# Area of a rectangle is 99 ft^2 , and the length of the rectangle is 7 ft more than twice the width, how do you find the dimensions of the rectangle?

Feb 9, 2016

Width $= \frac{11}{2}$ ft and Length $= 18$ ft

#### Explanation:

Let width of the rectangle be $x$ ft
Given that length is $7$ ft more than the twice the width.
$\therefore$ length $= 2 \times \text{width} + 7$ ft
or Length $= 2 x + 7$ ft

Area of the rectangle = Length $\times$ Width
Inserting given and above values

$99 = \left(2 x + 7\right) \times x$
$\implies 2 {x}^{2} + 7 x - 99 = 0$
Solve by finding roots $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- 7 \pm \sqrt{{7}^{2} - 4 \cdot 2 \cdot \left(- 99\right)}}{2 \cdot 2}$
or $x = \frac{- 7 \pm \sqrt{49 + 792}}{4}$
or $x = \frac{- 7 \pm \sqrt{841}}{4}$
or $x = \frac{- 7 \pm 29}{4}$
or $x = - 9 , \frac{11}{2}$
Ignoring $- v e$ root as width can not be negative.
Width $= \frac{11}{2} \mathmr{and}$ Length $2 \times \frac{11}{2} + 7 = 18$