# Solve the cubic equation:

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

**Quick Answer**

Since the discriminant $$ \Delta=0$$, the cubic equation has three real roots, of which two are equal.

$$ \Delta=0$$

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

In decimals,

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

**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 $$f(x) = 2x³ - 3x² + 1$$ 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 $$2x³ - 3x² + 1$$ 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.

$$f(1) = 2(1)³ - 3(1)² + 1 = 0$$

then, $$x = 1$$ is one of roots of the cubic equaiton $$2x³ - 3x² + 1 = 0$$. 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 $$2x³ - 3x² + 1$$ by $$x - 1$$

2x² | -x | -1 | |||

x - 1 | 2x³ | -3x² | 0x | +1 | |

2x³ | -2x² | ||||

-x² | 0x | ||||

-x² | +x | ||||

-x | +1 | ||||

-x | +1 | ||||

0 |

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

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

### Find factors of trinomial

Create a table of all factors for constant of $$-1$$

$$\begin{array}{|c|} \hline1\times( -1)\\ \hline(-1)\times 1\\ \hline\end{array}$$

Create another table of all factors for coeffcient of quadratic term $$2$$

$$\begin{array}{|c|} \hline1\times2\\ \hline(-1)\times (-2)\\ \hline\end{array}$$

### Determine the roots by factorizing

Cross multiply the pairs from constant and leading term to find pairs of which the sum of products is equal to the coefficient of linear term 2

$$\begin{array}{|c|} \hline1\times1 + 2\times(-1) = -1\\ \hline\end{array}$$

The pair $$1\times1 + 2\times(-1) = -1$$ is found to satisfy the rule.

Then, the quadratic equaiton is factored as

$$(x - 1)(2x + 1) = 0$$

which results in another two roots

$$x_2 = 1$$

$$x_3 = -\dfrac{1}{2}$$

That is,

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

Another method to find the roots of a general cubic equation is simplifing it to a depressed form. This method is applicable to all cases, especially to those difficult to find factors.

## 2. 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

$$p = \dfrac{3\cdot 2\cdot 0-(-3)^2}{3\cdot 2^2}=-\dfrac{3}{4}$$

$$q = \dfrac{2\cdot (-3)^3-9\cdot2\cdot (-3)\cdot 0+27\cdot 2^2\cdot1}{27\cdot 2^3}=\dfrac{1}{4}$$

### 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 $$x=t+\dfrac{1}{2}$$

The cubic equation $$2x³ - 3x² + 1=0$$ is transformed to

$$t^3 -\dfrac{3}{4}t+\dfrac{1}{4}=0$$

## 3. 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)

$$\begin{cases} 3uv=-\dfrac{3}{4}\quad\text{or}\quad v=-\dfrac{1}{4u}\\ v^3-u^3=\dfrac{1}{4}\\ \end{cases}$$

$$v=-\dfrac{1}{4u}$$ gives relationship between the two variables. Substitute the value of $$v$$ to the second equation

$$\Big(-\dfrac{1}{4u}\Big)^3-u^3=\dfrac{1}{4}$$

Simplifying gives,

$$u^3+\dfrac{1}{64}\dfrac{1}{u^3}+\dfrac{1}{4}=0$$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 $$v^3=\dfrac{1}{4}+u^3$$.

$$m^2+\dfrac{1}{4}m+\dfrac{1}{64}=0$$

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.

$$\begin{aligned} \\u^3=m&=-\dfrac{1}{8}+\dfrac{1}{2}\sqrt{\Big(\dfrac{1}{4}^2\Big)-4\cdot \dfrac{1}{64}}\\ & =-\dfrac{1}{8}+\dfrac{1}{2}\sqrt{\dfrac{1}{16}-\dfrac{1}{16}}\\ & =-\dfrac{1}{8}\\ \end{aligned}$$

$$\begin{aligned} \\v^3&=\dfrac{1}{4}+u^3\\ & =\dfrac{1}{4}-\dfrac{1}{8}\\ & =\dfrac{1}{8}\\ \end{aligned}$$

Taking cubic root offor $$u^3$$ and $$v^3$$ gives 3 roots.

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

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

Subtracting $$v$$ from $$u$$ yields the roots of the equation.

$$\begin{aligned} \\t_1&=u_1-v_1\\ & =-\dfrac{1}{2}-\dfrac{1}{2}\\ & =-1\\ \end{aligned}$$

$$\begin{aligned} \\t_2&=u_2-v_2\\ & =\dfrac{-1+i\sqrt{3}}{2}\cdot -\dfrac{1}{2}-\dfrac{-1-i\sqrt{3}}{2}\cdot \dfrac{1}{2}\\ & =\dfrac{1}{2}\\ \end{aligned}$$

$$\begin{aligned} \\t_3&=u_3-v_3\\ & =\dfrac{-1-i\sqrt{3}}{2}\cdot -\dfrac{1}{2}-\dfrac{-1+i\sqrt{3}}{2}\cdot \dfrac{1}{2}\\ & =\dfrac{1}{2}\\ \end{aligned}$$

