Question

find two consecutive positive integers sum of whose square is 365​

Answer 1

Answer:

It is given that the sum of squares of two consecutive positive integers is 365. Let, the first positive integer is m then the second positive integers is m+1. Since it is given that the consecutive numbers are positive integers. Hence, two consecutive positive integers are 13&14.

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Answer 2

Answer:

The two consecutive positive integers are 13 & 14.

Step-by-step-explanation:

Let the first positive integer be x.

And the consecutive positive integer be ( x + 1 ).

From the given condition,

The sum of squares of the integers is 365.

∴ x² + ( x + 1 )² = 365

⇒ x² + x² + 2x + 1² = 365

⇒ 2x² + 2x + 1 = 365

⇒ 2x² + 2x = 365 - 1

⇒ 2x² + 2x = 364

⇒ x² + x = 182

⇒ x² + x - 182 = 0

⇒ x² + 14x - 13x - 182 = 0

⇒ x ( x + 14 ) - 13 ( x + 14 ) = 0

⇒ ( x + 14 ) ( x - 13 ) = 0

⇒ ( x + 14 ) = 0 OR ( x - 13 ) = 0

⇒ x + 14 = 0 OR x - 13 = 0

⇒ x = - 14 OR x = 13

But the given integer is positive.

∴ x = - 14 is unacceptable.

∴ x = 13

The first positive integer = 13

And the consecutive positive integer = 13 + 1 = 14

∴ The two consecutive positive integers are 13 & 14.