Respuesta :
Given:
The area of the rectangle, A=65ft^2.
Let l be the length of the rectangle and w be the width of the rectangle.
It is given that the length of the rectangle is two feet less than 3 times the width.
Hence, the expression for the length of the rectangle is,
[tex]l=3w-2\text{ ----(A)}[/tex]Now, the expression for the area of the rectangle can be written as,
[tex]\begin{gathered} A=\text{length}\times width \\ A=l\times w \\ A=(3w-2)\times w \\ A=3w^2-2w \end{gathered}[/tex]Since A=65ft^2, we get
[tex]\begin{gathered} 65=3w^2-2w \\ 3w^2-2w-65=0\text{ ---(1)} \end{gathered}[/tex]Equation (1) is similar to a quadratic equation given by,
[tex]aw^2+bw+c=0\text{ ---(2)}[/tex]Comparing equations (1) and (2), we get a=3, b=-2 and c=-65.
Using discriminant method, the solution of equation (1) is,
[tex]\begin{gathered} w=\frac{-b\pm\sqrt[]{^{}b^2-4ac}}{2a} \\ w=\frac{-(-2)\pm\sqrt[]{(-2)^2-4\times3\times(-65)}}{2\times3} \\ w=\frac{2\pm\sqrt[]{4^{}+780}}{2\times3} \\ w=\frac{2\pm\sqrt[]{784}}{6} \\ w=\frac{2\pm28}{6} \end{gathered}[/tex]Since w cannot be negative, we consider only the positive value for w. Hence,
[tex]\begin{gathered} w=\frac{2+28}{6} \\ w=\frac{30}{6} \\ w=5\text{ ft} \end{gathered}[/tex]Now, put w=5 in equation (A) to obtain the value of l.
[tex]\begin{gathered} l=3w-2 \\ =3\times5-2 \\ =15-2 \\ =13ft \end{gathered}[/tex]Therefore, the length of the rectangle is l=13 ft and the width is w=5 ft.
