Solve the cubic equation:

$$x^3-5x^2+3x+2=0 $$

Quick Answer

Since the discriminant $$ \Delta <0$$, the cubic equation has three distinct real roots.

$$ \Delta=-4.3425925925926$$

$$\begin{cases} x_1=\dfrac{8}{3}\cos \bigg(\dfrac{1}{3}\cdot \arccos\dfrac{61}{128}\bigg)+\dfrac{5}{3} \\ x_2=\dfrac{8}{3} \cos \bigg( \dfrac{1}{3}\cdot \arccos\dfrac{61}{128}+\dfrac{2\pi}{3}\bigg)+\dfrac{5}{3} \\ x_3=\dfrac{8}{3} \cos \bigg( \dfrac{1}{3}\cdot \arccos\dfrac{61}{128}+\dfrac{4\pi}{3} \bigg)+\dfrac{5}{3} \end{cases}$$

In decimals,

$$\begin{cases} x_1=4.1642479384602 \\ x_2=-0.3913823806309 \\ x_3=1.2271344421707 \end{cases}$$

Detailed Steps on Solution

1. Convert to depressed cubic equation

The idea is to convert general form of cubic equation

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

to the form without quadratic term.

$$t^3+pt+q = 0$$

By substituting $$x$$ with $$t - \dfrac{b}{3a}$$, the general cubic equation could be transformed to

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

Compare with the depressed cubic equation. Then,

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

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

Substitute the values of coefficients, $$p, q$$ is obtained as

$$p = \dfrac{3\cdot 1\cdot 3-(-5)^2}{3\cdot 1^2}=-\dfrac{16}{3}$$

$$q = \dfrac{2\cdot (-5)^3-9\cdot1\cdot (-5)\cdot 3+27\cdot 1^2\cdot2}{27\cdot 1^3}=-\dfrac{61}{27}$$

Use the substitution to transform

Let $$p$$ and $$q$$ being the coefficient of the linean and constant terms, the depressed cubic equation is expressed as.

$$t^3 +pt+q=0$$

Let $$x=t+\dfrac{5}{3}$$

The cubic equation $$x³ - 5x² + 3x + 2=0$$ is transformed to

$$t^3 -\dfrac{16}{3}t-\dfrac{61}{27}=0$$

2. Cardano's solution

Let $$t=u-v$$

Cube both sides and extract common factor from two middle terms after expanding the bracket.

$$\begin{aligned} \\t^3&=(u-v)^3\\ & =u^3-3u^2v+3uv^2-v^3\\ & =-3uv(u-v)+u^3-v^3\\ \end{aligned}$$

Since $$u-v=t$$, substitution gives a linear term for the equation. Rearrange terms.

$$x^3+3uvx-u^3+v^3=0$$

Compare the cubic equation with the original one (1)

$$\begin{cases} 3uv=-\dfrac{16}{3}\quad\text{or}\quad v=-\dfrac{16}{9u}\\ v^3-u^3=-\dfrac{61}{27}\\ \end{cases}$$

$$v=-\dfrac{16}{9u}$$ gives relationship between the two variables. Substitute the value of $$v$$ to the second equation

$$\Big(-\dfrac{16}{9u}\Big)^3-u^3=-\dfrac{61}{27}$$

Simplifying gives,

$$u^3+\dfrac{4096}{729}\dfrac{1}{u^3}-\dfrac{61}{27}=0$$2

Let $$m=u^3$$, then the equation is transformed to a quadratic equation in terms of $$m$$. Once the value of $$m$$ is determined, $$v^3$$ could be determined by $$v^3=-\dfrac{61}{27}+u^3$$.

$$m^2-\dfrac{61}{27}m+\dfrac{4096}{729}=0$$

Sovling the quadratic euqation will give two roots (some may be equal). Here we only cosider one case with positive sign before the square root radical since the negative case will produce the same result.

$$\begin{aligned} \\u^3=m&=\dfrac{61}{54}+\dfrac{1}{2}\sqrt{\Big(-\dfrac{61}{27}^2\Big)-4\cdot \dfrac{4096}{729}}\\ & =\dfrac{61}{54}+\dfrac{1}{2}\sqrt{\dfrac{3721}{729}-\dfrac{16384}{729}}\\ & =\dfrac{61}{54}+\dfrac{1}{2}\sqrt{\dfrac{469}{27}}i\\ & =\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i\\ \end{aligned}$$

$$v^3$$ can be determined by the equation we deduced $$v^3-u^3=-\dfrac{61}{27}$$. Then,

$$\begin{aligned} \\v^3&=-\dfrac{61}{27}+u^3\\ & =-\dfrac{61}{27}+\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i\\ & =-\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i\\ \end{aligned}$$

Now we have,

$$u^3=\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i$$ and $$v^3=-\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i$$

Evaluating the simplest cubic equation $$x^3-A=0$$, it has 3 roots, in which the first root is a real number or a complex number. The second and third are expressed in the product of cubic root of unity and the first one.

