# Why is the “solving for cubic equation roots general rule” sometimes not applicable while the equation obviously has roots?

the general rule:

we have $$ax^3+bx^2+cx+d=0$$

$$\Delta_0=b^2-3ac$$

$$\Delta_1=2b^3-9abc+27a^2d$$

$$C=\sqrt{\Delta_1^2-4\Delta_0^3}$$

$$D=(\frac{\Delta_1+C}{2})^\frac{1}{3}$$

$$x=-\frac{1}{3a}(b+D+\frac{\Delta_0}{D})$$

imagine $$x^3-6x^2+11x-6=0$$

we know its roots are $$x=1$$, $$x=2$$ and $$x=3$$.

but when you use the general rule, you will find a negative $$\Delta_1^2-4\Delta_0^3$$ and thus; you can't continue the process!

Also, when you use this rule, you will just find ONE real root (remember the last process to find final $$x$$); while that equation has 3 real roots and no imaginary roots. so how to find the other real roots using the general rule?!

• You encountered the same problem as Cardano and other mathematicians who first studied cubic equations. Yes, sometimes this formula allows only to express real roots through complex radicals. This was one of the many reasons people started to take complex numbers seriously – Yuriy S Mar 7 '20 at 15:15

You use the Cardano formula to solve a cubic equation. Your formula for $$x$$ contains real numbers $$C$$ and $$D$$ if $$\Delta_1^2-4\Delta_0^3 \ge 0$$ and thus obviously gives you a real root. If $$\Delta_1^2-4\Delta_0^3 > 0$$, there are two additional non-real complex roots which are complex conjugate. If $$\Delta_1^2-4\Delta_0^3 = 0$$, there is one additional real root of multiplicity two.

The case $$\Delta_1^2-4\Delta_0^3 < 0$$ is known as the casus irreducibilis. In that case there exist three distinct real roots, but your formula represents them via non-real complex numbers $$C$$ and $$D$$. See my answer to Is there really analytic solution to cubic equation? where the complete story is told.

Edited:

In your example $$x^3- 6x^2 + 11x - 6 = 0$$ we have the three real roots $$1,2,3$$. Transformation as in the above link with $$x = y - \frac{1}{3}(-6) = y + 2$$ yields $$y^3 - y = 0$$. This shows that $$y = 0$$ is a solution and reduces the problem to $$y^2 - 1 = 0$$ which gives $$y = \pm 1$$.

Nevertheless we are in the casus irreducibilis: We have $$a = -1, b = 0$$, thus $$R = -1/27 < 0$$. Thus $$\sqrt{R} = i\sqrt{1/27}$$ and $$w = \eta /\sqrt{3}$$, where $$\eta$$ is one of the three complex third roots of $$i$$. We may take $$\eta = \sqrt{3}/2 + i/2$$ which gives $$w = 1/2 + i/2\sqrt{3}$$. Thus $$w' = 1/(3w) = 1/2 - i/2\sqrt{3}$$ and $$y = w + w'= 1$$ is a solution found by Cardano's formula.

Your formula yields $$\Delta_0 = 3, \Delta_1 = 0$$. Similar computations as above give a solution of your equation: $$C = i\sqrt{27}$$, $$D = \eta \sqrt{3}/\sqrt[3]{2}$$.

• in the case of $\Delta_1^2-4\Delta_0^3=0$, again you obtain one real root from the final expression $x=-\frac{1}{3a}(b+D+\frac{\Delta_0}{D})$. how the other real root can be obtained? – aminabzz Mar 7 '20 at 15:46
• This is explained in my linked answer. We can reduce the general case to $x^3 +A x = B$, where you have to express $A, B$ in terms of $a,b,c,d$ . In this special case, if your formula yields the root $x_0$, then $-x_0/2$ is the other root of multiplicity two. Then transform $x_0$ and $-x_0/2$ to the solutions of the original equation. You can also make a polynomial division by $(x-x_0)$ to obtain a quadratic equation. – Paul Frost Mar 7 '20 at 16:00
• thank you so much – aminabzz Mar 7 '20 at 16:05
• The formula works; you just need to carry the calculation through the complex domain (and take care to use the right selection of values since both complex square and complex cube roots most naturally have multiple values - 2 for square, 3 for cube.). This is why complex numbers were created, after all! – The_Sympathizer Mar 8 '20 at 6:40
• @The_Sympathizer You are right, I edited my answer to clarify this point. – Paul Frost Mar 8 '20 at 9:21

Yuriy's comment is right. The solutions exist, and the results you get with those complex radicals are such that all the imaginary parts cancel and, as if by magic, they equal the real roots you knew were there all along.

There are two ways to get around the casus irreducibilis. One way is to cheat and find a rational root (if one exists), allowing you to factor the cubic and get the remaining roots from the quadratic quotient. A variation on that is in some applications, the cubic equation is constructed in a way where you know one root, and you can factor with that root. The second approach is to introduce non-algebraic functions, which in this case means trigonometric functions, as described first by Vieta. Details are in the link given above.