Question

as seen in the diagram below, camila is building a walkway with a width of x feet to go around a swimming pool that measures 13 feet by 10 feet. if the total area of the pool and the walkway will be 304 square feet, how wide should the walkway be?

Explanation:

Step1: Find the dimensions of the outer - rectangle

The length of the outer - rectangle (pool + walkway) is $13 + 2x$ feet and the width is $10+2x$ feet.

Step2: Set up the area equation

The area of a rectangle is $A = l\times w$. So, $(13 + 2x)(10 + 2x)=304$.
Expand the left - hand side using FOIL:
\[

$$\begin{align*} 13\times10+13\times2x+2x\times10 + 4x^{2}&=304\\ 130+26x + 20x+4x^{2}&=304\\ 4x^{2}+46x+130 - 304&=0\\ 4x^{2}+46x - 174&=0 \end{align*}$$

\]
Divide the entire equation by 2 to simplify: $2x^{2}+23x - 87 = 0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 2$, $b = 23$, $c=-87$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(23)^{2}-4\times2\times(-87)=529 + 696 = 1225$.
Then, $x=\frac{-23\pm\sqrt{1225}}{4}=\frac{-23\pm35}{4}$.
We have two solutions for $x$:
$x_1=\frac{-23 + 35}{4}=\frac{12}{4}=3$ and $x_2=\frac{-23 - 35}{4}=\frac{-58}{4}=-14.5$.
Since the width cannot be negative, we discard $x_2$.

Answer:

3 feet