Roots of Cubic Equations

A cubic equation is a polynomial equation of third-degree. The general form of a cubic function is

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

(1)

in which a is nonzero number, that is a\ne0 .

A cubic equation normally has three roots (incl. equal roots). If the coefficients of each term and constant term are real numbers, then the cubic equation has at least one real root. Depending on different coefficients and constant term, the real root could be rational or irrational. A rational root implies the root could be found by factorization. Nevertheless, all of the roots could be determined by the root formula of a cubic equation.

The idea to solve a general cubic equation algebraically is to transform it to the form without quadratic term, which is often called a depressed cubic equation. By algebraic manipulation, the solution of a depressed cubic equation could be obtained by classic Cardano method, Vieta's Substitution and factorization method. So we will focus on roots of a general cubic equations and factors to determine the distribution of the roots.

Transform to a depressed cubic equation

By algebraic manipulation, a cubic equation of general form can be transformed to the depressed form.

x^3+px+q=0

(2)

Let

x = t - \dfrac{b}{3a}

(3)

Substitute x variable in (1) with the expression in terms t in (3).

Then,

a( t - \dfrac{b}{3a})^3+b( t - \dfrac{b}{3a})^2+c( t - \dfrac{b}{3a})+d = 0

(4)

Expand the first term.

a( t - \dfrac{b}{3a})^3 = a(t^3-\dfrac{b^3}{27a^3}+3\dfrac{b^2}{9a^2}t - 3\dfrac{b}{3a}t^2 )

=at^3-\dfrac{b^3}{27a^2}+\dfrac{b^2}{3a}t - bt^2

Expand the second quadratic term.

b( t - \dfrac{b}{3a})^2 = b(t^2-2 \dfrac{b}{3a}t+\dfrac{b^2}{9a^2} )

= bt^2- \dfrac{2b^2}{3a}t+\dfrac{b^3}{9a^2}

The third term

c( t - \dfrac{b}{3a}) = ct - \dfrac{bc}{3a}

Plug in the expanded terms to the equation (4) and combine the like terms. And then, we get a depressed for of cubic equation.

at^3-\dfrac{b^3}{27a^2}+\dfrac{b^2}{3a}t -\cancel{ bt^2 + bt^2}- \dfrac{2b^2}{3a}t+\dfrac{b^3}{9a^2}+ct - \dfrac{bc}{3a}+d = 0

at^3+(\dfrac{b^2}{3a}-\dfrac{2b^2}{3a}+c)t+(-\dfrac{b^3}{27a^2}+\dfrac{b^3}{9a^2} - \dfrac{bc}{3a}+d) =0

at^3+(-\dfrac{b^2}{3a}+c)t+(\dfrac{2b^3}{27a^2} - \dfrac{bc}{3a}+d) =0

at^3+\dfrac{3ac-b^2}{3a}t+\dfrac{2b^3-9abc+27a^2d}{27a^2} = 0

t^3+\dfrac{3ac-b^2}{3a^2}t+\dfrac{2b^3-9abc+27a^2d}{27a^3} = 0

(5)

Let

p = \dfrac{3ac-b^2}{3a^2}

(6)

q = \dfrac{2b^3-9abc+27a^2d}{27a^3}

(7)

Then, the equation (5) becomes exact form of equation (2), which has roots represented by the following expressions.

t_{1,2,3} =\begin{cases} \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } +\sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}& \\ \\ ω\cdotp \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + \overline{ω} \sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}&\ \\ \\ \overline{ω}\cdotp \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + ω\cdotp \sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }} \end{cases}

(8)

in which, ω = \dfrac{-1+i\sqrt{3}}{2} and \overline{ω} =\dfrac{-1-i\sqrt{3}}{2}

The discriminant of a cubic equation

The radicant under square root symbol is the discriminant of the cubic equation, which is determined by the coefficients of the cubic equation. By substitution of (6) and (7), the discriminant for a general cubic equation could be obtained.

