﻿ Solve 2x^3-6x^2+5x+3=0 | Uniteasy.com

# Solve the cubic equation:

## $$2x^3-6x^2+5x+3=0$$

Since the discriminant $$\Delta >0$$, the cubic equation has one real root and two conjugate complex roots.

$$\Delta=0.99537037037037$$

$$\begin{cases} x_1=\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}+\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}+1 \\ x_2=\dfrac{1}{2}\Big(-\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)+\dfrac{\sqrt{3}}{2}\Big(\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)i+1 \\ x_3=\dfrac{1}{2}\Big(-\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)-\dfrac{\sqrt{3}}{2}\Big(\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)i+1 \end{cases}$$

In decimals,

$$\begin{cases} x_1=-0.39176877223553 \\ x_2=1.6958843861178+0.97609694012813i \\ x_3=1.6958843861178-0.97609694012813i \end{cases}$$

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

$$p = \dfrac{3\cdot 2\cdot 5-(-6)^2}{3\cdot 2^2}=-\dfrac{1}{2}$$

$$q = \dfrac{2\cdot (-6)^3-9\cdot2\cdot (-6)\cdot 5+27\cdot 2^2\cdot3}{27\cdot 2^3}=2$$

### 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 $$x=t+1$$

The cubic equation $$2x³ - 6x² + 5x + 3=0$$ is transformed to

$$t^3 -\dfrac{1}{2}t+2=0$$

## 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)

$$\begin{cases} 3uv=-\dfrac{1}{2}\quad\text{or}\quad v=-\dfrac{1}{6u}\\ v^3-u^3=2\\ \end{cases}$$

$$v=-\dfrac{1}{6u}$$ gives relationship between the two variables. Substitute the value of $$v$$ to the second equation

$$\Big(-\dfrac{1}{6u}\Big)^3-u^3=2$$

Simplifying gives,

$$u^3+\dfrac{1}{216}\dfrac{1}{u^3}+2=0$$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 $$v^3=2+u^3$$.

$$m^2+2m+\dfrac{1}{216}=0$$

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&=-1+\dfrac{1}{2}\sqrt{\Big(2^2\Big)-4\cdot \dfrac{1}{216}}\\ & =-1+\dfrac{1}{2}\sqrt{4-\dfrac{1}{54}}\\ & =-1+\dfrac{1}{2}\sqrt{\dfrac{215}{54}}\\ & =-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}\\ \end{aligned}

$$v^3$$ can be determined by the equation we deduced $$v^3-u^3=2$$. Then,

\begin{aligned} \\v^3&=2+u^3\\ & =2-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}\\ & =1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}\\ \end{aligned}

Now we have,

$$u^3=-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}$$ and $$v^3=1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}$$

Evaluating the simplest cubic equation $$x^3-A=0$$, it has 3 roots, in which the first root is a real 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, $$\dfrac{1}{ω}=\overline{ω}$$.

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

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

$$\begin{cases} u_1=\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\\ u_2=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\\ u_3=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\\ \end{cases}$$

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 $$v=-\dfrac{1}{6u}$$. 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.

$$\begin{cases} v_1=\sqrt{1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\\ v_2=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt{1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\\ v_3=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt{1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\\ \end{cases}$$

Verification for the redicand in $$v$$.

\begin{aligned} \\v_1&=-\dfrac{1}{6u_1}\\ & =-\dfrac{1}{6}\cdot \dfrac{1}{\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}}\\ & =-\dfrac{1}{6}\cdot \dfrac{1}{\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}}\cdot \dfrac{\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}}{\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}}\\ & =-\dfrac{1}{6}\cdot \dfrac{\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}}{\sqrt{\Big(-1\Big)^2-\Big(\dfrac{1}{6}\sqrt{\dfrac{215}{6}}\Big)^2}}\\ & =-\dfrac{1}{6}\cdot \dfrac{\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}}{\sqrt{1-\dfrac{215}{216}}}\\ & =-\dfrac{1}{6}\cdot \dfrac{\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}}{\sqrt{\dfrac{1^3}{6^3}}}\\ & =-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\\ \end{aligned}

Since $$x=u-v$$, combining the real and imaginary parts gives 3 results for $$t$$

\begin{aligned} \\t_1&=u_1-v_1\\ & =\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}+\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\\ \end{aligned}

\begin{aligned} \\t_2&=u_2-v_2\\ & =\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\Big(\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt{1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)\\ & =\dfrac{1}{2}\Big(-\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)+\dfrac{\sqrt{3}}{2}\Big(\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)i\\ \end{aligned}

\begin{aligned} \\t_3&=u_3-v_3\\ & =\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\Big(\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt{1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)\\ & =\dfrac{1}{2}\Big(-\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)+\dfrac{\sqrt{3}}{2}\Big(-\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}+\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)i\\ \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. $$t=u+\dfrac{1}{6u}$$. And then substitute the equation to the cubic equation $$t^3-\dfrac{1}{2}t+2=0$$. 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 $$t=u+\dfrac{1}{6u}$$ to the cubic equation

$$\Big(u+\dfrac{1}{6u}\Big)^3-\dfrac{1}{2}\Big(u+\dfrac{1}{6u}\Big)+2=0$$

Expand brackets and cancel the like terms

$$u^3+\cancel{\dfrac{1}{2}u^2\dfrac{1}{u}}+\cancel{\dfrac{1}{12}u\dfrac{1}{u^2}}+\dfrac{1}{216}\dfrac{1}{u^3}-\cancel{\dfrac{1}{2}u}-\cancel{\dfrac{1}{12}\dfrac{1}{u}}+2=0$$

