Question

The product of two consecutive natural numbers is 12. The equation form of this statement is
$\begin{align}
  & a){{x}^{2}}+2x-12=0 \\
 & b){{x}^{2}}+x-12=0 \\
 & c){{x}^{2}}+x+12=0 \\
 & d){{x}^{2}}+2x+12 \\
\end{align}$

Answer 1

Hint: Now we are given that the product of two consecutive natural numbers is 12. Now let us know that the two consecutive terms can be written in the form of x and x + 1 as well as x and x – 1. Hence we will consider these two terms as two consecutive natural numbers and use the required condition to form an equation.

Complete step-by-step answer:
Now we know that natural numbers are 1, 2, 3, ……
Now two consecutive numbers means the numbers next to each other. Hence we have 2 and 3 are consecutive numbers as well as 4 and 5 are consecutive terms and so on.
Now we are given that the product of two consecutive numbers is 12.
Let x and x + 1 be the respective consecutive terms.
Now using the given condition we get,
x(x + 1) = 12
Now opening the brackets we get,
${{x}^{2}}+x=12$
Rearranging we have ${{x}^{2}}+x-12=0$
Now there is also the possibility that the numbers are x and x – 1.
$x\left( x-1 \right)=12$
Now again opening the bracket we have ${{x}^{2}}-x=12$
Again rearranging the terms we get, ${{x}^{2}}-x-12=0$ .
Hence the equation representing the given condition is ${{x}^{2}}+x-12=0$ or ${{x}^{2}}-x-12=0$

So, the correct answer is “Option (b)”.

Note: Now note that if we are ${{x}^{2}}+x-12=0$ and ${{x}^{2}}-x-12=0$ both are different equation but they represent same numbers. For the equation ${{x}^{2}}+x-12=0$ we have the numbers are x and x + 1 and for ${{x}^{2}}-x-12=0$ the numbers are x and x – 1. The value of these numbers in both cases will be the same.