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# Solve the cubic equation:

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

Quick Answer

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

$$\Delta=0.0625$$

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

In decimals,

$$\begin{cases} x_1=-0.7937005259841 \\ x_2=0.5+0.86602540378444i \\ x_3=0.5-0.86602540378444i \end{cases}$$

Detailed Steps on Solution

## 1. Use the sum of cubes identity

This cubic equation $$2x³ + 1=0$$ contains only a cubic term and a constant term. It might be the simplest form of a cubic equation.

To simplify, divide both sides of the equation by $$2$$ so that the coefficient of the leading term is equat to $$1$$.

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

Isolate the leading term by moving the constant term to the right hand side.

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

and then take cubic roots on the both sides, which results in

$$x =\sqrt{\dfrac{1}{2}}$$

If the solution is to find only real root, that's all. However, if it is to find all solutions, need more steps since a cubic eqaution normally has three roots.(Some may have duplicate ones)

### Factorization by sum of cubes identity

To find complete roots, use the sum of cubes formula to factorize the binomial

$$\Big(x+\sqrt{\dfrac{1}{2}}\Big)\Big[(x)^2-x\cdot \sqrt{\dfrac{1}{2}}+(\sqrt{\dfrac{1}{2}})^2\Big] = 0$$

Now the binomial on the left hand side is transformed to a product of two expressions. If the above equation is to hold true, either of the expressions is equal to zero.

Then we get a linear equation

$$x+\sqrt{\dfrac{1}{2}}=0$$

and a quadratic equation.

$$(x)^2-x\cdot \sqrt{\dfrac{1}{2}}+(\sqrt{\dfrac{1}{2}})^2=0$$

### Solve the linear equation $$x+\sqrt{\dfrac{1}{2}}=0$$

Subtract the value of the constant term from both sides yields the first solution.

$$x_1=-\sqrt[3]{\dfrac{1}{2}}\approx-0.7937005259841$$

### Solve the quadratic equation $$(x)^2-x\cdot \sqrt{\dfrac{1}{2}}+(\sqrt{\dfrac{1}{2}})^2$$

Let $$m = -\sqrt[3]{\dfrac{1}{2}}$$, then the equation is simplified as

$$x^2+mx+m^2= 0$$

Using the root identity for a quadratic equation, $$x$$ is found to be the product of $$m$$ and a complex number.

\begin{aligned}\\ x&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a} \\ &=\dfrac{-m\pm m\sqrt{3}i}{2} \\ \end{aligned}

Substitute the value of $$m$$, that is, $$m=-\sqrt[3]{\dfrac{1}{2}}$$, then we get another two complex roots

$$x_2 =\dfrac{1}{2}\sqrt[3]{\dfrac{1}{2}}\Big(1+ \sqrt{3}i\Big)$$

$$x_3 =\dfrac{1}{2}\sqrt[3]{\dfrac{1}{2}}\Big(1- \sqrt{3}i\Big)$$

## 2. Cubic Root Formula

The equation $$2x³ + 1=0$$ has no quadratic term, compared with the general cubic equation. We can use the root formula to calculate the roots direvtly.

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

For the equation , 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}

### 2.1 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,

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

\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[0-\Big(-\sqrt[3]{\dfrac{1}{2}}\Big)\Big]+\dfrac{\sqrt{3}}{2}\Big[0-\Big(-\sqrt[3]{\dfrac{1}{2}}\Big)\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[0-\Big(-\sqrt[3]{\dfrac{1}{2}}\Big)\Big]-\dfrac{\sqrt{3}}{2}\Big[0-\Big(-\sqrt[3]{\dfrac{1}{2}}\Big)\Big]i\\ & =\dfrac{1}{2}\sqrt[3]{\dfrac{1}{2}}-\dfrac{1}{2}\sqrt[3]{\dfrac{1}{2}}\sqrt{3}i\\ \end{aligned}

## 3. Summary

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

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

in decimal notation,

$$\begin{cases} x_1=-0.7937005259841 \\ x_2=0.5+0.86602540378444i \\ x_3=0.5-0.86602540378444i \end{cases}$$

## 4. Graph for the function $$f(x) = 2x³ + 1$$

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

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