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Example 2 Write a Pythagorean triplet whose smallest member is 8. We know 2m, π‘š^2βˆ’1 and π‘š^2+1 form a Pythagorean triplet. We know 2m, π’Ž^πŸβˆ’πŸ and π’Ž^𝟐+𝟏 form a Pythagorean triplet. Given, Smallest member of the triplet = 8. Let 2m = 8 2m = 8 m = 8/2 m = 4 Let π’Ž^πŸβˆ’πŸ" = 8" π‘š^2 = 8 + 1 π‘š^2 = 9 ∴ m = 3 Let π’Ž^𝟐+𝟏 = 8 π‘š^2 = 8 βˆ’ 1 π‘š^2 = 7 Since, 7 is not a square number, ∴ π‘š^2+1 β‰  8 It is not possible. Therefore, m = 4 or m = 3 Finding Triplets for m = 3 1st number = 2m 2nd number = π‘š^2βˆ’1 3rd number = π‘š^2+1 ∴ 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 = π‘š^2βˆ’1 3rd number = π‘š^2+1 As 8 is the smallest member of this triplet. ∴ 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 12 years. He provides courses for Maths and Science at Teachoo.