Chapter 3.1, Problem 68E

To determine

To calculate: .The equation of patio and solve the value for $x$ .

Answer to Problem 68E

The required value of $x$ is $7$ .

Explanation of Solution

Given Information:

The rectangular patio whose length is $25\text{ft}$ and width is $15\text{ft}$ . Also, the owner of patio is add $329\text{}\text{ft}$ to the area of patio.

Formula Used:

Quadratic formula:

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

Where $a,b,c$ are constant and $a\ne 0$ .

Calculation:

Consider the length of patio is $25\text{ft}$ and width of patio is $15\text{ft}$ .

Let, $x$ be the extended distance of patio.

Now, the equation of patio is:

  $\left(25+x\right)\left(15+x\right)-\left(15\times 25\right)=329$

Simplify the above equation.

  $\begin{array}{c}\left(25+x\right)\left(15+x\right)-\left(15\times 25\right)=329\\ x^2+40x+375-375=329\\ x^2+40x-329=0\end{array}$

Now, use quadratic formula in the above equation.

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $1$ for $a$ , $40$ for $b$ and $-329$ for $c$ in the above equation.

  $\begin{array}{c}x=\frac{-40\pm \sqrt{{40}^2-4\left(1\right)\left(-329\right)}}{2\left(1\right)}\\ =\frac{-40\pm \sqrt{1600+1316}}{2}\\ =\frac{-40\pm \sqrt{2916}}{2}\\ =\frac{-40\pm 54}{2}\end{array}$

Simplify the equation.

  $\begin{array}{c}x=\frac{-40\pm 54}{2}\\ =\frac{-40+54}{2},\frac{-40-54}{2}\\ =\frac{14}{2},\frac{-94}{2}\\ =7,-47\end{array}$

Since, $x$ is measurement and it is known that the measurement cannot be negative.

Thus, the solution of $x$ is $7$ .

Hence, the required solution of $x$ is $7$ .