RS Aggarwal Solutions for Class 8 Maths Chapter 6 Exercise 6.4: RS Aggarwal Solutions for Class 8 Maths Chapter 6 Exercise 6.4 prepared by subject experts of Physics Wallah provide detailed guidance and solutions for a range of problems related to operations on algebraic expressions.
By working through these solutions, students can gain a deeper understanding of algebraic expressions, clarify any doubts, and improve their problem-solving skills.RS Aggarwal Solutions for Class 8 Maths Chapter 6 Exercise 6.4 PDF
(Question 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(Question 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(Question 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(Question 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(Question 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(Question 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(Question 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(Question 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(Question 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(Question 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(Question 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(Question 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(Question 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.