# The length of a rectangle is 3ft more than twice its width, and the area of the rectangle is 77ft^2, how do you find the dimensions of the rectangle?

Mar 26, 2016

Width = $\frac{11}{2} \text{ ft = 5 foot 6 inches}$

Length = $14 \text{ feet}$

#### Explanation:

Breaking the question down into its component parts:

Let length be $L$
Let width be $w$
Let area be $A$

Length is 3 ft more than: L=" "?+3
twice" "L=2?+3
its width$\text{ } L = 2 w + 3$

Area $= A = 77 = \text{width "xx" Length}$

$A = 77 = w \times \left(2 w + 3\right)$

$2 {w}^{2} + 3 w = 77$

$2 {w}^{2} + 3 w - 77 = 0$ This is a quadratic equation
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Standard form $y = a {x}^{2} + b x + c$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = 2 \text{; "b=3"; } c = - 77$

$x = \frac{- \left(3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(2\right) \left(- 77\right)}}{2 \left(2\right)}$

$x = \frac{- \left(3\right) \pm 25}{4} = - 7 \mathmr{and} \frac{11}{2}$

As we can not have a negative area in this context the answer for $x$ is $\frac{11}{2}$

But $\textcolor{b l u e}{x = w \text{ so the width is } \frac{11}{2}}$

$\textcolor{b l u e}{L = 2 w + 3 = 11 + 3 = 14}$