Q. 460 - Intermediate Algebra

Question

The product of two consecutive integers is $156$. Find the integers.

Step-by-Step Solution

Answer

The two required integers are $\left\{-13,-12\right\},\{12,13\}$.

Step 1. Assume the integers.

Assume the first integer to be $x$ and the other consecutive integer to be $x+1$.

From the given question,

The product of two consecutive integers is $156$. Then,

$$\begin{gathered} x\left(x+1\right)=156 \\ {x}^2+x=156 \end{gathered}$$

Bring all the terms to one side,

$x^2+x-156=0$

Step 2. Factor the equation.

On factoring,

$$\begin{gathered} x^2+13x-12x-156=0 \\ x\left(x+13\right)-12\left(x+13\right)=0 \\ \left(x+13\right)\left(x-12\right)=0 \end{gathered}$$

Then,

$$\begin{gathered} x+13=0 \\ x=-13 \end{gathered}$$

Also,

$$\begin{gathered} x-12=0 \\ x=12 \end{gathered}$$

Step 3. Find the two integers.

When $x=-13$ then the other consecutive number,

$$\begin{gathered} x+1=-13+1 \\ =-12 \end{gathered}$$

When $x=12$ then the other consecutive number,

$$\begin{gathered} x+1=12+1 \\ =13 \end{gathered}$$

Step 4. Check the answers.

Substitute $-13,-12$ in the original equation,

$$\begin{gathered} \left(-13\right)\times \left(-12\right)=156 \\ 156=156 \end{gathered}$$

This is true.

Substitute $12,13$ in the original equation,

$$\begin{gathered} 12\times 13=156 \\ 156=156 \end{gathered}$$

This is also true.