Find two consecutive odd positive integers, sum of whose squares is 970.

Given:

Sum of the squares of two consecutive odd positive integers is 970. 

 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,

$(2x-1)^2+(2x+1)^2=970$

$4x^2-4x+1+4x^2+4x+1=970$

$8x^2+2=970$

$8x^2=970-2$

$8x^2=968$

$x^2=\frac{968}{8}$

$x^2=121$

$x^2-121=0$

$x^2-(11)^2=0$

$(x+11)(x-11)=0$

$x+11=0$ or $x-11=0$

$x=-11$ or $x=11$

We need only odd positive integer. Therefore, the value of $x$ is $11$.

$2x-1=2(11)-1=22-1=21$

$2x+1=2(11)+1=22+1=23$

The required odd positive integers are $21$ and $23$. 

Tutorialspoint

Updated on: 10-Oct-2022

49 Views

Kickstart Your Career

Get certified by completing the course

Get Started