Question

In the following exercises, solve by using methods of factoring, the square root principle, or the quadratic formula.

The front walk from the street to Pam’s house has an area of 250 square feet. Its length is two less than four times its width. Find the length and width of the sidewalk. Round to the nearest tenth.

Short answer

The width of the sidewalk is$8.2ft.$and the length is$30.8ft.$

Step-by-step solution

Given information.

The given statement is:

The front walk from the street to Pam’s house has an area of 250 square feet. Its length is two less than four times its width. Find the length and width of the sidewalk. Round to the nearest tenth.

Set up an equation such that the length is two less than four times its width.  

Let the width be $x.$

$$\begin{gathered} Length=4\times width-2 \\ Length=4x-2 \end{gathered}$$

Use the formula of the area of the rectangle.

$$\begin{gathered} Area=length\times breadth \\ Area=x\left(4x-2\right) \\ 250=4x^2-2x \\ 4x^2-2x-250=0 \end{gathered}$$

Use the quadratic formula. 

The quadratic formula for the equation

$a{x}^2+bx+c=0isx=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$

Now we compare the given equation with the standard form we get

$$\begin{gathered} a=4,b=-2andc=-250. \\ x=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(4\right)\left(-250\right)}}{2\left(4\right)} \\ x=\frac{2\pm \sqrt{4004}}{8} \\ x=-7.7,8.2 \end{gathered}$$

Because the width cannot be negative. So, the width is $8.2ft.$and the length is$4\left(8.2\right)-2=30.8ft.$