Question

Find two consecutive positive integers,the sum of whose square is 365​

Answer 1

let a and a+1 be two consecutive positive integers

Answer 2

Two consecutive positive integers are 13 and 14

Solution:

Let the two consecutive positive integer be x and x + 1

Sum of whose square is 365​

Therefore,

$x^2 + (x+1)^2 = 365\\\\x^2 + x^2 + 2x + 1 = 365\\\\2x^2 + 2x -364=0\\\\x^2 + x - 182 = 0\\\\Split\ the\ middle\ term\\\\x^2+14x-13x - 182= 0\\\\Group\ the\ expressions\\\\(x^2 + 14x) - (13x + 182) = 0\\\\Factor\ out\ x\ from\ x^2 + 14x\\\\x(x + 14) - (13x+182) = 0\\\\Factor\ out\ 13\ from\ 13x + 182\\\\x(x + 14) - 13(x + 14) = 0\\\\Factor\ out\ the\ common\ term\ x + 14\\\\(x-13)(x + 14) = 0\\\\x = 13 \text{ and } x = -14\\\\Ignore\ negative\ value\\\\Thus\\\\x = 13\\\\x + 1 = 13 + 1 = 14$

Thus two consecutive positive integers are 13 and 14

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