What is difference between number of roots of a polynomial equation and number of solutions to a polynomial equation?

Do they mean same thing OR can an equation have different number of roots and different number of solutions?

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- Aug 1st 2009, 08:34 AMpankajRoots and Solutions
What is difference between number of roots of a polynomial equation and number of solutions to a polynomial equation?

Do they mean same thing OR can an equation have different number of roots and different number of solutions? - Aug 1st 2009, 08:39 AMAlephZero
They are the same...

Unless they specify, in terms of solutions, that they need "real solutions," or "nontrivial solutions," etc. The roots are always the roots, but what is or is not an acceptable "solution" may depend upon how the question is asked.

EDIT: Also to consider: The above discussion presupposes that by "polynomial equation" you mean one that is given in the format $\displaystyle a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0$." - Aug 1st 2009, 08:47 AMyeongil
In my precalc book, if you have a polynomial function f(x), the following are equivalent:

* a root of the polynomial f(x),

* a solution to the polynomial equation f(x) = 0,

* a zero of the polynomial f(x), and

* an x-intercept of the polynomial f(x) (assuming that we're only talking about real roots).

01 - Aug 1st 2009, 08:51 AMVonNemo19
I agree with Aleph, because the differences between the two are subtle at best. A solution is finding the values in an equation that satisyfy the given condition, and finding roots generally implies finding the zeros (a type of solution) that satisfy the given conditions. I would maybe say that the word "solution" is a bit more broad than "root".

Read the first few lines of this:

Roots or zeros of a polynomial - Topics in precalculus - Aug 1st 2009, 08:54 AMyeongil
- Aug 1st 2009, 08:57 AMVonNemo19
- Aug 1st 2009, 09:08 AMAlephZero
One other consideration comes to mind here. Sometimes roots are counted with multiplicity, and sometimes without. So for instance

$\displaystyle (x-2)^5=0$

has 5 roots counting multiplicity, and 1 root, not counting multiplicity. In either case, there's really only 1 "solution." - Aug 1st 2009, 09:18 AMVonNemo19
- Aug 1st 2009, 05:28 PMpankaj
- Aug 1st 2009, 07:13 PMAlephZero
OK, well, any polynomial of the form $\displaystyle a_nx^n + a_{n-1}x^{n-1}+\ldots+a_1x+a_0=0$ can be factored in terms of its roots $\displaystyle r_1, r_2, \ldots, r_k$, with $\displaystyle k\leq n$, and corresponding multiplicities $\displaystyle m_1, m_2, \ldots, m_k$ thusly:

$\displaystyle c_0(x-r_1)^{m_1}(x-r_2)^{m_2}\cdots(x-r_k)^{m_k}=0.$

Here $\displaystyle c_0$ is just a constant. There will be $\displaystyle k$*distinct*roots, and therefore $\displaystyle k$ "solutions." But if we include multiplicity, there will be $\displaystyle n$ roots.

So in general, the amount of "solutions" is less than or equal to the amount of roots, depending on whether or not we count repeated roots.

I hope that answers your question.