Cardano's Method

The Cardano's formula (named after Girolamo Cardano 1501-1576), which is similar to the perfect-square method to quadratic equations, is a standard way to find a real root of a cubic equation like

$$ax^3+bx^2+cx+d=0.$$

We can then find the other two roots (real or complex) by polynomial division and the quadratic formula. The solution has two steps. We first "depress" the cubic equation and then solve the depressed equation.

To depress the cubic equation, we substitute

$$x=y-\frac{b}{3a}.$$

Then we get

$$\begin{aligned} a\left(y-\frac{b}{3a}\right)^3 +b \left(y-\frac{b}{3a}\right)^2 + c\left(y-\frac{b}{3a}\right) + d &=0\\ ay^3-by^2+\frac{b^2}{3a}y-\frac{b^3}{27a^2}+by^2-\frac{2b^2}{3a}y+\frac{b^3}{9a^2}+cy-\frac{bc}{3a} + d &= 0\\ ay^3+\left( c-\frac{b^2}{3a} \right) y + \left( d + \frac{2b^3}{27a^2}-\frac{bc}{3a} \right) &= 0. \end{aligned}$$

Note that the above equation has been depressed in such a way that it is without the $y^2$ term. Let's try this on an example.

Depress the following equation: $2x^3-30x^2+162x-350=0.$

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Substituting $x = y - \frac {-30}{6} = y + 5,$ the equation is depressed to

$$\begin{aligned} 2\left( y^3+15y^2+75y+125\right)-30\left(y^2+10y+25\right)+162(y+5)-350&=0\\ 2y^3 +(150-300+162)y + 250-750+810-350&=0\\ 2y^3+12y-40&=0\\ y^3+6y-20&=0. \ _\square \end{aligned}$$

The next step is to solve the depressed equation of the form $y^3+Ay=B$.

We first substitute

$$\begin{aligned} 3st & =A\\ s^3-t^3 & =B. \end{aligned}$$

Then we solve for

$$y = s -t.$$

Find the real root of the cubic equation $2x^3-30x^2+162x-350=0.$

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From the above example, we know the given equation is depressed into $y^3+6y=20.$ Then we substitute

$$\begin{aligned} 3st &= 6 &&\qquad (1)\\ s^3-t^3 &= 20. &&\qquad (2) \end{aligned}$$

From $(1)$ we have $s = \frac {2}{t},$ substituting which into $(2)$ gives

$$\dfrac {8}{t^3} - t^3=20. \qquad (2a)$$

Multiplying $(2a)$ throughout by $t^3$, we obtain the quadratic equation for $t^3$ as follows:

$$t^6 +20t^3-8=0.$$

Solving for $t^3$ and then $t$, we have

$$t^3 = \frac {-20\pm 12\sqrt{3}}{2}\implies t = \sqrt[3]{-10\pm 6\sqrt{3}}.$$

It can be shown that, whether we take the positive or negative root of $t,$ we will get the same value for $y = s-t$. We shall only consider the positive $t = \sqrt[3]{-10+6\sqrt{3}}$ here.

From $(2)$ we have

$$s = \sqrt [3] {20-10+6\sqrt{3}} = \sqrt [3] {10+6\sqrt{3}},$$

which implies

$$y=s-t=\sqrt [3] {10+6\sqrt{3}}-\sqrt [3] {-10+6\sqrt{3}} = 2.$$

Therefore, the real root of the original cubic equation $2x^3-30x^2+162x-350=0$ is

$$x = y +5 = 2+5 = 7. \ _\square$$

Note: We check that $x=7$ is the only real root of the cubic equation $2x^3-30x^2+162x-350=0$ because $2x^3-30x^2+162x-350$ can be factorized as $2(x-7)(x^2-8x+25).$ $_\square$

The cubic polynomial $P: ax^3 + bx^2 + cx + d =0$ has solutions

\[ \begin{align} x_1 & = S + T - \frac b {3 a} \\
x_2 & = - \frac {S + T} 2 - \frac b {3 a} + \frac {i \sqrt 3} 2 \left({S - T}\right) \\ x_3 & = - \frac {S + T} 2 - \frac b {3 a} - \frac {i \sqrt 3} 2 \left({S - T}\right), \\ \end{align} \]

where $S = \sqrt [3] {R + \sqrt{Q^3 + R^2}}, T = \sqrt [3] {R - \sqrt{Q^3 + R^2}},$ and in turn $Q = \frac {3 a c - b^2} {9 a^2}, R = \frac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}$.

