# RS Aggarwal Class 8 Math Sixth Chapter Operation on Algebric Expressions Exercise 6D Solution

## EXERCISE 6D

**(1) Find each of the following products:**

**(i) (x + 6) (x + 6) **

= x^{2} + 6x + 6x + 36

= x^{2 }+ 12x + 36

= x^{2} + (2 × x × 6) + 6^{2}

= (x + 6)^{2}

**(ii) (4x + 5y) (4x + 5y)**

= 16x^{2} + 20xy + 20xy + 25y^{2}

= (4x)^{2} + (2 × 4x × 5y) + (5y)^{2}

= (4x + 5y)^{2}

**(iii) (7a + 9b) (7a + 9b)**

= 49a^{2} + 63ab + 63ab + 81b^{2}

= (7a)^{2} + (2 × 7a × 9b) + (9b)^{2}

= (7a + 9b)^{2}

**(v) (x ^{2} + 7) (x^{2} + 7)**

= x^{4} + 7x^{2} + 7x^{2} + 49

= (x^{2})^{2} + (2 × x^{2} × 7) + 7^{2}

= (x^{2} + 7)^{2}

**(2) Find each of the following products:**

**(i) (x – 4) (x – 4) **

= x^{2} – 4x – 4x + 16

= x^{2} – 8x + 16

= x^{2} – (2 × x × 4) + 4^{2}

= (x – 4)^{2}

**(ii) (2x – 3y) (2x – 3y)**

= 4x^{2} – 6xy – 6xy + 9y^{2}

= (2x)^{2} – 12xy + (3y)^{2}

= (2x)^{2} – (2 × 2x × 3y) + (3y)^{2}

= (2x – 3y)^{2}

**(3) Expand:**

**(i) (8a + 3b) ^{2}**

= (8a)^{2} + 2 × 8a × 3b + (3b)^{2}

= 64a^{2} + 48ab + 9b^{2}

**(ii) (7x + 2y) ^{2}**

= (7x)^{2} + (2 × 7x × 2y) + (2y)^{2}

= 49x^{2} + 28xy + (2y)^{2}

**(iii) (5x + 11) ^{2}**

= (5x)^{2} + (2 × 5x × 11) + (11)^{2}

= 25x^{2} + 110x + 121

**(vi) (9x – 10) ^{2}**

= (9x)^{2} – (2 × 9x × 10) + (10)^{2}

= 81x^{2} – 180x + 100

**(vii) (x ^{2}y – yz^{2})^{2}**

= (x^{2}y)^{2} – (2 × x^{2}y × yz^{2}) + (yz^{2})^{2}

= x^{4}y^{2} – 2x^{2}y^{2}z^{2} + y^{2}z^{4}

**(4) Find each of the following products:**

**(i) (x + 3) (x – 3) **

= x^{2} + 3x – 3x – 9

= x^{2} – 3^{2}

**(ii) (2x + 5) (2x – 5)**

= 4x^{2} + 10x – 10x – 25

= (2x)^{2} – 5^{2}

**(iii) (8 + x) (8 – x)**

= 64 + 8x – 8x – x^{2}

= 4^{2} – x^{2}

**(iv) (7x + 11y) (7x – 11y)**

= 49x^{2} + 77xy – 77xy – 121y^{2}

= (7x)^{2} – (11y)^{2}

**(5) Using the formula for squaring a binomial, evaluate the following:**

**(i) (54) ^{2}**

= (50 + 4)^{2}

^{ }= (50)^{2} + (2 × 50 × 4) + (4)^{2}

= 2500 + 400 + 16

= 2916

**(ii) (82) ^{2}**

= (80 + 2)^{2}

= (80)^{2 }+ (2 × 80 × 2) + (2)^{2}

= 6400 + 320 + 4

= 6724

**(iii) (103) ^{2}**

= (100 + 3)^{2}

= (100)^{2} + (2 × 100 × 3) + (3)^{2}

= 10000 + 600 + 9

= 10609

**(iv) (704) ^{2}**

= (700 + 4)^{2}

= (700)^{2} + (2 × 700 × 4) + (4)^{2}

= 490000 + 5600 + 16

= 495616

**(6) Using the formula for squaring a binomial, evaluate the following:**

**(i) (69) ^{2}**

= (70 – 1)^{2}

= (70)^{2} – (2 × 70 × 1) + 1

= 4900 – 140 + 1

= 4761

**(ii) (78) ^{2}**

= (80 – 2)^{2}

= (80)^{2} – (2 × 80 × 2) + (2)^{2}

= 6400 – 320 + 4

= 6084

**(iii) (197) ^{2}**

= (200 – 3)^{2}

= (200)^{2} – (2 × 200 × 3) + (3)^{2}

= 40000 – 1200 + 9

= 38809

**(iv) (999) ^{2}**

= (1000 – 1)^{2}

= (1000)^{2} – (2 × 1000 × 1) + 1

= 1000000 – 2000 + 1

= 998001

**(7) Find the value of:**

**(i) (82) ^{2} – (18)^{2}**

= [(80+2)^{2}] – [(20 – 2)^{2}]

