MATH · Mathematics

The Relationship Between the Roots and Coefficients of a Cubic Equation

A complete guide to Vieta's relations for cubic equations, the reduction to a depressed cubic, and the derivation of Cardano's formula for all three roots.

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From a cubic polynomial to its three roots

Consider the general cubic equation

\[ ax^3+bx^2+cx+d=0,\qquad a\ne 0. \]

Over the complex numbers, it has three roots when multiplicity is counted. Call them \(\alpha\), \(\beta\), and \(\gamma\). The polynomial can then be factored as

\[ ax^3+bx^2+cx+d =a(x-\alpha)(x-\beta)(x-\gamma). \]

Expanding the product gives

\[ a\left[x^3-(\alpha+\beta+\gamma)x^2 +(\alpha\beta+\beta\gamma+\gamma\alpha)x -\alpha\beta\gamma\right]. \]

Matching coefficients with the original polynomial produces Vieta's relations:

\[ \boxed{\alpha+\beta+\gamma=-\frac{b}{a}},\qquad \boxed{\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}},\qquad \boxed{\alpha\beta\gamma=-\frac{d}{a}}. \]

The alternating signs are not accidental. They come directly from choosing either \(x\) or a negative root from each factor \((x-\alpha)(x-\beta)(x-\gamma)\).

How the relations can be used

Vieta's relations often answer questions about the roots without requiring the roots themselves. For example,

\[ \alpha^2+\beta^2+\gamma^2 =(\alpha+\beta+\gamma)^2 -2(\alpha\beta+\beta\gamma+\gamma\alpha) =\frac{b^2-2ac}{a^2}. \]

They also reconstruct a monic cubic from its roots. If the sum of the roots is \(S_1\), the sum of their pairwise products is \(S_2\), and their product is \(S_3\), then

\[ x^3-S_1x^2+S_2x-S_3=0. \]

As a quick example, the roots \(1\), \(2\), and \(3\) have \(S_1=6\), \(S_2=11\), and \(S_3=6\), so their cubic is \(x^3-6x^2+11x-6=0\).

Step 1: normalize the equation

To derive a formula for the roots, divide by \(a\) and introduce shorter coefficient names:

\[ x^3+Ax^2+Bx+C=0, \qquad A=\frac{b}{a},\quad B=\frac{c}{a},\quad C=\frac{d}{a}. \]

The quadratic term makes the algebra difficult. We remove it by shifting the variable:

\[ x=y-\frac{A}{3}. \]

Substitution and collection of like powers of \(y\) transform the equation into the depressed cubic

\[ y^3+py+q=0, \]

where

\[ p=B-\frac{A^2}{3} =\frac{3ac-b^2}{3a^2}, \]

\[ q=\frac{2A^3}{27}-\frac{AB}{3}+C =\frac{2b^3-9abc+27a^2d}{27a^3}. \]

The choice \(-A/3\) is exactly what makes the coefficient of \(y^2\) vanish.

Step 2: Cardano's substitution

Now write the unknown as a sum:

\[y=u+v.\]

Substituting into \(y^3+py+q=0\) gives

\[ u^3+v^3+(3uv+p)(u+v)+q=0. \]

There is freedom in the choice of \(u\) and \(v\), so impose

\[ 3uv+p=0 \qquad\Longrightarrow\qquad uv=-\frac{p}{3}. \]

The remaining equation is \(u^3+v^3=-q\). Cubing the product condition also gives

\[ u^3v^3=-\frac{p^3}{27}. \]

Therefore \(u^3\) and \(v^3\) are two numbers with known sum and product. They are the roots of

\[ z^2+qz-\frac{p^3}{27}=0. \]

The quadratic formula now gives

\[ u^3=-\frac{q}{2}+\sqrt{\Delta_C}, \qquad v^3=-\frac{q}{2}-\sqrt{\Delta_C}, \]

with Cardano's quantity

\[ \Delta_C=\left(\frac{q}{2}\right)^2 +\left(\frac{p}{3}\right)^3. \]

