Understanding Cubic Equation Formula for Polynomials of Degree Three

[gevni]

Hi, Can someone please help me in understanding few parameters of cubic equation formula for solving polynomial of degree three. I attached the formula in the screenshot.
My questions are:
(1) what is ". " dot in the end of the formula and what does it mean?
(2) I want to use it only for real number not complex, so do the formula remain the same?
(3) It states the solution is for this kind of equation\[ ax^3+ax^2+cx+d=0 \] what if I have equation like that \[ ax^3+ax^2-cx+d=0 \]? Do i have to put negative value with c parameter in the below formula or whatever the case formula remain same?

(4) It is just one formula then how I can get three roots by using this?
(5) I want to use this formula for getting the minimum possible value of x and if it results in 3 roots then which root show minimum value of x.

I am very thankful if someone help me understanding this formula or any resource that can help me understanding this.

 

[cbarker1]

Hi, Can someone please help me in understanding few parameters of cubic equation formula for solving polynomial of degree three. I attached the formula in the screenshot.
My questions are:
(1) what is ". " dot in the end of the formula and what does it mean?
(2) I want to use it only for real number not complex, so do the formula remain the same?
(3) It states the solution is for this kind of equation\[ ax^3+ax^2+cx+d=0 \] what if I have equation like that \[ ax^3+ax^2-cx+d=0 \]? Do i have to put negative value with c parameter in the below formula or whatever the case formula remain same?

(4) It is just one formula then how I can get three roots by using this?
(5) I want to use this formula for getting the minimum possible value of x and if it results in 3 roots then which root show minimum value of x.

I am very thankful if someone help me understanding this formula or any resource that can help me understanding this.

View attachment 9772

The "." means that the sentence is finished. I have been studying the cubic formula for several years. So the formula remain the same. To see what is happening in the formula, it is recommended to derive the formula. Good news, I have a derivation on the cubic formula. I will provide the derivation to you:

Let $ax^3+bx^2+cx+d=0$ where $a\neq0$

 

[gevni]

The "." means that the sentence is finished. I have been studying the cubic formula for several years. So the formula remain the same. To see what is happening in the formula, it is recommended to derive the formula. Good news, I have a derivation on the cubic formula. I will provide the derivation to you:

Let $ax^3+bx^2+cx+d=0$ where $a\neq0$

Thank you for reply. The part of your post is not displayed which shows the derivation steps. Can you please post them again

 

[cbarker1]

I was in the middle of that. However, it was deleted. So sorry for that.

 

[gevni]

I was in the middle of that. However, it was deleted. So sorry for that.

No problem but can you please let me know that the same formula that was shown for \[ ax^3+bx^2+cx+d=0 \] we can use for \[ ax^3+bx^2-cx+d=0 \]

 

[cbarker1]

The "." means that the sentence is finished. I have been studying the cubic formula for several years. So the formula remain the same. To see what is happening in the formula, it is recommended to derive the formula. Good news, I have a derivation on the cubic formula. I will provide the derivation to you:

