# The length of a rectangle is 5 cm more than 4 times its width. If the area of the rectangle is 76 cm^2, how do you find the dimensions of the rectangle to the nearest thousandth?

Jul 6, 2016

Width $w \cong 3.7785 c m$

Length $l \cong 20.114 c m$

#### Explanation:

Let length $= l$, and, width $= w .$

Given that, length=5+4(width) $\Rightarrow l = 5 + 4 w \ldots \ldots \ldots . . \left(1\right)$.

Area = 76 $\Rightarrow$ length x width=76 $\Rightarrow l \times w = 76. \ldots \ldots . \left(2\right)$

Sub.ing for$l$ from $\left(1\right)$ in $\left(2\right)$, we get,

$\left(5 + 4 w\right) w = 76 \Rightarrow 4 {w}^{2} + 5 w - 76 = 0.$

We know that the Zeroes of Quadratic Eqn. $: a {x}^{2} + b x + c = 0$, are

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

Hence, $w = \frac{- 5 \pm \sqrt{25 - 4 \cdot 4 \cdot \left(- 76\right)}}{8} = \frac{- 5 \pm \sqrt{25 + 1216}}{8}$
$= \frac{- 5 \pm \sqrt{1241}}{8} \cong \frac{- 5 \pm 35.2278}{8}$

Since $w$, width, can not be $- v e$, we can not take $w = \frac{- 5 - 35.2278}{8}$

Therefore, width $w = \frac{- 5 + 35.2278}{8} = = \frac{30.2278}{8} \cong 3.7785 c m$

$\left(1\right)$ then, gives us, length $l = 5 + 4 \left(3.7785\right) \cong 20.114 c m$

With these dimensions, Area $= 3.7785 \times 20.114 = 76.000749 s q . c m$.

Hence, the roots satisfy the eqns.