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

1.	Find the two positive integers whose sum of the squares is 365.
Answer» 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 x² +( x + 1)² = 365 x² + (x² + 2x + 1) = 365 [ ∵ (a + b)² = a² + 2ab + b²] 2x² + 2x + 1 = 365 2x² + 2x + 1- 365 = 0 2x² + 2x - 364 = 0 2(x² + x - 182) = 0 x² + x - 182 = 0 x² + 14x - 13x - 182 = 0 x (x + 14) - 13 (x + 14) = 0 (x - 13) (x + 14) = 0 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

Answer»

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

x² +( x + 1)² = 365

x² + (x² + 2x + 1) = 365 [ ∵ (a + b)² = a² + 2ab + b²]

2x² + 2x + 1 = 365

2x² + 2x + 1- 365 = 0

2x² + 2x - 364 = 0

2(x² + x - 182) = 0

x² + x - 182 = 0

x² + 14x - 13x - 182 = 0

x (x + 14) - 13 (x + 14) = 0

(x - 13) (x + 14) = 0

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

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