Question

find two consecutive positive integers, sum of whose square is 365

Answer 1

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Answer 2

$\Large{\underline{\underline{\bf{Solution:-}}}}$

${\rm{\bold{Let\: the \: consecutive\: positive\: Integers\: be \: x \: and \: x+1}}}$

⟹${\rm{ {x}^{2} + {x + 1}^{2} = 365 }}$

⟹ ${\rm{{x}^{2} + {x}^{2} + 1 + 2x = 365 }}$

⟹${\rm{ 2{x}^{2} + 2x - 364 = 0}}$

⟹ ${\rm{ {x}^{2} + x - 182 = 0 }}$

⟹ ${\rm{ {x}^{2} + 14x - 13x - 182 = 0 }}$

⟹${\rm{x(x + 14) - 13(x + 14) = 0 }}$

⟹${\rm{ (x + 14)(x - 13) = 0}}$

⟹${\rm{ x = - 14 \: or \: x = 13}}$

${\underline{\underline{\bf{Integers\:are \: positive }}}}$

$\rm{So, \: x\: can \: be \: only \: 13}$

$\rm{x+1=13+1=14}$

${\boxed{\boxed{\rm{First\: consecutive\: positive\: integer \: be \: 13}}}}$

${\boxed{\boxed{\rm{Other\: consecutive\: positive\: integer \: be \: 14}}}}$