We have a reactanlge whose width and length are defined as follows:
Width ( w ) = x
Length ( L ) = 2x - 6
We have defined the unknown dimension of the rectangle width as ( x ). Then length of the rectangle is expressed in terms of the width as 6 meters less than twice of width.
The two dimensions ( width and length ) have been defined above. We will go ahead and express the area of a rectangle using the two dimension defined as follows:
![\begin{gathered} \text{Area = w}\cdot L \\ \text{Area = }x\cdot(2x-6) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/77v4r3dasg7uxur81sfcggtusd8ci0x2eh.png)
The area of the rectangle is given to us as:
![\begin{gathered} \text{Area = 308 = 2x}^2\text{ - 6x} \\ 2x^2-6x\text{ - 308 = 0} \\ x^2\text{ -3x - 154 = 0} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/u1qaw5umudk56wwmdewob8duan4by7vgdi.png)
What we expressed above is a quadratic equation for the width ( w ) of the rectangle that satisfies the corresponding relationship with length ( L ) and Area ( A = 308 m^2 ).
To determine the width ( w ) with above conditions, we will have to solve the derived quadratic equation in ( x ).
The general formula for solving a quadratic equation is expressed for a form:
![\begin{gathered} ax^2+bx\text{ + c = 0} \\ a,b,c\text{ are constants} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ztrwflswvf3pyy2lu3j4o366jka43u4d9v.png)
The general formula gives us:
![x\text{ = }(-b\pm√(b^2-4\cdot a\cdot c))/(2\cdot a)](https://img.qammunity.org/2023/formulas/mathematics/college/9bu7oawfzcogiq7nj4pfmh18x7xp3y5rpf.png)
So for our quadratic equation we have:
![\begin{gathered} a\text{ = 1, b = -3, c = -154} \\ x\text{ = }(3\pm√((-3)^2-4\cdot(1)\cdot(-154)))/(2\cdot(1))\text{ = }\frac{3\pm\sqrt{9\text{ +616 }}}{2} \\ x\text{ = }(3\pm√(625))/(2)\text{ = }(3\pm25)/(2) \\ x\text{ = }(3+25)/(2)\text{ , }(3-25)/(2) \\ x\text{ = }(28)/(2)\text{ , }(-22)/(2) \\ x\text{ = 14 , -11} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cv2jfexgw28a2gg8d3dgdzhqtckapsjp4z.png)
We have two values from the quadratic equation for the width of the rectangle. However, all dimensions only perceives positive values as its a length which is only defined by positive real numbers. Hence,
![\textcolor{#FF7968}{w}\text{\textcolor{#FF7968}{ = x = 14 m}}](https://img.qammunity.org/2023/formulas/mathematics/college/8qx4ji3vucskhvmxyz865xk0a27a9z8jf6.png)
We will use the relationship between width ( w ) and length ( L ) to determine the length wise dimension of the rectangle as follows:
![\begin{gathered} L\text{ = 2}\cdot(14)\text{ - 6} \\ L\text{ = 28 - 6} \\ \textcolor{#FF7968}{L}\text{\textcolor{#FF7968}{ = 22 m}} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kr2wxdjmy3wwbpo7tikvhb8hwlejtfm8a6.png)
Therefore, with the given condtions for the rectangle the width x length dimension of the trianglw would be expressed as:
![\begin{gathered} w\text{ x L} \\ \textcolor{#FF7968}{(14}\text{\textcolor{#FF7968}{ x 22 ) m }}\textcolor{#FF7968}{\ldots}\text{\textcolor{#FF7968}{ Option ( D )}} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6xraoia24f0uq96c5iblnlh2mzy7oflmap.png)