1. Jun 1, 2014

### Bluskyz

Just over two years ago, I was introduced to the process of completing the square as a way to solve the roots in a quadratic equation. More recently, I've thought about how I could go about extending this to completing the cube. The following short story/proof is the result of during just that and then going even further. As always, criticism and comment is welcome. I've tried looking around to see if anybody has published anything similar to this but haven't been able to find anything.

I'll begin by reviewing the method of completing the square with the general quadratic $\text{ax}^2+\text{bx}+c$ and thus deriving the quadratic formula.
We then set the equation equal to zero (because we are finding its zeros) and divide everything by the first term's coefficient a: $x^2+\frac{\text{bx}}{a}=-\frac{c}{a}$
At this point, we see that we must look at the expanded form of the general squared binomial $\text{(x+n)}^2$ where $\text{n}$ is some constant which gives $x^2+2 \text{nx}+n^2$
In order to complete the square, we need to get $x^2+\frac{\text{bx}}{a}$ to look like that expanded form. As the first term, $x^2$, in both equations is exactly the same, nothing must be done to them. However, in the $x^2+\frac{\text{bx}}{a}$, we are missing the constant term $n^2$ and as such, we must find this value of $n$. By comparing the second terms of both equations, $\frac{\text{bx}}{a}$ and $2 \text{nx}$, we can set these equal to each other and solve for $\text{n}$: $\frac{\text{bx}}{a}=2 \text{nx}$. Through algebraic manipulation, $\text{n}$ can be found to equal $\frac{\text{b}}{\text{2a}}$.

Now going back to the original $x^2+\frac{\text{bx}}{a}=-\frac{c}{a}$, we can add the square of $\frac{\text{b}}{\text{2a}}$ to both sides yielding $x^2+\frac{\text{bx}}{a}+\frac{b^2}{4 a^2}=-\frac{c}{a}+\frac{b^2}{4 a^2}$.
Next, we collapse the left side of the equation into $\left(x+\frac{b}{2 a}\right)^2$ and add the two terms on the right side of the equation yielding $\frac{b^2-4 \text{ac}}{4 a^2}$. By taking the square root of both sides, we get $x+\frac{b}{2 a}=\pm \sqrt{\frac{b^2-4 \text{ac}}{4 a^2}}$. The $4a^2$ in the fraction can be taken out of the square root to give $x+\frac{b}{2 a}=\pm \frac{\sqrt{b^2-4 \text{ac}}}{2 a}$. Finally, we can subtract $\frac{\text{b}}{\text{2a}}$ giving us the ever famous quadratic equation $x=\frac{-b\pm \sqrt{b^2-4 \text{ac}}}{2 a}$

Next, I moved onto a cubic equation in an attempt to "complete the cube" and find a cubic equivalent quadratic equation. I started with the general form of a cubic: $\text{ax}^3+\text{bx}^2+\text{cx}+d$
As with completing the square, I continued to move the constant, $d$, to the other side of the equation yielding $\text{ax}^3+\text{bx}^2+\text{cx}=-d$
I then divided everything by the coefficient of the $x^3$ term, $a$, and got $x^3+\frac{\text{bx}^2}{a}+\frac{\text{cx}}{a}=-\frac{d}{a}$. Now, this time I looked at the expanded form of $\text{(x+n)}^2$ which works out to be $x^3+3 x^2 n+3 \text{xn}^2+n^3$ As with the quadratic, we can look at the second and third terms of both equations to find the value of $n$ for the cubic equation. Setting both equal to each other, we get:
$\frac{\text{bx}^2}{a}=3 x^2 n$

