# How to determine if a polynomial is of a particular order: 3rd degree (cubic), 4th degree (quartic) etc.

I am working on a math puzzle that results in the answer setting up a pair of equations for corresponding sides of similar triangles, then solving the first for y and substituting in the second that gives an equation with a single unknown, like this:

$$x - \frac {x} {\sqrt {16 - x^2}} - \frac {x} {\sqrt {9 - x^2}} = 0$$

Now the trick is to solve for $x$. But it has been pointed out to me that this equation is a quartic. OK, there are lots of places on the 'net that can solve the roots of quartics, no problem, as computers have come a long way. But how do I, when given a polynomial of this type, deduce that it is a cubic or a quartic or even a quintic, and solve for it but not have the coefficients of the general form? Since this equation has no $x^4$th term in it, how do I know that I'm dealing with a quartic? How can I manipulate this equation to get to the general form of $ax^4 + bx^3 + cx^2 +dx + e = 0$, thus having numbers for $a, b, c, d$, and $e$? (My TI-89 using nSOLVE gives the answer as $\pm 2.60328775442$, and $0$ for $x$, thus giving me only 3 solutions, not 4, making me think that it is a cubic)

If anyone would like the complete puzzle to see what I am working on, please ask, I am happy to supply the puzzle!

Thanks for any help!!

• $x=0$ is a solution. Setting this possibility aside and dividing through by $x$ you have two square roots to eliminate, and doing this gives you (at most) a quartic. You can see that you will get an even function (if $x$ is a solution then so is $-x$) but your solutions may not all be real. You will also have to check that the solutions of the quartic do solve the original equation, as squaring can add additional solutions, which would be picked out by choosing a negative opposite sign for the square root. – Mark Bennet Jun 22 '12 at 7:12
• I don't have time to write a post but it seems that there is only one real solution: the trivial one $x=0$. All the others are complex and if my calculation are correct you reach a quartic after substituting $x^2$. To visualize solutions you could also plot your equation. – rubik Jun 22 '12 at 7:24
• I'd love to see the original puzzle, certainly! – Steven Stadnicki Jun 22 '12 at 7:25
• @rubik, after dividing through by $x$ the left side is positive for $x=0$ and negative for $x$ just less than 3, so surely there's a real solution between 0 and 3. – Gerry Myerson Jun 22 '12 at 7:29
• Oh my bad sorry. I had copied the problem when it was still non-latex and I copied it wrong! I had $x$ instead $x^2$ inside the square roots... – rubik Jun 22 '12 at 9:18

First of all, it's not really correct to say that your equation 'is' a quartic, or a polynomial at all; 'polynomial' is reserved for terms of the form $\sum_{i=0}^n a_ix^i$ — that is, those that involve only positive whole powers of the variable. It would be more accurate to say that the roots of your equation are also roots of a polynomial equation, and that they can be found that way.

Your equation does in fact lead to a quartic, but the root at 0 you're seeing is 'spurious' in some sense, and it's possible that the other two roots of the quartic aren't actually roots of your initial equation. The most straightforward of finding out what polynomial equation your roots satisfy is to clear the square roots out of your expression:

• Firstly, $x=0$ is obviously a solution, since all three terms have an x in them; dividing out by this $x$ (and thus eliminating the zero solution) yields $1-1/\sqrt{16-x^2}-1/\sqrt{9-x^2}=0$.
• Now, we can set $y=x^2$ to get the equation $1-1/\sqrt{16-y}-1/\sqrt{9-y}=0$.
• Move the last term to the right to get $1-1/\sqrt{16-y} = 1/\sqrt{9-y}$.
• Now square both sides: $\left(1-1/\sqrt{16-y}\right)^2 = 1/(9-y)$.
• Expand the square on the left side: $1-2/\sqrt{16-y}+1/(16-y) = 1/(9-y)$.
• Shuffle around terms so there's just the remaining square root term on one side: $1-1/(9-y)+1/(16-y) = 2/\sqrt{16-y}$.
• Square both sides: $\left(1-1/(9-y)+1/(16-y)\right)^2 = 4/(16-y)$.
• Multiply out the left side: $\displaystyle{1+\frac{1}{(9-y)^2}+\frac{1}{(16-y)^2}-\frac{2}{9-y}-\frac{2}{16-y}-\frac{2}{(9-y)(16-y)} = \frac{4}{16-y}}$.
• Multiply both sides by $(9-y)^2(16-y)^2$: $(9-y)^2(16-y)^2+(16-y)^2+(9-y)^2-2(9-y)(16-y)^2-2(9-y)^2(16-y)-2(9-y)(16-y) = 4(9-y)^2(16-y)$.

I'll let you do the rest of the algebra from here, but you can see that this gives you a quartic in $y$ (which, if you can recall, is $x^2$). Not all of the roots of this quartic will necessarily be roots of the original equation, because we had to square our equation multiple times, so be careful! Of course, just because the equation is a quartic doesn't mean it's a general quartic, and it's very possible that some alternate formulation of the problem will give a much easier solution...

