﻿ Solve 2x^3+6x^2+6x+1=0 | Uniteasy.com

Solve the cubic equation:

$$2x^3+6x^2+6x+1=0$$

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

$$\Delta=0.0625$$

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

In decimals,

$$\begin{cases} x_1=-0.2062994740159 \\ x_2=-1.5+0.86602540378444i \\ x_3=-1.5-0.86602540378444i \end{cases}$$

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

$$p = \dfrac{3\cdot 2\cdot 6-6^2}{3\cdot 2^2}=0$$

$$q = \dfrac{2\cdot 6^3-9\cdot2\cdot 6\cdot 6+27\cdot 2^2\cdot1}{27\cdot 2^3}=-\dfrac{1}{2}$$

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 $$2x³ + 6x² + 6x + 1=0$$ is transformed to

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

For the equation $$t^3 -\dfrac{1}{2}$$, we have $$p=0$$ and $$q = -\dfrac{1}{2}$$

Calculate the discriminant

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

\begin{aligned} \\\Delta&=\dfrac{q^2}{4}+\dfrac{p^3}{27}\\ & =\dfrac{\Big(-\dfrac{1}{2}\Big)^2}{4}+0\\ & =\dfrac{1}{16}\\ & =0.0625\\ \end{aligned}

1.1 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.

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

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[3]{\dfrac{1}{4}+\sqrt{\dfrac{1}{16}}}+\sqrt[3]{\dfrac{1}{4}-\sqrt{\dfrac{1}{16}}}\\ & =\sqrt[3]{\dfrac{1}{4}+\dfrac{1}{4}}+\sqrt[3]{\dfrac{1}{4}-\dfrac{1}{4}}\\ & =\sqrt[3]{\dfrac{1}{2}}\\ \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,

$$\sqrt[3]{R} = \sqrt[3]{\dfrac{1}{2}}$$, $$\sqrt[3]{\overline{R}} =0$$

\begin{aligned} \\t_2&= ω\cdotp \sqrt[3]{R}+ \overline{ω} \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\\ & =\dfrac{1}{2}\Big(-\sqrt[3]{\dfrac{1}{2}}-0\Big)+\dfrac{\sqrt{3}}{2}\Big(\sqrt[3]{\dfrac{1}{2}}-0\Big)i\\ & =-\dfrac{1}{2}\sqrt[3]{\dfrac{1}{2}}+\dfrac{1}{2}\sqrt[3]{\dfrac{1}{2}}\sqrt{3}i\\ \end{aligned}

\begin{aligned} \\t_3&= \overline{ω}\cdotp \sqrt[3]{R}+ ω\cdotp \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 \\ & =\dfrac{1}{2}\Big(-\sqrt[3]{\dfrac{1}{2}}-0\Big)-\dfrac{\sqrt{3}}{2}\Big(\sqrt[3]{\dfrac{1}{2}}-0\Big)i\\ & =-\dfrac{1}{2}\sqrt[3]{\dfrac{1}{2}}-\dfrac{1}{2}\sqrt[3]{\dfrac{1}{2}}\sqrt{3}i\\ \end{aligned}

Roots of the general cubic equation

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

$$x_1 = t_1-1$$

$$x_2 = t_2-1$$

$$x_3 = t_3-1$$

2. Summary

In summary, we have tried the method of cubic root formula to explore the solutions of the equation. The cubic equation $$2x³ + 6x² + 6x + 1=0$$ is found to have one real root and two complex roots. Exact values and approximations are given below.

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

in decimal notation,

$$\begin{cases} x_1=-0.2062994740159 \\ x_2=-1.5+0.86602540378444i \\ x_3=-1.5-0.86602540378444i \end{cases}$$

3. Graph for the function $$f(x) = 2x³ + 6x² + 6x + 1$$

Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = 2x³ + 6x² + 6x + 1$$ has one intersection point with the x-axis.

Scroll to Top