Question

Find two consecutive positive integers sum of whose square is 365 ?
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Answer 1

Let first number = x

Then second number will one more so that next number wil x+1

Given that sum of whose squares is 365.

x2+ ( x + 1)2 = 365

use formula of (a +b)2 = a2 + 2ab +b2

x2+ x2+ 2x + 1 – 365 = 0

2x2+ 2x – 364 = 0

Divide by 2 to simplify it

x2+ x – 182 = 0

factorize

13 and -14

so the answer is 13 and 14

mark as brainliest

Answer 2

Sorry for late reply....