If $$ω = \dfrac{-1+i\sqrt{3}}{2}$$, then its reciprocal is equal to its conjugate, $$\dfrac{1}{ω}=\overline{ω}$$.

$$\begin{cases} r_1=\sqrt[3]{A}\\ r_2=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{A}\\ r_3=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{A}\\ \end{cases}$$

Similary, taking cubic root for $$u^3$$ and $$v^3$$ also gives 3 roots.

$$\begin{cases} u_1=\sqrt[3]{\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i}\\ u_2=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i}\\ u_3=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i}\\ \end{cases}$$

For $$v_2$$ and $$v_3$$, the complex numbers before radicals are the conjugates of those for $$u_2$$ and $$u_3$$, which can be verified by the reciprocal property of the cubic root of unity from the equation $$v=-\dfrac{16}{9u}$$. The radicand can be taken as the negative conjugate of that in $$u_1$$, $$u_2$$ and $$u_3$$, which is the same in value.

$$\begin{cases} v_1=\sqrt[3]{-\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i}\\ v_2=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{-\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i}\\ v_3=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{-\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i}\\ \end{cases}$$

Since $$t=u-v$$, the firt root $$t_1$$ can be expressed as the sum of cubic root of two conjugate complex numbers

$$t_1=\sqrt[3]{\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i}+\sqrt[3]{\dfrac{61}{54}-\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i}$$

Let $$u^3_1=z=\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i$$, and $$z$$ can be expressed in trigonomic form $$r(\cos θ + i \sin θ)$$, where $$r$$ and $$θ$$ are the modulus and principle argument of the complex number.

Then $$-v^3_1$$ is the conjugate $$\overline{z}=\dfrac{61}{54}-\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i$$, or $$\overline{z} = r(\cos θ - i \sin θ)$$

Now let's calculate the value of $$r$$ and $$ θ$$.

$$\begin{aligned} \\r&=\sqrt{\Big(\dfrac{61}{54}\Big)^2+\Big(\dfrac{1}{6}\sqrt{\dfrac{469}{3}}\Big)^2}\\ & =\dfrac{64}{27}\\ \end{aligned}$$

$$\cosθ=\dfrac{\dfrac{61}{54}}{\dfrac{64}{27}}=\dfrac{61}{128}$$

The argument is the inverse function of the cosθ

$$θ=\arccos\Big(\dfrac{61}{128}\Big)$$

Using de Moivre’s formula, the cubic root of $$z$$ could be determined.

$$\begin{aligned} \\u_1=\sqrt[3]{z}&=\sqrt[3]{r}(\cos\dfrac{θ}{3}+i\sin\dfrac{θ}{3})\\ & =\sqrt[3]{\dfrac{64}{27}}\Big[\cos\dfrac{1}{3}\arccos\Big(\dfrac{61}{128}\Big)+\sin\dfrac{1}{3}\arccos\Big(\dfrac{61}{128}\Big)\Big]\\ & =\dfrac{4}{3}\Big[\cos\dfrac{1}{3}\arccos\Big(\dfrac{61}{128}\Big)+\sin\dfrac{1}{3}\arccos\Big(\dfrac{61}{128}\Big)\Big]\\ \end{aligned}$$

Since $$-v^3$$ is the conjugate of $$z$$ as we mentioned above,

$$\begin{aligned} \\v_1=-\sqrt[3]{\overline{z}}&=-\sqrt[3]{r}(\cos\dfrac{θ}{3}-i\sin\dfrac{θ}{3})\\ & =\dfrac{4}{3}\Big[-\cos\dfrac{1}{3}\arccos\Big(\dfrac{61}{128}\Big)+\sin\dfrac{1}{3}\arccos\Big(\dfrac{61}{128}\Big)\Big]\\ \end{aligned}$$

