# The length of a rectangle is 1 foot less than 4 times its width. How do you find the dimensions of this rectangle if the area is 60 square feet?

Feb 10, 2016

Dimensions of rectangle are $4$ and $15$.

#### Explanation:

Assume that the width in feet is $x$. Hence, length of rectangle, which is one less than four times, will be $4 x - 1$.

As area of a rectangle (which is $60$) is length multiplied by width, it is given by $x \cdot \left(4 x - 1\right)$ i.e. $4 {x}^{2} - x$.

Hence $4 {x}^{2} - x = 60$ or

$4 {x}^{2} - x - 60 = 0$.

As this is a quadratic equation of type $a \cdot {x}^{2} + b \cdot x + c$, to factorize this as $a \cdot c$ is negative (it is $- 240$), we should factorize $240$ in two factors whose difference is $1$ note that $b = - 1$. Hit and trial shows these are 16 and 15.

Hence equation can be written as

$4 {x}^{2} - 16 x + 15 x - 60 = 0$ or $4 x \left(x - 4\right) + 15 \left(x - 4\right) = 0$

or $\left(4 x + 5\right) \cdot \left(x - 4\right) = 0$ i.e.

solution is $x = - \frac{5}{4}$ or $x = 4$.

As width cannot be $- \frac{5}{4}$, it is $4$ and length is $\left(4 \cdot 4 - 1\right)$ or $15$.

Hence dimensions of rectangle are $4$ and $15$.