\Delta = \dfrac{q^2}{4}+\dfrac{p^3}{27}

(9)

Roots of a cubic equation

A cubic equation has roots of not only in real numbers but also imaginary number. The nature of the roots is determined by the sign of the discriminant.

Roots when discriminant is less than zero

Looking at the roots formula of a cubic equation (8), if \Delta < 0, square root of a negative number gives a complex number. However, the radicants under two cubic root symbol are a pair of conjugate complex numbers, which are still conjugate after taking cubic root. Addition of conjugate complex numbers gives a real number. Therefore, a negative discriminant of a cubic equation implies three distinct real roots. Here are the steps to prove the statement.

Let's denote the radicant under the first cubic root symbol as z_1, then

z_1 = r_1(\cos θ_1 + i \sin θ_1)

where r_1 = |-\dfrac{q}{2}+i\sqrt{-\dfrac{q^2}{4}-\dfrac{p^3}{27} } | is the modulus of z_1, and θ_1 is the argument of z_1.

Let's denote the radicant under the second cubic root symbol as z_2, then

z_2 = r_2(\cos θ_2 + i \sin θ_2)

where r_2 = |-\dfrac{q}{2}-i\sqrt{-\dfrac{q^2}{4}-\dfrac{p^3}{27} } | is the modulus of z_2, and a . θ_2 is the argument of z_2.

Then,

r_1 = r_2 = \sqrt{(-\dfrac{q}{2})^2+(-\dfrac{q^2}{4}-\dfrac{p^3}{27} )} = \sqrt{-\dfrac{p^3}{27} }

\tan θ_1= \dfrac{\sqrt{-\dfrac{q^2}{4}-\dfrac{p^3}{27} } }{-\dfrac{q}{2}} = -\sqrt{\dfrac{-\dfrac{q^2}{4}-\dfrac{p^3}{27}}{\dfrac{q^2}{4} } }=-\sqrt{1-\dfrac{4p^3}{27q^2} }

\tan θ_2 = \dfrac{-\sqrt{-\dfrac{q^2}{4}-\dfrac{p^3}{27} } }{-\dfrac{q}{2}} = \sqrt{\dfrac{-\dfrac{q^2}{4}-\dfrac{p^3}{27}}{\dfrac{q^2}{4} } }=\sqrt{1-\dfrac{4p^3}{27q^2} }

It turns out z_1 and z_2 are a pair of conjugate complex numbers which have the same modulus and argument in length and symmetric about x axis.

Therefore we can rewrite z_1 as

z= r(\cos θ + i \sin θ)

(10)

where

r = \sqrt{-\dfrac{p^3}{27} }

(11)

\cos \theta = \dfrac{-\dfrac{q}{2}}{\sqrt{-\dfrac{p^3}{27} }} = \dfrac{-\dfrac{-q}{2}}{\dfrac{-p}{3}\sqrt{\dfrac{-p}{3} } } = \dfrac{3q}{2p}\sqrt{-\dfrac{3}{p} }

(12)

\theta = \arccos\dfrac{-q}{2r} =\arccos(\dfrac{3q}{2p}\sqrt{-\dfrac{3}{p} } )

(13)

Then, z_2 can be expressed as

\overline{z}= r(\cos θ - i \sin θ)

(14)

t_{1,2,3} =\begin{cases} \sqrt[3]{ z}+\sqrt[3]{ \overline{z}} \\ω\sqrt[3]{z} + \overline{ω} \sqrt[3]{ \overline{z}}\\ \overline{ω} \sqrt[3]{ z}+ω \sqrt[3]{ \overline{z}} \end{cases}

\sqrt[3]{z} =\sqrt[3]{r} ( \cos\dfrac{ θ}{3} + i \sin \dfrac{ θ}{3})

\sqrt[3]{ \overline{z}} = \sqrt[3]{r} (\cos\dfrac{ θ}{3} - i \sin \dfrac{ θ}{3} )

where

\sqrt[3]{r} = \sqrt{\dfrac{-p}{3} }

(15)

\dfrac{\theta}{3} =\dfrac{1}{3} \arccos\dfrac{-q}{2r} =\dfrac{1}{3} \arccos(\dfrac{3q}{2p}\sqrt{-\dfrac{3}{p} } )

