# The area of a rectangle is 12 square inches. The length is 5 more than twice it’s width. How do you find the length and width?

Mar 12, 2018

Using the positive root in the quadratic equation, you find that $w = 1.5$, which means $l = 8$

#### Explanation:

We know two equations from the problem statement. First is that the area of the rectangle is 12:

$l \cdot w = 12$

where $l$ is the length, and $w$ is the width. The other equation is the relationship between $l$ and $w$. It states that 'The length is 5 more than twice it's width'. This would translate to:

$l = 2 w + 5$

Now, we substitute the length to width relationship into the area equation:

$\left(2 w + 5\right) \cdot w = 12$

If we expand the left-hand equation, and subtract 12 from both sides, we have the makings of a quadratic equation:

$2 {w}^{2} + 5 w - 12 = 0$

where:
$a = 2$
$b = 5$
$c = - 12$

plug that into the quadratic equation:

$w = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} \Rightarrow w = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \left(2 \cdot - 12\right)}}{2 \cdot 2}$

$w = \frac{- 5 \pm \sqrt{25 - \left(- 96\right)}}{4} \Rightarrow w = \frac{- 5 \pm \sqrt{121}}{4}$

$w = \frac{- 5 \pm 11}{4}$

we know that the width must be a positive number, so we only worry about the positive root:

$w = \frac{- 5 + 11}{4} \Rightarrow w = \frac{6}{4} \Rightarrow \textcolor{red}{w = 1.5}$

now that we know the width ($w$), we can solve for the length ($l$):
$l = 2 w + 5 \Rightarrow l = 2 \left(1.5\right) + 5$

$l = 3 + 5 \Rightarrow \textcolor{red}{l = 8}$