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adratic Equations to Fi... m You Try! As a Visual Example C: A rectangle has a length that is 3 meters more than twice the width. The area of the rectangle is 44 square meters. Find the dimensions of the rectangle. Sketch a visual of this problem below. 1

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Let l be the length and w be the width of the rectangle.

Then the statement: "length that is 3m more than twice the width" becomes

l = 2w + 3

The area of a rectangle is length times width

Therefore


\begin{gathered} l* w=44m^2 \\ \Rightarrow w(2w+3)=44 \\ \Rightarrow2w^2+3w-44=0 \\ \end{gathered}

Factorising the quadratic equation, we have:


\begin{gathered} 2w^2+11w-8w-44=0 \\ \Rightarrow w(2w+11)-4(2w+11)=0 \\ \Rightarrow(w-4)(2w+11)=0 \\ \Rightarrow w=4\text{ or -5.5} \end{gathered}

Since w cannot be negative, then the only possibility is:

w=4

since l = 2w+3, then

l=2(4)+3=8+3=11

Hence the length is 11m and the width is 4m

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