# The area of a rectangle is 21 ft^2,and the length of the rectangle is 1 ft less than twice the...

## Question:

The area of a rectangle is 21 {eq}ft^2 {/eq}, and the length of the rectangle is 1 ft less than twice the width. Find the dimensions of the rectangle.

## Application of Quadratic Root Formula for the Rectangular Dimensions:

Sometimes, when finding the roots for the dimensions, we need to apply a quadratic root formula. This can be applied if the equation is a quadratic polynomial. The resulting roots could be either positive or negative numbers. When solving and dealing with real-life applications like a rectangle, we should always use a positive number.

The area of a rectangle is {eq}21 \ \text{ft}^2 {/eq}.

The length of the rectangle is {eq}1 \ \text{ft} {/eq} less than twice the width. This can be expressed as {eq}L_{rect}=W_{rect}-1 \ \text{ft} {/eq} as the length of the rectangle.

Therefore, we use the area formula of the rectangule:

\begin{align} A_{rect}&=L_{rect} \times W_{rect} && \left [ \text{Area formula of the rectangular} \right ]\\[0.2cm] 21 \ \text{ft}^2&=\left ( W_{rect}-1 \ \text{ft} \right )\left ( W_{rect} \right ) && \left [ \text{Substitute the values for length, width and area} \right ]\\[0.2cm] 21 \ \text{ft}^2&=\left ( W_{rect} \right )^{2} -1 \ \text{ft}\left ( W_{rect} \right ) \\[0.2cm] \left ( W_{rect} \right )^{2} -1 \ \text{ft}\left ( W_{rect} \right )-21 \ \text{ft}^2&=0 && \left [ \text{Quadratic Equation 1} \right ]\\[0.2cm] \end{align}

To solve for the roots of the quadratic equation:

$$\displaystyle Roots_{W_{rect}}=\frac{-b\pm \sqrt{b^{2}-4ac} }{2a}$$

Where the main quadratic equation formula is:

$$\displaystyle a\left ( W_{rect} \right )^{2} -b \ \text{ft}\left ( W_{rect} \right )-c \ \text{ft}^2=0$$

Therefore, we have:

\begin{align} \displaystyle Roots_{W_{rect}}&=\frac{-b\pm \sqrt{b^{2}-4ac} }{2a} \ \text{ft}\\[0.2cm] \displaystyle Roots_{W_{rect}}&=\frac{-\left (-1 \right )\pm \sqrt{\left (-1 \right )^{2}-4\left (1 \right )\left (-21 \right )} }{2\left (1 \right )} \ \text{ft} && \left [ \text{Substitute the values for a, b, and c from the quadratic equation 1} \right ]\\[0.2cm] \displaystyle Roots_{W_{rect}}&=\frac{1\pm \sqrt{1+84}}{2} \ \text{ft}\\[0.2cm] \displaystyle Roots_{W_{rect}}&=\frac{1\pm \sqrt{85}}{2} \ \text{ft}\\[0.2cm] \displaystyle Roots_{+W_{rect}}&=\frac{1+ \sqrt{85}}{2} \ \text{ft} && \left [ \text{Addition Condition} \right ]\\[0.2cm] \displaystyle Roots_{+W_{rect}}&\approx5.1 \ \text{ft}\\[0.2cm] \displaystyle Roots_{-W_{rect}}&=\frac{1- \sqrt{85}}{2} \ \text{ft} && \left [ \text{Minus Condition} \right ]\\[0.2cm] \displaystyle Roots_{-W_{rect}}&\approx-4.1 \ \text{ft} \end{align}

Since we are dealing with a rectangle, we will select the addition (+) condition of: {eq}\displaystyle Roots_{+W_{rect}}=\frac{1+ \sqrt{85}}{2} \ \text{ft}\approx5.1 \ \text{ft} {/eq}. Thus, The length of the rectangle is 1 ft less than twice the width. This can be expressed as {eq}\displaystyle L_{rect}=W_{rect}-1 \ \text{ft}=\frac{1+ \sqrt{85}}{2} \ \text{ft}-1 \ \text{ft}\approx4.1 \ \text{ft} {/eq}.

Therefore, the length and width of the rectangle is given by {eq}\approx4.1 \ \text{ft} \ \text{and} \ \approx5.1 \ \text{ft} {/eq}, respectively. 