# What is the dimensions of a rectangle with an area of 10 square feet if its length is 8 ft more than twice the width?

Sep 18, 2017

$10 \text{ ft."xx1" ft.}$

#### Explanation:

Let the length be $L$ feet
and the width be $W$ feet

We are told
$\textcolor{w h i t e}{\text{XXX}} L = 2 W + 8$

so the area, $A$, is
$\textcolor{w h i t e}{\text{XXX}} A = L \times W = \left(2 W + 8\right) \cdot W = 2 {W}^{2} + 8 W$
but we are also told that the area is $10 \text{ sq.ft.}$

So
$\textcolor{w h i t e}{\text{XXX}} 2 {W}^{2} + 8 W = 10$

$\textcolor{w h i t e}{\text{XXX}} {W}^{2} + 4 W = 5$

$\textcolor{w h i t e}{\text{XXX}} {W}^{2} + 4 W - 5 = 0$

$\textcolor{w h i t e}{\text{XXX}} \left(W + 5\right) \left(W - 1\right) = 0$

$W = - 5 \textcolor{w h i t e}{\text{xxx")"or"color(white)("xxx}} W = 1$

A negative length is not possible, so the only valid possibility is $W = 1$

and, since $L = 2 W + 8$
$\textcolor{w h i t e}{\text{XXX}} L = 10$