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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 $23$.
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