1. ## Roots 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?

2. 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 $a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0$."

3. 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).

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4. Originally Posted by pankaj
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?
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

5. Originally Posted by VonNemo19
Okay, that was scary -- I was looking at the same exact site just minutes ago!

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6. Originally Posted by yeongil
Okay, that was scary -- I was looking at the same exact site just minutes ago!

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7. One other consideration comes to mind here. Sometimes roots are counted with multiplicity, and sometimes without. So for instance

$(x-2)^5=0$

has 5 roots counting multiplicity, and 1 root, not counting multiplicity. In either case, there's really only 1 "solution."

8. Originally Posted by AlephZero
One other consideration comes to mind here. Sometimes roots are counted with multiplicity, and sometimes without. So for instance

$(x-2)^5=0$

has 5 roots counting multiplicity, and 1 root, not counting multiplicity. In either case, there's really only 1 "solution."
Good example, man!

9. Originally Posted by AlephZero
One other consideration comes to mind here. Sometimes roots are counted with multiplicity, and sometimes without. So for instance

$(x-2)^5=0$

has 5 roots counting multiplicity, and 1 root, not counting multiplicity. In either case, there's really only 1 "solution."
Yes.This is the part I wanted to confirm.Could you provide more examples like this this.

10. Originally Posted by pankaj
Yes.This is the part I wanted to confirm.Could you provide more examples like this this.
OK, well, any polynomial of the form $a_nx^n + a_{n-1}x^{n-1}+\ldots+a_1x+a_0=0$ can be factored in terms of its roots $r_1, r_2, \ldots, r_k$, with $k\leq n$, and corresponding multiplicities $m_1, m_2, \ldots, m_k$ thusly:

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

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