Question

Find two consecutive positive integers, sum of whose squares is 365.​

Answer 1

Answer:

let one number be x so other is x+1

atq:

x²+(x+1)²=365

2x²+2x–364=0

x²+x–182=0

182=2×13×7

so,

x²+14x–13x–182=0

x(x+14)–13(x+14)=0

x=13 or –14(rejected)

Thus the two consecutive poisitive integers are 13 and 14

Answer 2

$\huge\underline\mathbb{SOLUTION:-}$

$\mathsf {Let,\: first\:number\:be\:x}$

$\mathsf {And,\:Let\:second\:number\:be\:(x + 1)}$

$\underline\mathsf {According\:to\:given\:condition,}$

$\mathsf {x^2 + (x + 1)^2 = 365\: {(a + b)^2 = a^2 + b^2 + 2ab}}$

$\implies \mathsf {x^2 + x^2 + 1 + 2x = 365}$

$\implies \mathsf {2x^2 + 2x - 354 = 0}$

$\underline\mathsf {Dividing\:equation\:by\:2}$

$\implies \mathsf {x^2 + x - 182 = 0}$

$\implies \mathsf {x^2 + 14x - 13x - 182 = 0}$

$\implies \mathsf {x(x + 14) - 13(x + 14) = 0}$

$\implies \mathsf {(x + 14) (x - 13) = 0}$

$\implies \mathsf {x = 13,\: -14}$

$\therefore \mathsf {First\:number = 13\:(We\:discard\: -14\:because}$

$\mathsf {\:\:\:it\:is\:negative\:number.)}$

$\mathsf {Second\:number = x + 1 = 13 + 1= 14}$

$\therefore \mathsf {Two\:consecutive\:positive\:integers}$

$\mathsf {\:\:\:are\:13\:and\:14\:whose\:sum\:of\:squares\:is\:equal\:to\:365.}$