After depressing the equation and making it monic by dividing by $a,$ we get

$$\begin{aligned} y^3+\left(\dfrac{3ac-b^2}{3a^2}\right)y+\dfrac{2b^3-9abc+27a^2d}{27a^3}&=0 \\ y^3+3Qy-2R&=0. \end{aligned}$$

Now consider the identity $(S+T)^3-3ST(S+T)-(S^3+T^3)=0$. If we make it match with the equation, we get

$$\begin{aligned} y&=S+T\\ ST&=-Q \\ S^3+T^3 &=2R. \end{aligned}$$

Cube both sides of the second equation to get $S^3T^3=-Q^3$. Now, by Vieta's formula, the polynomial $P(z)=z^2-2Rz-Q^3$ will have roots $S^3$ and $T^3$, so let's solve it using the quadratic formula:

$$z=R \pm \sqrt{R^2+Q^3}.$$

Notice that the system of equations is symmetric in $S$ and $T$, so the order we choose doesn't matter, and the value of $y$ will be the same. So

$$\begin{aligned} S&=w^m\sqrt[3]{R + \sqrt{R^2+Q^3}}\\ T&=w^n\sqrt[3]{R - \sqrt{R^2+Q^3}}, \end{aligned}$$

where $0 \leq m,n \leq 2$ and $w$ is any $3^\text{rd}$ primitive root of unity. We see that then we have 9 combinations for the value of $S+T$, but only 3 of them work. By looking at the second equation, we see that $m+n$ must be a multiple of 3, so $(m,n)=(0,0),(1,2),(2,1)$ and our solutions are

$$\begin{aligned} y_1&=S+T \\ y_2&=Sw+Tw^2\\ y_3&=Sw^2+Tw. \end{aligned}$$

We choose $w=\frac{-1+\sqrt{3}i}{2}$ and $w^2=\frac{-1-\sqrt{3}i}{2}$, so

$$\begin{aligned} y_2&=\dfrac{-(S+T)+\sqrt{3}(S-T)i}{2} \\ y_3&=\dfrac{-(S+T)-\sqrt{3}(S-T)i}{2}. \end{aligned}$$

Undo the change $x=y-\frac{b}{3a},$ and we get our desired solutions. $_\square$

Find the roots of the following polynomial equation using Cardano's method:

$$x^3+2x^2+3x+4=0.$$

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It is given that

$$a=1, \quad b=2, \quad c=3, \quad d=4.$$

Compute $Q$ and $R:$

$$\begin{aligned} Q &= \frac {3 a c - b^2} {9 a^2} =\frac{5}{9} \\ \\ R &= \frac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3} = -\frac{35}{27}. \end{aligned}$$

Compute $S$ and $T:$

$$\begin{aligned} S &= \sqrt [3] {R + \sqrt{Q^3 + R^2}} = \frac{\sqrt[3]{-35+15\sqrt{6}}}{3} \\ \\ T &= \sqrt [3] {R - \sqrt{Q^3 + R^2}} = -\frac{\sqrt[3]{35+15\sqrt{6}}}{3}. \end{aligned}$$

Compute the roots:

$$\begin{aligned} x_1 &= S + T - \frac b {3 a} \\ \\ &= \frac{\sqrt[3]{-35+15\sqrt{6}}-\sqrt[3]{35+15\sqrt{6}}-2}{3} \\ \\ &\approx -1.651 \\ \\ \\ x_2 & = - \frac {S + T} 2 - \frac b {3 a} + \frac {i \sqrt 3} 2 \left({S - T}\right) \\ &= \frac{-\sqrt[3]{-35+15\sqrt{6}}+\sqrt[3]{35+15\sqrt{6}}-4}{6} + i\frac{\sqrt{3}\left(\sqrt[3]{-35+15\sqrt{6}}+\sqrt[3]{35+15\sqrt{6}}\right)}{6} \\ \\ &\approx -0.175+1.547i \\ \\ \\ x_3 & = - \frac {S + T} 2 - \frac b {3 a} - \frac {i \sqrt 3} 2 \left({S - T}\right) \\ \\ &= \frac{-\sqrt[3]{-35+15\sqrt{6}}+\sqrt[3]{35+15\sqrt{6}}-4}{6} - i\frac{\sqrt{3}\left(\sqrt[3]{-35+15\sqrt{6}}+\sqrt[3]{35+15\sqrt{6}}\right)}{6} \\ \\ &\approx -0.175-1.547i.\ _\square \end{aligned}$$