Question 5 - Figure it out - Page 154-156 - Chapter 6 Class 8 - We Distribute yet things Multiply (Ganita Prakash)

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Question 5 Verify which of the following statements are true. (i) (k + 1) (k + 2) – (k + 3) is always 2.Solving (𝑘+1)(𝑘+2)−(𝑘+3) = [𝒌(𝒌+𝟐)+𝟏(𝒌+𝟐)] −(𝑘+3) = [𝑘^2+2𝑘+𝑘+2] −(𝑘+3) = [𝒌^𝟐+𝟑𝒌+𝟐] −(𝒌+𝟑) = 𝑘^2+3𝑘+2−𝑘−3 = 𝑘^2+(3𝑘−𝑘)+(2−3) = 𝒌^𝟐+𝟐𝒌−𝟏 Since the value is not 2. ∴ The statement is false Question 5 Verify which of the following statements are true. (ii) (2q + 1) (2q – 3) is a multiple of 4.Solving (2𝑞+1)(2𝑞−3) = 𝟐𝒒(𝟐𝒒−𝟑)+𝟏(𝟐𝒒−𝟑) = 2𝑞 × 2𝑞−2 × 3 × 𝑞+2𝑞−3 =4𝑞^2−6𝑞+2𝑞−3 =𝟒𝒒^𝟐−𝟒𝒒−𝟑 Since we cannot take out 4 common, the expression is not a multiple of 4 ∴ The statement is false Question 5 Verify which of the following statements are true. (iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8.Any even number can be written as 2n Now, Square of even number = (2n)2 = 4n2 Since 4 is multiplied to the number, it means it is a multiple Therefore, Squares of even numbers are multiples of 4 Any odd number can be written as 2n + 1 Now, Square of odd number = (2n + 1)2 = (2n)2 + 2 × 2n × 1 + 12 = 4n2 + 4n + 1 = 4(n2 + n) + 1 Thus, squares of odd numbers are 1 more than multiple of 4 We can also say that squares of odd numbers are 1 more than multiple of 8 Example: 32 = 9 = 8 + 1 Thus, the given statement is True Question 5 Verify which of the following statements are true. (iv) (6n + 2)2 – (4n + 3)2 is 5 less than a square number.Now, (6n+2)^2−(4n+3)^2 Using (a + b)2 = a2 + 2ab + b2 = [(𝟔𝒏)^𝟐+𝟐 × 𝟔𝒏 × 𝟐+𝟐^𝟐 ]−[(𝟒𝒏)^𝟐+𝟐 × 𝟒𝒏 × 𝟑+𝟑^𝟐 ] = [36𝑛^2+24𝑛+4]−[16𝑛^2+24𝑛+9] = 36𝑛^2+24𝑛+4−16𝑛^2−24𝑛−9 = (36𝑛^2−16𝑛^2)+(24𝑛−24𝑛)+(4−9) = 20𝑛^2−5 Thus, the equation is 5 less than square number Hence, the statement is true