Question

The length of a rectangle is 3 M less than twice the width, and the area of the rectangle is 14m^2. Find the dimensions of the rectangle.Length= MWidth= M

Answer 1

Let's take l and w as the length and the width of the rectangle.

According to the given information the length is 3 less than twice the width:

`l=2w-3`

And the area, which is the product of the length times the width is 14. Using the equation above, we can re write the formula for the area, this way:

`\begin{gathered} A=w\cdot l \\ 14=w\cdot(2w-3) \\ 14=2w^2-3w \\ 0=2w^2-3w-14 \end{gathered}`

Now, we have to solve the quadratic equation using the quadratic formula:

`\begin{gathered} w=\frac{-(-3)\pm\sqrt[]{(-3)^2-4(2)(-14)}}{2(2)} \\ w=\frac{3\pm\sqrt[]{9+112}}{4} \\ w=\frac{3\pm\sqrt[]{121}}{4} \\ w=(3\pm11)/(4) \\ w1=(14)/(4)=(7)/(2) \\ w2=-(8)/(4)=-2 \end{gathered}`

w has 2 values, but for this context, the logic value is 7/2. It means that the width of the rectangle is 7/2.

Now, we can use this value and the first equation to find the length:

`\begin{gathered} l=2w-3 \\ l=2((7)/(2))-3 \\ l=7-3 \\ l=4 \end{gathered}`

The length of the rectangle is 4 and the width is 7/2.