Question 188524
<font face="Garamond" size="+2">


Let <i>x</i> be the number of shares purchased.  Let <i>p</i> be the price of each share at the time of the initial purchase.  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  xp = 720] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  p = \frac {720}{x} ]


The number of shares sold was <i>x</i> - 20 and the price at sale was <i>p</i> + 8, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - 20)(p + 8) = 800]


Multiply the binomials:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ xp + 8x - 20p - 160 = 800]


Make substitutions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 720 + 8x - 20\left(\frac {720}{x}\right) - 160 = 800]


Simplify:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x - \frac{14400}{x} = 240]


Multiply by *[tex \Large \frac{x}{8}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 1800 = 30x]


Standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 30x - 1800 = 0]


This factors.  Exclude the extraneous negative root.


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