# The length of a rectangle is 5 yd less than double the width, and the area of the rectangle is 52 yd^2 . How do you find the dimensions of the rectangle?

Aug 18, 2016

Width = 6.5 yds, length = 8 yds.

#### Explanation:

Define the variables first.

We could use two different variables, but we have been told how the length and width are related.

Let the width be $x \text{ width is the smaller side}$
The length = $2 x - 5$

"Area = l x w" and the area is given to be 52 squ yards.

$A = x \left(2 x - 5\right) = 52$

$2 {x}^{2} - 5 x = 52 \text{ quadratic equation}$

$2 {x}^{2} - 5 x - 52 = 0$

To factorise, find factors of 2 and 52 which cross-multiply and subtract to give 5.

$\textcolor{w h i t e}{\times x} \left(2\right) \text{ } \left(52\right)$
$\textcolor{w h i t e}{\times . x} 2 \text{ 13 } \Rightarrow 1 \times 13 = 13$
$\textcolor{w h i t e}{\times . x} 1 \text{ 4 "rArr2xx4 = 8" } 13 - 8 = 5$

We have the correct factors, now fill in the signs. We need -5.

$\textcolor{w h i t e}{\times x} \left(2\right) \text{ } \left(- 52\right)$
$\textcolor{w h i t e}{\times . x} 2 \text{ - 13 } \Rightarrow 1 \times - 13 = - 13$
$\textcolor{w h i t e}{\times . x} 1 \text{ +4 "rArr2xx+4 = +8" } - 13 + 8 = - 5$

$\left(2 x - 13\right) \left(x + 4\right) = 0$

Each factor could be equal to 0

$x = 6.5 \mathmr{and} x = - 4$ (reject)

The width = 6.5 yards. Now find the length: 6.5 x 2 -5 = 8 yards

Check:
Width = 6.5yds, length = 8yds
Area = 6.5 x 8 = 52

Aug 18, 2016

Length$= 8 y d$
Width $= 6.5 y d$.

#### Explanation:

Let width be $= x$
Therefore, length $= 2 x - 5$

We know that
$\text{Area" = "Length" xx "Width}$
Inserting given and assumed numbers we get

$52 = \left(2 x - 5\right) \times x$
rearranging we obtain

$2 {x}^{2} - 5 x - 52 = 0$

To factorize we use split the middle term method. We have two parts of middle term as $- 13 x \mathmr{and} 8 x$. The equation becomes

$2 {x}^{2} - 13 x + 8 x - 52 = 0$
Paring and taking out common factors we have
$x \left(2 x - 13\right) + 4 \left(2 x - 13\right) = 0$
$\implies \left(2 x - 13\right) \left(x + 4\right) = 0$

Setting each factor equal to $0$, we have two roots
$\left(2 x - 13\right) = 0 \mathmr{and} \left(x + 4\right) = 0$
$x = \frac{13}{2} = 6.5$
$x = - 4$, rejected as width can not be a $- v e$ value

$\therefore$Width $= 6.5 y d$. And length$= 2 \times 6.5 - 5 = 8 y d$

Check:
Area $= 8 \times 6.5 = 52 y {d}^{2}$