Then we get the same equation as (2)

$$u^3+\dfrac{1}{216}\dfrac{1}{u^3}+2=0$$

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

## $$t^3-\dfrac{1}{2}t+2=0$$

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

$$t^3=\dfrac{1}{2}t-2$$3

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

$$t=\sqrt{r_1}+\sqrt{r_2}$$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)

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

Substituting, the equation is simplified to

$$t^3=3\sqrt{β}t+\alpha$$

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

$$\begin{cases} 3\sqrt{β}=\dfrac{1}{2}\\ \alpha=-2\\ \end{cases}$$

Solving for $$β$$ gives

$$β=\dfrac{1}{216}$$

So the quadratic equation (5) is determined as

$$z^2+2z+\dfrac{1}{216}=0$$6

$$\begin{cases} r_1=-1+\dfrac{\sqrt{1290}}{36}\approx-0.0023175002184461\\ r_2=-1-\dfrac{\sqrt{1290}}{36}\approx-1.9976824997816\\ \end{cases}$$

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

$$t_1=\sqrt{-1+\dfrac{\sqrt{1290}}{36}}+\sqrt{-1-\dfrac{\sqrt{1290}}{36}}$$

in decimals,

$$t_1=-1.3917687722355$$

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

The other two roots could be determined as,

$$t_2=\dfrac{-1+i\sqrt{3}}{2}\sqrt{-1+\dfrac{\sqrt{1290}}{36}}+\dfrac{-1-i\sqrt{3}}{2}\sqrt{-1-\dfrac{\sqrt{1290}}{36}}$$

$$t_3=\dfrac{-1-i\sqrt{3}}{2}\sqrt{-1+\dfrac{\sqrt{1290}}{36}}+\dfrac{-1+i\sqrt{3}}{2}\sqrt{-1-\dfrac{\sqrt{1290}}{36}}$$

Combining the real and imaginary parts results in the same result as that obtained by Cardano's solution.

For the equation $$t^3 -\dfrac{1}{2}t+2$$, we have $$p=-\dfrac{1}{2}$$ and $$q = 2$$

### 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. However, there's the scenario like this one the discriminant is greater than zero when $$p$$ is negative.

\begin{aligned} \\\Delta&=\dfrac{q^2}{4}+\dfrac{p^3}{27}\\ & =\dfrac{2^2}{4}+\dfrac{\Big(-\dfrac{1}{2}\Big)^3}{27}\\ & =1-\dfrac{1}{216}\\ & =\dfrac{1\cdot 216-1\cdot 1}{216}\\ & =0.99537037037037\\ \end{aligned}

### 4.1 Use the root formula directly

If the discriminant is greater than zero, we can use the root formula to determine the roots of the cubic equation.

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

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

Substitute the values of $$p, q$$ and $$\Delta$$ which we have calculated. Then,

\begin{aligned} \\t_1&=\sqrt{-1+\sqrt{\dfrac{215}{216}}}+\sqrt{-1-\sqrt{\dfrac{215}{216}}}\\ & =\sqrt{-1+\sqrt{\dfrac{215}{6\cdot 6^2}}}+\sqrt{-1-\sqrt{\dfrac{215}{6\cdot 6^2}}}\\ & =\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}+\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\\ \end{aligned}

If we denote

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

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

then,

$$\sqrt{R} = \sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}$$, $$\sqrt{\overline{R}} =\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}$$

\begin{aligned} \\t_2&= ω\cdotp \sqrt{R}+ \overline{ω} \sqrt{\overline{R} }\\ & =\dfrac{-\sqrt{R}-\sqrt{\overline{R} }}{2} +\dfrac{\sqrt{3}( \sqrt{R} - \sqrt{\overline{R} }) }{2} i\\ & =\dfrac{1}{2}\Big(-\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)\\&+\dfrac{\sqrt{3}}{2}\Big(\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)i\\ \end{aligned}

\begin{aligned} \\t_3&= \overline{ω}\cdotp \sqrt{R}+ ω\cdotp \sqrt{\overline{R}}\\ & =\dfrac{-\sqrt{R}-\sqrt{\overline{R} }}{2} +\dfrac{\sqrt{3}(- \sqrt{R} + \sqrt{\overline{R} }) }{2}i \\ & =\dfrac{1}{2}\Big(-\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)\\&-\dfrac{\sqrt{3}}{2}\Big(\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)i\\ \end{aligned}

## Roots of the general cubic equation

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

$$x_1 = t_1+1$$

$$x_2 = t_2+1$$

$$x_3 = t_3+1$$

## 5. Summary

In summary, we have tried the method of cubic root formula to explore the solutions of the equation. The cubic equation $$2x³ - 6x² + 5x + 3=0$$ is found to have one real root and two complex roots. Exact values and approximations are given below.

$$\begin{cases} x_1=\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}+\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}+1 \\ x_2=\dfrac{1}{2}\Big(-\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)+\dfrac{\sqrt{3}}{2}\Big(\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)i+1 \\ x_3=\dfrac{1}{2}\Big(-\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)-\dfrac{\sqrt{3}}{2}\Big(\sqrt{-1+\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}-\sqrt{-1-\dfrac{1}{6}\sqrt{\dfrac{215}{6}}}\Big)i+1 \end{cases}$$

in decimal notation,

$$\begin{cases} x_1=-0.39176877223553 \\ x_2=1.6958843861178+0.97609694012813i \\ x_3=1.6958843861178-0.97609694012813i \end{cases}$$

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

Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = 2x³ - 6x² + 5x + 3$$ has one intersection point with the x-axis.

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