(16)

t_1 = \sqrt[3]{r} ( \cos\dfrac{ θ}{3} + i \sin \dfrac{ θ}{3}) + \sqrt[3]{r} (\cos\dfrac{ θ}{3} - i \sin \dfrac{ θ}{3} ) = 2\sqrt[3]{r} \cos\dfrac{ θ}{3}

(17)

Apply the rule for complex multiplication to the two terms

\begin{aligned} t_2 &= ω\sqrt[3]{z} + \overline{ω} \sqrt[3]{ \overline{z}} \\ & = \sqrt[3]{r}[ \dfrac{-1+i\sqrt{3}}{2} (\cos\dfrac{ θ}{3} + i \sin \dfrac{ θ}{3} ) + \dfrac{-1-i\sqrt{3}}{2}(\cos\dfrac{ θ}{3} - i \sin \dfrac{ θ}{3} )] \\& = \sqrt[3]{r}[-\dfrac{1}{2} \cos\dfrac{ θ}{3} - \dfrac{\sqrt{3}}{2} \sin \dfrac{ θ}{3} + ( -\dfrac{1}{2}\sin \dfrac{ θ}{3} +\dfrac{\sqrt{3} }{2} \cos\dfrac{ θ}{3} ) i \\&\quad -\dfrac{1}{2} \cos\dfrac{ θ}{3} - \dfrac{\sqrt{3}}{2}\sin \dfrac{ θ}{3}+ ( \dfrac{1}{2}\sin \dfrac{ θ}{3} -\dfrac{\sqrt{3} }{2} \cos\dfrac{ θ}{3} ) i ] \\ & = -\sqrt[3]{r}( \cos\dfrac{ θ}{3} + \sqrt{3} \sin\dfrac{ θ}{3} ) \end{aligned}

\begin{aligned} t_3&= \sqrt[3]{r} [ \dfrac{-1-i\sqrt{3}}{2} (\cos\dfrac{ θ}{3} + i \sin \dfrac{ θ}{3} ) + \dfrac{-1+i\sqrt{3}}{2}(\cos\dfrac{ θ}{3} - i \sin \dfrac{ θ}{3} )] \\& = \sqrt[3]{r} [-\dfrac{1}{2} \cos\dfrac{ θ}{3} + \dfrac{\sqrt{3}}{2}\sin \dfrac{ θ}{3} + ( -\dfrac{1}{2}\sin \dfrac{ θ}{3} -\dfrac{\sqrt{3} }{2} \cos\dfrac{ θ}{3} ) i \\&\quad -\dfrac{1}{2} \cos\dfrac{ θ}{3} + \dfrac{\sqrt{3}}{2}\sin \dfrac{ θ}{3}+ ( \dfrac{1}{2}\sin \dfrac{ θ}{3} +\dfrac{\sqrt{3} }{2} \cos\dfrac{ θ}{3} ) i] \\ & = \sqrt[3]{r} (-\cos\dfrac{ θ}{3}+ \sqrt{3} \sin \dfrac{ θ}{3} ) \end{aligned}

which shows a negative discriminant results in three distinct real roots.

\begin{cases} t_1 &= 2\sqrt[3]{r} \cos\dfrac{ θ}{3} \\ t_2 &=-\sqrt[3]{r}( \cos\dfrac{ θ}{3} + \sqrt{3} \sin\dfrac{ θ}{3} ) \\ t_3 &= \sqrt[3]{r} (-\cos\dfrac{ θ}{3}+ \sqrt{3} \sin \dfrac{ θ}{3} ) \end{cases}

(18)

Substituting t_{1,2,3} to (3) gives the roots for the general cubic equation when \Delta < 0.

\begin{cases} x_1 &= 2\sqrt[3]{r} \cos\dfrac{ θ}{3}- \dfrac{b}{3a} \\ x_2 &=-\sqrt[3]{r}( \cos\dfrac{ θ}{3} + \sqrt{3} \sin\dfrac{ θ}{3} )- \dfrac{b}{3a} \\ x_3 &= \sqrt[3]{r} (-\cos\dfrac{ θ}{3}+ \sqrt{3} \sin \dfrac{ θ}{3} ) - \dfrac{b}{3a} \end{cases}

