Q 6.105 - Intermediate Algebra

Question

A rectangular sign has area $30$ square feet. The length of the sign is one foot more than the width. Find the length

and width of the sign.

Step-by-Step Solution

Answer

The length of the rectangular sign is $6 feet$ and the width of the rectangular sign is $5 feet.$

Step 1. Given Information

There is a rectangular sign whose area is $30$ square feet and the length of the sign is one foot more than the width.

We have to find the length

and width of the sign.

Step 2. Assume the length and width of the sign

Let the width of the sign be x

As we know the length of the sign is one foot more than the width,

so the length of the sign will be $x+1$

Step 3. Summarize into an equation

The area of the rectangular sign is $30$ square feet.

By using the formula of area of a rectangle

$$\begin{gathered} \left(x\right)(x+1)=30 \\ {x}^2+x-30=0 \\ {x}^2+6x-5x-30=0 \\ x(x+6)-5(x+6)=0 \\ (x-5)(x+6)=0 \end{gathered}$$

Step 4. Use the Zero Product Property

The equation we get is

$(x-5)(x+6)=0$

So,

$$\begin{gathered} x-5=0 \\ x=5 \end{gathered}$$ and $$\begin{gathered} x+6=0 \\ x=-6 \end{gathered}$$

Step 5. Substitute the value of x in assumed width and length to find the exact width and length

Since x is the width of the rectangular sign, it does not be negative therefore, we will eliminate the value $-6.$

If $x=5$

So, the width of the rectangular sign is $5 feet.$

The length of the rectangular sign is $5+1=6 feet$