Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If not, is there only a limited rangle of polynomials for which the root can be found?

Also, if $u=x^{\frac{3}{2}}+x$, is $x$ expressible in terms of $u$?

share|improve this question
    
What do you exactly mean by solvable? Expressible as radicals? –  Aryabhata Oct 8 '10 at 18:35
4  
It is important that you are very precise about what you mean by solvable. In your case write y = x^{1/2} and use the cubic formula. –  Qiaochu Yuan Oct 8 '10 at 18:37
7  
And also note that $x^{\frac{3}{2}}+x$ is not a polynomial in $x$ (though it is a polynomial in $\sqrt{x}$). –  Arturo Magidin Oct 8 '10 at 19:43
    
I am wondering if he is asking about the existence of the splitting fields of polynomials? –  awllower Mar 6 '11 at 2:54

3 Answers 3

up vote 12 down vote accepted

The answer to your question depends entirely on what "can be found" means. I will note, before we start, that the expression you have with fractional exponents is not generally considered a polynomial: a polynomial in the variable $x$, with real coefficients, is an expression of the form $$a_0 + a_1x + a_2x^2+ \cdots + a_nx^n$$ where $a_0,a_1,\ldots,a_n$ are real numbers, and $n$ is a nonnegative integer. Note that all the powers of the variable are integral powers. If $a_n\neq 0$, we say the polynomial has degree $n$.

When $n\leq 4$, then there are formulas that express all the roots of the polynomial in terms of the coefficients ($a_0$, $a_1,\ldots,a_n$). For $n=1$, that is a polynomial $a_0+a_1x$ with $a_1\neq 0$, the solution is simply $x=-\frac{a_0}{a_1}$. For $n=2$, that is a polynomial $a_0 + a_1x + a_2x^2$ with $a_2\neq 0$, you get the well-known quadratic formula that gives the two roots: $$ r_1 = \frac{-a_1 + \sqrt{a_1^2 - 4a_0a_2}}{2a_2}\quad\text{and}\quad r_2 = \frac{-a_1-\sqrt{a_1^2-4a_0a_2}}{2a_2}.$$ When $n=3$ and when $n=4$ there are also formulas to express all the roots of the polynomial in terms of the coefficients and using only the operations of addition, subtractions, multiplication, division, and root extraction. For $n=3$, these are the Cardano formulae, and for $n=4$ the solution is due to Ferrari.

There are no similar formulas when $n\geq 5$; this is the celebrated Abel-Ruffini theorem. This does not mean we have no way of finding the roots, just that there is no formula that applies to all polynomials that gives the roots in terms of the coefficients, using only certain kinds of operations. For example, quintic equations (degree 5 polynomials) can be solved using other more complicated kinds of functions and operations (theta functions).

On the other hand, there are plenty of methods for finding approximate values for roots of polynomials, or in fact values that are "as close as you want" to the roots of the polynomial. For example, Sturm's theorem can help you locate the roots approximately, and then you can combine it with Newton's method to find very good approximations to the roots of the polynomial. See for example the Wikipedia page on root-finding algorithms.

But of course, this turns on whether "finding an approximation" qualifies as "can be found" (or even if "can be found, theoretically, given enough time to work" qualifies as "can be found").

For your "also" question, you want to express $x$ in terms of $u$ if $u=x^{3/2}+x$. This is slightly different from "solving a polynomial"; you are really trying to find a formula for the inverse of $x^{3/2}+x$ (assuming it has one, which it does because the function is increasing). You can turn it into a problem of solving a polynomial because if you let $z=x^{1/2}$, then you have $u=z^3+z^2$, which is equivalent to $z^3+z^2-u=0$; this is a cubic polynomial, so the roots of the polynomial can be expressed in terms of the coefficients (in this case, $1$, $1$, $0$, and $-u$) using Cardano's formulae. Then you would replace $x^{1/2}$ for $z$ in the case in which $z$ is nonnegative and real, and squaring gives you the answer. So, yes, it can be done.

share|improve this answer

Yes, there is just a limited amount of polynomials for which we can find the exact, i.e. algebraic roots by some general formula (like it's possible for quadratic polynomials).

Particularly, the Abel-Ruffini-theorem states that there is no such general solution for polynomials of degree five or higher in terms of radicals.

This doesn't mean they don't have roots (in fact they always have), but one has to write them down or approximate them in other ways.


As to your equation (note that this is no longer a polynomial equation, which would require it's exponents to be integers, but a general power sum)

$$ u = x ^ {\frac{3}{2}} + x$$

We substitute $k = x^{\frac{1}{2}}$ and now have to solve a cubic (polynomial) equation

$$ k^3 + k^2 - u = 0 $$

For degree three, we can use a well-known but kinda unwieldy general solution formula and finally obtain zero to three results.

share|improve this answer
1  
@Dario: I disagree with your use of the phrase "finite representation." What you mean is that they don't have roots which can be expressed as radicals, but it is unclear to me why this counts as a "finite representation" while expressing roots in terms of, say, Bring radicals is not. This is a prejudice which has no basis in mathematical reality. –  Qiaochu Yuan Oct 8 '10 at 20:07
1  
@Dario: so as I see it, the only sensible definition of what it means to "know" the root of a polynomial is that one has an algorithm for computing any finite sequence of its digits in finite time, preferably an efficient one. And the existence of such an algorithm is not limited by the Abel-Ruffini theorem and has nothing to do with our narrow conception of what constitutes a "closed form number." –  Qiaochu Yuan Oct 8 '10 at 20:22
1  
@Dario: in fact, have you ever tried to use the cubic formula on an actual cubic polynomial? The first or second time I tried to do this I found I had to take the cube root of an imaginary number and I could only write the roots of the polynomial in terms of cos pi/9, which is itself the root of an irreducible cubic polynomial; so I end up not much better off than when I started! To find a representation of the roots of a real cubic polynomial (such as cos pi/9) in terms of real radicals is a much less trivial problem. –  Qiaochu Yuan Oct 8 '10 at 20:25
1  
@Dario: and regarding my cubic formula comment, which I can no longer edit, see sbseminar.wordpress.com/2008/03/30/… . The point I am trying to make here is that the cubic formula should not be seen as an analogue of the quadratic formula in that it does not give you the kind of answers you think it does. –  Qiaochu Yuan Oct 8 '10 at 20:34
1  
As a commenter mentioned in the blog post Qiaochu linked to, for the cubic case, whether a solution to the cubic can be expressed as radicals of real quantities depends on the discriminant of the cubic equation; for the so-called casus irreduciblis, you are essentially forced to use trigonometric functions (alternatively, hyperbolic functions) if you do not want to stray into the realm of $\mathbb{C}$, even if all the solutions are real. –  J. M. Oct 8 '10 at 23:03

Your equation can be solved: http://www.wolframalpha.com/input/?i=solve+u%3Dx%5E%283%2F2%29%2Bx+for+x

The mathematics that decides when a polynomial is solvable by radicals is called Galois Theory and is one of the hallmarks of modern algebra.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.