and

$\frac{\text{cx}}{a}=3 \text{xn}^2$


Solving for $n$ in both equations, we find that

$n=\frac{b}{3 a}$

and

$n=\sqrt{\frac{c}{3 a}}$

With this, we have two different answers for $n$ and this is where the problems for completing the cube start to arise. If we were to pick just one of two values for $n$ and add it's cube to both sides (just as we added $n^2$ to both sides of the quadratic) and attempt to solve for $x$, we would find that the $x$ can't be isolated to just one side and thus a simple solution to the completing the cube can not be found. I will quickly attempt using $n=\frac{b}{3 a}$ to demonstrate the problem. After some algebraic manipulation, I found that the equation simplifies to $\left(x+\frac{b}{3 a}\right)^3=\frac{x \left(b^2 x-3 \text{ac}\right)}{3a^2}+\frac{b^3}{27 a^3}-\frac{d}{a}$. Because we have $x$'s on both sides, a nice solution can not be found. This is due to the fact that picking and using just one of the values for $n$ keeps the other term that needs the other value for $n$ unsatisfied. In the last example, I had to add $\frac{x \left(b^2 x-3 \text{ac}\right)}{3a^2}$ to both sides in order to complete the $x$ term and get the left to collapse into $\left(x+\frac{b}{3 a}\right)^3$. In order to eliminate this conflict, we need to find some value of $n$ that satisfies both the $x^2$ term and the $x$ term. To achieve this, the value of $n$ must equal both $\frac{b}{3 a}$ and $\sqrt{\frac{c}{3 a}}$ at the same time. Because $n$ must be equal to both of those expressions, the two expressions must be equal to each other: $\frac{b}{3 a}=\sqrt{\frac{c}{3 a}}$ I decided to solve for the variable $c$ and found that $c=\frac{b^2}{3 a}$. This ratio that we just found tells us the ratio the variables $a$, $b$ and $c$ must have between one another in order to allow for this method of completing the cube to satisfy both terms and work correctly. If I plug this ratio into $\text{ax}^3+\text{bx}^2+\text{cx}+d$ for $c$ and try to complete the cube now, the solution works perfectly. After solving for x, I found that the cubic version of the quadratic equation is $x=\frac{-b+ \sqrt[3]{b^3-27 a^2 d}}{3 a}$. And thus, this equation is guaranteed to give you one real root of a cubic equation so long as the coefficients of the equation satisfy to ratio $c=\frac{b^2}{3 a}$.

And with this, I still wasn't content. I decided to attempt to find the ultimate "quadratic equation" that works for all polynomials to the $f$th degree. Before doing any further math, I sat down and thought about the problem. Just as the quadratic had one value for $n$ and the cubic equation had two possible values for $n$, I knew that an equation of the $f$th degree would have $f-1$ possible values for $n$, only further restricting the domain of equations for which the ultimate equation would work correctly. Because of this fact, the final generalized equation won't be very practical as it will only find a maximum of two roots for any polynomial and it will only work for polynomials that satisfy the required ratios of coefficients. Nevertheless, I kept working towards an answer as i knew that it would work for polynomials over the 5th degree and I thought it would be interesting to know the generalized extension of the quadratic equation.

And so I continued toward the answer by attempting to find the generalized "quadratic equation" for a polynomial to the $n$th degree ( From now on, the portions of my work that were labeled with $n$ i.e. $n=\frac{b}{3a}$, will instead be labeled with $p$. Moreover, $n$ will now assume the variable that denotes the degree of a polynomial) The required ratios that the coefficients of the polynomials must hold will be found later in the post. By using the same process that I used to derive the quadratic and cubic equations above, I was able to also derive an equation that works for quartics. It is as follows: $x=\frac{-b\pm \sqrt[4]{b^4-256 a^3 e}}{4 a}$. I encourage you to work this out on your own if you are curious, however, you might want to wait until I explain how to find the required ratios later in the post. By examining the patterns that exist between the three equations that we have found, I've found that the final generalized equation for a polynomial to the $n$th degree is $x=\frac{-b+ \sqrt[n]{b^n-(n^n) (a^(n-1)) c }}{n a}$ where $c$ is the constant that exists in the polynomial. The constant in the equation $\text{ax}^3+\text{bx}^2+\text{cx}+d$, for example, would be the variable $d$. Keep in mind that if $n$ were an even number, you would need to $\pm$ the $\sqrt[n]{}$. Now, this equation, as I explained earlier, only works for polynomials whose coefficients satisfy certain ratios. I will now begin to explain the final bit of my work in which I explain how to find these required ratios.

