MATH 110-02 – Algebra Through History The Tartaglia-Cardano Formula for Cubics November 11, 2019

Background

We just saw that the formula Tartaglia found and Cardano printed in his Ars Magna can be written as follows. For the “cubes and roots equals numbers” case x^{3} + px = q, (one) solution is

x = ^{3} \sqrt{\frac{q}{2} + \sqrt{(\frac{q}{2})^{2} + (\frac{p}{3})^{3}}} + ^{3} \sqrt{\frac{q}{2} − \sqrt{(\frac{q}{2})^{2} + (\frac{p}{3})^{3}}}

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(!)

Questions

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??

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.)