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

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13, 14

-13,-14

-13,14

13,-14

answer is A.

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Detailed Solution

Let the first integer be x.
The next consecutive positive integer will be x + 1.
According to the given question,
x² + ( x + 1)² = 365
Or, x² +( x + 1)² = 365
Or, x² + (x² + 2x + 1) = 365 [ ∵ (a + b)² = a² + 2ab + b²]
Or, 2x² + 2x + 1 = 365
Or, 2x² + 2x + 1- 365 = 0
Or, 2x² + 2x - 364 = 0
Or, 2(x² + x - 182) = 0
Or, x² + x - 182 = 0
Or, x² + 14x - 13x - 182 = 0
Or, x (x + 14) - 13 (x + 14) = 0
Or, (x - 13) (x + 14) = 0
Or, x - 13 = 0 and x + 14 = 0
x = 13 and x = - 14
The value of x cannot be negative (because it is given that the integers are positive).
∴ x = 13 and x + 1 = 14
Hence, Option 1 is the correct option.

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