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

Given:

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

 To do:

We have to find the numbers.

Solution:

Let the two consecutive odd positive integers be $2x-1$ and $2x+1$.

According to the question,

$(2x-1)^2+(2x+1)^2=394$

$4x^2-4x+1+4x^2+4x+1=394$

$8x^2+2=394$

$8x^2=394-2$

$8x^2=392$

$x^2=\frac{392}{8}$

$x^2=49$

$x^2-49=0$

$x^2-(7)^2=0$

$(x+7)(x-7)=0$

$x+7=0$ or $x-7=0$

$x=-7$ 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 $15$.

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Updated on: 10-Oct-2022

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