Question

The sum of the squares of the two consecutive odd positive integers as 394. Find them.

Answer 1

Given: The sum of the squares of the two consecutive odd positive integers is 394.

Let the consecutive odd positive integers are 2x – 1 and 2x + 1

According to the question,

(2𝑥 - 1)² + (2𝑥 + 1)² = 394

`4x^2 − 4x + 1 + 4x^2 + 4x + 1 = 394  ...{((a + b)^2 = a^2 + 2ab + b^2), ((a - b)^2 = a^2 - 2ab + b^2):}`

`4x^2 − cancel(4x) + 1 + 4x^2 + cancel(4x) + 1` = 394

8𝑥² + 2 = 394

8𝑥² = 394 - 2

8𝑥² = 392

𝑥² = `392/8`

𝑥² = 49

Taking square root both the sides,

x = `+-`7

x = 7 or x = - 7

We need only odd positive integer. Therefore, the value of x is 7.

2x − 1 = 2(7) − 1 = 14 − 1 = 13

2x + 1 = 2(7) − 1 = 14 + 1 = 15

The required odd positive integers are 13 and 15.