Question 231604
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The volume of a rectangular solid is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_{rect}\ =\ lwh]


So for your first box:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_1\ =\ x\cdot x\cdot(x+5)\ =\ x^3\ +\ 5x^2]


And for the second box:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V_2\ =\ (x+1)(x+1)(x+2)\ =\ x^3\ +\ 4x^2\ +\ 5x\ +\ 2]


(verification left as an exercise for the student.)


Set these two volumes equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \ x^3\ +\ 5x^2\ =\ x^3\ +\ 4x^2\ +\ 5x\ +\ 2]


Collect like terms:


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


Leaving a quadratic that unfortunately does not factor.  Use the quadratic formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-b \pm sqrt{b^2 - 4ac}}{2a} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-(-5) \pm sqrt{(-5)^2 - 4(1)(-2)}}{2(1)} ] 


Of the two roots, one will be negative -- finding them is left as an arithmetic exercise for the student.  Discard the negative root because you are looking for a positive measure of length.  Hint:  Leave your answer in radical form for the exact answer -- at least until you have verified that the root you obtained is the correct answer by using it to calculate the volume of each of the boxes and making sure they are equal quantities -- another exercise for the student.


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