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• RD Sharma Class 8 Solutions for Maths

# Class 8 RD Sharma Solutions – Chapter 6 Algebraic Expressions And Identities – Exercise 6.6 | Set 2

### Question 11. If x – y = 7 and xy = 9, find the value of x2+y2

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 9x2 + 25y2

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

### (i) 16x2 + 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) 64x2 + 81y2 + 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) 81x2 + 16y2 – 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

4×2 + 9y2 + 12xy – 4×2 – 9y2 + 12xy = 196 – 4

24 xy = 192

xy = 8

### Question 17. If x2 + y2 = 29 and xy = 2, find the value of

(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

### (i) 4x2 – 12x + 7

(2x)2 – 2 (2x) (3) + 32 – 32 + 7

(2x – 3)2 – 9 + 7

(2x – 3)2 – 2

### (ii) 4x2 – 20x + 20

(2x)2 – 2 (2x) (5) + 52 – 52 + 20

(2x – 5)2 – 25 + 20

(2x – 5)2 – 5

### i) (x – y) (x + y) (x2 + y2) (x4 + y4)

(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) (4x2 + 1) (16x4 + 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) (m2 – n2m)2 + 2m3n2

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

### (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 = 16m2/9 + 9n2/16

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 = 48pq2

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