Question 1018183
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You are given that *[tex \Large l\ =\ 2w\ +\ 1]


The perimeter of a rectangle is *[tex \Large P\ =\ 2l\ +\ 2w].  Substituting for *[tex \Large l] you get the perimeter as a function of the width:  *[tex \Large P(w)\ =\ 2(2w\ +\ 1)\ +\ 2w\ =\ 6w\ +\ 2]


The area of a rectangle is *[tex \Large A\ =\ lw].  Substituting for *[tex \Large l] you get the area as a function of the width:  *[tex \Large A(w)\ =\ (2w\ +\ 1)w\ =\ 2w^2\ +\ w]


Since the value of the area is 5 greater than the perimeter, you can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w^2\ +\ w\ =\ 6w\ +\ 2\ +\ 5]


which simplifies to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w^2\ -\ 5w\ -\ 7\ =\ 0]


Solve the factorable quadratic for the positive root.  Then calculate *[tex \Large 2w\ +\ 1].


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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