Question

A rectangular room is 9 feet longer than it is wide. The area of the room is 360 square feet. How many feet long is the room?

Answer 1

we make a drawing

where w is width

area of a rectangle is

$A=l\times w$

where a is tha area, l the length and w the width

then replacing the area and values of length and width

$\begin{gathered} 360=(w+9)\times(w) \\ 360=w^2+9w \end{gathered}$

we rewrite the expression equaling 0 to find w

$\begin{gathered} w^2+9w=360 \\ w^2+9w-360=0 \end{gathered}$

factor by quadratic formula

$w=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

where a is 1, b is 9 and c -360

$\begin{gathered} w=\frac{-(9)\pm\sqrt[]{(9)^2-4(1)(-360)}}{2(1)} \\ \\ w=\frac{-9\pm\sqrt[]{81+1440}}{2} \\ \\ w=\frac{-9\pm\sqrt[]{1521}}{2} \\ \\ w=\frac{-9\pm39}{2} \end{gathered}$

we have two solutions to w

$\begin{gathered} w_1=\frac{-9+39}{2}=15 \\ \\ w_2=\frac{-9-39}{2}=-24 \end{gathered}$

but we use the possitive number because is a measure and the measure are not negatives

the long of the room is

$w+9$

replacing w

$\begin{gathered} 15+9 \\ =24 \end{gathered}$

tha