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# Solve the cubic equation:

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

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

$$\Delta=0.05787037037037$$

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

In decimals,

$$\begin{cases} x_1=-1 \\ x_2=0.5+0.5i \\ x_3=0.5-0.5i \end{cases}$$

Detailed Steps on Solution

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.

## 1. Factorization Method

Find all possible factors for constant $$1$$, which are,

$$1$$

$$-1$$

and all possible factors of the coefficient of the leading term $$2$$,

$$1, 2$$

$$-1, -2$$

Dividing factors of constant term by those of the cubic term one by one gives the following fractions.

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

Substitute the fractions to the function $$f(x) = 2x³ - x + 1$$ 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 $$2x³ - x + 1$$ has a factor $$x-n$$, that is, $$x=n$$ is the root of the equation.

Fortunetely, one of the numbers is found to make the equation equal.

$$f(-1) = 2(-1)³ - (-1) + 1 = 0$$

then, $$x = -1$$ is one of roots of the cubic equaiton $$2x³ - x + 1 = 0$$. And, the equation can be factored as

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

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

### Long division

Divide the polynomial $$2x³ - x + 1$$ by $$x + 1$$

 2x² -2x +1 x + 1 2x³ 0x² -x +1 2x³ +2x² -2x² -x -2x² -2x x +1 x +1 0

Now we get another factor of the cubic equation $$2x² - 2x + 1$$

## Solve the quadratic equation: $$2x² - 2x + 1 = 0$$

Given $$a =2, b=-2, c=1$$,

Use the root solution formula for a quadratic equation, the roots of the equation are given as

\begin{aligned} \\t&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}\\ & =\dfrac{-(-2)\pm\sqrt{(-2)^2-4\cdot 2\cdot 1}}{2 \cdot 2}\\ & =\dfrac{2\pm2\sqrt{-1}}{4}\\ & =\dfrac{1}{2}\pm\dfrac{1}{2}i\\ \end{aligned}

Since the discriminat is less than zero, we get another two complex roots:

That is,

$$\begin{cases} t_2 =\dfrac{1}{2}+\dfrac{1}{2}i \\ t_3=\dfrac{1}{2}-\dfrac{1}{2}i \end{cases}$$

## 2. Cardano's solution

Divide each term by $$2$$, the coefficient of the cubic term.

Then the cubic equation  is converted to,

Let $$x=u-v$$

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

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

Since $$u-v=x$$, 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)

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

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

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

Simplifying gives,

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

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

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&=-\dfrac{1}{4}+\dfrac{1}{2}\sqrt{\Big(\dfrac{1}{2}^2\Big)-4\cdot \dfrac{1}{216}}\\ & =-\dfrac{1}{4}+\dfrac{1}{2}\sqrt{\dfrac{1}{4}-\dfrac{1}{54}}\\ & =-\dfrac{1}{4}+\dfrac{1}{2}\sqrt{\dfrac{25}{108}}\\ & =-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}\\ \end{aligned}

$$v^3$$ can be determined by the equation we deduced $$v^3-u^3=\dfrac{1}{2}$$. Then,

\begin{aligned} \\v^3&=\dfrac{1}{2}+u^3\\ & =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}\\ & =\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}\\ \end{aligned}

Now we have,

$$u^3=-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}$$ and $$v^3=\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}$$

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.

$$\begin{cases} u_1=\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\\ u_2=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\\ u_3=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\\ \end{cases}$$

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.

$$\begin{cases} v_1=\sqrt{\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\\ v_2=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt{\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\\ v_3=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt{\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\\ \end{cases}$$

Verification for the redicand in $$v$$.

\begin{aligned} \\v_1&=-\dfrac{1}{6u_1}\\ & =-\dfrac{1}{6}\cdot \dfrac{1}{\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}}\\ & =-\dfrac{1}{6}\cdot \dfrac{1}{\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}}\cdot \dfrac{\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}}{\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}}\\ & =-\dfrac{1}{6}\cdot \dfrac{\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}}{\sqrt{\Big(-\dfrac{1}{4}\Big)^2-\Big(\dfrac{5}{12}\sqrt{\dfrac{1}{3}}\Big)^2}}\\ & =-\dfrac{1}{6}\cdot \dfrac{\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}}{\sqrt{\dfrac{1}{16}-\dfrac{25}{432}}}\\ & =-\dfrac{1}{6}\cdot \dfrac{\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}}{\sqrt{\dfrac{1^3}{6^3}}}\\ & =-\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\\ \end{aligned}

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

\begin{aligned} \\x_1&=u_1-v_1\\ & =\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}+\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\\ \end{aligned}

\begin{aligned} \\x_2&=u_2-v_2\\ & =\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}-\Big(\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt{\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)\\ & =\dfrac{1}{2}\Big(-\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}-\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)+\dfrac{\sqrt{3}}{2}\Big(\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}-\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)i\\ \end{aligned}

\begin{aligned} \\x_3&=u_3-v_3\\ & =\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}-\Big(\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt{\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)\\ & =\dfrac{1}{2}\Big(-\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}-\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)+\dfrac{\sqrt{3}}{2}\Big(-\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}+\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)i\\ \end{aligned}

## 3. Vieta's Substitution

In Cardano' solution, $$x$$ is defined as the difference of $$u$$ and $$v$$. If we substitute the value of $$v$$ (4) into (2), we get the equation. $$x=u+\dfrac{1}{6u}$$. And then substitute the equation to the cubic equation $$x^3-\dfrac{1}{2}x+\dfrac{1}{2}=0$$. 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.

