# Number of roots

1. Jun 3, 2014

### Jhenrique

If x^5 has 5 roots, if x^3 has 3 roots and if x^10 has 10 roots, so how many roots has x^3.14 ?

2. Jun 3, 2014

### pwsnafu

It has one root with multiplicity 5.

It has one root with multiplicity 3.

It has one root with multiplicity 10.

One, and the root is also a branch point.

3. Jun 3, 2014

### Jhenrique

Why one and what's a branch point?

4. Jun 3, 2014

### symbolipoint

Jhenrique is referring most likely to the degree of a function and not to just specific functions.

5. Jun 3, 2014

### pwsnafu

The only solution to $x^{3.14}=0$ is $x=0$. There is no other choice.

A branch point is a point where, on the complex plane, traveling around that point is dependent on the direction of travel. Visually, the function $x^{3.14}$ "tears" the complex plane from the point $x=0$.

Even then the statement "a polynomial of degree 10 has 10 roots" is still wrong. The correct form is "a polynomial with degree 10 has 10 roots counting multiplicity", which is a sloppy way of saying "the sum of the multiplicities of the roots of a polynomial of degree 10 is equal to 10".

Also Jhenrique, the fundamental theorem of algebra is a theorem about polynomials, $x^{3.14}$ is not a polynomial. You can't expect the pattern to continue.

Last edited: Jun 3, 2014
6. Jun 3, 2014

### pwsnafu

Let me explain at length.

Suppose we have a polynomial $p(x)$. Suppose that the degree is $n$. This means
$p(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$.

Now suppose that the roots of $p(x)$ are $x_1, x_2, \ldots, x_j$, each one different. Now the "number of roots" is $j$. The fundamental theorem does not say $j = n$.

Each root has what we call "multiplicity", I'm going to denote as $m_1, m_2, \ldots, m_j$.
This means when we factor
$p(x) = (x-x_1)^{m_1} (x-x_2)^{m_2} \ldots (x-x_j)^{m_j}$
The fundamental theorem states that
1. $j \neq 0$, and
2. $m_1 + m_2 + \ldots+ m_j = n$.
This is what "the number of roots counting multiplicity" means.

7. Jun 4, 2014

### HallsofIvy

For n a positive integer, and a non zero, $$x^n= a$$ has n distinct roots. If n is NOT an integer it has infinitely many roots.

(don't know what you mean by "x^5 has 5 roots". Equations have roots, not polynomials.)

8. Jun 4, 2014

### gopher_p

Are you saying that there are infinitely many complex numbers satisfying $x^\frac{1}{2}=1$?

9. Jun 4, 2014

### Mentallic

Yep!

$$x = e^{4i\pi n }$$ for all integers n. Granted, they are all the same complex numbers.

10. Jun 4, 2014

### jbunniii

Isn't that rather like saying that there are infinitely many numbers equal to zero, since $n - n = 0$ for all integers $n$?

11. Jun 4, 2014

### Mentallic

I honestly don't understand it either

12. Jun 4, 2014

### gopher_p

I think it's more like saying that every linear polynomial $mx+n$ with integer coefficients has infinitely many rational roots, $\{-\frac{nk}{mk}\}_{k\in\mathbb{Z}\setminus\{0\}}$.

13. Jun 6, 2014

### Boorglar

Actually, infinitely many roots occur only when the exponent is irrational.
For example, $x^{√2} = 1$ has infinitely many (complex) solutions.

Indeed, let $x = e^{i\pi√2n}$ for any integer n.
All those solutions are distinct, because otherwise $i\pi√2m = i\pi√2n + 2k\pi i$ or
$√2=\frac{2k}{m-n}$ would be rational, which is false.

Rational exponents have finitely many solutions because then the sequence becomes periodic, when n is a multiple of the denominator.

14. Jun 7, 2014

### HallsofIvy

Yes, as many have pointed out, I misspoke. With rational power, $x^{m/n}= a$, with m/n reduced to lowest terms, has m roots. It is only if the power is irrational, $x^\alpha= a$, that there are an infinite number of roots,

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