Question

The product of two consecutive odd integers is equal to $675$. Find the two integers.

Answer 1

Hint: Consecutive odd integers have a difference of $2$ between them. So, consider the two consecutive odd integers to be ${\text{x}}$ and ${\text{x + 2}}$. Put the product of these two integers equal to $675$ and then simplify the equation to find the value of ${\text{x}}$. Put the value of ${\text{x}}$ in the consecutive numbers you considered and you’ll get the answer.

Complete step-by-step answer:
Given, the product of two consecutive integers is equal to $675$. We have to find two integers. Since there is a difference of $2$ between two consecutive odd integers so we can consider the first integer to be ${\text{x}}$ and the second integer to be ${\text{x + 2}}$. Then according to question,
\[
   \Rightarrow {\text{x}} \times \left( {{\text{x + 2}}} \right) = 675 \\
   \Rightarrow {{\text{x}}^2} + 2{\text{x = 675}} \\
 \]
If we add one on both sides we can make a perfect square which will make solving the equation easier. So add $1$ on both the sides-
$
   \Rightarrow {{\text{x}}^2} + 2{\text{x + 1 = 675 + 1}} \\
   \Rightarrow {\left( {{\text{x}} + 1} \right)^2} = 676 \\
 $
Now remove the square from ${\text{x + 1}}$
$ \Rightarrow {\text{x + 1 = }}\sqrt {676} = 26$
Now transfer 1 from left side to the right side of equation and we have,
$ \Rightarrow {\text{x = 26 - 1 = 25}}$
Now we have the value of${\text{x}}$, so on putting the value in the considered integers we get-
$
  {\text{x}} = 25 \\
  {\text{x}} + 2 = 27 \\
 $
Here since the product of these two integers is positive so both integers can either be positive or negative as $\left( - \right) \times \left( - \right) = + $ and $\left( + \right) \times \left( + \right) = + $
Hence the two integers are either $25{\text{and}}27$ or$ - 25{\text{ and - 27}}$.

Note: Consecutive integers are the integers that are continuous like one after other in a series. Consecutive odd integers are continuous integers which are odd and they have a difference of 2 between them. They are of form $2{\text{n + 1,2n + 3}}$ and so on.