Q. 323 - Intermediate Algebra

Question

The product of two consecutive even integers is $168$. Find the integers.

Step-by-Step Solution

Answer

The two required integers are $\{-14,-12\}$ and $\{12,14\}$.

Step 1. Assumption of integers.

Assume the first integer to be x.

Next consecutive even integer$=x+2$

From the question,

$$\begin{gathered} x\left(x+2\right)=168 ...... \left(i\right) \\ {x}^2+2x=168 \\ {x}^2+2x-168=0 \end{gathered}$$

Step 2. Factor the equation.

On factoring,

$$\begin{gathered} x^2+14x-12x-168=0 \\ x\left(x+14\right)-12\left(x+14\right)=0 \\ \left(x+14\right)\left(x-12\right)=0 \end{gathered}$$

Use the zero product property,

$$\begin{gathered} x+14=0 \\ x=-14 \end{gathered}$$

Then,

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

Step 3. Find out the integers.

If the first integer $x=-14$, then the next even integer,

$$\begin{gathered} =x+2 \\ =-14+2 \\ =-12 \end{gathered}$$

Therefore, the sets of two consecutive even integers is $-14,-12$.

If the first integer $x=12$, then the next even integer,

$$\begin{gathered} =n+2 \\ =12+2 \\ =14 \end{gathered}$$

Therefore, the sets of two consecutive even integers is $12,14$.

Step 4. Check the answers.

Substitute the consecutive integers $-14,-12$ in equation (i),

$$\begin{gathered} \left(-14\right)\left(-12\right)=168 \\ 168=168 \end{gathered}$$

This is true.

Substitute the consecutive integers $12,14$ in equation (i),

$$\begin{gathered} 12\times 14=168 \\ 168=168 \end{gathered}$$

This is also true.

Hence, the sets of consecutive integer are $\{-14,-12\},\{12,14\}$.