Let $ax^3+bx^2+cx+d=0$ where $a\neq0$ and $a,b,c,d$ are arbitrary real numbers. Divide $a$ through the previous equation (since the roots of the previous equation does not change if it is modified by the division): $$x^3+b'x^2+c'x+d'=0 \, \text{where} \, b'=\frac{b}{a}, c'=\frac{c}{a}, \text{and}\,\, d'=\frac{d}{a} \, \text{(1).}$$ By introduction of a new unknown this equation can be simplified, moreover, so that (1) will not have a second power of the unknown. To do what I have described I need to set $x=y+k$ with $k$ still arbitrary. By Taylor's formula (look up in a calculus textbook if you don't know), I will use the first four terms of the formula to determine what the value of $k$: So let $f(x)=x^3+b'x^2+c'x+d'$. $f(y+k)=f(k)+f'(k)y+\frac{f''(k)}{2}y^2+\frac{f'''(k)}{6}y^3$ and $f(k)=k^3+b'k^2+c'k+d'$, $f'(k)=3k^2+2b'k+c'$, $\frac{1}{2}f''(k)=3k+b'$, $\frac{1}{6} f'''(k)=1$. To get rid of the term involving $y^2$, it is suffices to choose $k$ so that $3k+b'=0 \implies k=\frac{-b'}{3}$. We will plugin the value of $k$ into the following functions $f$ and $f'$ and it yields the following values:
$$f\left(\frac{-b'}{3}\right)=d'-\frac{b'c'}{3}+\frac{2{b'}^2}{27} \, \, \text{and} \, \, f'\left(\frac{-b'}{3}\right)=c'-\frac{{b'}^2}{3}.$$ So the substitution for the elimination of $y^2$ is $x=y-\frac{b'}{3}.$ Then the equation (1) is transformed into the following equation through the substitution: $$y^3+py+q=0\, \text{(2)} \, \text{where} \, \, p=c'-\frac{{b'}^2}{3} \, \text{and} \, q= d'-\frac{b'c'}{3}+\frac{2{b'}^2}{27}.$$ A cubic equation of the form (2) can be solved by means of the following device: We seek to satisfy it by setting $y=u+v$, thus introducing two variables $u$ and $v$. On substituting this expression into (2) and arranging terms in a proper way, $u$ and $v$ have to satisfy the equation (let's called it (3): $$u^3+v^3+(p+3uv)(u+v)+q=0,$$ with two variables. This problem is indeterminate unless we find another relationship between $u$ and $v$. This is the relationship that we will be using $3uv+p=0 \implies uv=\frac{-p}{3}$. Then, it follows from (3) that $u^3+v^3=-q$, so that the solution of the cubic (2) can be obtained by solving the system of two equation:$$\begin{cases} u^3+v^3=-q \\ uv=\frac{-p}{3} \end{cases}$$ Taking the cube to the second equation, we will have $u^3v^3=\frac{-p^3}{27}$ and so, from the system of equations and the previous equation, we know the sum and product of the two unknown quantities $u^3$ and $v^3$. (How? There is a formula called the Viete's formula that describes how the roots of a polynomial is related to the coefficient of the same polynomial). These quantities are the roots of the quadratic equation: $t^2+qt-\frac{p^3}{27}=0$. We will separate the roots of the quadratic into $A$ and $B$, respectively: $$A=-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}} \,\text{and}\, B=-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}} $$ where we at liberty to the square root as we please. Now owing to the symmetry between the terms $u^3$ and $v^3$ in the systems of equation, we can set $u^3=A$ and $v^3=B$. If some determined value of the cube root of $A$ is denoted $\sqrt[3]{A}$, the three possible values of $u$ will be $u=\sqrt[3]{A}$, $u=\omega\sqrt[3]{A}$, $u={\omega}^2\sqrt[3]{A}$, where $\omega=\frac{-1+i\sqrt{3}}{2}$ is an imaginary cube root of unity. As to $v$, it will have also three values: $v=\sqrt[3]{B}$, $v=\omega\sqrt[3]{B}$, $v={\omega}^2\sqrt[3]{A}$ but not every one of them can be associated with three possible values of $u$, since $u$ and $v$ must satisfy the relation $uv=-\frac{p}{3}$. If $\sqrt[3]{B}$ stands for that cube root of $B$ which satisfies the relation $\sqrt[3]{A}\sqrt[3]{B}=-\frac{p}{3}$, then the values of $v$ that can be associated with $u=\sqrt[3]{A}$, $u=\omega\sqrt[3]{A}$, $u={\omega}^2\sqrt[3]{A}$ will be $v=\sqrt[3]{B}$, $v=\omega\sqrt[3]{B}$, $v={\omega}^2\sqrt[3]{A}$. Hence, equation (2) will have the following roots: $y_1=\sqrt[3]{A}+\sqrt[3]{B}$, $y_2=\omega\sqrt[3]{A}+{\omega}^2\sqrt[3]{B}$, and $y_3=\omega\sqrt[3]{B}+{\omega}^2\sqrt[3]{A}$. After you find what is the value $A$ and $B$, you back-substitute the $y$s into $x=y-\frac{b'}{3}$ in order to fine the value of $x$.

I will post the discussion about the solutions (Discriminant and how to calculate a special case of the sum of two cube roots that leads to an integer answer as well as irreducible case and how to find the roots in that case.) from the book listed below.
Theory of Equation by Uspensky

 

[cbarker1]

This is from Theory of Equations by J.V Upsensky. Copyright in 1948 by Mcgraw-Hill Book Company Inc in New York City. Page 86 starts at the second to the bottom to 87 to the bottom then 88 and so on.

 

[topsquark]

No problem but can you please let me know that the same formula that was shown for \[ ax^3+bx^2+cx+d=0 \] we can use for \[ ax^3+bx^2-cx+d=0 \]

In a word, just replace c with -c in the formula.

-Dan

 

[gevni]

