Respuesta :
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:
[tex]\begin{gathered} \text{Area = w}\cdot L \\ \text{Area = }x\cdot(2x-6) \end{gathered}[/tex]The area of the rectangle is given to us as:
[tex]\begin{gathered} \text{Area = 308 = 2x}^2\text{ - 6x} \\ 2x^2-6x\text{ - 308 = 0} \\ x^2\text{ -3x - 154 = 0} \end{gathered}[/tex]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:
[tex]\begin{gathered} ax^2+bx\text{ + c = 0} \\ a,b,c\text{ are constants} \end{gathered}[/tex]The general formula gives us:
[tex]x\text{ = }\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}[/tex]So for our quadratic equation we have:
[tex]\begin{gathered} a\text{ = 1, b = -3, c = -154} \\ x\text{ = }\frac{3\pm\sqrt{(-3)^2-4\cdot(1)\cdot(-154)}}{2\cdot(1)}\text{ = }\frac{3\pm\sqrt{9\text{ +616 }}}{2} \\ x\text{ = }\frac{3\pm\sqrt{625}}{2}\text{ = }\frac{3\pm25}{2} \\ x\text{ = }\frac{3+25}{2}\text{ , }\frac{3-25}{2} \\ x\text{ = }\frac{28}{2}\text{ , }\frac{-22}{2} \\ x\text{ = 14 , -11} \end{gathered}[/tex]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,
[tex]\textcolor{#FF7968}{w}\text{\textcolor{#FF7968}{ = x = 14 m}}[/tex]We will use the relationship between width ( w ) and length ( L ) to determine the length wise dimension of the rectangle as follows:
[tex]\begin{gathered} L\text{ = 2}\cdot(14)\text{ - 6} \\ L\text{ = 28 - 6} \\ \textcolor{#FF7968}{L}\text{\textcolor{#FF7968}{ = 22 m}} \end{gathered}[/tex]Therefore, with the given condtions for the rectangle the width x length dimension of the trianglw would be expressed as:
[tex]\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}[/tex]