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

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#### Solution

Let the consecutive positive integers be *x* and *x* + 1.

Therefore, *x*^{2} + (*x* + 1)^{2} = 365

⇒ *x*^{2 }+ *x*^{2 }+ 1 + 2*x* = 365

⇒ 2*x*^{2} + 2x - 364 = 0

⇒ *x*^{2 }+ *x *- 182* *= 0

⇒ *x*^{2 }+ 14*x* - 13*x* - 182 = 0

⇒ *x*(*x* + 14) -13(*x* + 14) = 0

⇒ (*x* + 14)(*x* - 13) = 0

Either *x* + 14 = 0 or *x* - 13 = 0,

⇒

*x*= - 14 or*x*= 13Since the integers are positive, x can only be 13.

∴ x + 1 = 13 + 1 = 14

Therefore, two consecutive positive integers will be 13 and 14.

Therefore, two consecutive positive integers will be 13 and 14.

Concept: Solutions of Quadratic Equations by Factorization

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