(19)

Addition of the three roots yields a constant number.

x_1+x_2+x_3 = -\dfrac{b}{a}

(20)

Roots when discriminant is equal to zero

If \Delta = 0, then 27 q^2+4p^3 = 0. If p=q = 0, then the cubic equation has a triple zero root. If p, q \ne 0, the cubic equation has 3 real roots, of which 2 are equal roots. The three roots could be determined by constructing the special equation below.

The discriminant is zero, that is, 27 q^2+4p^3 = 0,

Then, -\dfrac{27q^2}{4p^2} = p and -\dfrac{27q^3}{4p^2}=q.

\Big( t - \dfrac{3q}{p} \Big) \Big( t+\dfrac{3q}{2p} \Big) ^2

= ( t - \dfrac{3q}{p})(t^2+\dfrac{3q}{p}t+\dfrac{9q^2}{4p^2} )

= t^3+\cancel{\dfrac{3q}{p}t^2}+\dfrac{9q^2}{4p^2}t- \cancel{\dfrac{3q}{p}t^2} - \dfrac{9q^2}{p^2}t-\dfrac{27q^3}{4p^2}

=t^3 -\dfrac{27q^2}{4p^2}t -\dfrac{27q^3}{4p^2}

=t^3+pt+q =0

which shows that the first root is \dfrac{3q}{p} and the second root is equal to the third one, that is

\begin{cases} t_1 &= \dfrac{3q}{p} \\ t_2 &=t_3 = - \dfrac{3q}{2p} \end{cases}

(21)

Substituting t_{1,2,3} to (3) gives the roots for the general cubic equation when \Delta =0.

\begin{cases} x_1 &= \dfrac{3q}{p}- \dfrac{b}{3a} \\ x_2 &=x_3 = - \dfrac{3q}{2p}- \dfrac{b}{3a} \end{cases}

(22)

And the statement (20) about addition of three roots holds true when \Delta =0.

Roots when discriminant is greater than zero

Looking at the roots formula of a cubic equation (8), if \Delta > 0,the radicants under two cubic root symbol are real numbers. Therefore, the first root is a real root. However, the addition of two cubic that are multiplied by a complex number results in two distinct complex numbers.

Considering the symmetric properties of the expression under two cubic root symbol in equation (8), lets denote the first radicant as R, that is

R = -\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }

Then, the second radicant as \overline{R}

\overline{R} = -\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }

Then,

\begin{cases} t_1 & =\sqrt[3]{R}+ \sqrt[3]{\overline{R} } & \\ t_2 &= ω\cdotp \sqrt[3]{R}+ \overline{ω} \sqrt[3]{\overline{R} } \\ t_3 &= \overline{ω}\cdotp \sqrt[3]{R}+ ω\cdotp \sqrt[3]{\overline{R} } \end{cases}

Obviously, t_1 is a real number. However, t_2 and t_3 are two distinct complex numbers

\begin{aligned} t_2 &= ω\cdotp \sqrt[3]{R}+ \overline{ω} \sqrt[3]{\overline{R} } \\ & = \dfrac{-1+i\sqrt{3}}{2} \sqrt[3]{R}+ \dfrac{-1-i\sqrt{3}}{2} \sqrt[3]{\overline{R} } \\& =\dfrac{-\sqrt[3]{R}-\sqrt[3]{\overline{R} }}{2} +\dfrac{\sqrt{3}( \sqrt[3]{R} - \sqrt[3]{\overline{R} }) }{2} i \end{aligned}

\begin{aligned} t_3&= \overline{ω}\cdotp \sqrt[3]{R}+ ω\cdotp \sqrt[3]{\overline{R} }\\& = \dfrac{-1-i\sqrt{3}}{2} \sqrt[3]{R} + \dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{\overline{R} } \\ & = \dfrac{-\sqrt[3]{R}-\sqrt[3]{\overline{R} }}{2} +\dfrac{\sqrt{3}(- \sqrt[3]{R} + \sqrt[3]{\overline{R} }) }{2}i \end{aligned}

