RS Aggarwal Solutions for Class 8 Maths Chapter 6 Exercise 6.4 (Exercise 6D)
Ananya Gupta31 Jul, 2024
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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 Operations on Algebraic Expressions Overview
Exercise 6.4 of RS Aggarwal Solutions for Class 8 Maths Chapter 6 focuses on complex operations involving algebraic expressions, including advanced techniques for multiplication and division. This exercise includes problems where students must apply the distributive property, combine like terms, and simplify expressions with multiple variables. The problems challenge students to work with more intricate algebraic structures and practice their skills in managing complex equations.RS Aggarwal Solutions for Class 8 Maths Chapter 6 Exercise 6.4 PDF
RS Aggarwal Solutions for Class 8 Maths Chapter 6 Exercise 6.4 are available in the PDF link below. This resource provides detailed solutions and explanations for complex operations on algebraic expressions. By using this PDF, students can access thorough, step-by-step guidance to tackle challenging problems helping them enhance their understanding of algebraic expressions. This will support effective study, clarify any doubts, and improve exam preparation and performance.RS Aggarwal Solutions for Class 8 Maths Chapter 6 Exercise 6.4 PDF
RS Aggarwal Solutions for Class 8 Maths Chapter 6 Exercise 6.4 (Exercise 6D)
RS Aggarwal Solutions for Class 8 Maths Chapter 6 Exercise 6.4 are available below. This resource provide detailed solutions and explanations for problems related to operations on algebraic expressions.(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.Benefits of RS Aggarwal Solutions for Class 8 Maths Chapter 6 Exercise 6.4
- Detailed Explanations : These solutions provide clear and comprehensive explanations for each problem, making it easier for students to grasp complex concepts and operations on algebraic expressions.
- Step-by-Step Guidance : Each solution is broken down into detailed steps, which helps students understand the methodology and apply it to similar problems, boosting their problem-solving skills.
- Concept Reinforcement : By working through these exercises, students reinforce their understanding of algebraic expressions and operations, leading to a stronger grasp of the material.
- Error Correction : The solutions help identify and correct common mistakes, allowing students to learn from their errors and improve their accuracy.
- Efficient Exam Preparation : With these solutions, students can efficiently prepare for exams by practicing and reviewing key concepts, ultimately leading to better performance.
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