The first root is the difference of $$u_1$$ and $$v_1$$.

$$\begin{aligned} \\t_1&=u_1-v_1\\ & =2\cdot \dfrac{4}{3}\cos\dfrac{1}{3}\arccos\Big(\dfrac{61}{128}\Big)\\ & =\dfrac{8}{3}\cos\dfrac{1}{3}\arccos\Big(\dfrac{61}{128}\Big)\\ \end{aligned}$$

The second root is the difference of $$u_2$$ and $$v_2$$, in which $$u_2$$ is the product of the cubic root of $$z$$ and the cubic root of unity, $$v_2$$ is the product of the negative of the conjugate of $$z$$ and the conjugate of the cubic root of unity.

$$\begin{aligned} \\t_2&=u_2-v_2\\ & =\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i}+\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{\dfrac{61}{54}-\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i}\\ & =\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{r}(\cos\dfrac{θ}{3}+i\sin\dfrac{θ}{3})+\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{r}(\cos\dfrac{θ}{3}-i\sin\dfrac{θ}{3})\\ & =-\sqrt[3]{r}(\cos\dfrac{ θ}{3} + \sqrt{3} \sin\dfrac{ θ}{3} ) \\ & =2\sqrt[3]{r}\cos\Big(\dfrac{θ}{3}+ \dfrac{4\pi}{3}\Big)\\ & =\dfrac{8}{3}\cos\Big(\dfrac{1}{3}\arccos\Big(\dfrac{61}{128}\Big)+ \dfrac{4\pi}{3}\Big)\\ \end{aligned}$$

The third root is the difference of $$u_3$$ and $$v_3$$, in which $$u_3$$ is the product of the cubic root of $$z$$ and the conjugate of the cubic root of unity, $$v_3$$ is the product of the negative of the conjugate of $$z$$ and the cubic root of unity.

$$\begin{aligned} \\t_3&=u_3-v_3\\ & =\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{\dfrac{61}{54}+\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i}+\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{\dfrac{61}{54}-\dfrac{1}{6}\sqrt{\dfrac{469}{3}}i}\\ & =\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{r}(\cos\dfrac{θ}{3}+i\sin\dfrac{θ}{3})+\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{r}(\cos\dfrac{θ}{3}-i\sin\dfrac{θ}{3})\\ & =\sqrt[3]{r}(-\cos\dfrac{θ}{3}+\sqrt{3} \sin \dfrac{ θ}{3})\\ & =2\sqrt[3]{r}\cos\Big(\dfrac{θ}{3}+ \dfrac{2\pi}{3}\Big)\\ & =\dfrac{8}{3}\cos\Big(\dfrac{1}{3}\arccos\Big(\dfrac{61}{128}\Big)+ \dfrac{2\pi}{3}\Big)\\ \end{aligned}$$

3. Vieta's Substitution

In Cardano' solution, $$t$$ is defined as the difference of $$u$$ and $$v$$. If we substitute the value of $$v$$ (4) into (2), we get the equation. $$t=u+\dfrac{16}{9u}$$. And then substitute the equation to the cubic equation $$t^3-\dfrac{16}{3}t-\dfrac{61}{27}=0$$. This method is called Vieta's Substitution for solving a cubic equation, which simplied the Cardano' solution. The substitution expression can be obtained by the following formula directly.

$$t=u-\dfrac{p}{3u}$$

Substitute the expression $$t=u+\dfrac{16}{9u}$$ to the cubic equation

$$\Big(u+\dfrac{16}{9u}\Big)^3-\dfrac{16}{3}\Big(u+\dfrac{16}{9u}\Big)-\dfrac{61}{27}=0$$

Expand brackets and cancel the like terms

$$u^3+\cancel{\dfrac{16}{3}u^2\dfrac{1}{u}}+\cancel{\dfrac{256}{27}u\dfrac{1}{u^2}}+\dfrac{4096}{729}\dfrac{1}{u^3}-\cancel{\dfrac{16}{3}u}-\cancel{\dfrac{256}{27}\dfrac{1}{u}}-\dfrac{61}{27}=0$$

Then we get the same equation as (2)

$$u^3+\dfrac{4096}{729}\dfrac{1}{u^3}-\dfrac{61}{27}=0$$

The rest of the steps will be the same as those of Cardano's solution

4. Euler's Solution

$$t^3-\dfrac{16}{3}t-\dfrac{61}{27}=0$$

Move the linear term and constant of (1) to its right hand side. We get the following form of the equation.

