Question

The product of two consecutive integers is $156$, find the numbers.A) $\text{10 }and\text{ 13}$B) $\text{12 }and\text{ 13}$C) $\text{12 }and\text{ 11}$D) $\text{1 }and\text{ 13}$

Hint: Here we assume variable to one of the consecutive integers to get the next one. Algebraic equation is framed and solved to get the unknown values.
Consecutive integers are integers whose difference value is $1$ .
We need to find the numbers as per the condition in the question.

Let the two consecutive numbers be $x$ and $x+1$ respectively.
The product of two consecutive integers is $156$ $\Rightarrow x\times (x+1)=156$
\begin{align} & \Rightarrow {{x}^{2}}+x=156 \\ & \Rightarrow {{x}^{2}}+x-156=0 \\ & \Rightarrow {{x}^{2}}+13x-12x-156=0 \\ & \Rightarrow x(x+13)-12(x+13)=0 \\ & \Rightarrow (x+13)(x-12)=0 \\ & \Rightarrow x=-13,12 \\ \end{align}
Therefore, If $x=-13$ the consecutive integers are $-13,-12$
Therefore, If $x=12$the consecutive integers are $12,13$
Note: The two consecutive integers can be even taken as $x-1$ and $x$ respectively. Even then we get the values to be $12$ and $13$. Upon performing the above calculation $x$ turns to be $13$ and therefore $x-1$ becomes $12$.