Question

find two consecutive positive integers,sum of whose square is 613

Answer 1

$\mathfrak{\huge{\pink{\underline{\underline{Question:-}}}}}$

find two consecutive positive integers,sum of whose square is 613

$\mathfrak{\huge{\pink{\underline{\underline{AnswEr:-}}}}}$

let two consecutive positive integers be X, x + 1

according to the question

x² + (x + 1)² = 613

x² + x² + 1 + 2x = 613

2x² + 2x = 613 - 1

2x² + 2x - 612 = 0

2 ( x² + x - 306 ) = 0

x² + x - 306 = 0

x² - 18x + 17x - 306 = 0

x (x - 18) + 17 (x - 18) = 0

(x - 18) (x + 17) = 0

x = 18

x = -17

Hope my answer helps you!!

Answer 2

$x^{2} + {x - 1}^{2} = 613 \\ x^{2} + x^{2} + 1 - 2x = 613 \\ 2x^{2} + 2x = 612 \\ 2x^{2} - 2x - 612 = 0 \\ x^{2} - x - 306 = 0$

X=18 OR X= -17