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  1. Chapter 6 Class 8 Squares and Square Roots
  2. Serial order wise
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Example 2 Write a Pythagorean triplet whose smallest member is 8. We know 2m, ๐‘š^2โˆ’1 and ๐‘š^2+1 form a Pythagorean triplet. Given, Smallest member of the triplet = 8. Let 2m = 8 2m = 8 m = 8/2 m = 4 Let ๐’Ž^๐Ÿโˆ’๐Ÿ" = 8" ๐‘š^2 = 8 + 1 ๐‘š^2 = 9 โˆด m = 3 Let ๐’Ž^๐Ÿ+๐Ÿ = 8 ๐‘š^2 = 8 โˆ’ 1 ๐‘š^2 = 7 Since, 7 is not a square number, โˆด ๐‘š^2+1 โ‰  8 It is not possible. Therefore, m = 4 or m = 3 Finding Triplets for m = 3 1st number = 2m 2nd number = ๐‘š^2โˆ’1 3rd number = ๐‘š^2+1 โˆด The required triplet is 6, 8, 10 But 8 is not a smallest member of this triplet. So, lets try for m = 4 Finding Triplets for m = 4 1st number = 2m 2nd number = ๐‘š^2โˆ’1 3rd number = ๐‘š^2+1 As 8 is the smallest member of this triplet. โˆด The required triplet is 8, 15, 17

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.