Question

write a pythagorean triplet whose one member is 14

Answer 1

2m=28
m^2-1=14^2-1=195
So,m=14
m^2-1=195
2m=28

Answer 2

HEY MATE
HERE IS YOUR ANSWER

$let \: 2m = 14 \\ m = 7 \\ {m}^{2} + 1 = {7}^{2} + 1 = 49 + 1 = 50 \\ {m}^{2} - 1 = {7}^{2} - 1 = 49 - 1 = 48 \\ \\ check. \\ {14}^{2} + {48}^{2} = 196 + 1304 = 2500 = {50}^{2} \\ \\ hence \\ the \: triplet \: is \: 14 \: \: \: 48 \: \: \: and \: \: \: 50$

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