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)
= x2 + 6x + 6x + 36
= x2 + 12x + 36
= x2 + (2 × x × 6) + 62
= (x + 6)2
(ii) (4x + 5y) (4x + 5y)
= 16x2 + 20xy + 20xy + 25y2
= (4x)2 + (2 × 4x × 5y) + (5y)2
= (4x + 5y)2
(iii) (7a + 9b) (7a + 9b)
= 49a2 + 63ab + 63ab + 81b2
= (7a)2 + (2 × 7a × 9b) + (9b)2
= (7a + 9b)2
(v) (x2 + 7) (x2 + 7)
= x4 + 7x2 + 7x2 + 49
= (x2)2 + (2 × x2 × 7) + 72
= (x2 + 7)2
(2) Find each of the following products:
(i) (x – 4) (x – 4)
= x2 – 4x – 4x + 16
= x2 – 8x + 16
= x2 – (2 × x × 4) + 42
= (x – 4)2
(ii) (2x – 3y) (2x – 3y)
= 4x2 – 6xy – 6xy + 9y2
= (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
= 64a2 + 48ab + 9b2
(ii) (7x + 2y)2
= (7x)2 + (2 × 7x × 2y) + (2y)2
= 49x2 + 28xy + (2y)2
(iii) (5x + 11)2
= (5x)2 + (2 × 5x × 11) + (11)2
= 25x2 + 110x + 121
(vi) (9x – 10)2
= (9x)2 – (2 × 9x × 10) + (10)2
= 81x2 – 180x + 100
(vii) (x2y – yz2)2
= (x2y)2 – (2 × x2y × yz2) + (yz2)2
= x4y2 – 2x2y2z2 + y2z4
(4) Find each of the following products:
(i) (x + 3) (x – 3)
= x2 + 3x – 3x – 9
= x2 – 32
(ii) (2x + 5) (2x – 5)
= 4x2 + 10x – 10x – 25
= (2x)2 – 52
(iii) (8 + x) (8 – x)
= 64 + 8x – 8x – x2
= 42 – x2
(iv) (7x + 11y) (7x – 11y)
= 49x2 + 77xy – 77xy – 121y2
= (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 (9x2 + 24x + 16), when x = 12.
Solution: 9x2 + 24x + 16
= [9 × (12)2] + (24 × 12) + 16
= (9 × 144) + 288 + 16
= 1296 + 288 + 16
= 1600
(9) Find the value of the expression (64x2 + 81y2 + 144xy), when x = 11 and y = 4/3.
Solution: 64x2 + 81y2 + 144xy
(10) Find the value of the expression (36x2 + 25y2 – 60xy), when x =2/3 and y = 1/5.
Solution: (36x2 + 25y2 – 60xy)
= (6x)2 – (2 × 6x × 5y) + (5y)2
= (6x – 5y)2
(13) Find the continued product:
(i) (x + 1) (x – 1) (x2 + 1)
= (x2 – 1) (x2 + 1)
= x4 – 1
(ii) (x – 3) (x + 3) (x2 + 9)
= [(x)2 – (3)2] (x2 + 9)
= (x2 – 9)(x2 + 9)
= x2 – 92 = x2 – 81
(iii) (3x – 2y) (3x + 2y) (9x2 + 4y2)
= [(3x)2 – (2y2)] (9x2 + 4y2)
= (9x2 – 4y2) (9x2 + 4y2)
= (9x2)2 – (4y2)2
= 81x4 – 16y4
(iv) (2p + 3) (2p – 3) (4p2 + 9)
= [(2p)2 – (3)2] (4p2 + 9)
= (4p2 – 9) (4p2 + 9)
= (4p2)2 – (9)2
= 16p4 – 81
(14) If x + y = 12 and xy = 14, find the value of (x2 + y2).
Solution: x + y = 12
⇒ (x + y)2 = (12)2
⇒ x2 + 2xy + y2 = 144
⇒ (x2 + y2) + (2 × 14) = 144
⇒ (x2 + y2) = 144 – 28
⇒ (x2 + y2) = 116
(15) If x – y = 7 and xy = 9, find the value of (x2 + y2).
Solution: x – y = 7
⇒ (x – y)2 = (7)2
⇒ x2 – 2xy + y2 = 49
⇒ (x2 + y2) – (2 × 9) = 49
⇒ (x2 + y2) = 49 – 18
⇒ (x2 + y2) = 31.
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