I'm trying to find all real roots of a cubic. I wanted to use Cardando's Method but I'm not sure I'm correctly understanding how to obtain all 3 roots given the depressed cubic: $$t^3 + pt + q = 0$$
What I'm understanding is that unless $q = 0$ and $p \neq 0$ there are at most 2 roots?
Generally these two roots should be obtained by finding the $u$ and $v$ from $t = u + v$ and then:
- The first root of $t$ is: $u(\frac{-1 + i\sqrt 3}{2}) + v(\frac{-1 - i\sqrt 3}{2})$
- And the second root of $t$ is: $u(\frac{-1 - i\sqrt 3}{2}) + v(\frac{-1 + i\sqrt 3}{2})$
So a couple questions here. How do we know there are only 2 roots in these cases, and where did those two magic coefficients come from that I'm multiplying $u$ and $v$ by?
When a cubic equation has $3$ real roots, Cardano's formulae require square roots of the negative discriminant of an auxiliary quadratic equation – it is even for this case what is now known as complex numbers were invented.
So it is better to use a trigonometric method: set $x=A\cos\theta, \enspace A>0$. The equation $x^3+px+q=0$ becomes $$A^3\cos^3\theta+pA\cos\theta+q=0$$ Now remember the formula $\;\cos3\theta=4\cos^3\theta-3\cos\theta$, and choose $A$ so that $A^3\cos^3\theta+pA\cos\theta$ is proportional to this formula, i. e. $$\frac{A^3}4=\frac{pA}{-3}\iff A^2=-\frac{4p}3\iff A=2\sqrt{-\frac p3}$$ (since we want to have $A>0$). This results in the trigonometric equation: $$-\frac{2p}3\sqrt{-\frac p3}(4\cos^3\theta-3\cos\theta)+q=-\frac{2p}3\sqrt{-\frac p3}\cos3\theta+q=0,$$ whence $\;\cos3\theta=…$