## 4. 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. $$t=u+\dfrac{1}{4u}$$. And then substitute the equation to the cubic equation $$t^3-\dfrac{3}{4}t+\dfrac{1}{4}=0$$. 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 $$t=u+\dfrac{1}{4u}$$ to the cubic equation

$$\Big(u+\dfrac{1}{4u}\Big)^3-\dfrac{3}{4}\Big(u+\dfrac{1}{4u}\Big)+\dfrac{1}{4}=0$$

Expand brackets and cancel the like terms

$$u^3+\cancel{\dfrac{3}{4}u^2\dfrac{1}{u}}+\cancel{\dfrac{3}{16}u\dfrac{1}{u^2}}+\dfrac{1}{64}\dfrac{1}{u^3}-\cancel{\dfrac{3}{4}u}-\cancel{\dfrac{3}{16}\dfrac{1}{u}}+\dfrac{1}{4}=0$$

Then we get the same equation as (2)

$$u^3+\dfrac{1}{64}\dfrac{1}{u^3}+\dfrac{1}{4}=0$$

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

## 5. Euler's Solution

## $$t^3-\dfrac{3}{4}t+\dfrac{1}{4}=0$$

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

$$t^3=\dfrac{3}{4}t-\dfrac{1}{4} $$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

$$\begin{cases} 3\sqrt[3]{β}=\dfrac{3}{4}\\ \alpha=-\dfrac{1}{4}\\ \end{cases}$$

Solving for $$β$$ gives

$$β=\dfrac{1}{64} $$

So the quadratic equation (5) is determined as

$$z^2+\dfrac{1}{4}z+\dfrac{1}{64}=0$$6

Solving the quadratic equation yields

$$\begin{cases} r_1=-\dfrac{1}{8}\approx-0.125\\ r_2=-\dfrac{1}{8}\approx-0.125\\ \end{cases}$$

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

$$t_1=\sqrt[3]{-\dfrac{1}{8}}+\sqrt[3]{-\dfrac{1}{8}} $$

in decimals,

$$t_1=-1 $$

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

The other two roots could be determined as,

$$t_2=\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{-\dfrac{1}{8}}+\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{-\dfrac{1}{8}} $$

$$t_3=\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{-\dfrac{1}{8}}+\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{-\dfrac{1}{8}} $$

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

For the equation $$t^3 -\dfrac{3}{4}t+\dfrac{1}{4}$$, we have $$p=-\dfrac{3}{4}$$ and $$q = \dfrac{1}{4}$$

### 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.

$$\begin{aligned} \\\Delta&=\dfrac{q^2}{4}+\dfrac{p^3}{27}\\ & =\dfrac{\Big(\dfrac{1}{4}\Big)^2}{4}+\dfrac{\Big(-\dfrac{3}{4}\Big)^3}{27}\\ & =\dfrac{1}{64}-\dfrac{1}{64}\\ & =0\\ \end{aligned}$$

### 5.1 Use the root formula directly

If $$\Delta = 0$$, then the cubic equation has three real roots, of which two are equal.

Substitute the values of p and q to the root formula, then we get,

$$\begin{aligned} \\t_1&=\dfrac{3q}{p}\\ & =\dfrac{3\cdot \dfrac{1}{4}}{-\dfrac{3}{4}}\\ & =-1\\ \end{aligned}$$

$$\begin{aligned} \\t_2=t_3&=-\dfrac{3q}{2p}\\ & =\dfrac{3\cdot \dfrac{1}{4}}{2\cdot -\dfrac{3}{4}}\\ & =\dfrac{1}{2}\\ \end{aligned}$$

## Roots of the general cubic equation

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

$$\begin{aligned} \\x_1&=t_1-\dfrac{b}{3a}\\ & =-1+\dfrac{1}{2}\\ & =-\dfrac{1}{2}\\ \end{aligned}$$

$$\begin{aligned} \\x_2&=t_2-\dfrac{b}{3a}\\ & =\dfrac{1}{2}+\dfrac{1}{2}\\ & =1\\ \end{aligned}$$

$$\begin{aligned} \\x_3&=t_3-\dfrac{b}{3a}\\ & =\dfrac{1}{2}+\dfrac{1}{2}\\ & =1\\ \end{aligned}$$

## 6. Summary

In summary, we have tried the method of factorization, cubic root formula to explore the solutions of the equation. The cubic equation $$2x³ - 3x² + 1=0$$ is found to have three real roots, of which two are equal roots . Exact values and approximations are given below.

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

Convert to decimals,

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

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

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

Since the discriminat is equal to zero, the curve of the cubic function $$f(x) = 2x³ - 3x² + 1$$ has one intersection point with the x-axis.