= [(80)^{2} + (2 × 80 × 2) + 4] – [(20)^{2} – (2 × 20 × 2) + 4]

= (6400 + 320 + 4) – (400 – 80 + 4)

= 6724 – 324

= 6400

**(ii) (128) ^{2} – (72)^{2}**

= [(130 – 2)^{2} – (70 + 2)^{2}]

= [(130)^{2} – (2 × 130 × 2) + 4] – [(70)^{2} + (2 × 70 × 2) + 4]

= (16900 – 520 + 4) – (4900 + 280 + 4)

= 16384 – 5184

= 11200

**(iii) 197 × 203**

= (200 – 3) × (200 + 3)

= (200)^{2} – (3)^{2}

= 40000 – 9

= 39991

**(v) (14.7 ×**** 15.3)**

= (15 – 0.3) × (15 + 0.3)

= (15)^{2} – (0.3)^{2}

= 225 – 0.09

= 224.91

**(vi) (8.63) ^{2} – (1.37)^{2}**

= (8.63 + 1.37) (8.63 – 1.37)

= 10 × 7.26

= 72.6

**(8) Find the value of the expression (9x ^{2} + 24x + 16), when x = 12.**

Solution: 9x^{2} + 24x + 16

= [9 × (12)^{2}] + (24 × 12) + 16

= (9 × 144) + 288 + 16

= 1296 + 288 + 16

= 1600

**(9) Find the value of the expression (64x ^{2} + 81y^{2} + 144xy), when x = 11 and y = 4/3.**

Solution: 64x^{2} + 81y^{2} + 144xy

**(10) Find the value of the expression (36x ^{2} + 25y^{2} – 60xy), when x =2/3 and y = 1/5.**

Solution: (36x^{2} + 25y^{2} – 60xy)

= (6x)^{2} – (2 × 6x × 5y) + (5y)^{2}

= (6x – 5y)^{2}

**(13) Find the continued product: **

**(i) (x + 1) (x – 1) (x ^{2} + 1)**

= (x^{2} – 1) (x^{2} + 1)

= x^{4} – 1

**(ii) (x – 3) (x + 3) (x ^{2} + 9)**

= [(x)^{2} – (3)^{2}] (x^{2} + 9)

= (x^{2} – 9)(x^{2} + 9)

= x^{2} – 9^{2} = x^{2} – 81

**(iii) (3x – 2y) (3x + 2y) (9x ^{2} + 4y^{2})**

= [(3x)^{2} – (2y^{2})] (9x^{2} + 4y^{2})

= (9x^{2} – 4y^{2}) (9x^{2} + 4y^{2})

= (9x^{2})^{2} – (4y^{2})^{2}

= 81x^{4} – 16y^{4}

**(iv) (2p + 3) (2p – 3) (4p ^{2} + 9)**

= [(2p)^{2} – (3)^{2}] (4p^{2} + 9)

= (4p^{2} – 9) (4p^{2} + 9)

= (4p^{2})^{2} – (9)^{2}

= 16p^{4} – 81

**(14) If x + y = 12 and xy = 14, find the value of (x ^{2} + y^{2}).**

Solution: x + y = 12

⇒ (x + y)^{2} = (12)^{2}

⇒ x^{2} + 2xy + y^{2} = 144

⇒ (x^{2} + y^{2}) + (2 × 14) = 144

⇒ (x^{2} + y^{2}) = 144 – 28

⇒ (x^{2} + y^{2}) = 116

**(15) If x – y = 7 and xy = 9, find the value of (x ^{2} + y^{2}).**

Solution: x – y = 7

⇒ (x – y)^{2} = (7)^{2}

⇒ x^{2} – 2xy + y^{2} = 49

⇒ (x^{2} + y^{2}) – (2 × 9) = 49

⇒ (x^{2} + y^{2}) = 49 – 18

⇒ (x^{2} + y^{2}) = 31.

Thank you R S Aggarwal ji

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