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Sagot :
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:
[tex]L=W+2[/tex]Then, if the area is equal to 15 in² and the area is given by the product of length and width, we have:
[tex]\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}=\frac{-2+8}{2}=3 \\ W_2=\frac{-2-8}{2}=-5 \end{gathered}[/tex]Since the width can't be negative, we have W = 3.
Calculating L, we have:
[tex]\begin{gathered} L=W+2 \\ L=3+2 \\ L=5 \end{gathered}[/tex]Therefore the length of the rectangle is 5 inches, while the width is 3 inches.
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