# Class 9 RD Sharma Solutions – Chapter 4 Algebraic Identities- Exercise 4.1 | Set 2

### Question 8. If x2 +1/x2 = 79, find the value of x +1/x

Solution:

Given, x2 +1/x2 = 79

Let us take the square of x + 1/x

So, (x + 1/x)2 = (x)2 + (1/x)2 + 2 Ã— (x) Ã— (1/x)

Since, (a + b)2 = a2 + b2 + 2ab

So,

(x + 1/x)2 = 79 + 2

(x + 1/x)2 = 81

x + 1/x = Â±9

Hence, the value of x + 1/x is Â±9

### Question 9. If 9x2 + 25y2 = 181 and xy = -6, find the value of 3x + 5y

Solution:

Given,

9x2 + 25y2 = 181 and xy = -6

Let us take a square of 3x + 5y

(3x + 5y)2 = (3x)2 + (5y)2 + 2 Ã— (3x) Ã— (5y)

Since, (a + b)2 = a2 + b2 + 2ab

So,

(3x + 5y)2 = 9x2 + 25y2 + 30xy

(3x + 5y)2  = 181 + 30(-6)

(3x + 5y)2 = 181 â€“ 180 = 1

So, 3x + 5y = Â±1

Hence, the value of 3x + 5y is Â±1

### Question 10. If 2x + 3y = 8 and xy = 2, find the value of 4x2 + 9y2

Solution:

Given,

2x + 3y = 8 and xy = 2

Let us take a square of 2x + 3y

(2x + 3y)2 = (2x)2 + (3y)2 + 2 Ã— (2x) Ã— (3y)

Since, (a + b)2 = a2 + b2 + 2ab

So,

(2x + 3y)2 = 4x2 + 9y2 + 12xy

(8)2  = 4x2 + 9y2 + 12(2)

64 = 4x2 + 9y2 + 24

4x2 + 9y2 = 40

Hence, the value of 4x2 + 9y2 is 40

### Question 11. If 3x -7y = 10 and xy = -1, find the value of 9x2 + 49y2

Solution:

Given,

3x -7y = 10 and xy = -1

Let us take a square of 3x -7y

(3x -7y)2 = (3x)2 + (7y)2 – 2 Ã— (3x) Ã— (7y)

Since, (a – b)2 = a2 + b2 – 2ab

So,

(3x -7y)2 = 9x2 + 49y2 – 42xy

(10)2  = 9x2 + 49y2 – 42(-1)

100 = 9x2 + 49y2 + 42

9x2 + 49y2 = 58

Hence, the value of 9x2 + 49y2 is 58

### Question 12. Simplify each of the following products:

(i) (1/2a – 3b)(3b +1/2a)(1/4a2 + 9b2)
(ii) (m +n/7)3 (m-n/7)
(iii) (x/2 -2/5)(2/5 -x/2) -x2 + 2x
(iv) (x2 + x -2)(x2 -x + 2)
(v) (x3 -3x2 -x)(x2 -3x + 1)
(vi) (2x4 -4x2 +1)(2x4 -4x2 -1)

Solution:

i) (1/2a -3b)(3b +1/2a)(1/4a2 + 9b2)

The above expression can be written as, (1/2a -3b)(1/2a +3b)(1/4a2 + 9b2)

We know that, (a + b)(a â€“ b) = a2 â€“ b2

So, [(1/2a)2 -(3b)2](1/4a2 + 9b2) = (1/4a2 -9b2)(1/4a2 + 9b2)

We now conclude it as, (1/4a2)2 – (9b2)2

= 1/16a4 -81b4

Hence, (1/2a -3b)(3b +1/2a)(1/4a2 + 9b2) = 1/16a4 -81b4

ii) (m +n/7)3 (m-n/7)

The above expression can be written as, (m +n/7)2 (m +n/7)(m-n/7)

We know that, (a + b)(a â€“ b) = a2 â€“ b2

So, (m +n/7)2 [(m)2 -(n/7)2]

= (m +n/7)2 (m2 -n2/49)

Hence, (m +n/7)3 (m-n/7) = (m +n/7)2 (m2 -n2/49)

iii) (x/2 -2/5)(2/5 -x/2) -x2 + 2x

The above expression can be written as, [-(x/2 -2/5)(x/2 -2/5)] -x2 + 2x

= [-(x/2 -2/5)2] -x2 + 2x

We know that, (a -b)2 = a2 â€“ 2ab + b2

So, -(x2/4 + 4/25 -2x/5) -x2 + 2x

= -x2/4 -4/25 + 2x/5 -x2 + 2x = -5x2/4 + 12x/5 -4/25

Hence, (x/2 -2/5)(2/5 -x/2) -x2 + 2x = -5x2/4 + 12x/5 -4/25

iv) (x2 + x -2)(x2 -x + 2)

The above expression can be written as, [x2 + (x -2)][x2 -(x -2)]

We know that, (a + b)(a â€“ b) = a2 â€“ b2

So, (x2)2 -(x -2)2

= x4 -(x2 + 4 -4x)

= x4 -x2 -4 + 4x

Hence, (x2 + x -2)(x2 -x + 2) = x4 -x2 -4 + 4x

v) (x3 -3x2 -x)(x2 -3x + 1)

The above expression can be written as, x(x2 -3x -1)(x2 -3x + 1)

= x[(x2 -3x) -1][(x2 -3x) + 1]

We know that, (a + b)(a â€“ b) = a2 â€“ b2

= x [(x2 -3x)2 -12]

= x[x4 + 9x2 -6x3 -1] = x5 + 9x3 -6x4 -x

Hence, (x3 -3x2 -x)(x2 -3x + 1) = x5 + 9x3 -6x4 -x

vi) (2x4 -4x2 +1)(2x4 -4x2 -1)

The above expression can be written as, [(2x4 -4x2)+1][(2x4 -4x2) -1]

We know that, (a + b)(a â€“ b) = a2 â€“ b2

So, (2x4 -4x2)2 -12

= 4x8 + 16x4 -16x6 -1

Hence, (2x4 -4x2 +1)(2x4 -4x2 -1) = 4x8 + 16x4 -16x6 -1

### Question 13. Prove that a2 + b2 + c2 â€“ ab â€“ bc â€“ ca is always non-negative for all values of a, b and c.

Solution:

In order to prove the given expression,

First Multiply and Divide a2 + b2 + c2 â€“ ab â€“ bc â€“ ca by 2

= 1/2[2a2 + 2b2 + 2c2 â€“ 2ab â€“ 2bc â€“ 2ca]

= 1/2[a2 + b2 + c2 +a2 + b2 + c2 â€“ 2ab â€“ 2bc â€“ 2ca]

= 1/2[a2 + b2 -2ab + b2 + c2 -2bc + c2 +a2 -2ac]

= 1/2[(a -b)2 + (b -c)2 + (c -a)2]

As we can clearly see the above expression is sum of square terms which will always be positive.

Hence, a2 + b2 + c2 â€“ ab â€“ bc â€“ ca is always non-negative for all values of a, b and c is proved

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