# The product of two consecutive positive odd numbers is 195. Find the numbers.

Answer

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Hint: Assume the numbers as some variables. Frame the equation using the condition given in the question and solve it.

As per the question, let the two positive consecutive odd numbers be $x - 1$ and $x + 1$.

The product of the numbers is given as 196. So, we have:

$ \Rightarrow \left( {x - 1} \right)\left( {x + 1} \right) = 195$

As we know that $\left( {x - a} \right)\left( {x + a} \right) = {x^2} - {a^2}$. Using this formula, we’ll get:

$

\Rightarrow {x^2} - 1 = 195, \\

\Rightarrow {x^2} = 196, \\

\Rightarrow x = \pm \sqrt {196} , \\

\Rightarrow x = \pm 14 \\

$

But we have to consider only positive integers, so we will ignore negative values. $x = 14$ is the valid solution.

$x - 1 = 13{\text{ and }}x + 1 = 15$

Therefore our numbers are 13 and 15.

Note: We can also assume numbers to be $x$ and $x + 2$. In that case, we will get a different quadratic equation but the end result will be the same.

As per the question, let the two positive consecutive odd numbers be $x - 1$ and $x + 1$.

The product of the numbers is given as 196. So, we have:

$ \Rightarrow \left( {x - 1} \right)\left( {x + 1} \right) = 195$

As we know that $\left( {x - a} \right)\left( {x + a} \right) = {x^2} - {a^2}$. Using this formula, we’ll get:

$

\Rightarrow {x^2} - 1 = 195, \\

\Rightarrow {x^2} = 196, \\

\Rightarrow x = \pm \sqrt {196} , \\

\Rightarrow x = \pm 14 \\

$

But we have to consider only positive integers, so we will ignore negative values. $x = 14$ is the valid solution.

$x - 1 = 13{\text{ and }}x + 1 = 15$

Therefore our numbers are 13 and 15.

Note: We can also assume numbers to be $x$ and $x + 2$. In that case, we will get a different quadratic equation but the end result will be the same.

Last updated date: 27th Sep 2023

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