This last bit requires us to look back at our method for determining the coefficients for the second and third degree polynomials. With the cubic, for example, we would look at our two equations $x^3+\frac{\text{bx}^2}{a}+\frac{\text{cx}}{a}=-\frac{d}{a}$ and $x^3+3 x^2 n+3 \text{xn}^2+n^3$. In order to determine the coefficient ratio for the third term, the $x$ term, we set both the second and third terms equal to their corresponding terms in the other equation. Thus:

$\frac{\text{bx}^2}{a}=3 x^2 n$

and

$\frac{\text{cx}}{a}=3 \text{xn}^2$


Solving for $n$, we get

$n=\frac{b}{3 a}$

and

$n=\sqrt{\frac{c}{3 a}}$

respectively. At this point, we took the first equation $n=\frac{b}{3 a}$ and plugged it in for $n$ in the second equation $n=\sqrt{\frac{c}{3 a}}$. We now solved for $c$ to yield the required ratio for the coefficient of the third term in terms of $a$ and $b$ : $c=\frac{b^2}{3 a}$. We can now do the same for a quartic equation in order to gather more information to help in discovering the patterns that lie within the equations.

With a quartic equation we have our generalized quartic $x^4+\frac{\text{bx}^3}{a}+\frac{\text{cx}^2}{a}+\frac{\text{dx}}{a}=-\frac{e}{a}$ and our expanded binomial to the 4th power $x^4+4 x^3 n+6 x^2 n^2+4 x n^3+n^4$. Just as before, we set their corresponding terms equal and solve for $n$ to give us

$n=\frac{b}{4 a}$

$n=\sqrt{\frac{c}{6 a}}$

and

$n=\sqrt[3]{\frac{d}{4 a}}$

Just as before, we take the first equation $n=\frac{b}{4 a}$ and plug that in for the next two equations. After solving for $c$ and $d$ separately in the two equations we find that

$c=\frac{6 b^2}{16 a}$

and

$d=\frac{4 b^3}{64 a^2}$

These two ratios we just found now tell us the required ratios that $c$ and $d$ must hold in a quartic equation in order to work correctly in our generalized quadratic equation that we found above. Notice that I did not reduce the numbers in the fractions for they will later help us to find patterns in the ratios. If you were to sit down and do the same for a polynomial to the 5th degree, you would find that

$c=\frac{10 b^2}{25 a}$

$d=\frac{10 b^3}{125 a^2}$

and

$e=\frac{5 b^4}{625 a^3}$

Now, arranging all the ratios that we have found so far into a table, we get this:

Now, we can finally begin finding patterns between the equations in an attempt to find a generalized ratio requirement equation. First, notice that $b$ is always raised to the power equal to the degree of the polynomial, $n$, minus the exponent of the term it is describing (we will call this value $k$). For the $x$ term of a polynomial of the 5th degree, for example, the degree is 5 and the $x$ is raised to the 1st power. Thus $b$ should be raised to the $5-1$ power which gives us the $b^4$ we see in the table. This process can be described as $b^{n-k}$. Also take note that the exponent to which the variable $a$ is raised is very similar. The only difference is that $a$ is always raised to $b$'s exponent minus 1. It then follows that the variables $b$ and $a$ in the equations can be described as $a \left(\frac{b}{a}\right)^{n-k}$. Now, we will attempt to integrate the coefficient in the numerator into the equation that is taking form. In short, the coefficient of the numerator can always be described as the binomial coefficient indexed by $n$ and $k$. Thus we obtain $a \binom{n}{k} \left(\frac{b}{a}\right)^{n-k}$ by added that into our equation. All that is left is to describe the coefficient in the denominator. If we look back at one of our equations and work backwards, we can better understand how to describe this coefficient. Take $d=\frac{4 b^3}{64 a^2}$ for example. This was found by solving $\frac{b}{4 a}=\sqrt[3]{\frac{d}{4 a}}$ for $d$. The coefficient in the denominator for $\frac{b}{4 a}$ is 4. It just so happens that this coefficient will always be equivalent to degree of the polynomial you are attempting to solve. The equation $\frac{b}{5 a}=\sqrt[3]{\frac{d}{10 a}}$ is used to find the ratio $d=\frac{10 b^3}{125 a^2}$. In that equation, the coefficient of $\frac{b}{5 a}$ is 5, the degree of a fifth degree polynomial. As such, we can add the single variable $n$ into our equation $a \binom{n}{k} \left(\frac{b}{a}\right)^{n-k}$ in the denominator to obtain the final equation $a \binom{n}{k} \left(\frac{b}{na}\right)^{n-k}$. And with that, we can see that this equation precisely describes the ratios that must exist in the polynomial in order for the generalized quadratic equation to work. Given the degree of the polynomial $n$ you are working with and the exponent $k$ to which the $x$ in the term is raised, you can now find the exact ratio between $a$ and $b$ that must accompany that term. Lets test the equation with an example we have already done, a polynomial of the forth degree and the $x^2$ term. Plugging in 4 for $n$ and 2 for $k$, we get $a \binom{4}{2} \left(\frac{b}{4a}\right)^{4-2}$. This then simplifies to
$6 a \left(\frac{b}{4a}\right)^{2}$
$6 a \left(\frac{b^2}{16 a^2}\right)$

