Question 708486
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Let *[tex \LARGE w] represent the width and *[tex \LARGE l] represent the length.


The perimeter of a rectangle is given by *[tex \LARGE P\ =\ 2l\ +\ 2w]


So we know that *[tex \LARGE 2l\ +\ 2w\ =\ 40] which is to say *[tex \LARGE l\ +\ w\ =\ 20], or, better for our purposes, *[tex \LARGE l\ =\ 20\ -\ w]


The area of a rectangle is the length times the width, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w(20\ -\ w)\ =\ 90]


Putting the quadratic into standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ -\ 20w\ +\ 90\ =\ 0]


Solve the quadratic for *[tex \LARGE w].  Discard the value that produces a negative result when you calculate *[tex \LARGE 20\ -\ w].  Note:  This quadratic DOES NOT factor over the rationals.  Use the quadratic formula, but leave your answer in simplest radical form.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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