Question 188229
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If <i>x</i> is the measure of the side of one of the cut out squares, then the length of the bottom of the box must be 14 - 2<i>x</i>, and the width must be 11 - 2<i>x</i>, so the area of the bottom of the box must be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (14 - 2x)(11 - 2x) = 154 - 50x + 4x^2 = 80]


Putting the quadratic into standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2 - 50x + 74 = 0]


Divide by 2 to eliminate a common factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2 - 25x + 37 = 0]


Use the quadratic formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-(-25) \pm sqrt{(-25)^2 - 4(2)(37)}}{2(2)}]


You can do your own arithmetic from here, but you need to exclude one of the roots because it is too large.  The larger answer is an extraneous root.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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