Slide9.JPG

Slide10.JPG

Go Ad-free

Transcript

Ex 5.2, 2 Write a Pythagorean triplet whose one member is. (ii) 14We know 2m, 𝑚^2−1 and 𝑚^2+1 form a Pythagorean triplet. Given, One member of the triplet = 14. Let 2m = 14 2m = 14 m = 14/2 m = 7 Let 𝒎^𝟐−𝟏" = 14" 𝑚^2 = 14 + 1 𝑚^2 = 15 Since, 15 is not a square number, ∴ 𝑚^2−1 ≠ 14 It is not possible. Let 𝒎^𝟐+𝟏 = 14 𝑚^2 = 14 − 1 𝑚^2 = 13 Since, 13 is not a square number, ∴ 𝑚^2+1 ≠ 14 It is not possible. Therefore, m = 7 Finding Triplets for m = 7 1st number = 2m 2nd number = 𝑚^2−1 3rd number = 𝑚^2+1 ∴ The required triplet is 14, 48, 50

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo