# A function f:R→R satisfies f(5−x)=f(5+x). If f(x)=0 has 5 distinct real roots, what is the sum of all of the distinct real roots?

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I spoke to pramodpandey and now I know why I was confused. He defines "root" as a zero of a *polynomial* function and I define root as a zero of *any *function. The Wikipedia page uses my definition, but Wikipedia isn't exactly the authority on math. I searched around and can't really find a clear definition. For this reason, I prefer to use the term "zero of a function", since that seems more clear.

My first response was that this is a nice problem, and one feature of nice problems is that they generate discussion. Thank you pramodpandey for your response and proof that no polynomial can satisfy the conditions, and thank you nice- for posting this problem (I hope you're still reading this).

This is a nice problem. Let `r` be a root, so that `f(r)=0.` It follows from what is given that

`0=f(r)=f(5-(5-r))=f(5+(5-r))=f(10-r),` so `10-r` is also a root. Thus any root `r` is naturally associated with another root `10-r.` If none of the roots `r` is equal to 5, then its associated root `10-r` is also not equal to 5 and the *associated roots are distinct and form a pair*. However, this can only result in an even number of distinct roots, so one root must be 5 in order for the number of distinct roots to be 5 (and odd number).

So the roots must look like `5,r_1,10-r_1,r_2,10-r_2`. Add these up and you get 25.

**The sum of the roots is 25.**

I don't understand your objection. The question can be reworded to say something like "Suppose a function `f` exists such that..."

So even if there can be no such function, we can still assume it can exists and say what must be true of the roots. Of course, if we reach a contradiction, we can always say that `f` can't exist.

However, unless I made a silly mistake, I found such a function `f.` I think you're imppsing unnecessary restrictions on the function. It's quite possible that a polynomial can't satisfy the conditions, although I'm sure some sort of piecewise polynomial that's not differentiable everywhere can be made to work.

The problem doesn't specify that `f` is a polynomial. The function `f` defined by

`f(x)=0` if `x=-5,0,5,10,15` and `f(x)=1` otherwise

seems to work. I get 5 distinct zeros, and it can be seen that `f(5-x)=f(x+5)` for all `x` (make sure to check my work, of course). And indeed, the zeros add to `-5+0+5+10+15=25.`

Here it is not condition that f(5-x)=f(5+x)=0,if this then x=+5 ,x=-5 are two roots it self.

Under this assumption f(5-x)=f(5+x), problem has no solution at all.

let f(x)=ax^5+bx^4+cx^3+dx^2+ex+f be given polynomial of degree five ,having five distinct real roots.

f(5-x)=a{.......+c(5,1)5(-x)^4+c(5,5)(-x)^5}+b{........+c(4,4)(-x)^4}+......

f(5+x)=a{.......+c(5,1)5x^4+c(5,5)x^5}+b{........+c(4,4)x^4}

f(5-x)=f(5+x)

comparing coeffients of x^4 and x^5 ,we have

ac(5,1)5+bc(4,4)=ac(5,1)5+bc(4,4) coefficient of x^4

-ac(5,5)=ac(5,5)

2 a c(5,5)=0

a can not be zero because of degree restriction,but c(5,5) can not be zero c(5,5)=1,

Here is contradiction

our assumption that f(5-x)=f(5+x) is wrong

so problem has no solution.

In question as well it is never mentioned about function is constant.

when condition applyied it is not for a particular function.