and finally,

$\frac{6 b^2}{16 a}$

Thus, we have obtained the required ratio that must exist for the $x^2$ term, a ratio that we had previously found and confirmed by manipulating our equations. Our final completely generalized quadratic formula is as follows:

$x=\frac{-b+ \sqrt[n]{b^n-(n^n) (a^(n-1)) c }}{n a}$

where the coefficient of the $x^k$ term must satisfy $a \binom{n}{k} \left(\frac{b}{na}\right)^{n-k}$.

And with that, I have finished explaining my little discovery. If you were able to sit through my paragraphs of explanation to get to this point, I applaud you. I can only hope now that you will comment on my work and perhaps take it even farther. If this has already been done and documented, then I apologize for attempting to take credit, please let me know. I plan on doing a bit more with the problem, specifically with the ratios that must exist. I have a feeling that there are further options for the ratios other than simply the relationship between $a$ and $b$. Notice that if you attempt to use this equation with a normal quadratic, the generalized quadratic formula, $x=\frac{-b+ \sqrt[n]{b^n-(n^n) (a^(n-1)) c }}{n a}$, turns simply into the regular quadratic formula. Also note that if you were to attempt to find the ratios for the terms $x^n$ and $x^{n-1}$, they end up being just $a$ and $b$ respectively, thus having no restrictions for the values of $a$ and $b$; they can be chosen arbitrarily. $a$ can not be 0, of course, as that would make the generalized quadratic formula undefined. This solution doesn't have much practicality but I find it interesting none the less. It completely describes the workings of the normal quadratic equation but also much more. Furthermore, it works for polynomials to the $n$th degree but can only find up to a maximum of two roots. If you find any mistakes in my work, don't hesitate to let me know. Thank you.

2. Jun 1, 2014

### Matterwave

I didn't get through your whole post, but there is, in fact a cubic formula, it's just really annoying:

http://en.wikipedia.org/wiki/Cubic_equation#General_formula_for_roots

For higher than cubic equations, I don't think there are general formulas for the roots anymore. I think the fact that that no such formula exists for n>3 (for maybe some other integer higher than 3, I can't remember) is proven.

It's possible that you found a way to find 2 roots, provided the coefficients satisfy some complicated ratios among themselves, but this should be quite a small subset of all potential equations of degree n.

3. Jun 1, 2014

### pwsnafu

This forum is not the place to post original research.

Further, there does not exist a general solution to quintics or higher due to the Abel-Ruffini theorem. Eg $x^5+x-1=0$
The necessary and sufficient requirement for solving in radicals is that the Galois group is solvable, but I don't believe you are at a level where we can explain what that entails.

Last edited: Jun 1, 2014
4. Jun 1, 2014

### micromass

Staff Emeritus
Indeed.