$$t^3=\dfrac{16}{3}t+\dfrac{61}{27} $$3

Let the root of the cubic equation be the sum of two cubic roots

$$t=\sqrt[3]{r_1}+\sqrt[3]{r_2} $$4

in which $$r_1$$ and $$r_2$$ are two roots of a quadratic equation

$$z^2-\alpha z+ β=0 $$5

Using Vieta's Formula, the following equations are established.

$$r_1+r_2 = \alpha \quad \text{and} \quad r_1r_2 = β $$

To determine $$\alpha$$, $$β$$, cube both sides of the equation (4)

$$t^3=3\sqrt[3]{r_1r_2}(\sqrt[3]{r_1}+\sqrt[3]{r_2})+r_1+r_2 $$

Substituting, the equation is simplified to

$$t^3=3\sqrt[3]{β}t+\alpha $$

Compare the cubic equation with (3), the following equations are established

$$\begin{cases} 3\sqrt[3]{β}=\dfrac{16}{3}\\ \alpha=\dfrac{61}{27}\\ \end{cases}$$

Solving for $$β$$ gives

$$β=\dfrac{4096}{729} $$

So the quadratic equation (5) is determined as

$$z^2-\dfrac{61}{27}z+\dfrac{4096}{729}=0$$6

Solving the quadratic equation yields

$$\begin{cases} r_1=\dfrac{61}{54}+\dfrac{\sqrt{1407}}{18}i\approx1.1296296296296+2.0838888148346i\\ r_2=\dfrac{61}{54}-\dfrac{\sqrt{1407}}{18}i\approx1.1296296296296-2.0838888148346i\\ \end{cases}$$

Therefore, one of the roots of the cubic equation could be obtained from (4).

$$t_1=\sqrt[3]{\dfrac{61}{54}+\dfrac{\sqrt{1407}}{18}i}+\sqrt[3]{\dfrac{61}{54}-\dfrac{\sqrt{1407}}{18}i} $$

in decimals,

$$t_1=2.4975812717935 $$

However, since the cube root of a quantity has triple values,

The other two roots could be determined as,

$$t_2=\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{\dfrac{61}{54}+\dfrac{\sqrt{1407}}{18}i}+\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{\dfrac{61}{54}-\dfrac{\sqrt{1407}}{18}i} $$

$$t_3=\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{\dfrac{61}{54}+\dfrac{\sqrt{1407}}{18}i}+\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{\dfrac{61}{54}-\dfrac{\sqrt{1407}}{18}i} $$

Since the expression involes cubic root of complex number, the final result can be deduced by using trigonometric method as shown in Cardano's solution.

For the equation $$t^3 -\dfrac{16}{3}t-\dfrac{61}{27}$$, we have $$p=-\dfrac{16}{3}$$ and $$q = -\dfrac{61}{27}$$

Calculate the discriminant

The nature of the roots are determined by the sign of the discriminant.

Since $$p$$ is negative, the discriminant will be less than zero if the absolute value of $$p$$ is large enough.

$$\begin{aligned} \\\Delta&=\dfrac{q^2}{4}+\dfrac{p^3}{27}\\ & =\dfrac{\Big(-\dfrac{61}{27}\Big)^2}{4}+\dfrac{\Big(-\dfrac{16}{3}\Big)^3}{27}\\ & =\dfrac{3721}{2916}-\dfrac{4096}{729}\\ & =\dfrac{3721\cdot 1-4096\cdot 4}{2916}\\ & =-4.3425925925926\\ \end{aligned}$$

4.1 Use the root formula directly

If $$\Delta < 0$$, then there are 3 distinct real roots for the cubic equation

$$\begin{cases} t_1 &= 2\sqrt[3]{r} \cos\dfrac{ θ}{3} \\ t_2 & = 2\sqrt[3]{r}\cos\Big( \dfrac{ θ}{3}+\dfrac{2\pi}{3} \Big) \\ t_3&= 2\sqrt[3]{r}\cos\Big( \dfrac{ θ}{3}+\dfrac{4\pi}{3} \Big) \end{cases}$$

