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The length of a rectangle is 2 inches more than the width. The area is 15 square inches. Find the dimensions.The length of the rectangle is ? inches, while the width is ? inches.

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Let's use the variable L to represent the length and W to represent the width.

If the length is 2 inches more than the width, we can write the following equation:


L=W+2

Then, if the area is equal to 15 in² and the area is given by the product of length and width, we have:


\begin{gathered} A=L\cdot W \\ A=(W+2)\cdot W \\ 15=W^2+2W \\ W^2+2W-15=0 \\ W_1=\frac{-2+\sqrt[]{4+60}}{2}=(-2+8)/(2)=3 \\ W_2=(-2-8)/(2)=-5 \end{gathered}

Since the width can't be negative, we have W = 3.

Calculating L, we have:


\begin{gathered} L=W+2 \\ L=3+2 \\ L=5 \end{gathered}

Therefore the length of the rectangle is 5 inches, while the width is 3 inches.

User Robbie Dee
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