Let $ax^3+bx^2+cx+d=0$ where $a\neq0$ and $a,b,c,d$ are arbitrary real numbers. Divide $a$ through the previous equation (since the roots of the previous equation does not change if it is modified by the division): $$x^3+b'x^2+c'x+d'=0 \, \text{where} \, b'=\frac{b}{a}, c'=\frac{c}{a}, \text{and}\,\, d'=\frac{d}{a} \, \text{(1).}$$ By introduction of a new unknown this equation can be simplified, moreover, so that (1) will not have a second power of the unknown. To do what I have described I need to set $x=y+k$ with $k$ still arbitrary. By Taylor's formula (look up in a calculus textbook if you don't know), I will use the first four terms of the formula to determine what the value of $k$: So let $f(x)=x^3+b'x^2+c'x+d'$. $f(y+k)=f(k)+f'(k)y+\frac{f''(k)}{2}y^2+\frac{f'''(k)}{6}y^3$ and $f(k)=k^3+b'k^2+c'k+d'$, $f'(k)=3k^2+2b'k+c'$, $\frac{1}{2}f''(k)=3k+b'$, $\frac{1}{6} f'''(k)=1$. To get rid of the term involving $y^2$, it is suffices to choose $k$ so that $3k+b'=0 \implies k=\frac{-b'}{3}$. We will plugin the value of $k$ into the following functions $f$ and $f'$ and it yields the following values:
$$f\left(\frac{-b'}{3}\right)=d'-\frac{b'c'}{3}+\frac{2{b'}^2}{27} \, \, \text{and} \, \, f'\left(\frac{-b'}{3}\right)=c'-\frac{{b'}^2}{3}.$$ So the substitution for the elimination of $y^2$ is $x=y-\frac{b'}{3}.$ Then the equation (1) is transformed into the following equation through the substitution: $$y^3+py+q=0\, \text{(2)} \, \text{where} \, \, p=c'-\frac{{b'}^2}{3} \, \text{and} \, q= d'-\frac{b'c'}{3}+\frac{2{b'}^2}{27}.$$ A cubic equation of the form (2) can be solved by means of the following device: We seek to satisfy it by setting $y=u+v$, thus introducing two variables $u$ and $v$. On substituting this expression into (2) and arranging terms in a proper way, $u$ and $v$ have to satisfy the equation (let's called it (3): $$u^3+v^3+(p+3uv)(u+v)+q=0,$$ with two variables. This problem is indeterminate unless we find another relationship between $u$ and $v$. This is the relationship that we will be using $3uv+p=0 \implies uv=\frac{-p}{3}$. Then, it follows from (3) that $u^3+v^3=-q$, so that the solution of the cubic (2) can be obtained by solving the system of two equation:$$\begin{cases} u^3+v^3=-q \\ uv=\frac{-p}{3} \end{cases}$$ Taking the cube to the second equation, we will have $u^3v^3=\frac{-p^3}{27}$ and so, from the system of equations and the previous equation, we know the sum and product of the two unknown quantities $u^3$ and $v^3$. (How? There is a formula called the Viete's formula that describes how the roots of a polynomial is related to the coefficient of the same polynomial). These quantities are the roots of the quadratic equation: $t^2+qt-\frac{p^3}{27}=0$. We will separate the roots of the quadratic into $A$ and $B$, respectively: $$A=-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}} \,\text{and}\, B=-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}} $$ where we at liberty to the square root as we please. Now owing to the symmetry between the terms $u^3$ and $v^3$ in the systems of equation, we can set $u^3=A$ and $v^3=B$. If some determined value of the cube root of $A$ is denoted $\sqrt[3]{A}$, the three possible values of $u$ will be $u=\sqrt[3]{A}$, $u=\omega\sqrt[3]{A}$, $u={\omega}^2\sqrt[3]{A}$, where $\omega=\frac{-1+i\sqrt{3}}{2}$ is an imaginary cube root of unity. As to $v$, it will have also three values: $v=\sqrt[3]{B}$, $v=\omega\sqrt[3]{B}$, $v={\omega}^2\sqrt[3]{A}$ but not every one of them can be associated with three possible values of $u$, since $u$ and $v$ must satisfy the relation $uv=-\frac{p}{3}$. If $\sqrt[3]{B}$ stands for that cube root of $B$ which satisfies the relation $\sqrt[3]{A}\sqrt[3]{B}=-\frac{p}{3}$, then the values of $v$ that can be associated with $u=\sqrt[3]{A}$, $u=\omega\sqrt[3]{A}$, $u={\omega}^2\sqrt[3]{A}$ will be $v=\sqrt[3]{B}$, $v=\omega\sqrt[3]{B}$, $v={\omega}^2\sqrt[3]{A}$. Hence, equation (2) will have the following roots: $y_1=\sqrt[3]{A}+\sqrt[3]{B}$, $y_2=\omega\sqrt[3]{A}+{\omega}^2\sqrt[3]{B}$, and $y_3=\omega\sqrt[3]{B}+{\omega}^2\sqrt[3]{A}$. After you find what is the value $A$ and $B$, you back-substitute the $y$s into $x=y-\frac{b'}{3}$ in order to fine the value of $x$.

I will post the discussion about the solutions (Discriminant and how to calculate a special case of the sum of two cube roots that leads to an integer answer as well as irreducible case and how to find the roots in that case.) from the book listed below.
Theory of Equation by Uspensky

Thank you for clarifying it.

 

[Theia]

(2) I want to use it only for real number not complex, so do the formula remain the same?

You can use the same formula, but you probably can't avoid using complex numbers with this formula. If the cubic equation has 3 real roots, the formula requires taking cube root of a complex number. If you want to use only real numbers, you'll need other approach to solve those cubic equations.

(4) It is just one formula then how I can get three roots by using this?

After you've found a root, you can divide the cubic and then use quadratic formula to solve the rest 2 roots.

(5) I want to use this formula for getting the minimum possible value of x and if it results in 3 roots then which root show minimum value of x.

Unfortunately it's not that straightforward. You need to do some kind of analysis beforehand to find out, which method is the most effective. I suggest you to take a look into the English wikipedia. Link here.