where

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

Substitute the values of $$p$$ and $$q$$ to determine the value of $$\theta$$

$$\begin{aligned} \\\theta&= \arccos(\dfrac{3q}{2p}\sqrt{-\dfrac{3}{p} } )\\ & =\arccos\Big(\dfrac{3\cdot -\dfrac{61}{27}}{2 \cdot -\dfrac{16}{3}}\sqrt{-\dfrac{3}{-\dfrac{16}{3}} }\Big)\\ & = \arccos\Big(\dfrac{1}{2}\cdot 3\cdot \dfrac{61}{27}\cdot \dfrac{3}{16}\sqrt{3\cdot \dfrac{3}{16}}\Big)\\ & =\arccos\Big(\dfrac{1}{2}\cdot 3\cdot \dfrac{61}{27}\cdot \dfrac{3}{16}\cdot \dfrac{3}{4}\Big)\\ & = \arccos\Big(\dfrac{61}{128}\Big)\\ \end{aligned}$$

Then we can determine one third of $$\theta$$

$$\dfrac{\theta}{3} = \dfrac{1}{3} \arccos\dfrac{61}{128} $$

Substitute the value of $$p$$ to determine the cubic root of $$r$$

$$\begin{aligned} \\\sqrt[3]{r}& =\sqrt{\dfrac{-p}{3}}\\ & =\sqrt{\frac{-(-16)}{3 \cdot 3}} \\ & =\sqrt{\dfrac{16}{9}}\\ & =\dfrac{4}{3}\\ \end{aligned}$$

Substitute the value of $$\dfrac{\theta}{3}$$ and $$\sqrt[3]{r}$$ to the root formulas. Then we get

$$\begin{aligned} \\t_1&= 2\sqrt[3]{r} \cos\dfrac{ θ}{3}\\ & =2\cdot \dfrac{4}{3}\cdot \cos \bigg( \dfrac{1}{3}\cdot\arccos\dfrac{61}{128}\bigg)\\ & =\dfrac{8}{3}\cos \bigg(\dfrac{1}{3}\cdot \arccos\dfrac{61}{128}\bigg)\\ & \approx \dfrac{8}{3} \cos 0.35801861519573\\ & \approx 2.4975812717935\\ \end{aligned}$$

$$\begin{aligned} \\t_2&=\dfrac{8}{3} \cos \bigg( \dfrac{1}{3}\cdot \arccos\dfrac{61}{128}+\dfrac{2\pi}{3}\bigg)\\ & \approx -2.0580490472976\\ \end{aligned}$$

$$\begin{aligned} \\t_3&=\dfrac{8}{3} \cos \bigg( \dfrac{1}{3}\cdot \arccos\dfrac{61}{128}+\dfrac{4\pi}{3} \bigg)\\ & \approx -0.43953222449598\\ \end{aligned}$$

4.2 Trigonometric Method

For a depressed cubic equation, another method is to convert the equation to the form of triple angle identity. By comparing the trigonometric value of the triple angle, the value of the single angle is determined. And then the roots of the cubic equation can be found.

Compare the given equation with the standard depressed cubic equation $$t^3 +pt+q=0$$, we get $$p=-\dfrac{16}{3}$$ and $$q = -\dfrac{61}{27}$$

Using the formula

$$t= 2\sqrt{\dfrac{-p}{3}}u $$

to introduce an intermediate variable $$u$$ so that the given equation is transformed to a new equation in terms of $$u$$, which is analogous to a trignometric triple angle identity.

Then,

$$t= \dfrac{8}{3}u $$

Substitute to the cubic equation and simplify.

$$27\Big(\dfrac{8}{3}u\Big)^3 -144\Big(\dfrac{8}{3}u\Big)-61=0$$

Simply the equation

$$27\cdot \dfrac{8}{3}\cdot \Big(\dfrac{8}{3}\Big)^2 u^3-144\cdot \dfrac{8}{3}u-61=0$$

$$9\cdot 8\cdot \dfrac{64}{9} u^3-144\cdot \dfrac{8}{3}u-61=0$$

Mutiply each term by $$9$$

$$4608 u^3-3456u-549=0$$

Divide each term by $$1152$$

$$4u^3-3u-\dfrac{61}{128}=0$$

The equation becomes the form of a triple angle identity for cosine function.

$$4\cos^3θ-3\cos θ-\cos3θ =0$$

Comparing the equation with triple angle identity gives

$$u =\cos θ$$

and

$$\cos3θ =\dfrac{61}{128}$$

The fact that the cosine function is periodic with period 2π implies the following equation.

$$\cos(3θ-2πk) =\dfrac{61}{128}$$

Solving for θ yields

$$ θ =\dfrac{1}{3}\cdot \arccos\dfrac{61}{128}+\dfrac{2πk}{3} , $$

in witch $$k=0,1,2$$.

