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. The followingfollowing 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_21, 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 simplify 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 on solving cubic equations

Question 1: Find solutions of the cubic equation: x³ - 6x² + 11x - 6 = 0

A cubic equation has at least one real root. If the coefficient of leading term is 1, one of solutions could be a factor of the constant term.

Find all possible factors for constant -6

1, -1

2, -2

3, -3

6, -6

Substitute the factors to the function f(x) = x³ - 6x² + 11x - 6 and find the one that makes f(x) = 0.

According to factor theorem, f(n) = 0, if and only if the polynomial x³ - 6x² + 11x - 6 has a factor x-n, that is, x=n is a root of the equation.

f(1) = 1³ - 61² + 111 - 6 = 0

then, x = 1 is one of roots of the cubic equaiton, which can be factored as

(x -1)(ax^2+bx+c) = 0

Then we can use either long division or synthetic division to determine the expression of trinomial

Dividing the polynomial x³ - 6x² + 11x - 6 by x - 1 by Long division, we get another factor of the cubic equation x² - 5x + 6 and a quadratic equation

x² - 5x + 6 = 0

Further factoring the quadratic equation gives

(x - 3)(x - 2) = 0

which results in another two roots

x_2 = 3

x_3 = 2

Finally, we have obtained the three roots for the cubic equation x³ - 6x² + 11x - 6=0, which are

\begin{cases} x_1 =1 \\ x_2 = 3 \\ x_3 = 2 \end{cases}

Question 2: Find solutions of the cubic equation: 4x³ - 5x² - 5x + 4 = 0

If a cubic equation has a rational root, the root could be a fracion number with a factor of constant term as the numerator and a factor of the coefficient of leading term as the denumerator.

Find all possible factors for constant 4, which is donated as N

1, -1

2, -2

4, -4

Find all possible factors of the coefficient of the leading term 4, which is donated as D

1, -1

2, -2

4, -4

N over D in each combinations gives the following fractions, which could be a root of the cubic equation.

1, -1, 2, -2, 4, -4, \dfrac{1}{2}, -\dfrac{1}{2}, \dfrac{1}{4}, -\dfrac{1}{4},

Substitute the fractions to the function f(x) = 4x³ - 5x² - 5x + 4 one by one to find out the one that makes f(x) = 0.

According to the factor theorem, f(n) = 0, if and only if the polynomial 4x³ - 5x² - 5x + 4 has a factor x-n, that is, x=n is the root of the equation.

f(1) = 4(1)³ - 5(1)² - 5(1) + 4 = -2,

f(-1) = 4(-1)³ - 5(-1)² - 5(-1) + 4 = 0,

then, x = -1 is one of roots of the cubic equation 4x³ - 5x² - 5x + 4 = 0. And, the equation can be factored as

(x +1)(ax^2+bx+c) = 0

Then we can use either long division or synthetic division to determine the expression of trinomial.


Dividing the polynomial 4x³ - 5x² - 5x + 4 by x + 1 by long division gives another factor of the cubic equation 4x² - 9x + 4 and a quadratic equation

4x² - 9x + 4 = 0

Use root solution formula for a quadratic equation

\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}

= \dfrac{-(-9)\pm\sqrt{(-9)^2-4\cdot 4\cdot 4} }{2 \cdot 4}

= \dfrac{9\pm\sqrt{17} }{8}

Finally we have obtained 3 roots for the cubic equation 4x³ - 5x² - 5x + 4=0 , which are

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

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 Steven Zheng posted 3 hours ago

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