Algebra Precalculus – How to Find the Dimensions of a Rectangle Where Length Is 6m Longer than the Width and Area Is $84^2$m?

algebra-precalculusfactoringquadratics

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:

$l = w + 6$

$84 = lw$

Best Answer

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*}

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