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

Substitute the expression $$x=u+\dfrac{1}{6u}$$ to the cubic equation

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

Expand brackets and cancel the like terms

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

Then we get the same equation as (2)

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

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

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

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

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

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

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

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

Substituting, the equation is simplified to

$$x^3=3\sqrt{β}x+\alpha$$

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

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

Solving for $$β$$ gives

$$β=\dfrac{1}{216}$$

So the quadratic equation (5) is determined as

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

$$\begin{cases} r_1=-\dfrac{1}{4}+\dfrac{5}{36}\sqrt{3}\approx-0.009437387837656\\ r_2=-\dfrac{1}{4}-\dfrac{5}{36}\sqrt{3}\approx-0.49056261216234\\ \end{cases}$$

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

$$x_1=\sqrt{-\dfrac{1}{4}+\dfrac{5}{36}\sqrt{3}}+\sqrt{-\dfrac{1}{4}-\dfrac{5}{36}\sqrt{3}}$$

in decimals,

$$x_1=-1$$

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

The other two roots could be determined as,

$$x_2=\dfrac{-1+i\sqrt{3}}{2}\sqrt{-\dfrac{1}{4}+\dfrac{5}{36}\sqrt{3}}+\dfrac{-1-i\sqrt{3}}{2}\sqrt{-\dfrac{1}{4}-\dfrac{5}{36}\sqrt{3}}$$

$$x_3=\dfrac{-1-i\sqrt{3}}{2}\sqrt{-\dfrac{1}{4}+\dfrac{5}{36}\sqrt{3}}+\dfrac{-1+i\sqrt{3}}{2}\sqrt{-\dfrac{1}{4}-\dfrac{5}{36}\sqrt{3}}$$

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

## 5. Cubic Root Formula

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

$$x^3 +px+q=0$$

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

### 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{\Big(\dfrac{1}{2}\Big)^2}{4}+\dfrac{\Big(-\dfrac{1}{2}\Big)^3}{27}\\ & =\dfrac{1}{16}-\dfrac{1}{216}\\ & =\dfrac{1\cdot 27-1\cdot 2}{432}\\ & =0.05787037037037\\ \end{aligned}

### 5.1 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} \\x_1&=\sqrt{-\dfrac{1}{4}+\sqrt{\dfrac{25}{432}}}+\sqrt{-\dfrac{1}{4}-\sqrt{\dfrac{25}{432}}}\\ & =\sqrt{-\dfrac{1}{4}+\sqrt{\dfrac{5^2}{3\cdot 12^2}}}+\sqrt{-\dfrac{1}{4}-\sqrt{\dfrac{5^2}{3\cdot 12^2}}}\\ & =\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}+\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\\ \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,

$$\sqrt{R} = \sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}$$, $$\sqrt{\overline{R}} =\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}$$

\begin{aligned} \\x_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{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}-\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)\\&+\dfrac{\sqrt{3}}{2}\Big(\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}-\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)i\\ \end{aligned}

\begin{aligned} \\x_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{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}-\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)\\&-\dfrac{\sqrt{3}}{2}\Big(\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}-\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)i\\ \end{aligned}

## 6. Summary

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

$$\begin{cases} x_1=\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}+\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}} \\ x_2=\dfrac{1}{2}\Big(-\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}-\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)+\dfrac{\sqrt{3}}{2}\Big(\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}-\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)i \\ x_3=\dfrac{1}{2}\Big(-\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}-\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)-\dfrac{\sqrt{3}}{2}\Big(\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}-\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)i \end{cases}$$

in decimal notation,

$$\begin{cases} x_1=-1 \\ x_2=0.5+0.5i \\ x_3=0.5-0.5i \end{cases}$$

Using the method of factorization, the roots are derived to the following forms

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

The decimal results by using factorization method are

$$\begin{cases} x_1=-1 \\ x_2=0.5+0.5i \\ x_3=0.5-0.5i \end{cases}$$

## 7. Math problems derived by the cubic equation

It is found that the methods of factorizaiton and using cubic root formula yield the roots for the cubic equation in different forms. However, their deicmal results show that they are equal in values. Comparing the real roots, we get the equation.

$$\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}+\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}=-1$$

Similarly, comparison of the imaginary parts of complex roots yields the following equation.

$$\dfrac{\sqrt{3}}{2}\Big(\sqrt{-\dfrac{1}{4}+\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}-\sqrt{-\dfrac{1}{4}-\dfrac{5}{12}\sqrt{\dfrac{1}{3}}}\Big)=\sqrt{\dfrac{1}{2}}$$

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

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

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