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 × (x) × (1/x)

Since, (a + b)² = a² + b² + 2ab

So,

(x + 1/x)² = 79 + 2

(x + 1/x)² = 81

x + 1/x = ±9

Hence, the value of x + 1/x is ±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 × (3x) × (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 – 180 = 1

So, 3x + 5y = ±1

Hence, the value of 3x + 5y is ±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 × (2x) × (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 × (3x) × (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 – b) = a² – 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 – b) = a² – 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 – 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 – b) = a² – 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 – b) = a² – 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 – b) = a² – 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 – ab – bc – 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 – ab – bc – ca by 2

= 1/2[2a² + 2b² + 2c² – 2ab – 2bc – 2ca]

= 1/2[a² + b² + c² +a² + b² + c² – 2ab – 2bc – 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 – ab – bc – ca is always non-negative for all values of a, b and c is proved