Substituting t_{1,2,3} to (3) yields three roots for a general cubic equation.

\begin{cases} x_1 & =\sqrt[3]{R}+ \sqrt[3]{\overline{R} } - \dfrac{b}{3a} & \\ x_2 &= \dfrac{-\sqrt[3]{R}-\sqrt[3]{\overline{R} }}{2} +\dfrac{\sqrt{3}( \sqrt[3]{R} - \sqrt[3]{\overline{R} }) }{2} i - \dfrac{b}{3a} \\ x_3 &= \dfrac{-\sqrt[3]{R}-\sqrt[3]{\overline{R} }}{2} +\dfrac{\sqrt{3}(- \sqrt[3]{R} + \sqrt[3]{\overline{R} }) }{2}i - \dfrac{b}{3a} \end{cases}

And it is found that the statement (20) about addition of three roots holds true when \Delta > 0.

Examples of the roots of cubic equations

If a cubic equation has a rational root, which means it's reducible, then the root could be a fraction number with a factor of constant term as the numerator and a factor of the coefficient of leading term as the denominator. Factorization is always the easiest way to find the rational root. And then, find another factor by using polynomial long division.

For example, solve x³ - 6x² + 11x - 6 = 0. The first root is determined as x_1 = 1 by observation using factor theorem. Dividing the polynomial x³ - 6x² + 11x - 6 by x - 1 by Long division gives another factor x² - 5x + 6 . The next two roots are obtained by solving the quadratic equation. x_2 = 3 and x_3 = 2

More examples of reducible cubic equations are

4x³ - 5x² - 5x + 4 = 0

which has 1 integer root and 2 real roots expressed as radicals

\begin{cases} x_1 =-1 \\ x_2 = \dfrac{9+\sqrt{17} }{8} \\ x_3 = \dfrac{9-\sqrt{17} }{8} \end{cases}

For this type of equation, the discriminant is less than zero. If solved with trigonometric method or Cardano method, the last two roots could be represented by trigonometric functions.

\begin{cases} x_1=\dfrac{\sqrt{85}}{6}\cos \bigg[\dfrac{1}{3}\cdot \arccos\Big(-\dfrac{17}{5}\sqrt{\dfrac{1}{85}}\Big)\bigg]+\dfrac{5}{12} \\ x_2=-1 \\ x_3=\dfrac{\sqrt{85}}{6} \cos \bigg[ \dfrac{1}{3}\cdot \arccos\Big(-\dfrac{17}{5}\sqrt{\dfrac{1}{85}}\Big)+\dfrac{4\pi}{3} \bigg]+\dfrac{5}{12} \end{cases}

x^3−2x^2+8x−16=0

which has 1 integer root and 2 complex roots.

\begin{cases} x_1=2 \\ x_2=2\sqrt{2}i \\ x_3=-2\sqrt{2}i \end{cases}

2x^3−x^2+3x−9=0

which has 1 rational root and 2 complex roots.

\begin{cases} x_1=\dfrac{3}{2} \\ x_2=-\dfrac{1}{2}+\dfrac{\sqrt{11}}{2}i \\ x_3=-\dfrac{1}{2}-\dfrac{\sqrt{11}}{2}i \end{cases}

However, not all reducible equations give a rational root. Here is an example that the equation could be factored by using the sum of cubes identity.

