Question

# The sum of the squares of two consecutive odd numbers are 394. Find the product of two numbers.

Hint:- The odd consecutive numbers has a difference of two.

$\Rightarrow$Let two consecutive odd numbers be $x$ and $x + 2$
So, according to the given condition in question
$\Rightarrow {x^2} + {\left( {x + 2} \right)^2} = 394$ (1)
Now, for solving equation 1. We get,
$\Rightarrow {x^2} + {x^2} + 4x + 4 = 394 \\ \Rightarrow 2{x^2} + 4x + 4 = 394 \\$
Solving above equation. We get,
$\Rightarrow {x^2} + 2x - 195 = 0$
Now, for solving above equation we split 2x, it becomes
$\Rightarrow {x^2} + 15x - 13x - 195 = 0$
Taking common factors of the above equation to get the value of $x$.
$\Rightarrow x\left( {x + 15} \right) - 13\left( {x + 15} \right) = 0 \\ \Rightarrow \left( {x + 15} \right)\left( {x - 13} \right) = 0 \\$
Hence possible values of $x$ are -15 and 13.
But x cannot be negative.
$\Rightarrow So,{\text{ }}x = 13$
$\Rightarrow$So, the first odd number will be 13.
$\Rightarrow$And, another odd number will be $13 + 2 = 15$.
So, we have two find the product of these two odd numbers.
$\Rightarrow$Required product$= 13*15 = 195$.
Hence, the product of two consecutive numbers whose sum of squares is 394 is 195.

Note:- Whenever we come up with this type of problem then first assume two numbers such that they depend on each other (to make calculations easy) like here we assume x+2 which depends on x. And then find the value of the variable using the given condition. Then we should multiply two numbers to get the required answer.