Then we can determine the value of $$u$$, that is $$\cos θ$$

$$\begin{aligned} \\u&= \cos θ\\ & = \cos\Big(\dfrac{1}{3}\cdot \arccos\dfrac{61}{128}+\dfrac{2πk}{3}\Big)\\ \end{aligned}$$

and subsiquently,

$$\begin{aligned} \\t&= \dfrac{8}{3}u\\ & = \dfrac{8}{3} \cos\Big(\dfrac{1}{3}\cdot \arccos\dfrac{61}{128}+\dfrac{2πk}{3}\Big)\\ \end{aligned}$$

Substituting $$k=0,1,2$$ gives the roots of solution set.

$$\begin{aligned} \\t_1&= \dfrac{8}{3}\cos\Big(\dfrac{1}{3}\cdot \arccos\dfrac{61}{128}\Big)\\ & \approx2.4975812717935\\ \end{aligned}$$

$$\begin{aligned} \\t_2&= \dfrac{8}{3}\cos\Big(\dfrac{1}{3}\cdot \arccos\dfrac{61}{128}+\dfrac{2π}{3}\Big)\\ & \approx-2.0580490472976\\ \end{aligned}$$

$$\begin{aligned} \\t_3&= \dfrac{8}{3}\cos\Big(\dfrac{1}{3}\cdot \arccos\dfrac{61}{128}+\dfrac{4π}{3}\Big)\\ & \approx-0.43953222449598\\ \end{aligned}$$

which shows the trigonometric method gets the same solution set as that by using roots formula.

Roots of the general cubic equation

Since $$x = t - \dfrac{b}{3a}$$, substituting the values of $$t$$, $$a$$ and $$b$$ gives

$$\begin{aligned} \\x_1&=t_1-\dfrac{b}{3a}\\ & =\dfrac{8}{3}\cos \bigg(\dfrac{1}{3}\cdot \arccos\dfrac{61}{128}\bigg)+\dfrac{5}{3}\\ \end{aligned}$$

$$\begin{aligned} \\x_2&=t_2-\dfrac{b}{3a}\\ & =\dfrac{8}{3} \cos \bigg( \dfrac{1}{3}\cdot \arccos\dfrac{61}{128}+\dfrac{2\pi}{3}\bigg)+\dfrac{5}{3}\\ \end{aligned}$$

$$\begin{aligned} \\x_3&=t_3-\dfrac{b}{3a}\\ & =\dfrac{8}{3} \cos \bigg( \dfrac{1}{3}\cdot \arccos\dfrac{61}{128}+\dfrac{4\pi}{3} \bigg)+\dfrac{5}{3}\\ \end{aligned}$$

5. Summary

In summary, we have tried the method of cubic root formula, trigonometric to explore the solutions of the equation. The cubic equation $$x³ - 5x² + 3x + 2=0$$ is found to have three real roots . Exact values and approximations are given below.

$$\begin{cases} x_1=\dfrac{8}{3}\cos \bigg(\dfrac{1}{3}\cdot \arccos\dfrac{61}{128}\bigg)+\dfrac{5}{3} \\ x_2=\dfrac{8}{3} \cos \bigg( \dfrac{1}{3}\cdot \arccos\dfrac{61}{128}+\dfrac{2\pi}{3}\bigg)+\dfrac{5}{3} \\ x_3=\dfrac{8}{3} \cos \bigg( \dfrac{1}{3}\cdot \arccos\dfrac{61}{128}+\dfrac{4\pi}{3} \bigg)+\dfrac{5}{3} \end{cases}$$

Convert to decimals,

$$\begin{cases} x_1=4.1642479384602 \\ x_2=-0.3913823806309 \\ x_3=1.2271344421707 \end{cases}$$

6. Graph for the function $$f(x) = x³ - 5x² + 3x + 2$$

Since the discriminat is less than zero, the curve of the cubic function $$f(x) = x³ - 5x² + 3x + 2$$ has 3 intersection points with horizontal axis.

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