 4x** 2 – 12x + 7
(2x)2 – 2 (2x) (3) + 32 – 32 + 7
(2x – 3)2 – 9 + 7
(2x – 3)2 – 2
****(ii) 4x** 2 – 20x + 20
(2x)2 – 2 (2x) (5) + 52 – 52 + 20
(2x – 5)2 – 25 + 20
(2x – 5)2 – 5
**Question 19. Simplify:
**i) (x – y) (x + y) (x 2 + y 2 ) (x 4 + y 4 )
(x2 – y2) (x2 + y2) (x4 + y4)
[(x2)2 – (y2)2] (x4 + y4)
(x4 – y4) (x4 – y4)
[(x4)2 – (y4)2]
x8 – y8
****(ii) (2x – 1) (2x + 1) (4x** 2 + 1) (16x 4 + 1)
[(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
****(iii) (7m – 8n)** 2 + (7m + 8n) 2
(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
****(iv) (2.5p – 1.5q)** 2 – (1.5p – 2.5q) 2
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)
****(v) (m** 2 – n 2 m) 2 + 2m3n 2
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
**Question 20. Show that:
****(i) (3x + 7)** 2 – 84x = (3x – 7) 2
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
****(ii) (9a – 5b)** 2 + 180ab = (9a + 5b) 2
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
****(iii) (4m/3 – 3n/4)** 2 + 2mn = 16m 2 /9 + 9n 2 /16
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
****(iv) (4pq + 3q)** 2 – (4pq – 3q) 2 = 48pq 2
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
****(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0**
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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