Question:
The sum of squares of two consecutive odd positive integers is 394. Find them.
Solution:
Let two consecutive odd positive integer be $(2 x-1)$ and other $(2 x+1)$
Then according to question
$(2 x+1)^{2}+(2 x-1)^{2}=394$
$8 x^{2}+2=394$
$8 x^{2}=394-2$
$x^{2}=\frac{392}{8}$
$x^{2}=49$
$x=\sqrt{49}$
$=\pm 7$
Since, x being a positive number, so x cannot be negative.
Therefore,
When $x=7$ then odd positive
$2 x-1=2 \times 7-1$
$=13$
And
$2 x+1=2 \times 7+1$
$=15$
Thus, two consecutive odd positive integer be 13,15
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.