Cardano's formula for one root

Choosing compatible cube roots so that \(uv=-p/3\), one solution of the depressed cubic is

\[ y= \sqrt[3]{-\frac{q}{2}+\sqrt{\Delta_C}} + \sqrt[3]{-\frac{q}{2}-\sqrt{\Delta_C}}. \]

Returning to \(x=y-A/3\) gives the familiar cubic formula

\[ \boxed{ x=-\frac{b}{3a} +\sqrt[3]{-\frac{q}{2}+\sqrt{\Delta_C}} +\sqrt[3]{-\frac{q}{2}-\sqrt{\Delta_C}} }, \]

where \(p\), \(q\), and \(\Delta_C\) are the coefficient expressions derived above. For real coefficients with \(\Delta_C\geq 0\), taking the real cube roots gives a real solution directly.

Generating all three roots

A nonzero complex number has three cube roots. Let

\[ \omega=e^{2\pi i/3}=-\frac12+\frac{\sqrt3}{2}i, \qquad \omega^3=1, \qquad 1+\omega+\omega^2=0. \]

Choose cube roots \(u\) and \(v\) of the two Cardano expressions with the compatibility condition \(uv=-p/3\). Then all three roots are

\[ \boxed{ x_k=-\frac{A}{3}+\omega^k u+\omega^{-k}v }, \qquad k=0,1,2. \]

The compatibility condition matters: independently taking principal complex cube roots can pair the branches incorrectly and produce values that do not satisfy the original cubic.

A compact formula using only the original coefficients

Define

\[ \Delta_0=b^2-3ac, \qquad \Delta_1=2b^3-9abc+27a^2d, \]

\[ \mathcal C= \sqrt[3]{\frac{\Delta_1\pm \sqrt{\Delta_1^2-4\Delta_0^3}}{2}}. \]

Choose the sign and cube-root branch so that \(\mathcal C\ne0\) when possible. With \(\zeta_k\in\{1,\omega,\omega^2\}\), the three roots are

\[ \boxed{ x_k=-\frac{1}{3a} \left( b+\zeta_k\mathcal C+ \frac{\Delta_0}{\zeta_k\mathcal C} \right) }, \qquad k=0,1,2. \]

If \(\Delta_0=\Delta_1=0\), the equation has the triple root \(x=-b/(3a)\), and the displayed fraction involving \(\mathcal C\) should be replaced by that limiting case.

How the discriminant classifies the roots

The usual discriminant of \(ax^3+bx^2+cx+d\) is

\[ D=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd. \]

It is related to Cardano's quantity by \(D=-108a^4\Delta_C\). For real coefficients:

If \(D>0\), the cubic has three distinct real roots. Equivalently, \(\Delta_C<0\), so Cardano's radical expression passes through complex numbers even though every final root is real. This is the casus irreducibilis.
If \(D<0\), there is one real root and one nonreal complex-conjugate pair.
If \(D=0\), at least two roots coincide.

A short example

For \(x^3-3x-2=0\), the equation is already depressed, with \(p=-3\), \(q=-2\), and \(\Delta_C=0\). Cardano's formula gives

\[ u^3=v^3=1, \qquad u=v=1, \qquad x_0=u+v=2. \]

Using the cube roots of unity gives the other two roots, both equal to \(-1\). Thus

\[ x^3-3x-2=(x-2)(x+1)^2. \]

Vieta's relations verify the result immediately: \(2+(-1)+(-1)=0\), the pairwise-product sum is \(-3\), and the product is \(2\), matching the coefficients of the monic cubic.

The main idea

Vieta's relations describe symmetric information about the roots directly from the coefficients. Cardano's method goes further by converting the cubic into a depressed cubic, splitting its unknown into \(u+v\), and reducing the problem to a quadratic equation for \(u^3\) and \(v^3\). The resulting formula is more delicate than the quadratic formula because complex cube-root branches must be paired consistently, but it provides an exact algebraic expression for every cubic root.