Question

find two consecutive positive integers sum of whose squares is 365

Answer 1

 Let first number = xThen second number will one more so that next number wil x+1Given that sum of whose squares is 365.      x2+ (  x + 1)2 = 365use formula of (a +b)2 = a2 + 2ab +b2    x2+   x2+ 2x + 1 – 365 = 0  2x2+ 2x – 364 = 0Divide by 2 to simplify it    x2+   x – 182 = 0factorize it now

Answer 2

Let one number be x and other number be x+1

Therefore,
x+x+1=365
2x+1=365
2x=365-1
2x=364
x=364÷2
x=182

Hence, the first no. be x=182
and other number be
x+1
=182+1
=183