Question 1119313
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Let *[tex \Large x] represent the width.  Three times the width is *[tex \Large 3x].  5 less than 3 times the width is then *[tex \Large 3x\ -\ 5], which is the length.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(x)\ =\ (3x\ -\ 5)x\ =\ 3x^2\ -\ 5x]


But if *[tex \LARGE A(x)\ =\ 28\text{ m^2}], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2\ -\ 5x\ -\ 28\ =\ 0]


Solve the quadratic for *[tex \Large x].  Since we know by inspection (the sign on the constant term is the opposite of the sign on the lead coefficient) that there are two real number roots for this equation, and because the sign on the constant term is negative we know that these two roots must have opposite signs, we know that one of the roots will be less than zero.  Hence, only one of the roots of this equation will be valid because a negative value for length defies the laws of physics.  The width will be the positive root.  Then calculate *[tex \Large 3x\ -\ 5] to find the length.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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