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\centerline{MATH 110-02 -- Algebra Through History}
\centerline{The Tartaglia-Cardano Formula for Cubics}
\centerline{November 11, 2019}
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{\it Background}
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We just saw that the formula Tartaglia found and Cardano
printed in his {\it Ars Magna} can be written as follows.
For the ``cubes and roots equals numbers'' case
$x^3 + px = q$, (one) solution is
$$x = \root 3 \of{{q\over 2} + \sqrt{\left({q\over 2}\right)^2 + \left({p\over 3}\right)^3}}
+ \root 3 \of{{q\over 2} - \sqrt{\left({q\over 2}\right)^2 + \left({p\over 3}\right)^3}}$$
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Tartaglia and Cardano would have been considering the case $p, q > 0$
but we can see that the formula is actually valid when $p < 0$ or $q < 0$
as well(!)
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{\it Questions}
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A.  Use the formula to solve
$$x^3 + 3x = 36$$
with $p = 3$ and $q = 36$.  Note:  There is an ``obvious'' real solution $x = 3$ here.  Is it 
clear your answer is giving that same value??
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B.  What happens if you use the formula to solve:
$$x^3 - 6x - 2 = 0$$
where $p = -6$ and $q = 2$?  (There are actually 3
distinct real roots as you can see by plotting $y = x^3 - 6x - 2$
on a graphing calculator.)
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