The length of a rectangle is 6m longer than the width. If the area of a rectangle is $84^2$m, find the dimensions of the rectangle. I'm in grade 11 math.
The length of a rectangle is $6m$ longer than the width. If the area of a rectangle is $84~\text{m}^2$, find the dimensions of the rectangle. I don't know where to start other than:
$l = w + 6$
$84 = lw$
 A: From the given information, you have the following system of equations to solve:
\begin{align*}
\begin{cases}
L = 6 + W\\
LW = 84\\
\end{cases}
\end{align*}
Where $W$ denotes the width and $L$ denotes the length.
If we substitute the first relation into the second, it results into the following equation:
\begin{align*}
(6+W)W = 84 \Longleftrightarrow W^{2} + 6W - 84 = 0
\end{align*}
Can you proceed from here?
EDIT
Observe that $W^{2} + 6W = (W^{2} + 6W + 9) - 9 = (W+3)^{2} - 9$, from whence we obtain
\begin{align*}
W^{2} + 6W - 84 = 0 \Longleftrightarrow (W+3)^{2} - 93 = 0 \Longleftrightarrow (W+3)^{2} = 93 \Longleftrightarrow W = \pm\sqrt{93} - 3
\end{align*}
Since $W$ must be positive, we conclude that $W = \sqrt{93}-3$ and $L = \sqrt{93} + 3$.
Another possible approach is to apply the Bhaskara's formula, which is given by:
\begin{align*}
ax^{2} + bx + c = 0 \Longleftrightarrow x = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}
\end{align*}
A: Substitute $w=l-6$ into $lw=84$ we get
$$
l^2-6l=84\Leftrightarrow l=3\pm \sqrt{93}
$$
Since $l>0$, we get $ l=3+\sqrt{93}$. So $w=\sqrt{93}-3$
