Q. 324 - Intermediate Algebra

Question

The product of two consecutive even integers is $288$. Find the integers.

Step-by-Step Solution

Answer

The two required integers are $\{-18,-16\}$ and $\{16,18\}$.

Step 1. Assumption of integers.

Assume the first integer to be $x$.

Next consecutive even integer$=x+2$

From the question,

$$\begin{gathered} x\left(x+2\right)=288 ...... \left(i\right) \\ {x}^2+2x=288 \\ {x}^2+2x-288=0 \end{gathered}$$

Step 2. Factor the equation.

On factoring,

$$\begin{gathered} x^2+18x-16x-288=0 \\ x\left(x+18\right)-16\left(x+18\right)=0 \\ \left(x+18\right)\left(x-16\right)=0 \end{gathered}$$

Use the zero product property,

$$\begin{gathered} x+18=0 \\ x=-18 \end{gathered}$$

Then,

$$\begin{gathered} x-16=0 \\ x=16 \end{gathered}$$

Step 3. Find out the integers.

If the first integer $x=-18$, then the next even integer,

$$\begin{gathered} =x+2 \\ =-18+2 \\ =-16 \end{gathered}$$

Therefore, the sets of two consecutive even integers is $-18,-16$.

If the first integer $x=16$, then the next even integer,

$$\begin{gathered} =x+2 \\ =16+2 \\ =18 \end{gathered}$$

Therefore, the sets of two consecutive even integers is $16,18$.

Step 4. Check the answers.

Substitute the consecutive integers $-18,-16$ in equation (i),

$$\begin{gathered} \left(-18\right)\left(-16\right)=288 \\ 288=288 \end{gathered}$$

This is true.

Substitute the consecutive integers $16,18$ in equation (i),

$$\begin{gathered} 16\times 18=288 \\ 288=288 \end{gathered}$$

This is also true.

Hence, the sets of consecutive integer are $\{-18,-16\},\{16,18\}$.