8x^3+9x^2+9x+3=0

This types of cubic equation normally gives 1 real root in radicals and 2 complex roots.

\begin{cases} x_1=\dfrac{\sqrt[3]{45}}{8}-\dfrac{\sqrt[3]{75}}{8}-\dfrac{3}{8} \\ x_2=-\dfrac{1}{8}\sqrt[3]{\dfrac{45}{8}}+\dfrac{1}{8}\sqrt[3]{\dfrac{75}{8}}+\sqrt{3}\Big(\dfrac{1}{8}\sqrt[3]{\dfrac{45}{8}}+\dfrac{1}{8}\sqrt[3]{\dfrac{75}{8}}\Big)i-\dfrac{3}{8} \\ x_3=-\dfrac{1}{8}\sqrt[3]{\dfrac{45}{8}}+\dfrac{1}{8}\sqrt[3]{\dfrac{75}{8}}-\sqrt{3}\Big(\dfrac{1}{8}\sqrt[3]{\dfrac{45}{8}}+\dfrac{1}{8}\sqrt[3]{\dfrac{75}{8}}\Big)i-\dfrac{3}{8} \end{cases}

In fact, the majority of cubic equations are not special ones that could be transformed to lower degree. For general form of cubic equation, the roots could be represented in complex radicals if the discriminant is greater than zero or trigonometric form if the discriminant is less than zero.

For example,

x^3+2x+4=0

It has 1 real root and 2 complex roots. All in complex radicals.

\begin{cases} x_1=\sqrt[3]{-2+\dfrac{2}{3}\sqrt{\dfrac{29}{3}}}+\sqrt[3]{-2-\dfrac{2}{3}\sqrt{\dfrac{29}{3}}} \\ x_2=\dfrac{1}{2}\Big(-\sqrt[3]{-2+\dfrac{2}{3}\sqrt{\dfrac{29}{3}}}-\sqrt[3]{-2-\dfrac{2}{3}\sqrt{\dfrac{29}{3}}}\Big)+\dfrac{\sqrt{3}}{2}\Big(\sqrt[3]{-2+\dfrac{2}{3}\sqrt{\dfrac{29}{3}}}-\sqrt[3]{-2-\dfrac{2}{3}\sqrt{\dfrac{29}{3}}}\Big)i \\ x_3=\dfrac{1}{2}\Big(-\sqrt[3]{-2+\dfrac{2}{3}\sqrt{\dfrac{29}{3}}}-\sqrt[3]{-2-\dfrac{2}{3}\sqrt{\dfrac{29}{3}}}\Big)-\dfrac{\sqrt{3}}{2}\Big(\sqrt[3]{-2+\dfrac{2}{3}\sqrt{\dfrac{29}{3}}}-\sqrt[3]{-2-\dfrac{2}{3}\sqrt{\dfrac{29}{3}}}\Big)i \end{cases}

x^3−3x+1=0 gives three nice trigonometric roots.

\begin{cases} x_1=2\cos\dfrac{2}{9}\pi \\ x_2=2 \cos\dfrac{8}{9}\pi \\ x_3=2 \cos\dfrac{14}{9}\pi \end{cases}

while x^3-3x^2−4x+2=0 gives three roots that are in the form of radicals and trigonometric.

\begin{cases} x_1=2\sqrt{\dfrac{7}{3}}\cos \bigg[\dfrac{1}{3}\cdot \arccos\big(\dfrac{6}{7}\sqrt{\dfrac{3}{7}}\big)\bigg]+1 \\ x_2=2\sqrt{\dfrac{7}{3}} \cos \bigg[ \dfrac{1}{3}\cdot \arccos\big(\dfrac{6}{7}\sqrt{\dfrac{3}{7}}\big)+\dfrac{2\pi}{3}\bigg]+1 \\ x_3=2\sqrt{\dfrac{7}{3}} \cos \bigg[ \dfrac{1}{3}\cdot \arccos\big(\dfrac{6}{7}\sqrt{\dfrac{3}{7}}\big)+\dfrac{4\pi}{3} \bigg]+1 \end{cases}

For the above equations, if you'd like know the detailed steps , just copy the equation to the cubic equation solver find more about the solutions.

In summary, the roots of general cubic equation are derived based on the solution of a depressed cubic equation. Depending on different coefficients and constant term, the discriminant of the cubic equation varies, which ultimately determine the nature of the roots of the cubic equation. A rational root implies the root could be found by factorization. Nevertheless, all of the roots could be determined by the root formula of a cubic equation.

Collected in the board: Algebraic equation

Steven Zheng posted 4 months ago