• In fact the quartic in $y$ is ${y}^{4}-46\,{y}^{3}+763\,{y}^{2}-5374\,y+13585$, which is irreducible over the rationals. It has two positive real roots, but only one of them gives solutions of the original equation. – Robert Israel Jun 22 '12 at 7:30

Since the given equation

$$\begin{equation*} x-\frac{x}{\sqrt{16-x^{2}}}-\frac{x}{\sqrt{9-x^{2}}}=0 \end{equation*}\tag{A}$$

is equivalent to $$\begin{equation*} x\underset{R}{\underbrace{\left( 1-\frac{1}{\sqrt{16-x^{2}}}-\frac{1}{\sqrt{9-x^{2}}}\right) }}=0,\tag{B} \end{equation*}$$

$x=0$ is a root. The other roots are the roots of $R=0$. To illustrate the more general case we write it as $$\begin{equation*} 1-\frac{1}{\sqrt{P}}-\frac{1}{\sqrt{Q}}=0, \end{equation*}\tag{C}$$ where $$\begin{equation*} P=16-x^{2},\qquad Q=9-x^{2}. \end{equation*}\tag{D}$$

The usual method to solve (C) is

1. isolate one of the irrational terms on one side,

2. square both sides,$^1$

3. repeat until you have only rational terms,

4. transform the rational equation into a polynomial one.

For instance, starting with $-\frac{1}{\sqrt{P}}$ on the LHS and squaring, we get successively

$$\begin{eqnarray*}-\frac{1}{\sqrt{P}}&=&\frac{1}{\sqrt{Q}}-1\Rightarrow \frac{1}{P} =\frac{1}{Q}-\frac{2}{\sqrt{Q}}+1,\qquad\text{(steps 1 and 2)}\\&\Leftrightarrow &\frac{2}{ \sqrt{Q}}=\frac{1}{Q}-\frac{1}{P}+1, \qquad\text{(steps 3 and 1) }\\ &\Rightarrow &\frac{4}{Q}=\frac{\left( P-Q+PQ\right) ^{2}}{P^{2}Q^{2}}, \qquad\text{(step 2)}\\&\Leftrightarrow &\frac{\left( P-Q+PQ\right) ^{2}}{P^{2}Q}=4,\qquad Q\neq 0, \\ &\Leftrightarrow &\left( P-Q+PQ\right) ^{2}-4P^{2}Q=0, \qquad\qquad\text{(step 4)}\qquad P,Q\neq 0,\\ &\Leftrightarrow &P^{2}Q^{2}+P^{2}-2PQ+Q^{2}-2P^{2}Q-2PQ^{2}=0,\qquad P,Q\neq 0.\tag{E} \end{eqnarray*}$$

The degree of the last equation is defined by the 8th degree term: $$\begin{equation*} P^{2}Q^{2}=\left( 16-x^{2}\right) ^{2}\left( 9-x^{2}\right) ^{2}\end{equation*}\tag{F}$$

Substituting $P,Q$ and expanding yields the 8th. degree equation $$\begin{equation*} x^{8}-46x^{6}+763x^{4}-5374x^{2}+13\,585=0.\tag{G} \end{equation*}$$

However this particular equation can be reduced to a quartic (4th. degree) equation in $y=x^{2}$, which we could have expected because $R$ is a function of $x^2$
$$\begin{equation*} y^{4}-46y^{3}+763y^{2}-5374y+13\,585=0.\tag{H} \end{equation*}$$

This equation has two real roots $$\begin{equation*} y_{1}\approx 6.7771,\qquad y_{2}\approx 8.4627. \end{equation*}$$ The corresponding $x$ values are thus

$$x_{1,2}\approx \pm \sqrt{6.7771}\approx 2.603\,3,\qquad x_{3,4}\approx \pm \sqrt{8.4627}\approx \pm 2.9091.$$

Among these, $x_{3,4}$ are not solutions of the original equation (A) because

$$\begin{equation*} 1-\frac{1}{\sqrt{16-y_{2}^2}}-\frac{1}{\sqrt{9-y_{2}^2}}\ne 0. \end{equation*}$$

--

$^1$ Raising both sides of an equation $A=B$ to the $n^{\text{th}}$ power yields a new equation $A^n=B^n$, which has all the solutions of the given equation and may admit other solutions, the so called extraneous solutions.

Case $n=2$. Since $$A^{2}-B^{2}=(A-B)(A+B),$$ the equation $$A^{2}-B^{2}=0\Leftrightarrow A^{2}=B^{2}$$ means that $A=B$ or $A=-B.$ The roots of the equation $A^2=B^2$ are the roots of the equation $A=B$ and the roots of the equation $A=-B$. In general the roots of equation $A=-B$ are different from the roots of the equation $A=B$, so when we square $A=B$ we may generate other solutions.

• A fourth-degree equation should not be called "quadratic"... – GEdgar Jun 23 '12 at 15:32
• @GEdgar Thanks! I will fix it. – Américo Tavares Jun 23 '12 at 15:34